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Fiitjee Aits 2015 Papers

The document discusses a concept recapitulation test for JEE preparation. It provides instructions for the test, which covers physics, chemistry, and math, and will last 3 hours. It also lists useful physical constants and formulas for the test.

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Fiitjee Aits 2015 Papers

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FIITJEE JEE (Main), 2015

FIITJEE Students From All Programs have bagged 34 in Top 100, 66 in Top 200 and 174 in Top 500 All India Ranks. FIITJEE Performance in JEE (Advanced), 2014:
2521 FIITJEE Students from Classroom / Integrated School Programs & 3579 FIITJEE Students from All Programs have qualified in JEE (Advanced), 2014.
CONCEPT RECAPITULATION TEST - I

Time Allotted: 3 Hours Maximum Marks: 432

 Pl ea s e r ea d t h e i n s t r u c t i o n s c a r ef u l l y. Yo u a r e a l l o t t ed 5 m i n ut es
s p ec i f i c a ll y f o r t h i s p u r p o s e.
 Yo u a r e n o t a ll o wed t o l ea v e t h e E xa m i n at i o n Ha ll b ef o r e t h e en d o f
ALL INDIA TEST SERIES

t h e t es t .

B. Filling of OMR Sheet

1. Ensure matching of OMR sheet with the Question paper before you start marking your answers
on OMR sheet.
2. On the OMR sheet, darken the appropriate bubble with black pen for each character of your
Enrolment No. and write your Name, Test Centre and other details at the designated places.
3. OMR sheet contains alphabets, numerals & special characters for marking answers.

C. Marking Scheme For All Three Parts.

(i) Section-A (01 to 02 and 09 to 30) contains 24 multiple choice questions which have only one
correct answer. Each question carries +4 marks for correct answer and – 1 mark for wrong
answer.

Section-A (03 to 08) contains 6 multiple choice questions which have only one correct answer.
Each question carries +8 marks for correct answer and – 2 marks for wrong answer.

Name of the Candidate

Enrolment No.

FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942
website: www.fiitjee.com
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2

Useful Data

PHYSICS
Acceleration due to gravity g = 10 m/s2

Planck constant h = 6.6 1034 J-s

Charge of electron e = 1.6  1019 C

Mass of electron me = 9.1  1031 kg

Permittivity of free space 0 = 8.85  1012 C2/N-m2

3 3
Density of water water = 10 kg/m

Atmospheric pressure Pa = 105 N/m2

Gas constant R = 8.314 J K1 mol1

CHEMISTRY

Gas Constant R = 8.314 J K1 mol1

= 0.0821 Lit atm K1 mol1
= 1.987  2 Cal K1 mol1
Avogadro's Number Na = 6.023  1023
Planck’s constant h = 6.625  1034 Js
= 6.625  10–27 ergs
1 Faraday = 96500 coulomb
1 calorie = 4.2 joule
1 amu = 1.66  10–27 kg
1 eV = 1.6  10–19 J

Atomic No: H=1, He = 2, Li=3, Be=4, B=5, C=6, N=7, O=8,

N=9, Na=11, Mg=12, Si=14, Al=13, P=15, S=16,
Cl=17, Ar=18, K =19, Ca=20, Cr=24, Mn=25,
Fe=26, Co=27, Ni=28, Cu = 29, Zn=30, As=33,
Br=35, Ag=47, Sn=50, I=53, Xe=54, Ba=56,
Pb=82, U=92.
Atomic masses: H=1, He=4, Li=7, Be=9, B=11, C=12, N=14, O=16,
F=19, Na=23, Mg=24, Al = 27, Si=28, P=31, S=32,
Cl=35.5, K=39, Ca=40, Cr=52, Mn=55, Fe=56, Co=59,
Ni=58.7, Cu=63.5, Zn=65.4, As=75, Br=80, Ag=108,
Sn=118.7, I=127, Xe=131, Ba=137, Pb=207, U=238.

FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942
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Physics PART – I

SECTION – A
Single Correct Choice Type

This section contains 30 multiple choice questions. Each question has four choices (A), (B), (C)
and (D) out of which ONLY ONE is correct.

1. Number of amplitude modulation broadcast stations that can be accommodated in a 100

KHz bandwidth if the highest frequency modulating a carrier is 5 KHz
(A) 10 (B) 100
(C) 1000 (D) 10,000

2. Regarding transistor what is not correct.

(A) For transistor to act an amplifier EB junction should be forward biased and CB
junction should be reverse biased
(B) IE = IB + IC in any configuration and for any transistor.

(C)   where  and  transistor parameter
1 

(D)   .
1 

3. A particle of charge q and mass m is projected from a large distance towards another
identical charged particle at rest with velocity v0. The distance of closest approach will be
(Assume negligible gravitational attraction)
q2 q2
(A) (B)
20mv 02 0mv 02
2q2 q2
(C) (D)
0mv 02 40mv 02

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4. A thin equiconvex lens of focal length f is cut into two

symmetric plano-convex lenses, and the lenses are placed
with their curved faces towards each other, with a common
optic axis. If the refractive index  of the material of the lens
is 1.45, the focal length of the combination is (assuming no
reflection at the boundaries) :
(A) f (B) f
(C) 0.9 f (D) + f/0.9

5. In an npn transistor amplifier circuit when working in common emitter configuration when
the input signal changes by 0.02 V, the base current changes by 10 A and the collector
1 A. If the load in the collector circuit is 10 K for this the power given.
4 4
(A) 10 (B) 2  10
4
(C) 5  10 (D) 5  103

6. Two nicols A and B are placed in the path of a beam of unpolarised light. In between
these two a third nicol C is placed such that its principal section is at an angle of 30 with
that of A. The percentage of intensity of incident unpolarized light that emerges from C to
B.
(A) 2.8% (B) 9.4%
(C) 15.3% (D) 10.2%

7. Two cells are connected with 2 resistance and a E1 = 3V E2 = 2V

TG(tangent galvanometer ). If the deflection in TG is 30. 1 2
Then the resistance of the coil.
Given that coil has 5 turns and radius = 0.08 and
BH = 5  105T TG
2
(A) 1.5  (B) 3.5 
(C) 5.5  (D) 7.5 

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8. A conducting ring of radius a falls vertically downward with a  

velocity v in a magnetic field B. The potential difference  Q
between two points P and Q located symmetrically on both P
sides of the vertical has the value 60° 
Bv 0a 
(A) 0 (B) 
2 

  
3 B
(C) Bv 0a (D) Bv 0a
2
v

9. In problem 8, the current flowing in the upper arc PQ of the loop is given by (resistance
per unit length is )
3Bv 0a
(A) 0 (B)
2
3 3 3
(C) Bv 0 a (D) Bv 0a
2 
0
10. A parallel beam of light of all wavelengths greater than 3000 A falls on a double slit in a
Young's double slit experiment. It is observed that the wavelengths 3600 A0 and 6000 A0
are absent at a distance of 31.5 mm from the position of the central maximum, and the
orders of the interference at this point for the two wavelengths differ by 7. If the distance
between the slits and the screen is 1m, the separation between the two slits is
(A) 0.08 mm (B) 0.13 mm
(C) 0.2mm (D) 0.1mm

11. A certain radioactive material can undergo three different types of  B1

decay, each with a different decay constant , 2 and 3. Then, the 2 B2
effective decay constant eff is given by
3 B3
1 1 1 
(A)    (B) eff =  + 2 + 3 
 eff  2 
  2  3 1 1 1 1 1 
(C) eff = (D)   
3  eff 3   2 3 

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-3 -1 -3 -1
12. A thin walled cylindrical metal vessel ( = 10 K ) contains a liquid (r = 10 K ). If the
vessel and its contents are now heated by 10 K the pressure due to the liquid at the
bottom
(A) increases by 2 % (B) decreases by 1 %
(C) remains unchanged (D) decreases by 2%

13. A projectile projected tangentially from the surface of a planet of

radius R from a point A, follows the indicated trajectory during its
motion. If it is at a height of 3R at the farthest point of its trajectory,
then the velocity of projection at A, v0 is given by (acceleration due
to gravity on surface = g) v0
(A) v 0 = 1.5gR (B) v 0 = 0.5gR A
(C) v 0 = 1.6gR (D) v 0 = 2gR / 3

14. A small uniform sphere of diameter d is rolling without

slipping with in another static hollow sphere of inner
diameter 6d, the velocity of the centre of mass of the inner
sphere being v 0 in magnitude. The angular velocity of the
smaller sphere about an axis passing through its C.M. o
perpendicular to its direction of motion is
2v 0 2v 0
(A) (B)
5d d
12v 0 14v 0
(C) (D)
5d 5d

15. 1 mole of H2 and 1 mole of He at a pressure of P0 & 2P0 respectively and a temperature
T0 are kept in two different containers. The ratio of their rms velocities is CH2 : CHe =
(A) 10 : 3 (B) 1 : 1
(C) 5 : 3 (D) 2 : 2

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0
16. The wavelength of the K line for an element is measured to be  = 1.55 A . The
element has Z (atomic number) equal to
(A) 28 (B) 27
(C) 29 (D) 30
17. An equimolar mixture of a monoatomic and a diatomic ideal gas is suddenly compressed
1 th
to of its original volume. The ratio of final temperature to the initial temperature.
8
(A) 80.53 (B) 80.5
1.53
(C) 8 (D) 82/3

18. The material of an electromagnet should have following properties, except

(A) high retentivity
(B) high value of saturation magnetisation
(C) low coercivity
(D) low hystersis loss

19. According to electromagnetic theory statement that is not correct.

 
(A) E and B vectors vibrate sinusoidally in the same phase
 
(B) E and B vectors vibrate sinusoidally in the opposite phase
 
(C) E and B vectors contribute to energy equally
 
(D) E and B vectors vibrate orthogonally with the same frequencies

20. A current (I) carrying circular wire of radius R is placed in a magnetic x

field B perpendicular to its plane. The change in tension T of the I x x
wire is x R
(A) BIR (B) 2BIR
x x
(C) BIR (D) 2BIR x
B I
x

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21. The velocity of sound varies according to the Cliff

relation : B
c(x) = c0 (1 + 0.1x/d) between a source located at A
x
x = 0 and a cliff located at x = d. The time taken x=0 x=d
for sound to travel from A to B and return back to A
is given by
2d 2d
(A) (B)
c0 0.1
c 0 (1  )
2
2d 20d
(C) (D) ln(1.1)
c 0  1.1 c0

22. A charged particle of mass m and charge e moves in a circular path of radius r under the
action of a constant magnetic field B acting perpendicular to the plane of its orbit. If
Bohr's postulate is valid for this system, then the radius of the path, r is given by
(A) r  n2 (B) r  1/n
-n
(C) r  e (D) r  n
where n is the quantum number of the orbit.

23. The molar heat capacity of a diatomic ideal gas undergoing the process :
1/4
PV = constant is
17 2
(A) R (B) R
6 5
23
(C) R (D) 0
6

24. A wheel of radius r rolls without slipping on a flat surface, with its centre of mass
undergoing a linear deceleration a. Then,
(A) a non -zero toque is acting on the wheel about its CM.
(B) the net torque acting on the wheel may be zero or non-zero
(C) the net torque acting on the wheel is zero
(D) the net torque acting on the wheel cannot be calculated as the data is insufficient.

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25. A capacitor C and a resistance R are connected in a circuit in series with each other, so
that the capacitor discharges through R. If the temperature coefficient of the resistance is
positive, as the temperature increases the time constant of the circuit
(A) increases
(B) decreases
(C) remains uncharged
(D) may increase or decrease depending on other factors

26. A particle A having charge q is placed at a distance d from another particle B having an
equal charge q an dielectric slab of thickness t is placed midway between the two.
Electrostatic force on A due to B
(A) decreases
(B) increases
(C) remains constant
(D) may increase or decrease depending upon the magnitudes of the charges

27. A radioactive element A decays into another element B with a decay constant 4, which
then decays into a third element C (decay constant = 9). If all the atoms in the
beginning consisted of atoms of type A only, then the ratio of the number of atoms of A to
that of type B (i.e. NA/NB) when the number of atoms of B is at a maximum (NB is
maximum ) is
3 9
(A) (B)
2 4
1 ln3
(C) (D)
1 ln2

28. If planck's constant (h) were to be doubled while other fundamental quantities like and
charge mass electron. remaining constant the force acting on the electron in the nth orbit
of a hydrogen atom will
(A) increases 4 times (B) decrease 1/8 of its value
(C) remains unchanged (D) decreases 1/16 of its original value

29. A positron -electron pair form a hydrogen like system. Applying Bohr's theory to this
system and ignoring any relativistic effects, the distance (a) between the positron and
electron in the ground state is given by
(A) a > 0.53 A0 (B) a < 0.53 A0
0
(C) a = 0.53 A (D) a = 1.06 A0

30. Two identical spherical stars each having a mass equal to that of the sun move in a
common circular orbit of radius equal to the earth's orbital radius, under their mutual
gravitational interaction. Their time period of rotation equals
(A) 2yr (B) 1 yr
(C) 22 yr (D) 2 yr

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Chemistry PART – II

SECTION – A
Single Correct Choice Type

This section contains 30 multiple choice questions. Each question has four choices (A), (B), (C)
and (D) out of which ONLY ONE is correct.

1. A gas in an open container is heated from 27°C to 127°C. The fraction of the original
amount of remaining in the container will be:
3 1
(A) (B)
4 4
1 1
(C) (D)
2 8

2. Which one has peroxide linkage

2 2
(A) S2O8 (B) S2O6
2 2
(C) S2O7 (D) S2O3

2+
3. The number of electrons gained in the following reaction Fe2(SO4)3  Fe + SO2 is
(A) 2 (B) 1
(C) 6 (D) 8

4. The values of observed and calculated molecular weights of silver nitrate are 92.64 and
170 respectively. The degree of dissociation of silver nitrate is
(A) 60% (B) 83.5%
(C) 46.7% (D) 60.23%

5. The pH of pure water at 25°C and 35°C are 7 and 6 respectively. The heat of formation of
water from H+ and OH– are
–1 –1
(A) 84.55 kcal mol (B) – 84.55 kcal mole
–1
(C) 74.55 kcal mol (D) – 74.55 kcal mole–1

6. In a face centred cubic arrangement of A and B atoms whose A atoms are at the corner
of the unit cell and B atoms are at the face centres. One of the A atom is missing from
one corner in unit cell. The simplest formula of the compound is:
(A) A7B3 (B) AB3
(C) A7B24 (D) A7/8B3

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7. In which one of the following reactions, we will get only one crossed Aldol product,
(A) C6H5CHO, (CH3)2CO (B) C6H5CHO, CH3CHO
(C) C6H5CHO, (C2H5)2CO (D) CH3CHO, CH3CH2CHO

8. CH3 – CH – C – CH3. This compound

| ||
OH O
(A) Can react with Schiff’s Reagent (B) Can react with Tollen’s Reagent
(C) Both (A) and (B) (D) None

9. The isomeric monosubstitution products theoretically possible for the structure

CH2 = CH – CH2 – CH2 – CH = CH2 are
(A) 3 (B) 2
(C) 4 (D) 6

10. CH3

CH OH / OH
CH3 – C – CH2    A. A is
3

O
CH3
CH3
(A) CH3 –– C –– CH2 (B)
CH3OH + CH3 – C = CH2
OCH3OH
CH3
CH3
(C) CH3 –– C –– CH2 (D)
HCHO + CH3 – C = CH2
OH OCH3
– +
11. HO – CH2 – CH2 – X + HC  C Na  A + B
A & B can be
(A) HO – CH2 – CH2 – C  CH, NaX (B) Acetylene + NaO – CH2 – CH2 – X
+ +
(C) Ethanol + XC  CNa (D) Ethylhalide + HOC  CNa

12. Which of the following will be most reactive towards the nucleophilic substitution.
Br OTs

(A) (B)

OH OSO2CF3

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13. The correct order of hybridisation in the following species SO2, ClO3–, XeF4 is
3 2 3 2 2 3 2
(A) sp , sp , sp d (B) sp , sp , sp d
2 3 3 2 3 3 2
(C) sp , sp , sp d (D) sp, sp , sp d

14. The species with maximum magnetic moment is

(A) K4[Ni(CN)6] (B) K4[Fe(CN)6]
(C) K3[Co(CN)6] (D) [Cu(NH3)4]SO4

15. The increasing order of dipole moment is

(A) CH3I CH3Br  CH3Cl  CH3F (B) CH3F  CH3Cl  CH3Br  CH3I
(C) CH3I  CH3Br  CH3F  CH3Cl (D) CH3I  CH3F  CH3Br  CH3Cl

16. 17g of hydrogen peroxide is present in 1120 ml of solution. This solution is called
(A) 5 volume solution (B) 10 volume solution
(C) 15 volume solution (D) 16 volume solution

17. The density of a gas at 27°C and 1 atm is d. Pressure remaining constant, at which of the
following temperature will its density become 0.75d.
(A) 20°C (B) 30°C
(C) 36°C (D) 127°C

18. Radioactivity of radioactive element remains 1/10 of the original radioactivity after 2.303
seconds. The half life period (in seconds) is
(A) 2.303 (B) 0.693
(C) 0.2303 (D) 0.0693

19. Assuming complete dissociation predict which of the following will have pH = 12.
(A) 500 ml 0.005 M NaOH (B) 1000 ml 0.005 M Ca(OH)2
(C) 100 ml 0.05 M Ca(OH)2 (D) 50 ml 0.05 M NaOH

20. The percentage of metal in its oxide is 60. The percentage of bromine in metal bromide
(oxidation state of metal remains the same) will be
(A) ~ 87 (B) ~ 97
(C) ~ 77 (D) ~ 80

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21. 50g of CaCO3 when treated with 60 ml of 20 N HCl at 300 K and 4 atm produces
(A) 22 g CO2 (B) 44g CO2
(C) 11g CO2 (D) 88 g CO2
–5
22. A certain weak acid has a dissociation constant of 10 . The equilibrium constant for its
reaction with a strong base is
(A) 10–9 (B) 10+14
9
(C) 10 (D) 10–5
st
23. The set representing the correct order of 1 ionisation potential is
(A) Cu  Ag  Au (B) Au Ag  Cu
(C) Ag  Cu  Au (D) Au  Cu  Ag

24. Acetophenone on reaction with p -nitroperbenzoic acid gives

(A) phenyl propionate (B) phenyl acetate
(C) methyl benzoate (D) benzophenone

25. The number of monochloro derivaties of 2-methoxy propane are possible are
(A) 2 (B) 1
(C) 3 (D) 4

26. The increasing order of bond angle in the following species is

(A) F2O  Cl2O  H2O (B) Cl2O  H2O  F2O
(C) F2O  H2O  Cl2O (D) H2O  F2O  Cl2O

27. 0.10g of a sample containing CuCO3 and some inert impurity was dissolved in dilute
sulphuric acid and volume made up to 50 ml. This solution was added into 50 ml of 0.04
–
M KI solution where copper precipitates as CuI and I is oxidised into I3 . A 10 ml portion
of this solution is taken for analysis filtered and made up free I3 and then treated with
excess of acidic permanganate solution. Liberated iodine required 20 ml of 2.5 mM
sodium thiosulphate solution to reach the end point. Determine the wt. percentage of
CuCo3 in the original sample.
(A) 7.41 (B) 74.1
(C) 61.75 (D) None of these

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28. Urea (H2NCONH2) is manufactured by passing CO2(g) through ammonia solution

followed by crystallization. CO2 for the above reaction is prepared by combustion of
hydrocarbon. If combustion of 236 kg of a saturated hydrocarbon (CnH2n+2) produces as
much CO2 as required for production of 999.6 kg urea then M. F. of hydrocarbon is:
(A) C10H22 (B) C12H26
(C) C13H28 (D) C8H18

29. For the reaction at 300 K 0.8

(T = 300 K)


A  
B H  2400 cal mol1
0.2

Fraction of moleculess
Frequency factor for both forward and backward reaction
is equal. The [Keq]400 K is 4e – 0.5n.
A
The value of ‘n’ is (Assume H independent of
temperature)
B
(Ea)b (Ea)f

Speed
(A) 1 (B) 2
(C) 4 (D) 6

30. Which of the following statement is not true?

(A) O-F bond length in OF2 is less than O-F bond length in O2F2.
(B) In HCO3 , all C – O bond length are not identical.
(C) In diborane, two different B – H bond lengths are observed although the hybridization
of both boron atoms are same.
(D) In hydrazine, the N – N bond length is larger than normal N – N bond length.

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Mathematics PART – III

SECTION – A
Single Correct Choice Type

This section contains 30 multiple choice questions. Each question has four choices (A), (B), (C)
and (D) out of which ONLY ONE is correct.

1. Let y = f(x) be a curve passing through (e, ee), which satisfy the differential equation
e
(2ny + xy ln x) dx – x ln x dy = 0, x > 0, y > 0. It g(x) = lim f  x  then
n  g  x  dx equals to
1/e
(A) e (B) 1
(C) 0 (D) none of these

2. The smaller area enclosed by y = f(x), when f(x) is a polynomial of least degree satisfying
1/ x
 f x  2 2
lim  1  3   e and the circle x + y = 2 above the x-axis is
x 0
 x 
 3
(A) (B)
2 5
 3  3
(C)  (D) 
2 5 2 5

1 cos 2 t
3. For x  R and a continuous function f, let I1   x f x  2  x  dx ,
sin2 t
1 cos2 t
I1
I2   f x  2  x  dx . Then is
sin2 t
I2
(A) 0 (B) 1
(C) 2 (D) 3

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2 2x 2  12xf  x   f  x 
4. If xf(x) = 3(f(x)) + 2, then  2
dx equals to
 6f  x   x   x2  f  x  
1 1
(A) 2
c (B) 2
c
x  f x x  f x
1 1
(C) c (D) c
x  f x x  f x

 y y
5. If f  2x  , 2x    xy , then f(m, n) + f(n, m) = 0
 8 8
(A) only when m = n (B) only when m  n
(C) only when m = –n (D) for all m and n

6. If period of sin2m kx , m  N is  then lim k n is equal to

n
(A) 0 (B) 
(C) 1 (D) none of these

1 1
7. A function g(x) is defined as g  x  
4
   
f 2x 2  1  f 1  x 2 and f(x) is an increasing
2
function then g(x) is increasing in the interval
 2 2  2   2 
(A)   ,  (B)   , 0    , 
 3 3  3   3 
     
(C) (–1, 1) (D) none of these

8. If sin  and –cos  are the roots of the equation ax2 – bx – c = 0, where a, b, c are the
sides of a triangle ABC, then cos B is equal to
c c
(A) 1  (B) 1 
2a a
c c
(C) 1  (D) 1 
2a 3a

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x
a a
 t 
2
9. The equation  8t  13 dt  x sin   has a solution if sin is equal to
0 x 6
1
(A) 1 (B)
2
1
(C) (D) 0
2
 
10. If x and y are two non-collinear vectors and a triangle ABC with side lengths a, b, c
satisfying
    
 20a  15b  x  15b  12c  y  12c  20a  x  y   0 . Then triangle ABC is
(A) an acute angle triangle (B) an obtuse angle triangle
(C) a right angle triangle (D) an isosceles triangle

11. The Cartesian equation of the plane r  1      ˆi   2    ˆj   3  2  2  kˆ is
(A) 2x + y = 5 (B) 2x – y = 5
(C) 2x + z = 5 (D) 2x – z = 5

x2 y2 2
12. If the ellipse  y 2  1 meets the ellipse x 2   1 in four distinct points and a = b –
4 a
5b + 7, then b does not lie in the interval
(A) [4, 5] (B) (–, 2)  (3, )
(C) (–, 0) (D) [2, 3]

x2 y2
13. Portion of asymptote of hyperbola   1 (between centre and the tangent at vertex)
a2 b 2
in the first quadrant is cut by the line y + (x – a) = 0 ( is a parameter) then
(A)   R (B)   (0, )
(C)   (–, 0) (D)   R – {0}


14. A focal chord of parabola y2 = 4x is inclined at an angle of with positive x-direction,
4
then the slope of normal drawn at the ends of chord will satisfy the equation
2 2
(A) m – 2m – 1 = 0 (B) m + 2m – 1 = 0
2 2
(C) m – 1 = 0 (D) m + 2m – 2 = 0

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15. The range of values of a such that the angle  between the pair of tangents drawn from
2 2  
(a, 0) to the circle x + y = 1 lies in the interval  ,   is
3 
(A) (–2, –1)  (1, 2)    
(B)  2, 0  0, 2


(C)  3,  2  (D)   3,  2    2, 3 

16. The equation of the parabola to which the line m2(y – 10) – mx – 1 = 0 is a tangent for
any real value of m is
2 2
(A) x = –4y (B) x = –4(y – 10)
2 2
(C) y = 4(x – 4) (D) x = y – 10

17. Consider a curve ax 2 + 2hxy + by2 – 1 = 0 and a point P not on the curve. A line drawn
from the point P intersects the curve at the points Q and R. If the product PQ·PR is
independent of the slope of the line, then the curve is
(A) a pair of straight lines (B) a circle
(C) a parabola (D) an ellipse or a hyperbola

18. In a ABC if A = (1, 2) and internal angle bisectors through B and C are y = x and y = –
2x. The in radius r of ABC is
1 1
(A) (B)
3 2
2
(C) (D) none of these
3

19. If 0  [x] < 2; –1  [y] < 1 and 1  [z] < 3 (where [.] denotes the greatest integer function)
 x   1  y z
then the maximum value of determinant  x   y   1  z  is
 x  y  z  1
(A) 2 (B) 6
(C) 4 (D) none of these

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20. If min{x2 + (a – b)x + 1 – a – b} > max{–x2 + (a + b)x – (1 + a + b)}, then

2 2 2 2
(A) a + b < 2 (B) a + b < 4
2 2
(C) a + b > 4 (D) a + b > ab

m
1  xi  1 
21. If m and x are two real numbers, then the real value of e2mi cot x
 xi  1  is equal to
 
m
(A) cos x + i sin x (B)
2
m
(C) 1 (D) 1
2

200
 5 
22. If {x} represents the fractional part of x, then   is
 8 
1 1
(A) (B)
4 8
3 5
(C) (D)
8 8

a 2a3 a 2  a3  a  a3 
23. If   3 2  , then a1, a2, a3, a4 are in
a1a 4 a1  a4  a1  a 4 
(A) A.P. (B) G.P.
(C) H.P. (D) none of these

24. n-similar balls each of weight w when weighed in pairs the sum of the weights of all the
possible pairs is 120 when they are weighed in triplets the sum of the weights comes out
to be 480 for all possible triplets, then n is
(A) 20 (B) 10
(C) 16 (D) 28

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       
25. If a, b, c are unit vectors such that a  2b  3c  3  2 2 . Angle between a and b is
      2 
, between a and c is  and angle between b and c varies in  ,  , then the
2 3 
greatest value of 4 cos  + 6 cos  is
(A) 2 2  5 (B) 2 2  5
(C) 2 2  5 (D) 2 2

26. If two events A and B are such that P(Ac) = 0.3, P(B) = 0.4 and P(A  Bc) = 0.5 then
 B 
P C 
is equal to
 A B 
(A) 0.9 (B) 0.5
(C) 0.6 (D) 0.25

27. ABC is a triangular park with AB = AC = 100m. A block tower is situated at the midpoint
of BC. The angles of elevation of the top of the tower at A and B are cot1(3.2) and cosec
1
(2.6) respectively. The height of the tower is
(A) 16 m (B) 25 m
(C) 50 m (D) none of these

28. If aN = {ax | x  N} and bN  cN = dN, where b, c  N, then

(A) d = bc (B) c = bd
(C) b = cd (D) none of these

29. If the normal to the curve y = f(x) at x = 0, be given by the equation 3x – y + 3 = 0, then
1

x 0
   
the value of lim x 2 f x 2  5f 4x 2  4f 7x 2   is equal to
1 1
(A) (B) 
3 3
1 1
(C)  (D)
5 4

30. Let P (n) be the statement 2n < n ! where n is a natural number, then P (n) is true for
(A) all n (B) all n > 2
(C) all n > 3 (D) none of these

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S. No. PHYSICS CHEMISTRY MATHEMATICS

ALL INDIA TEST SERIES

1. A A C
2. D A D
3. B D B
4. A B A
5. C B D
6. B C C
7. B B B
8. D B C
9. A A A
10. C C C
11. B B C
12. D D D
13. C C B
14. B D B
15. D C A
16. C A B
17. B D B
18. A B B
19. B B C
20. A A B
21. D A C
22. D C B
23. C D C
24. A B B
25. A C C
26. C C D
27. B B B
28. D B D
29. A B B
30. A D C

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2

Physics PART – I

SECTION – A

Total Bandwidth 100  103

1. No. of stations = = = 10.
BW per statation 10  103

3. B,
v0 q
q
Mass=m m  
v v

Applying conservation of momentum : 2mv = mv 0 . .. (i)

1 1 q2 1
conservation of energy : 2  mv 2 + = m v20 . . . (ii)
2 40 d 2
(I)  v = v0/2, putting in (2), we get,
q2
d=
0mv 02

4. the powers of the lenses add, and the power of each half is one half of that of the entire lens.

IC V
5.   100 , Ri  B
B IB
RL
power gains = 2  5  104
Ri

I0
6. IA 
2
I0 I 3 3I
IC  cos 30  0  0
2 24 8
3I 3
IB  IC cos2 60  0 cos 2 60  I0
8 32
IB 3
  9.4%
IC 32

5
7. i
5R
B = BH tan 
0Ni
 BH tan 
2r
2rBH 5
i tan  
0N 5R

Solving R = 3.5 

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9. Since the current flowing through a loop is determined by the net emf acting in the loop, which
again is zero n this case - as the flux doesn't change.

10. At x = 31.5 mm, both 3600 A0 and 6000 A0 produce minima. If the slit separation is d,
then,
xd 1 xd 1
 n1  ,  n2 
D1 2 D 2 2
xd  1 1 
or ,     n1  n2  7
D  1 2 
0 0
Substituting the values, D = 1m, 1 = 3600 A , 2 = 6000 A
we get d = 0.2 mm.

dNA
11.  (  NA )  ( 2.NA )  ( 3 .NA )
dt
= 6 . NA

12. The mass of the liquid does not change, the area on which it exerts its weight increases by
2  10-3  10 or 2  10-2
or 2 %, the pressure decreases by 2 %.

13. At the farthest point X,

mv0R = mv x 4R
1 GMm 1 GMm
m v2x  = mv 02 -
2 4R 2 R
 v0 = 1.6 gR .

14. The angular velocity  of the sphere about an

v0 2v
axis through its C.M. is  0 , it does
d/ 2 d
not depend on whether the axis is
accelerated or not. o

3
15. Their average translational K.E. is kB T, kB  Boltzmann's constant
2
1 1 3
mH2 CH2 2  mHe CHe
2
= k BT
2 2 2
CH2 : CHe = mHe : mH2 = 4 : 2 =2: 2

3 12400 0
16. Using Ek = 13.6 eV (Z- 1)2 & k = A
4 Ek 
7 5
C(1)  C(2) R R
17. For the mixture,  = P(1) P
 2 2 3
(2) 5 3 2
CV  CV R R
2 2
Applying TV - 1 = constant, the temperature increases by a factor of 8.

20. (A), Consider a semicircular portion of the F=BI(2R)

wire
2T = BI. 2R
 T = BIR

T T

d
dx
21. Time to reach B from A = tAB = 
x 0
c(x)
d
= ln (1.1) ; total time = 2tAB
0.1c 0

22. For the system,

mv 2
 Bev
r
nh
mvr =
2
nh
 r= .
Be

23. pV1/4 = constant

RT 1/4
 v = constant
V
or TV-3/4 = constant
3
dT V-3/4 - dV.V 7/4 .T = 0
4
dV 4 V
or 
dt 3T
dQ dU dV 4V 5 4 23
C=  P  Cv  p.  R R  R.
dT dT dT 3T 2 3 6

25. Time constant = RC, increases with temperature

dNB
27.   A NA  BNB = 4NA - 9 NB = 0 (NB is maximum)
dt
NA 9
 
NB 4

29. For the position -electron system, the mass of the position = mass of the electron, much less
than that of a proton . the distance in the ground state > aB (hydrogen atom)
A careful calculation shows a  1.06 A0 ( = 2 aB)

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v 2 GM2
30. (A), M 
R 4R2 v
2 GM
v = M M
4R
2R 2R R
Time period, T = =
v GM / 4R v
2 3/2
= 2R
GM
rd
If M = Msun, R = Rearth - sun, Kepler's 3 law gives us,
2
Tearth =  R3/2
earth  sun = 1 yr.
GMsun
2
 T=  2  R3/2
earth  sun = 2  1 yr = 2yr .
GMsun

Chemistry PART – II
SECTION – A
2. O O
–
O—S—O—O—S—O–

normal molecular weight

4. i for AgNO3 = =1+
observed molecular weight
170
= – 1 = 0.835 = 83.5%
92.64

5. At 25°C [H+] = 10–7;  Kw = 10–14

At 35°C [H+] = 10–6;  Kw = 10–12
Now using,
KW H  T2  T1 
2.303 log 2
  
KW 1
R  T1T2 
We get, H = 84.55 kcal/mole
Thus H2O H+ + OH– H = 84.55
+ –
 H + OH H2O H = – 84.55 kcal/mole

7
6. No. of atoms of A from corners of unit cell =
8
No. of atoms of B from faces of unit cell = 3
7
Thus A : B : : 3 (or) 7 : 24
8
 (C)

O
||

12. O  S  CF3 is the best leaving group as it is the weakest base
||
O

14. Because it has maximum no. of unpaired electrons

17
16. Weight of H2O2 in 1 ml =
1120
17 1
Moles of H2O2 in 1 ml = 
1120 34
17 1 1
Moles of O2 liberated =  
1120 34 2
17 1 1
Volume of O2 liberated =    22400 = 5 ml
1120 34 2

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1
17. PM = dRT or d 
T
d1 T2 d T  273
   2
d2 T1 0.75d 27  273
T2 = 127°C

0.693 2.303 A
18.  log 0
t1 / 2 t At
Putting the values we get
t1/2 = 0.693

19. pH = 12, pOH = 2  [OH–] = 10–2 M

xM  100
20. % of M in MxOy = = 60  xM = 24y
xM  16y
xM 24 y
% M in MxBr2y =  100   100  13
xM  160y 24 y  160y
 % of bromine in MxBr2y  87

50  2
21. Equivalents of CaCO3 = =1
100
60  20
Equivalent of HCl = = 1.2
1000
 Equivalent of CO2 = 1, moles of CO2 = 0.5 (n-factor = 2)
 mass of CO2 = 22g

Ka
22. HA + OH– A– + H2O
Kh
K w 10 14
Kh =  5
= 10–9
Ka 10
Ka = Kh–1 = 109

23. Ag has a lower ionisation potential than Cu because of larger size. But in Au due poor shielding
effect of f-electrons the nuclear charge is not shielded and so I.P. is high
 Au  Cu  Ag

O O–
CH3 CH3
O
C = O + NO2Ph – C – O – O–  C
24. Ph Ph O – O – C – Ph – NO2

CH3 – C – O – Ph

27. m moles of KI = 50 × 0.04 = 2

-3
m.eq. of Na2S2O3 reacted with 10 ml solution = 20 × 2.5 × 10
-3
= 50 × 10
- 100
m.eq. of I left unreacted in 100 ml solution = 50  10 3   0.5
10
m. moles of I- reacted with CuCO3 = 2 – 0.5 = 1.5
 2Cu   5I  2CuI  I3
++2
m.moles of Cu =
 1.5  0.6
5
wt. of CuCO3 = 0.6 × 10-3 × 123.5
= 0.0741
0.0741
wt. of CuCO3 =  100  74.1
0.1

28. 2NH2  CO2  NH2  CO  NH2  H2O

999.6  103
moles of urea = = 16.66  103
60
 moles of CO2 = 16.66  103
CnH2n 2  O2  nCO2  H2O
16.66  103
 12n   2n  2    236  103
n
 n  12
so, M. F. of hydrocarbon is C12H26

  E /RT
Kf A.e a f 0.8
29. Keq(300 K)   4
K b A.e  Ea b /RT 0.2
k H  1 1
 In 2    
k1 R  T1 T2 
 4 
In    1
 k2 
4
 e  k 2  4e1
k2

30. In hydrazine (NH2 – NH2) bond order is one, whereas in nitrogen gas bond order is three.

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Mathematics PART – III

SECTION – A
1. f(x) = ex (ln x)2n
1 
lim f  x  = 0, if x   , e 
n e 
e
  g  x  dx  0
1/e

4
2. Clearly f(x) = x
1
 3
 Required area = 2
0
 
2  x2  x 4 dx = 
2 5

1 cos2 t
3. I1   x f x  2  x  dx = 2I2 – I1
sin2 t
I1
 1
I2

f  x
4. f ' x 
6f  x   x
2x  f '  x  1
 I   2
dx  2
c
x  f  x
x 2
 f x 
y y
5. 2x    , 2x   
8 8
 f(, ) = 2 – 2
 f(m, n) + f(n, m) = 0


6.    k  1  lim kn  1
k n 

7. g(x) = x(f(2x2 – 1) – xf(1 – x2))

 2 
If, x > 0  x   ,  
 3 
 2 
If, x < 0  x    , 0 
 3 
 2   2 
 x   , 0   , 
 3   3 
   

b
8. sin   cos  
a
c
sin  cos  
a
 b2 = a2 – 2ac
a2  c 2  b2 c
 cosB   1
2ac 2a

x3 a
9.  4x 2  13x  x sin  
3 x
1 a

3
 2

 x  6   3  sin
x
a
 sin  1
6
   
10.  x, y and x  y are linearly independent
 20a – 15b = 15b – 12c = 12c – 20a = 0
 c2 = a2 + b2

11.      
r  î  2jˆ  3kˆ   î  ˆj  2kˆ   î  2kˆ

a  î  2ˆj  3kˆ

b  î  ˆj  2kˆ

c  î  2kˆ
   

 Vector equation of plane is  r  a  b  c  0 
 2x + z = 5

12. a>1
 b2 – 5b + 7 > 1
 b  (–, 2)  (3, )

13. Clearly, – < 0,  > 0

2 y
14. m AB  1 A(t1)
t1  t 2
t1 + t2 = 2
t1t2 = –1
x
 Required equation is m2 + 2m – 1 = 0 (1, 0)

B(t2)

 M
15. 
3 1
   1 1 /2
     1  1< a< 2 P(a, 0)
6 2 2 2 a O
 a  (–2, –1)  (1, 2)

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1
16. x  m  y  10  
m
a
x  h  my  k 
m
 h = 0, k = 10, a = –1
 (x – 0)2 = –4(y – 10)

x  h' y  k ' R
17.   r
cos  sin 
Q
ah' 2  2hh'k ' bk ' 2
 PQ  PR  P(h, k)
acos2   2hsin  cos   b sin2 
 a = b, h = 0

18. A1 = (2, 1) A(1, 2)

 11 2 
A2    ,  y = –2x y=x
 5 5
Equation of BC: x – 7y + 5 = 0
5 1 I(0, 0)
 r  ID  
1  49 2 r
B A2 A1 C
19. R1  R1 – R2, R2  R2 – R3
 = [x] + [y] + [z] + 1
 max = 1 + 0 + 2 + 1 = 4

  a  b  2  4 1  a  b     a  b 2  4 1  a  b  
20.     
 4 1   4   1 
2 2
a +b <4

21. Put, cot–1 x = 

1 xi  1
 e2i cot x
 cos 2  isin2 =
xi  1
m
2mi cot 1 x  xi  1 
 e   1
 xi  1 

100
5200 1  24 
22. 
8 8
 5200  1
  
 8  8

1 1 1 1
23.    ….. (1)
a4 a3 a2 a1
 1 1  1 1
3    ….. (2)
 a3 a2  a 4 a1
1 1 1 1 1 1
 From (1) and (2), we get     
a2 a1 a3 a2 a 4 a3
 a1, a2, a3, a4 are in H.P.

n n  n  1
24. C2  = Number of possible pairs of n objects
2
n  n  1 n  n  1
Total weight of pairs are  2  w  n  n  1 w units
2 2
 n(n – 1)w = 120 ..... (1)
Similarly, total weight of all triplets = 480
n  n  1 n  2  3w
  480 ….. (2)
6
 From (1) and (2), we get n = 10

  2
25. a  2b  3c  3  2 2
 
4 cos  + 6 cos  = –11 + 2 2 – 12 cos  ( is angle between b and c )
 4 cos  + 6 cos    11  2 2,  5  2 2 

 B  P  A  B  0.2 1
26. P C 
=  
 A  B  P A  Bc  0.8 4 
1
27. Let   MAP  cot (3.2)
and   MBP  cosec 1(2.6)
 cot = 3.2 and cosec = 2.6
Let height of the tower = h P
 MP = h.
Now BM = MP cot = h cot h
= h cosec 2  1  h (2.6)2  1  (2.4)h 
and AM = MP cot = h cot = h(3.2) = 3.2 h B C
M
Hence, from ABM
AB2 = AM2 + BM2
(100)2 = (3.2h)
2
2
+ (2.4h)22 100 m  100 m
100 100
h2  
10.24  5.76 16
100
h  25 .
4 A
28. We have,
bN = {bx | x  N } = the set of positive integral multiples of b
cN = {cx | x  N } = the set of positive integral multiple of c
 bN  cN = the set of positive integral multiples of L.C.M of b and c
 d = L.C.M. of b and c

1
29. f ' 0  
3
2x 1
 Required limit = lim = 
x 0
   40xf '  4x   56xf ' 7x 
2xf ' x 2 2 2 3

30. P(1), P(2) and P (3) are not true

k
P (4) is true and 2 < k!
k
 2.2 < 2k!  (k + 1)k! for k  1

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FIITJEE JEE (Advanced), 2015
CONCEPT RECAPITULATION TEST - I
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Paper 1

Time Allotted: 3 Hours Maximum Marks: 180

 Pl ea s e r ea d t h e i n s t r u c t i o n s c a r ef u ll y . Yo u a r e a l l o t t ed 5 m i n u t es
s p ec i f i c a ll y f o r t h i s p u r p o s e.
ALL INDIA TEST SERIES

 Yo u a r e n o t a l l o wed t o l ea v e t h e E xa m i n at i o n Ha l l b ef o r e t h e en d o f
t h e t es t .

INSTRUCTIONS

C. Marking Scheme For All Three Parts.

(i) Section-A (01 – 10) contains 10 multiple choice questions which have one or more than one
correct answer. Each question carries +3 marks for correct answer. There is no negative
marking.

(ii) Section-C (01 – 10) contains 10 Numerical based questions with answers as numerical value
from 0 to 9 and each question carries +3 marks for correct answer. There is no negative
marking.

Name of the Candidate

Enrolment No.

Useful Data

PHYSICS
Acceleration due to gravity g = 10 m/s2

Planck constant h = 6.6 1034 J-s

Charge of electron e = 1.6  1019 C

Mass of electron me = 9.1  1031 kg

Permittivity of free space 0 = 8.85  1012 C2/N-m2

Density of water water = 103 kg/m3

Atmospheric pressure Pa = 105 N/m2

Gas constant R = 8.314 J K1 mol1

CHEMISTRY

Gas Constant R = 8.314 J K1 mol1

Atomic No: H=1, He = 2, Li=3, Be=4, B=5, C=6, N=7, O=8,

Physics PART – I

SECTION – A
(One or More than One Options Correct Type)
This section contains 10 multiple choice questions. Each question has four choices (A), (B), (C)
and (D) out of which ONE or MORE THAN ONE are correct.

1. Initially two particles A and B are present at (0, 0) and (d, 0) respectively. They start
moving with speed v A  viˆ  vˆj and v B  vˆj . If R is magnitude of relative separation
between them and T0 be the time when separation between them is minimum, then
d 2d
(A) T0  (B) R min 
5v 5
(C) Graph of R versus time is straight line (D) Graph of R versus time is circle.

2. Four forces act on a point object. The object will be in equilibrium if

(A) all of them are in the same plane
(B) they are opposite to each other in pair
(C) the sum of x, y and z components of force is zero separately
(D) they form a closed figure of 4 sides

3. A person of mass 70 kg jump from a 3.0 m height. Then

(A) impulse on the man by ground will be 539 N-s
(B) average force on man if he land with stiffed leg and body moves by 1.0 cm during
5
impact is 2.1 × 10 N
(C) average force on man feet by ground if he land with bent legs and body moves by
50 cm is 4.2 × 103 N
(D) average force on man feet by ground if he land with bent legs and body moves by
3
50 cm is (4.9 × 10 N)

4. A mass m of radius r is rolling horizontally without any slip with a linear speed v. It then
3 v2
rolls up to a height given by
4 g
(A) the body is identified to be a disc or a solid cylinder
(B) the body is a solid sphere
3
(C) MOI of the body about instantaneous axis of rotation is mr 2
2
2 2
(D) MOI of the body about instantaneous axis of rotation is mr
5
Space for Rough work

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AITS-CRT-I(Paper-1)-PCM-JEE(Advanced)/15

5. A particle initially at rest is displaced by applying a non conservative force F in a uniform

gravitational field. In the process following physical quantities associated with the particle
are measured
U  change in gravitational potential energy
K  change in kinetic energy
W1  work done by the force F
W 2  work done by the gravitational force
(A) W2  U (B) K  W1  W2
(C) K  U  W1  W2 (D) W1  W2

6. An object is floating in a liquid, kept in a container. The container is placed in a lift.

Choose the correct option(s)
(A) Buoyant force increases as lift accelerates up
(B) Buoyant force decreases as lift accelerates up
(C) Buoyant force remains constant as lift accelerates
(D) The fraction of solid submerged into liquid does not change

7. A source of sound moves along a circle of radius 2 m with S

constant angular velocity 40 rad/s. Frequency of the source is
300 Hz. A detector is kept at some distance from the circle in the D
same plane of the circle (as shown in figure). Which of the
following is not the possible value of frequency registered by the
detector? (Speed of sound = 320 m/s)
(A) 250 Hz (B) 360 Hz
(C) 410 Hz (D) 220 Hz

8. Two gases have same initial pressure, volume and temperature. They expand to same
final volume, one adiabatically and the other isothermally.
(A) the final temperature is greater for isothermal process
(B) the final temperature is lesser for isothermal process
(C) the work done by the gas is greater for isothermal process
(D) the work done by the gas is greater for adiabatic process

Space for Rough work

9. A charged particle q is placed at a distance d from the centre of conducting sphere of

radius R(<d), then in static condition at the centre of sphere
kq
(A) magnitude of electric field due to induced charge is
d2
kq
(B) magnitude of electric field due to induced charge is 2
R
(C) magnitude of electric field due to induced charge is zero
kq
(D) magnitude of electric field due to charge q is (where k = 1/40)
d2

10. Three identical small balls of same 9m / s

mass 1 kg are placed on a horizontal
smooth table. The velocity give to
middle ball parallel to the surface is 9
m/s. Then:
1m 1m
(A) the speed of colliding balls just before collision is 5 m/s
(B) the speed of any of the balls when they collide is 9 m/s
(C) the tension in the string connected between balls just before the two balls collide is 5
newton
(D) the tension in the string connected between balls just before the two balls collide is 9
newton

Space for Rough work

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AITS-CRT-I(Paper-1)-PCM-JEE(Advanced)/15

SECTION – C
(One Integer Value Correct Type)

This section contains 10 questions. Each question, when worked out will result in one integer
from 0 to 9 (both inclusive).

1. A ball translates on a horizontal circular path as shown in the figure. 2

When the ball reaches the diametrically opposite point. The position   
of point A in shown by point 1, 2, 3 and 4. Find the correct position 3 4
1
of point A: A

C

B

2. A rod of finite length having linear charge density A C


 subtends an angle of 15 at a point P which is also the 

centre of curvature of a circular arc having same linear  P
M 
charge density  as shown in figure. If the net electric field R

bK  
 R
at the origin is . Find the value of b: D
R B

3. A charge particle having specific charge  is initially

located at point P at a distance of X 0 from an infinite I0 P
current carrying wire as shown in figure. The maximum  v0
distance from the wire the charge will go, if it is given X0
b v0
velocity v0 perpendicular to the wire is X 0e . Find
0 I 0
the value of b:

4. An electrical network is formed by two hexagon and joining grid

points with wires. The resistance of each side and the
connecting wires is R. An ideal battery of emf E is joined as
shown in figure. Find the current passing through the battery.
Consider the wire connected with the cell is ideal. [take
E  11 Volt ; R  6 ohm ]


Space for Rough work

5. A point object is placed at the midpoint in water having  2   4/3

boundary width of 30 cm and other medium having Air Air

O
refractive index 2 is having boundary width 20 cm as
15cm
shown in figure. The position of image of object O as
seen from air through the curved part is 10k. Find the 20cm
value of k………. 18cm

6. An ideal gas undergoes a process shown in the P-V diagram. P

If the change in heat content of the gas in this process 2P0 1
K
is PV
0 0 . Then find the value of K: 2
2 P0
V
V0 2V0

7. A source of sound of frequency 1000 Hz is moving towards a stationary wall with velocity
30 m/s and observer is also moving behind the source with the same velocity. If the
2
frequency band width detected by the observer is K  10 . There find the value of k:
8. A vessel containing liquid of density  is accelerated
2
10 m / s in a gravity free
horizontally with acceleration
space. A small sphere of density 2  is released from rest
with respect to the vessel. Find the initial acceleration of the
sphere with respect to the vessel:

X L  50 2ohm
9. In an AC circuit capacitive reactance of capacitor is X C  50 2
X C  50 2ohm
and inductive reactance of inductor is X L  50 2ohm . If the source
voltage is 100 volts. Then find the current passing through the
source. 
100volts

10. Out of the three conducting identical plates, middle flat is given charge   3 C
3 C and other two plates are not having any charge. If switch S is
closed. Find the charge passing through switch S. (in C ):
 d 2d
S

Space for Rough work

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Chemistry PART – II

1. Dacron is a copolymer of ethylene glycol and

(A) Phthalic acid (B) Adipic acid
(C) Benzoic acid (D) Terephthalic acid

2. Select the correct statement/s to the following decay chain:

  
A   B   C  D
(A) A and D are isotopes (B) B, C, D are isobars
(C) A and B are isodiaphers (D) A and C are isotones

3. In isothermal expansion of a gas:

(A) Kinetic energy of gas molecules remain same
(B) Kinetic energy of gaseous molecule decreases
(C) Pressure of gas decrease
(D) Kinetic energy of gaseous molecule increases

4. 5.3% (w/v) Na2CO3 solution and 6.3%(w/v) H2C2O4.2H2O solution have same:
(A) Molality (B) Molarity
(C) Normality (D) Mole fraction

5. Given that
o o
ENi2
/Ni
 0.25V, ECu 2
/Cu
 0.34V
EoAg / Ag  0.80V, EoZn2  / Zn  0.76 V
Which of the following redox process will not take place in specified direction?
(A) Ni2   aq   Cu  s   Ni  s   Cu2   aq
(B) Cu  s   2Ag  aq  Cu2  aq   2Ag  s 

(C) Cu  s   2H  aq  Cu2  aq   H2  g

(D) Zn  s   2H  aq  Zn2  aq   H2  g 

Space for Rough work

6. 
H /
 A
OH OH
The product ‘A’ is
(A) (B)

O
O
(C) (D)

i CH MgBr H SO / 
HBO
7. CH3 CHO  3
 ii H O
  A  
 2 4
 B     C
2

(A) and (C) are related as:

(A) Identical (B) Position isomers
(C) Functional isomer (D) Optical isomers

8. Select molecules in which all -orbitals are in same plane

(A) C6H6 (B) O  C  C  C  O
(C) CH2  C  CH2 (D) CH2  CH  CH  CH2

9. Select correct statement (s)

(A) Bond length of NO   NO
(B) Bond order of NO  NO
(C) Bond energy of NO  NO 
(D) NO is paramagnetic but NO is diamagnetic

10. Which of the following properties is/are same in:

3 3 3
(I) CrCl6 3
(II) Cr H2O 6  (III) Cr NH3 6  (IV) Cr NH3 6 
(A) Magnetic moment (B) EAN of central metal
(C) Magnitude of splitting energy (D) Colour

Space for Rough work

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SECTION –C
Integer Answer Type

This section contains 10 questions. Each question, when worked out will result in one integer
from 0 to 9 (both inclusive).

1. For the complex compound Pt NH3 NO2  Py  ONO   , total number of geometrical
isomers will be:

2. Find the total number of acidic radicals which produce volatile product with dilute HCl:
SO24 ,I ,NO2 ,NO3 ,SO32 ,HCO3

3. Total number of stereoisomers of 1, 2-Dibromo-3-chlorocyclo propane.

4.
 i HNO conc.
2

 ii Cl /FeCl
 X,

2 3

The number of possible isomers of ‘X’ is?

5. Number of aromatic compounds among the following:

, graphite, diamond.
, N , , , , , , N N
N
H

6. Molality of a 5M solution of H2SO4 having density 1.115 gm/ml.

7. Total number of electron in phosphorus which have zero value of at least one quantum
number.

8. The pressure necessary to obtain 50% dissociation of PCl 5 at 400 K is numerically equal
to ……….times of Kp.

9. Number of S—S bonds in cyclic trimer of SO3

10. Number of unpaired electrons in compound [Mn(CN)6]3– is:

Space for Rough work

Mathematics PART – III

1. If x1, x2, x3, x4 are four positive real numbers such that
1 1 1 1
x1   4, x 2   1, x 3   4 and x 4   1 , then
x2 x3 x4 x1
(A) x1 = x3 (B) x2 = x4
(C) x1x2 = 1 (D) x3x4 = 1

2. If the normals at (xi, yi) i = 1, 2, 3, 4 to the rectangular hyperbola xy = 2 meet at the point
(3, 4), then
(A) x1 + x2 + x3 + x4 = 3 (B) y1 + y2 + y3 + y4 = 4
(C) x1x2x3x4 = –4 (D) y1y2y3y4 = 4

3. In a triangle ABC with sides a, b and c, a semicircle touching the sides AC and CB is
inscribed whose diameter lies on AB, then the radius of the semicircle is
a 2abc A B C
(A) (B) sin sin sin
2 s a  b 2 2 2
2 2abc A B C
(C) (D) cos cos cos
ab s a  b 2 2 2

4. An urn contains fair tickets with numbers 112, 121, 211, 222 and one ticket is drawn. Let
Ai (i = 1, 2, 3) be the event that the ith digit of the number of ticket drawn is 1 then
1
(A) P(A1) = P(A2) = P(A3) = (B) P(A1  A2) = P(A1) P(A2)
2
(C) A1, A 2, A3 are pair wise independent events (D) P(A1  A 2  A3)  P(A1) p(A2) P(A3)

Space for rough work

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z4
5. Suppose that the complex number z lies on the curve such that is purely
z  2i
imaginary. If the complex number z1 represents the mid-point of chord OA of this curve,
O being the origin, then z 1 necessarily satisfies
z 2 z1
(A) 1  ik , k  R – {0} (B)  ik , k  R – {0}
z1  i z1  2  i
z1  2 5
(C)  k , k  R – {0} (D) z1 
2z1  i 2


3
6. e

x

 cos x ln x  1  x 2 dx   2
, then  can be

(A) 1 (B) 2
(C) 3 (D) 4

3 2 max. f  t  ; 0  t  x ; 0  x  1
7. Let f(x) = x – x + x + 1 and g  x    then
3  x ; 1 x  2
(A) g(x) is discontinuous at x = 1 (B) g(x) is continuous at x = 1
(C) g(x) is differentiable at x = 1 (D) g(x) is non-differentiable at x = 1

8. Which of the following set of values of x satisfies the equation

2
2 sin 2
x – 3 sin x  1   22 – 2 sin 2
x  3 sin x  9
 
(A) x  n  ,nI (B) x  n  ,nI
6 3

(C) x = n, n  I (D) x  2n  , n  I
2

9. Consider all the strings of n-digits which can be formed using the numbers from the set
{0, 1, 2, 3} (may be starting with zero also). If En denotes the number of strings containing
even number of zeros and On denotes the strings containing odd number of zeros then
n 4n  2n
(A) En + On = 4 (B) En 
2
n n
4 2
(C) O n  (D) En = On
2

Space for rough work

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 dy 
10. 
The solution of   x 2 y 3  xy  1 is
 dx 

1 2 1 2
(A)  2  y 2  c  e  y /2 (B)  2  y 2  c  e  y /2
x x
2 1  2x 2
(C)  1  y 2  e y /2 (D)   y 2  c  e  y /2
x x

SECTION – C

(One Integer Value Correct Type)

This section contains 10 questions. Each question, when worked out will result in one integer
from 0 to 9 (both inclusive)

1. Three circles lie on a plane so that each of them externally touches the other two. Two of
them has radius 3, the third having radius unity. If A, B and C are the points of tangency
a b
of those circles and the area of the triangle ABC is then value of c – a – b is, where
c
a, b and c are positive integers _____
3 2 2
2. The number of polynomials of the form x + ax + bx + c which are divisible by x + 1
where a, b, c  {1, 2, 3, ….., 10} is 10k, then k is _____

3. OABC is a tetrahedron in which O is the origin and position vector of points A, B, C are
ˆi  2ˆj  3kˆ , 2iˆ  ˆj  kˆ and ˆi  3 ˆj  2kˆ respectively. An integral value of  for which

3
shortest distance between OA and BC is is _____
2

 1  2 1  2x 
4. If f  x   f   , where x is a real number x  0, x  1, then the value of f(2)
 1  x  x 1  x 
must be _____

Space for rough work

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5. If x5 – x3 + x = a, when x > 0, then the maximum value of 2a – x6 is equal to _____

6. If x4 + 3 cos(ax2 + bx + c) = 2(x2 – 2) has two solutions with a, b, c  (2, 5) then the

ac
maximum value of 2 is _____
b

100 2(99) 3(98) 100 f ' 101  k 

7. If f(x) = (x – 1) (x – 2) (x – 3) ….. (x – n) , let k  , then   97  is
f 101  50 
_____

(a2  b2  c 2 )
8. The minimum value of in any quadrilateral, where a, b, c and d are sides
d2
of quadrilateral, is k then 6k is equal to _____

9. The number of integral values of parameter , so that the point (, (1 + 2)–1) does not lie
outside the triangle formed by the lines L1 : 15y = x + 1, L2 : 78y = 118 – 23x and
L3 : y + 2 = 0, is equal to _____

1
10. Let f(x) = sin x – a sin 2x – sin 3x + 2ax, increases throughout the number line then
3
minimum prime value of a is equal to _____

Space for rough work

ANSWERS, HINTS & SOLUTIONS

CRT– I
(Paper-1)
ALL INDIA TEST SERIES

Q. No. PHYSICS CHEMISTRY MATHEMATICS

1. A, B D A, B, C, D

2. C, D A, B, C A, B, C

3. A, B, D A, C C, D

4. A, C B, C A, B, C, D

5. A, B A, C A, B

6. A, D B A, B, C, D

7. C, D B B, D

8. A, C A, D A, D

9. A, D B, D A, B, C

10. B, D A, B A, D

1. 2 3 0

2. 0 3 1

3. 2 4 3

4. 3 3 3
5. 4 7 1
6.
3 8 1
7. 2 9 4
8. 5 3 2
9. 0 0 4
10. 3 2 2

Physics PART – I

SECTION – A
1. Using concept of relative velocity we can
solve v 
vA

v B(d, )
(0, 0)

vB  vjˆ
R min

3. Use v 2  u2  2as
And impulse = change in momentum

2gh
4. Use v 

I
where,   1 
Mr 2

5. Use work energy theorem

6. Buoyant force FB = Vegeff

If lift accelerates
geff = g + a

 v  v0 
7. Use,  app   v  
 v  vs 
v = speed of sound

8. for isothermal process, PV = constant

For adiabatic process, PV = constant

9. Electric field at centre = 0 R

So, charge q will produce same electric
field as induced charge on sphere at centre d q
so A & D

10. Using conservation linear moment and conservation mechanical energy.

SECTION – C
1. For translation position of A will be 2.

2. formula for rod and are both are

k k
 sin   sin   ˆi   cos   cos   ˆj
d d
 
So, E  0 and b = 0

3
6. work done = P0 V0
2
P V  P1V1
U  2 2
 1
3
Q  P0 V0
2
k=3

7. f1 = f 30 m/s 30 m/s
C 330 O
fw  f  1000 Hz S
C  30 300
C  v0 360 330
f2  fw    1000  1200 Hz
C 330 300
Frequency Band width = f2  f1 = 200 Hz

8. Fnet  Fpseudo  FBuoyant  V  2  10  v  10  v  10

Fnet v  10
a   5 m/s
n v  2

100
9. IL   2 A
50 2
100
IC   2
50 2
Isource  IL  IC  0

Chemistry PART – II

SECTION – A
1. Dacron is a copolymer of ethylene glycol and terephthalic acid

A  A 4  A 4  A 4
2. ZA 
 B 
 C 
 D
Z 2 Z 1 Z
 - emission gives isodiapher.

3. From kinetic theory of gases

K. E.  T

WB  1000 WB  1000
MNa2CO3  MH2 C2O4 .2H2O 
mB  V mB  V
4.
5.3  1000 6.3  1000
  0.5   0.5
106  100 126  100
W  1000 N  1000
NNa2CO3  B NH2C2O4 .2H2O  B
EB  V EB  V
5.3  1000 6.3  1000
 1  1
53  100 63  100
n-factor =2 n-faction = 2

o
5. Eredox change will be negative for non-spontaneous reaction

Ni2  aq  Cu  s   Ni  s   Cu2  aq 

Eo  EoNi2  / Ni  ECu
o
2
/ Cu
 0.25  0.34  0.59V   ve, non  spon tan eous 

Cu  s   2H  aq  Cu2   aq  H2  g

Eo  EHo  /H  ECu
o
2
/Cu
 0  0.34  0.34V  ve, non  spon taneous 
2

6.

H Ring H
 
Exp
  
H O
2

OH OH OH OH
O

7. H
 i CH3MgBr H SO /  HBO
 CH3  CH  CH2  CH3  CH2  CH2  OH
CH3 CHO  
 H3C C CH3 
2 4
ii H2O B  C 
OH
B 
2
8. Compounds have all ‘c’ atom sp hybridised are coplanar.

9. B. O. of NO = 2.5, paramagnetic
NO+ = 3.0, diamagnetic

10. Cr 3  , d3 configuration shows same number paired e– and EAN but strength of all ligands are
different so does not slow same colour and splitting.

SECTION –C

1. [Mabcd] type square planar complex has 3 G. I.

2. NO2  HCl  HNO2  Cl

3HNO2  HNO3  2NO  H2 O

2NO  O2  2NO2 

SO23   HCl  H2O  2Cl  SO2 
HCO3  HCl  H2O  CO2  Cl

3. H Cl Cl Cl

H Cl H H H H H H Br Br H H

Br Br Br Br Br H H Br

meso Enatiomers

5. Aromatic compounds

, graphite
N , , , , , N N
N
A layer of graphite (graphene) contains infinite lattice of fused aromatic rings. All the valencies
are satisfied (except at the edges) and no bonds are needed between layers. (Planar layered
structure of fused aromatic rings).

6. 5 mol H2 SO 4  5  98  490 gm of H2 SO 4
5M
1000 ml of solution  1000  1.115  1115 gm solution
 wt. of solvent = 1115 – 490 = 625 gm
moles of solute 5
m   8 mol / kg
wt. of solvent  in kg  625  103

2
7. 15 P  1s ,2s2 ,2p6 ,3s2 3p3
For ‘s’  = 0, m = 0, e– = 6
-1
For ‘p’,  = 1, m = +1
0

-1 0  1 0 1

 2e  1e
Total e– = 6 + 2 + 1 = 9

8. 

PCl5 
 PCl3  Cl2
a 0 0
a ax a a  0.50

 2P Kp 0.5  0.5
Kp  2
 
1  P 0.75
P
 3
KP
 P = 3  KP

9. O O
S

O O

O O
S S
O O O

3
10. Mn  CN 
 6
x + 6(–1) = –3
 x = +3
Mn3    Ar  3d4 4S0

t 2g eg
As CN– is strong ligand hence due to splitting it will from low spin complex.

Mathematics PART – III

SECTION – A
1 x
1. 4  x1  2 1
x2 x2
1 x
1  x2  2 2
x3 x3
1 x
4  x3  2 3
x4 x4
1 x  1  1  1  1 4
1  x4   2 4  16   x1    x2    x3   x 4    2
x1 x1  x 2  x 3  x 4  x1

 2
2. Any point on xy = 2 is P  2t, 
 t 

 2 2
Normal at P is  y    t x  2t
t 
 


3.  ABC  BPC   APC C

1 1
 ar  br
2 2
2 A B
r P
ab
A B C s s  a s  s  b s s  c  s
cos cos cos    
2 2 2 bc ac ab abc

1 2
4. P(A1  A2) = , P(A1) =
4 4
1
P(A2) = . Thus P(A1  A2) = P(A1) P(A2)
2

z4 z4 2 2
5.   0 which is a circle x + y – 4x – 2y = 0
z  2i z  2i
xx1 + yy1 – 2(x + x1) – (y + y1) = x12  y12  4x1  2y1
2 2
It passes through (0, 0), so, the locus of (x1, y1) is x + y – 2x – y = 0.
So, z1 = x1 + iy1, lies on this circle for which the points (2, 0) and (0, 1) are extremities of a
diameter. Also (0, 0) and (2, 1) represent extremities of another diameter

 
6. I e

x
dx   cos x ln  x 

1  x 2 dx 
3
 e  e  
2
3 2
7. f(t) = t – t + t + 1
f(t) > 0
 max.{f(t): 0  t  x} = f(x)

8. Put 2
 2 sin 2
x  3 sin x 1t

n n n–2 n n–4
9. En = 3 + C2·3 + C4·3 + …..
On = nC1·3n–1 + nC3·3n–3 + …..
Adding and subtracting, we get En + On = (3 + 1)n = 4n, En – On = (3 – 1)2 = 2n
4n  2n
Adding and subtracting again, we get 2En = 4n + 2n  En 
2
n n
n n 4  2
2On = 4 – 2  On 
2

dx
10.  x 2 y 3  xy
dy
dx
 xy  x 2 y 3
dy
1 dx y
    y3
x 2 dy x
1
Put t
x
SECTION – C

3 x c3
1. cos  
4 1 1 1
3 3  M
 x  AB  A B p 7
4 2 x
7 3 3
p  4 2  3 2  7 and c 3M 
4 
c1 C c2
7 3 7
 CM  7   3 3
4 4

2. x2 + 1 = (x + i)(x – i)
b = 1, a = c
Number of ways of choosing a, b, c = 10 = 10  1
k=1
 
3. Unit vector  n̂  perpendicular to OA and CB
 
CB  OA  3  7  ˆi  4ˆj   5    kˆ
n̂    
CB  OA  3  7 2  16   5   2
ˆi ˆj kˆ
    7  3   4  2     2  5   
As CB  OA  1   3 1 = BA  nˆ 
1 2 3  3  7 2  16   5   2
 
Shortest distance = BA  n
1  1   1 2
4. Replacing x by , we obtain f   f 1    2x 
1  x   1 x   x x
1
Again, replacing x by 1 
x

5. We have a =

x x6  1 
2
x 1
 x 2  1
 x6 + 1 = a    2a
 x 
 2a – x6  1
2 2 2
6.  (x – 1) = –3{1 + cos(ax + bx + c)} solution exists when L.H.S. = R.H. S. = 0
 x = 1 and cos(ax2 + bx + c) = –1
When x = 1, x = –1
 a + b + c = 3, a – b + c = 
 b =  and a + c = 2

100
i101 i 
7. f  x     x  i
i 1
100
ln  f  x     i 101  i  ln  x  i 
i 1

1 
100 i 101  i
 f ' x  


f x i 1  x  i 

8. For any quadrilateral,

a + b + c > d.
Now, (a  b)2 + (b  c)2 + (c  a)2  0
 a2 + b2 + c2  (ab + bc + ca)
 3 (a2 + b2 + c2)  3 (ab + bc + ca)
 3 (a2 + b2 + c2)  a2 + b2 + c2 + 2 (ab + bc + ca)
 3 (a2 + b2 + c2)  (a + b + c)2
 3 (a2 + b2 + c2) > d2
(a2  b2  c 2 ) 1
  .
d2 3
 a2  b2  c 2  1
 The minimum value of   is
 d2  3

2 –1 1  1
9. As P(, (1 +  ) ) lie on y   2, 
1 x2  5
1  1 
 On solving y  with L1, we get 5,
2
x 1 P2  26 
P1
 1
 2, 5  ….. (1)
 
 1 
and with line L2, we get P2  5,  ….. (2)
 26  y+2=0
 Equation (1) and (2), we get 2    5 L1 = 0 L2 = 0

10. f(x) = cos x – 2a cos 2x – cos 3x + 2a  0  x  R

 2 sin2 x (cos x + a)  0
 a  – cos x  1

Paper 2

Time Allotted: 3 Hours Maximum Marks: 180

 Pl ea s e r ea d t h e i n s t r u c t i o n s c a r ef u l l y. Yo u a r e a l l o t t ed 5 m i n ut es
s p ec i f i c a ll y f o r t h i s p u r p o s e.

ALL INDIA TEST SERIES

Yo u a r e n o t a l l o wed t o l ea v e t h e E xa m i n at i o n Ha l l b ef o r e t h e en d o f
t h e t es t .

INSTRUCTIONS
A. General Instructions
1. Attempt ALL the questions. Answers have to be marked on the OMR sheets.
2. This question paper contains Three Parts.
3. Part-I is Physics, Part-II is Chemistry and Part-III is Mathematics.
4. Each part has only one section: Section-A.
5. Rough spaces are provided for rough work inside the question paper. No additional sheets will be
provided for rough work.
6. Blank Papers, clip boards, log tables, slide rule, calculator, cellular phones, pagers and electronic
devices, in any form, are not allowed.
B. Filling of OMR Sheet
1. Ensure matching of OMR sheet with the Question paper before you start marking your answers
on OMR sheet.
2. On the OMR sheet, darken the appropriate bubble with black pen for each character of your
Enrolment No. and write your Name, Test Centre and other details at the designated places.
3. OMR sheet contains alphabets, numerals & special characters for marking answers.
C. Marking Scheme For All Three Parts.
(i) Section-A (01 to 10) contains 10 multiple choice questions which have one correct answer.
Each question carries +3 marks for correct answer and – 1 mark for wrong answer.
Section-A (11 to 16) contains 3 paragraphs with each having 2 questions. Each question carries
+3 marks for correct answer and – 1 mark for wrong answer.
Section-A (17 – 20) contains 4 Matching Lists Type questions: Each question has four
statements in LIST I & 4 statements in LIST II. The codes for lists have choices (A), (B), (C), (D)
out of which only one is correct. Each question carries +3 marks for correct answer and
– 1 mark for wrong answer.

Name of the Candidate

Enrolment No.

Useful Data

PHYSICS
Acceleration due to gravity g = 10 m/s2

Planck constant h = 6.6 1034 J-s

Charge of electron e = 1.6  1019 C

Mass of electron me = 9.1  1031 kg

Permittivity of free space 0 = 8.85  1012 C2/N-m2

Density of water water = 103 kg/m3

Atmospheric pressure Pa = 105 N/m2

Gas constant R = 8.314 J K1 mol1

CHEMISTRY

Gas Constant R = 8.314 J K1 mol1

Atomic No: H=1, He = 2, Li=3, Be=4, B=5, C=6, N=7, O=8,

Physics PART – I

SECTION – A
(Only One Option Correct Type)
This section contains 10 multiple choice questions. Each question has four choices (A), (B), (C)
and (D) out of which ONLY ONE option is correct.

1. Which of the following is the most accurate instrument for measuring length?
(A) vernier calipers having 20 divisions on the sliding scale which coincide with 19
divisions on the main millimeter scale
(B) a screw gauge having pitch 1 mm and 50 divisions on the circular scale
(C) a vernier scale of least count 0.01 mm
(D) a screw gauge of least count 0.001 mm

2. A ball is thrown from the ground to clear a wall 3 m high at a distance of 6 m and falls 18
m away from the wall, the angle of projection of ball is
3  2
(A) tan1   (B) tan1  
2
  3
 1 3
(C) tan1   (D) tan1  
2
  4

3. A block of mass m moving with velocity v strikes elastically

another identical mass connected to a spring as shown in
k
figure. The maximum compression produced in the spring m v m
(Assume surfaces to be smooth) is

mv 2 mv 2
(A) (B)
2k k
mv 2
(C) (D) none of these
3k

Space for Rough work

4. A cube of side l and mass M is placed on rough horizontal surface and the friction is
sufficient so that it will not move, if a constant force F = Mg is applied horizontally l/4
above the surface. Then the torque due to normal force about center of the cube is equal
to
Mgl Mgl
(A) (B)
2 4
Mgl
(C) (D) zero
8

5. The work done to take a particle of mass m from surface of the earth to a height equal to
2R is (R is radius of earth)
mgR
(A) 2 mgR (B)
2
2mgR
(C) 3 mgR (D)
3

6. A spherical object of mass 1 kg and radius 1 m is falling vertically downward inside a

viscous liquid in a gravity free space. At a certain instant the velocity of the sphere is 2
1
m/s. If the coefficient of viscosity of the liquid is N-s/m2, then velocity of ball will
18
become 0.5 m/s after a time
(A) ln 4 s (B) 2 ln 4 s
(C) 3 ln 4 s (D) 2 ln 2 s

7. Two straight long conductors AOB and COD are perpendicular to each other and carry
currents i1 and i2. The magnitude of the magnetic induction at a point P at a distance a
from the point O in a direction perpendicular to the plane ACBD is
 
(A) 0  i1  i2  (B) 0  i1  i2 
2a 2a
 1/2  i1i2
2a

(C) 0 i12  i22  (D) 0
2a  i1  i2 

Space for Rough work

8. A non-conducting rod AB of length l has a positive linear + B

+
charge density . The rod is rotated about point A with an ++
angular velocity  in the plane of paper. The magnetic +
+
moment of the rod is + l
+
A


3 3
l 2l
(A) (B)
2 3
3l3 l3
(C) (D)
2 6

9. A bird is flying over a swimming pool at a height of 2m from the water surface. If the bottom is
perfectly plane reflecting surface and depth of swimming pool is 1 m, then the distance of final
image of bird from the bird itself is (  w  4 / 3 )
11 23
(A) m (B) m
3 3
11 11
(C) m (D) m
4 2

10. The mean lives of a radioactive sample are 30 years and 60 years for -emission and -
emission respectively. If the sample decays both by -emission and -emission
simultaneously, the time after which, only one-fourth of the sample remain is
(A) 10 years (B) 20 years
(C) 40 years (D) 45 years

Space for Rough work

Comprehension type (Only One Option Correct)

This section contains 3 paragraphs, each describing theory, experiments, data etc. Six questions
relate to the three paragraphs with two questions on each paragraph. Each question has only
one correct answer among the four given options (A), (B), (C) and (D).

Paragraph for Questions 11 & 12

 
The position vector of a body of mass m = 4 kg is given as r  ˆi(t 2  4t)  ˆj( t 3 ) , where r is in
metres and t in seconds.

11. The magnitude of torque with respect to origin, acting on the particle at t = 1s will be
(A) zero (B) 48 Nm
(C) 64 Nm (D) 80 Nm

12. Work done by force acting on it in first two seconds will be

(A) 32 J (B) 256 J
(C) 320 J (D) 288 J

Paragraph for Questions 13 & 14

A uniform ring of mass m and radius R can rotate freely about an axis E0
passing through centre C and perpendicular to plane of paper. Half of – + + y
ring is positively charge and other half is negatively charge. Uniform –– +
–
electric field E0 is switched on along –ve x-axis (Axis are shown in – +
– + x
figure) –
– C +
– +
[magnitude of charge density ] –– +
– + +

13. The dipole moment of ring is

(A) 2 R2 (B) 4 R2
2 2
(C) 2 R (D) 4  R

14. If ring is slightly disturb from given position, find the angular speed of ring when it rotate
by /2.
E0 E0
(A) 2 (B)
m m
8  E0
(C) (D) none
m

Space for Rough work

Paragraph for Questions 15 & 16

An ammeter and a voltmeter are connected in series to a battery with emf E = 6 volt and
negligible resistance. When a resistance R = 3 is connected in parallel to voltmeter, reading of
ammeter increases three times while that of voltmeter reduces to one third.

15. Reading of voltmeter after the connection of resistance is

(A) 1 Volt (B) 3 Volt
(C) 9/2 Volt (D) 3/2 Volt

16. Reading of ammeter before the connection of the resistance is

3 6
(A) A (B) A
4 7
3
(C) A (D) 1 A
16
(Matching List Type)

This section contains 4 multiple choice questions. Each question has matching lists. The
codes for the lists have choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

17. Line joining the centre of path and O


cylinder is having angular velocity  R
and angular acceleration  at the  y
given instant: Match the statements 
from List I with those in List II and x
select the correct answer using the C
code given below the lists.
List-I List-II
Rotational component of kinetic R  r   ˆi  R  r  2 ˆj
(P) 1.
energy of cylinder
Kinetic energy of cylinder 2R R  r  ˆ
(Q) 2. j
r
Acceleration of centre of cylinder 1 2
(R) 3. m  R  r  2
4
Acceleration of point of contact 3 2
(S) 4. m  R  r  2
4
Codes:
P Q R S
(A) 3 4 1 2
(B) 2 4 1 3
(C) 1 2 3 4
(D) 4 1 2 3

Space for Rough work

18. A disc of mass m and radius R is rolling without slipping on y

a horizontal fixed rough surface with tramlational velocity v
 
as shown in the figure. Here Lo ,Lp and k o ,k p represents
C
the angular momentum and kinetic energy about point O r  R/2 v
and P respectively: Match the physical quantities from List I P x
with those in List II and select the correct answer using the O
code given below the lists.
List-I List-II

Lo 3
(P) 1. mVR  R 
4

Lp 3
(Q) 2. mVR2
4
Ko 3
(R) 3.
2
mVR kˆ  
Kp 3
(S) 4. mv 2
8
Codes:
P Q R S
(A) 4 2 1 3
(B) 2 4 1 3
(C) 3 1 2 4
(D) 4 1 2 3

19. A block of mass 2kg is kept over a block of   0.6

mass 8 kg as shown in the figure. A force
2kg F  14 N
F  14N acts on the upper block. Match the
physical quantities from List I with those in List   0.1 8kg
II and select the correct answer using the code
given below the lists. All values are in S.I.
units:
List-I List-II
(P) Acceleration of block 2 kg 1. 10
(Q) Acceleration of block 8 kg 2. 1
(R) Friction action between the blocks 3. 12
Friction acting between lower block 1
(S) and ground 4.
4
Codes:
P Q R S
(A) 4 2 1 3
(B) 2 4 1 3
(C) 1 2 3 4
(D) 2 4 3 1

Space for Rough work

20. A thin biconvex lens of small aperture

and having focal length 30 cm is cut
tow ways as shown in the figure. The
focal length of different combination
given in List-I with their value in List-II.
Match the statements from List I with
those in List II and select the correct
answer using the code given below
the lists.
List-I List-II
Infinite
(P) 1.

30 cm
(Q) 2.

60 cm
(R) 3.

15 cm
(S) 4.

Codes:
P Q R S
(A) 4 2 1 3
(B) 2 3 4 1
(C) 1 2 3 4
(D) 2 4 3 1

Space for Rough work

Chemistry PART – II

SECTION – A
(Only One Option Correct Type)
This section contains 10 multiple choice questions. Each question has four choices (A), (B), (C)
and (D) out of which ONLY ONE option is correct.

1. Which of the following is not matched correctly?

(A) CH3
HCl/ 
H3C C CH2 OH   P; Reaction through SN1 and rearrangement.

CH3
(B) SOCl2
CH3  CH2  OH   P; Reaction through SNi.
(C) HI
CH3  CH2  OH   P;Reaction through SN 2 reaction.
Reflux
(D) C 2H 5
H3C C OH HI

Reflux
 P;
HC CH3
CH3

2. O
COOC2H5
i H
3 O

 ii 


The major product is

(A) O (B) O
O

CH2

Space for Rough work

3. Br 1eq 
KCN  H O / H
2
CH3  COOH 
Re d P
  A     B  
2
  C    D  , Pr oduct D is

(A) COOH (B) O

H2 C
H3C C COOH
COOH
(C) CH3COOH (D) CH4

4.
Et 2NH/DMF  i Re duction
dil. HBF4
F NO2  
 X 
ii  NaNO /HCl
 Y  
Z
2

What is ‘Z’?
(A) (B)
F NH2 Et 2N NH2

CH3
CH3
5. The monomer of the polymer: CH2 C CH2 C is
CH3
CH3

(A) CH3 (B) (H3C)2C C(CH3) 2

H2C C
CH3

(C) H3CHC CHCH3 (D) H3C CH CH2

6. State of hybridization of sulphur, carbon-1 and carbon-2 in F3SCCF3 respectively are”

1 2
(A) sp3, sp3, sp3 (B) sp3, sp2, sp3
(C) sp3d, sp, sp3 (D) sp3, sp, sp3

Space for Rough work

7. A XF4 type molecule have  = 0, which additional information is required to conform the
geometry of molecule
(A) All the X—F bond length are identical
(B) Molecule has same F—X—F bond angle between any two adjacent F
(C) Number of lone pair of electron on central atom ≤ 2
(D) Planarity of molecule

8. Solid AB haz ZnS type structure. If the radius of A+ ion is 22.5 pm then radius of B– ion
will be:
(A) 100 pm (B) 200 pm
(C) 150 pm (D) 95 pm

9. 2Zn  O2  2ZnO; Go  616 J

2Zn  S  2ZnS; Go  293 J

2S  2O 2  2SO2  g  ; Go  408 J

o
G for the following reaction:
2ZnS  3O2  2ZnO  2SO2 , would be:
(A) –731 J (B) –1317 J
(C) +731 J (D) +1317 J

10. A gas shows heating effect on sudden expansion. It shows that:

(A) the gas is nobel
(B) the gas is ideal
(C) the inversion temperature of the gas is very low
(D) its Vanders Waal’s constant ‘b’ is high

Space for Rough work

Comprehension Type (Only One Option Correct)

This section contains 3 paragraphs each describing theory, experiment, data etc. Six questions
relate to four paragraphs with two questions on each paragraph. Each question of a paragraph
has only one correct answer among the four choices (A), (B), (C) and (D).

Paragraph for Question Nos. 11 to 12

Hybridisation is a process of mixing of atomic orbitals to give mixed or hybrid orbitals. Hybrid
orbitals have equal energy and their hypothetical shape may be given as
Head Tail
Five d-orbitals are non-degenerate, they are divided into two different set of orbitals
eg  dx2  y2 ,dz2
t2g  dxy ,dy 2 ,dzx
Hybridisation involving d-orbitals are dsp2, sp3d, dsp3, sp3d2, d2sp3, sp3d3
3 2
11. Which of the following orbitals are involved in sp d hybridisation?
(A) dxy, dyz (B) dx2  y2 ,dxy
(C) dx2  y2 , dz2 (D) dz2 , dxy

12. Which of the following d – orbitals are involved in sp3d3 hybridisation?

(A) dx2  y2 ,dz2 ,dxy (B) dxy ,dyz ,dzx
(C) dx2  y2 ,dxy ,dxz (D) dz2 ,dy2 ,dzx

Space for Rough work

Paragraph for Question Nos. 13 to 14

If a cell has cell potential ‘E’ and standard cell potential ‘Eo’, then free energy change of cell
process may be calculated as,
o o
G = –W = –nFE and G = –W max = –nFE
Where ‘n’ is the number of electrons involved in overall all process . According to Gibs-Helmholtz
equation:
G =  H – TS
 d G 
G  H  T  
 dT P
 dE 
Temperature coefficient of cell ‘’ will be equal to  
 dT P

o
13. G for the Daniell cell
Zn(s)|ZnSO4||CuSO4|Cu(s)
EoZn2  / Zn  0.76V; EoCu2  / Cu  0.34 V
Will be:
(A) –312.3 KJ (B) –212.3 KJ
(C) –123.2 KJ (D) –323.1 KJ

14. The temperature coefficient of a cell whose cell reaction is

Pb  s   HgCl2  aq   PbCl2  aq   Hg   
 dE  4 1
 dT   1.5  10 vK at 298 K
 P
The change in entropy in JK–1 mol–1 for the given cell reaction will be
(A) 14.475 (B) 57.9
(C) 28.95 (D) 86.82

Space for Rough work

FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942
website: www.fiitjee.com
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15
AITS-CRT-I-(Paper-2)-PCM-JEE(Advanced)/15

Paragraph for Question Nos. 15 to 16

Methyl red is commonly used as indicator for acid base titrations. It is prepared by treating
NaNO2/HCl with anthranilic acid and the resulting solution is mixed with N, N-dimethyl aniline and
shaked well the solution for some minute to get ‘Methyl Red”.

Given
COOH
NH2

(Anthranilic Acid)

15. Which is most likely to be “Methyl Red”?

(A) O (B) COOH
C OH
N N NMe2
NH NMe2

(C) COOH Me2N (D)

HOOC N N NMe 2

N N

16. Which method can be prepare o-amino benzoic acid, the main raw material?
(A) NH2

CH COCl / Pyridine
Conc. H2SO 4 KMnO4 CH Cl / AlCl
dil. H2SO 4 / 

3

   
3 3
    

(B) NH2

3 CH  Cl / AlCl
3 4 KMnO
   

HNO / H
Sn / HCl

3
  

(D) NH2

CH COCl / Pyridine CH Cl / AlCl KMnO /  dil. H SO


3

 
3 3
 
4
 
2 4


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(Matching List Type)

This section contains 4 multiple choice questions. Each question has matching lists. The
codes for the lists have choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

17. Match the list- I with list – Ii

List– I List – II
(P) Calomel (1) Reference
(Q) Glass (2) Redox
(R) Hydrogen (3) Membrane
(S) Quinohydrone (4) Gas
Codes:
P Q R S
(A) 1 3 4 2
(B) 3 1 2 4
(C) 1 2 4 3
(D) 1 4 2 3

18. Match List – I with List – II.

List – I List – II
Transformation Name
O
(P) O (1) Hofmann
H 3C C Cl
  C CH3
Isocyanide
AlCl 3

(Q) O O (2) Wittin Reaction

MCPBA
O
 

(R) O (3) Bayer-Villiger

R
oxidation
C  Ph3P  CH2  C CH2
R R
R
(S) Ph  NH3 
CHCl3 / KOH
 Ph  NC (4) Friedel Craft’s
Acylation
Codes:
P Q R S
(A) 4 3 2 1
(B) 3 1 4 2
(C) 1 2 3 4
(D) 1 3 2 4

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19. Match the list- I with list – II

List– I List – II
(P) H3C C CH (1) Positive test with Fehling’s solution
(Q) HCOOH (2) Positive test with Tollen’s reagent
(R) NH2 (3) Decolourise Br2 water
(S) (4) Isocyanide test
H2N CHO

Codes:
P Q R S
(A) 2,3 1,2 4 2,4
(B) 1, 2, 3 1,2 2 4
(C) 2 1, 2 1, 2, 3, 4 4
(D) 1 4 2 3

20. Match the list- I with list – II

List– I List – II
(P) COOH (1) O, P – directing compound
CH COCl
3

AlCl

3

(Q) Cl (2) Activated compound

H3C CH3
KNH
 2
liq NH

3

(R) NMe 2 (3) No reaction

(S) NO 2 (4) Deactivated compound

Codes:
P Q R S
(A) 2 1, 2 3, 4 4
(B) 4 1, 2, 3 1, 2 4
(C) 1, 2 3, 4 1, 2 2, 3
(D) 1 2, 3 2 1, 4

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Mathematics PART – III

SECTION – A
(Only One Option Correct Type)
This section contains 10 multiple choice questions. Each question has four choices (A), (B), (C)
and (D) out of which ONLY ONE option is correct.

1. tan1  sin x   sin 1  tan x  holds true for

 
(A) x  R (B) 2n   x  2n  ; (n  z)
2 2
(C) x  n, (n  z) (D) x  {0, z+}

2. Let P(x) be a polynomial with degree 2009 and leading co-efficient unity such that
P(0) = 2008, P(1) = 2007, P(2) = 2006, ….. P(2008) = 0 and the value of P(2009) =
 n   a where n and a are natural number than value of (n + a) is
(A) 2008 (B) 2009
(C) 2010 (D) 2011

3 2 1 2 3
3. Let f(x) = x + x + 100x + 7 sin x, then equation    0 has
y  f 1 y  f  2  y  f  3 
(A) no real root (B) one real root
(C) two real roots (D) more than two real roots

n n
n r Cr
4. Let 1  x    Cr xr and un    1 at x = 2. Then the sum to infinity of
r 0 r 0 xr
u1 + u2 + u3 + ….. is
1
(A) 0 (B)
2
(C) 1 (D) 2

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5. Consider the system of equations ax + by = 0 and cx + dy = 0 where a, b, c, d  {1, 2}.

The probability that the system of equations has a unique solution is
3 5
(A) (B)
8 16
9 5
(C) (D)
16 8

x2 y2
6. If circumcentre of an equilateral triangle inscribed in   1 with vertices having
a2 b2
eccentric angles , ,  respectively is (x1, y1) then  cos  cos    sin  sin  is
9x12 9y12 3 x12 y12 5
(A) 2
 2
 (B) 2
 2

2a 2b 2 2a
2b 2
x12 y12 5 x12
y2 1
(C)   (D) 2  12 
9a2 9b2 9 a b 2

C 1
7. In a ABC, CD is the bisector of the angle C. If cos has the value and CD = 6, then
2 3
ab
is equal to
a b
(A) 3 (B) 6
(C) 9 (D) 18

dx dx
8. Let S  x    ex  8e x  4e3x , R  x    e3x  8ex  4e x and M(x) = S(x) – 2R(x).

1
If M  x   tan1  f  x    c where c is an arbitrary constant then f  loge 2  is equal to
2
1 3
(A) (B)
2 2
5 7
(C) (D)
2 2

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x
9. f(x) is a differentiable function satisfying the relation f  x   x 2   e  t f  x  t  dt , then
0
9
 f k  is equal to
k 1
(A) 960 (B) 1060
(C) 1224 (D) 1260

10. The solution of differential equation x 2(x dy + y dx) = (xy – 1)2 dx is (where c is an
arbitrary constant)
(A) xy – 1 = cx (B) xy – 1 = cx2
1 1
(C)  c (D) none of these
xy  1 x

Comprehension type (Only One Option Correct)

Paragraph for Question Nos. 11 to 12

Read the following write up carefully and answer the following questions:
Let each of the circles
S1 = x2 + y2 + 4y – 1 = 0
S2 = x2 + y2 + 6x + y + 8 = 0
2 2
S3 = x + y – 4x – 4y – 37 = 0
touches the other two. Let P1, P2, P3 be the point of contact of S1 and S2, S2 and S3, S3 and S1
respectively. Let T be the point of concurrence of the tangents at P1, P2, P3 to the circles. C1, C2,
C3 are the centres of S1, S2, S3 respectively

11. P2 and P3 are reflections of each other in the line

(A) y = x (B) y = x + 1
(C) 2x – y + 3 = 0 (D) 2x + y + 9 = 0

12. The area of the quadrilateral TP2C3P3 is

(A) 11 sq. units (B) 25 sq. units
(C) 15 sq. units (D) 9 sq. units

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Paragraph for Question Nos. 13 to 14

Read the following write up carefully and answer the following questions:
If f(x) is a differentiable function wherever it is continuous and f(c1) = f(c2) = 0,
f(c1)f(c2) < 0, f(c1) = 5, f(c2) = 0 and (c1 < c2)

13. If f(x) is continuous in [c1, c2] and f(c1) – f(c2) > 0, then minimum number of roots of f(x)
= 0 in [c1 – 1, c2 + 1] is
(A) 2 (B) 3
(C) 4 (D) 5

14. If f(x) is continuous in [c1, c2] and f(c1) – f(c2) > 0, then minimum number of roots of f(x)
= 0 in [c1 – 1, c2 + 1] is
(A) 2 (B) 3
(C) 4 (D) 5

Paragraph for Question Nos. 15 to 16

Read the following write up carefully and answer the following questions:
 
Define a function : N  N as follows:  1  1,  pn  pn1  p  1 if p is prime and n  N and
(mn) = (m) (n), if m and n are relatively prime natural numbers then

15. (8n + 4) where n  N is equal to

(A) 2(4n + 2) (B) (2n + 1)
(C) 2(2n + 1) (D) 4(2n + 1)

16. The number of natural numbers ‘n’ such that (n) is odd is
(A) 1 (B) 2
(C) 3 (D) 4

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(Match List Type)

This section contains 4 multiple choice questions. Each question has matching lists. The codes
for the lists have choices (A), (B), (C) and (D) out of which ONLY ONE is correct

17. Match the following List–I with List–II

List – I List – II
2 2
(P) Let f(x) = x + xg(1) + g(2) and g(x) = x + xf(2) + f(3), then
1. 2
f(1) – f(2) is equal to
(Q) If f(x – y), f(x)·f(y) and f(x + y) are in A.P., for all x, y and
2. 1
f(0)  0, then f(2) + f(–2) is equal to
(R) If f(x) = x3 + x2·f(1) + xf(2) + f(3) for all x, then f(0) + f(3) + 3. 4
1 is equal to
n
(S) Let f(x) = x , n being a positive integer, then value of n for
which the equality f(a + b) = f(a) + f(b) is valid for all a, b > 4. 0
0 is
Codes:
P Q R S
(A) 4 3 2 1
(B) 3 4 2 1
(C) 3 4 2 1
(D) 1 2 3 4

x
et
18. A function F is defined by F  x    dt  x > 0. Now express the functions in List–I in
1
t
terms of F, then match the following List–I with List–II
List – I List – II
x t
(P)
e ex
 t  2 dt 1. F x 
x
e
1

x
e3t 1
 1
(Q)  t dt 2. xe x  e  F  
1 x
x
et –2
(R)  t 2 dt 3. e [F(x + 2) – F(3)]
1

x 1
(S)  e t dt 4. F(3x) – F(3)
1

Codes:
P Q R S
(A) 4 3 2 1
(B) 3 4 1 2
(C) 3 4 2 1
(D) 1 2 3 4

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19. Match the following List–I with List–II

List – I List – II
(P) An urn contains five balls, two balls are drawn and are 31
found to be white. If probability of all the balls in urn are 1. k
white is k, then 91

(Q) Out of 15 consecutive integers three are selected at 3

random, then the probability of the sum is divisible by 3 2. k
is k, then 16

(R) If 3 cards are placed at random and independently in 4 7

boxes lying in a straight line. Then the probability of the 3. k
cards going into 3 adjacent boxes is k, then 8

(S) A box contains 4 balls which are either red or black, 2

balls are drawn and found to be red if these are 1
4. k
replaced, then the probability that next draw will result in 2
a red ball is k, then
Codes:
P Q R S
(A) 4 1 3 2
(B) 1 2 3 4
(C) 3 4 2 1
(D) 4 1 2 3

20. Match the following List–I with List–II

List – I List – II
   10 2 
(P) sin   
  r   r    is equal to 1. 0
 900  r 1 
(Q) if root of t2 + t + 1 = 0 be , , then 4 + 4 + –1–1 is equal to 2. 4
4
 1  cos   isin  
(R) if    cosn  isinn , then n is equal to 3. i
 sin   i 1  cos   

 
(S) if zr  cos  isin r (where r  1, 2, 3, .....), then value of
3r
3 4. 1
z 1z 2z3 …..  is equal to
Codes:
P Q R S
(A) 4 1 3 2
(B) 1 4 2 3
(C) 2 3 4 1
(D) 4 1 2 3

Space for rough work

ANSWERS, HINTS & SOLUTIONS

CRT– I
(Paper-2)
ALL INDIA TEST SERIES

Q. No. PHYSICS CHEMISTRY MATHEMATICS

1. D A C

2. B A C

3. B C C

4. B C B

5. D A D

6. C D A

7. C D C

8. D A B

9. D A A

10. C C C

11. D C A

12. B B C

13. B B C

14. C C A

15. D B C

16. C A B

17. A A C

18. C A B

19. D A D

20. B B D

Physics PART – I

SECTION – A

2. 6 = V cos  t ...(1)
1
3 = V sin  t  gt 2 ...(2)
2
V 2 sin2
24 = ...(3)
g
Solving the equation we will get the result

1 2 1
3. kx  mv 2
2 2
mv 2
x
k

4. F = mg use concept of toppling

5. Use concept of conservation of energy Ei = Ef

mg  F v
6. a
m
v  u  at

0 2i
7. Use B  
4 r
 +
8. M  iA B
+
dx 2
A  x2 ,i  ,t +
t 
+ dx
Place the value in M = iA
x
A


h
9. x  distance of final image of bird from x

bird itself
h

 ˆ 2
11. r  i (t  4t)  ˆj( t 3 )

Differentiation twice to get a  ˆi(2)  6t ˆj
  
  r F

12.   F  dr

 x x-volt
15. (6  x) i0RA and  6    (3i0 )R A A V
 3
i0
6x 1
 
x 3
6
3 6V
9
 x V
2
3
 after connection = V
2

 3R v  x
16. i0  RV = x and (3i0 )  
 3  Rv  3
3
 i0  A
16

17. (P) Vcm = (R  r)

1 2
 rotational E  m  R  r  2
4
1 2 1 3
(Q) total kinetic energy = mv cm  Icm 2  m(R  r)2 2
2 2 4
 ˆ 2ˆ
(R) acm  (R  r) i  (R  r) j
  
18. Lany point  Lcm, that point  L about CM
3
(P) 
2
mvr kˆ  
(Q)  (1)
1 1
K  mv 2cm  Icm 2
2 2
3
 (R)  mv 2
4

19. Slipping between 2 and 8

14  12
(P) a2 kg  1
2
12  10 1
(Q) a8 kg  
8 4
(R) 0.6  2  10 = 12

20. Basic combination of lenses.

Chemistry PART – II

SECTION – A

1. C2H5 H2C CH3 H3C CH2

C2H5
H3C C OH H3C C  H3C C I I C CH3
HI I
  
HC CH3
2H O2 CH CH3 CH CH
CH3 H3C CH3 H3C CH3
CH3
3o Racemic mixture

2. O O O O O
C O C2H5 C OH
H3O  



  CO 2

  ketoacid

3. Br2 KCN 
COOH
H3C COOH   H2C COOH   H2C H2O / H
COOH 
Re d P  H2C
Br CN COOH



 CO
 CH3COOH  CO2 

2

4.
Et2NH / DMF /  Reduction
F NO 2   Et 2N NO 2   Et 2N NH2
HF

NaNO /HCl + dil. HBF


2
 Et 2N N2 Cl 
4
Et 2N F


5. CH3 CH3
H2C C opens as as operating unit
CH2 C
CH3
CH3

6. F
F F
S
C1
C2
F F
F

7. XF4-type molecule have zero dipole moment when it is either tetrahedral or square planar.

8. rA  22.5
 0.225   0.225
rB rB
 rB  100 pm

9. Go  2GoZnO  2 GSO

o
2
 2GoZnS
= (–616–408)–(–293) = –731 J

10. Expansion of real gas above its inversion temperature shows heating effect.

11. sp3d2  it involves dx 2  y 2 and dz2 orbitals in hybridisation.

3 3
12. sp d  It involves dxy, dyz, dzx d-orbtials in hybridisation.

13. Go = –nFEo

= – 2  96500  1.10 = – 212.3 KJ

 dE  S
14.   
 dT P nF
S
 1.5  10–4 =  S = 28.95 JK–1 mol–1
2  96500

15. COOH COOH

COOH
NH2 N2 Cl NMe2
NaNO HCl

2

 
 N N NMe2

16. NHAc NHAc NHAc

NH2 CH3
3 CH COCl 2H SO
4 MeCl/ AlCl KMnO
4

   
3
  

SO3H SO3H
NHAc NH2
COOH COOH

H


SO 3H

17. Fact.

19. Read the text.

20. –COOH, –NO2 groups are deactivating groups and –CH3, –NMe2 are activating group. Friedel

Craft’s acylation not take place with the deactivated benzene compound NH2 is nucleophile.

Mathematics PART – III

SECTION – A

1. At x = 0, , 2
LHS = RHS
Now let f(x) = tan–1(sin x) and g(x) = sin–1(tan x)
2
f 'x 
cos x
and g'  x   sec x  cos x 
cos x
 f ' x 
2
1  sin x 2
1  tan x 2
1  tan x 1  sin2 x
Hence, f(x)  g(x)  x  domain. So f(x) and g(x) are equal only at x = n
g(x)

2 3
f(x) 4
 3
4 2    5
–
4 4 4
  
 –
– 2 2
4

2. P(x) – 2008 + x = x(x – 1)(x – 2)(x – 3) ….. (x – 2008)

Put x = 2009

3.  f(x) = 3x2 + 2x + 100 + 7 cos x

= 3x2 + 2x + 93 + 7(1 + cos x)
2
 1   278
= 3  x      7 1  cos x   0
 3   3
 f(x) is increasing function  f(1) < f(2) < f(3)
(Let a < b < c)
Now g(y) = (y – b)(y – c) + 2(y – c)(y – a) + 3(y – a)(y – b) = 0
 g(a) > 0, g(b) < 0 and g(c) > 0

C0 C C C n Cn n!
4.   1  2  3  .....   1 
x x 1 x  2 x  3 x  n x  x  1 x  2  .....  x  n 
C0 C1 C 2 C3 1 1 1
 At x = 2, un      .....   
2 3 4 5  
n  1 n  2  n  1 n  2
1 1 
Hence, sn = u1 + u2 + u3 + ….. + un =  
 2 n  2 

5. (1) ad = 1, bc = 4 (2) ad = 1, bc = 2
(3) ad = 2, bc = 1 (4) ad = 2, bc = 4
(5) ad = 4, bc = 1 (6) ad = 4, bc = 2

  a cos   b sin  
6. Centroid = circumcentre =  x1, y1    , 
 3 3 
 9x12 9y12 
 2  2  3   2   cos  cos    sin  sin  
 a b 

7.  = 1 + 2 C
1 C C
 = ab sin c = ab sin cos
2 2 2
1 C 1 C
1 + 2 = 6b sin  6a sin b a
2 2 2 2
 = 1 + 2
1 1 1 ab 1 2
   9
a b 9 ab D B
A

8. M x   

e x e2x  2  dx
4x 2x
e  8e 4
x
Put e = t

x x x
f  x   x 2   e t f  x  t  dt = x 2   e   f  t  dt  x 2  e x  e t f  t  dt
 x t
9.
0 0 0
x
 ex f(x) = x2e x + t
 e f  t  dt
0
Differentiating both sides w.r.t. x
ex f(x) + e x f1(x) = 2xe x + x2e x + e x f(x)
 f1(x) = 2x + x2
x3
 f  x   x2  c
3

10. Given differential equation can be written as

xdy  ydx dx
 2
(xy  1)2 x
Integrating both sides
1 1 1 1
   c  c
(xy  1) x xy  1 x

11.-12. Shown in figure

P1(–2, –1) C3(2, 2)
P2(–4, –1) S2 S1
P3(–1, –4)  1 P1
C 2  3,   C1(0, –2)
 2 P P3
2

T(–3, –3)

13. f(c1) > f(c2)

f(c1) > 0 and f(c2) < 0
 at x = c1, f has local minimum and at x = c2, f has local maximum

14. See above solution

2
15. (8n + 4) = (4(2n + 1)) = (4) (2n + 1) = (2 ) (2n + 1) = 2(2n + 1)

16. (pn) is odd  pn – 1(p – 1) is odd

 p is prime, the only value p can take is p = 2

 (2n) is odd

17. (P) f(x) = 2x + g(1), f(x) = 2

g(x) = 2x + f(2), g(x) = 2
at x = 1, f(1) = 2 + g(1) and g(1) = 2 + f(2)
 f(1) = 4 + f(2)
(Q)  2f(x) f(y) = f(x – y) + f(x + y) ….. (1)
Replacing x by y and y by x, then
2f(y) f(x) = f(y – x) + f(y + x) ….. (2)
From equation (1) and (2), we get f(x – y) = f(y – x)
Put y = 0, then differentiate
2
(R) f(x) = 3x + 2xf(1) + f(2)
 f(1) = 3 + 2f(1) + f(2)
 f(1) + f(2) = –3
 and f(x) = 6x + 2f(1)
 f(2) = 12 + 2f(1)
 –2f(1) + f(2) = 12
(S) n(a + b)n – 1 = nan – 1 + nbn – 1

18. (P) Put t + 2 = z

(Q) Put 3t = z
(R) Integrate by part
1
(S) Put  z
t
1
1
A  4 1
19. (P) P  4 
 1 1 1 3 1 6 1 
 B        1
2
4 10 4 10 4 10 4
3  5 C3  53 31
(Q) Required probability = 15

C3 91
3!  2  3
(R) Required probability = 
4 44 16
3
A 1 1 1 1 1 5
(S) P  A    P  Ei   P     1     
i 1 E
 i 3 3 2 3 6 9
1 1 1 1 1
1  
 E1  3 3  E2  3 2 3  E3  3 6 1
Thus, P     , P    and P    
A 5 5  A  5 10  A  5 10
9 9 9
10 10
20. (P)  r   r  2    r 2  r  1  1
r 1 r 1
4 4 –1 –1 4 8 –1 –2 2
(Q)  +  +   = + +  =+ +1=0
T    
(R) N = 1 + cos  + i sin  = 2cos cos  isin   2cos  ei/2
2 2 2 2
r 2 r 
D = –i sin  + i(1 + cos ) = i [conjugate of N ] = i2cos  e i /2
2
   
(S)    .....  
3 3 2 33 2

Time Allotted: 3 Hours Maximum Marks: 432

t h e t es t .

B. Filling of OMR Sheet

C. Marking Scheme For All Three Parts.

(i) Section-A (01 to 03 and 10 to 12) contains 6 multiple choice questions which have only one
correct answer. Each question carries +8 marks for correct answer and – 2 mark for wrong
answer.

Section-A (04 to 09 and 13 to 30) contains 24 multiple choice questions which have only one
correct answer. Each question carries +4 marks for correct answer and – 1 mark for wrong
answer.

Name of the Candidate

Enrolment No.

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Useful Data

PHYSICS
2
Acceleration due to gravity g = 10 m/s

Planck constant h = 6.6 1034 J-s

Charge of electron e = 1.6  1019 C

Mass of electron me = 9.1  1031 kg

Permittivity of free space 0 = 8.85  1012 C2/N-m2

Density of water water = 103 kg/m3

Atmospheric pressure Pa = 105 N/m2

Gas constant R = 8.314 J K1 mol1

CHEMISTRY

Gas Constant R = 8.314 J K1 mol1

Atomic No: H=1, He = 2, Li=3, Be=4, B=5, C=6, N=7, O=8,

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Physics PART – I

SECTION – A
Single Correct Choice Type

This section contains 30 multiple choice questions. Each question has four choices (A), (B), (C) and (D)
out of which ONLY ONE is correct.

1. There are some passengers inside a stationary railway compartment. The centre of mass of the
compartment itself (without the passengers) is C1, while the centre of mass of the ‘compartment
plus passengers’ system is C2. If the passengers move about inside the compartment,
(A) both C1 and C2 will move with respect to the ground
(B) neither C1 nor C2 will move with respect to the ground
(C) C1 will move but C2 will stationary with respect to the ground
(D) C2 will move but C1 will be stationary with respect to the ground

2. A false balance has equal arms. An object weighs x when placed in one pan and y in the other
pan. The true weight of the object is equal to
xy
(A) xy (B)
2
x2  y 2 x2  y2
(C) (D)
2 2

3. Two identical cylindrical vessels with their bases at the same level, each contains a liquid of
density . The height of the liquid in one vessel is h1 and that in the other h2. The area of either
base is A. The work done by gravity in equalizing the levels when the vessels are inter connected
is
2
 h  h2  h h 
(A) Ag  1  (B) Ag  1 2 
 2   2 
2
 h  h2  h h 
(C) Ag  1  (D) Ag  1 2 
 4   4 

4. Dimensional formula for Plank’s constant is identical to that of

(A) Torque (B) Power
(C) Linear momentum (D) Angular momentum

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5. A wheel of radius r rolls without slipping with a speed v on a horizontal road. When it is at a point
A on the road, a small blob of mud separates from the wheel at its highest point and lands at
point B on the road
r r
(A) AB  v (B) AB  2v
g g
r r
(C) AB  4v (D) AB  8v
g g

6. A particle moves along the parabolic path y = ax2 in such a way that the x-component of the
velocity remains constant, say c. The acceleration of the particle is
(A) ackˆ (B) 2ac 2 ˆj
(C) ac 2kˆ (D) a 2c 2 ˆj

7. A radioactive nucleus of mass number A, initially at rest, emits an  particle with speed v. The
recoil speed of the daughter nucleus is
4v 4v
(A) (B)
A4 A
(A  4)v (A  4)v
(C) (D)
A 4

8. A particle of mass m is fixed to one end of a light spring of force 

constant k and unstretched length l. The system is rotated
about the other end of the spring with an angular velocity , in k m
gravity free space. The increase in length of the spring will be
ml2 ml2
(A) (B)
k k  m2
2
ml
(C) (D) None of these
k  m2

9. A particle is moving in a circle of radius r centred at O with v2

constant speed v. The change in velocity in moving from P to Q
(  POQ = 40°) is Qv
1
(A) 2vcos40° (B) 2vsin40°
40
(C) 2vcos20° (D) 2vsin20° P

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10. The electric potential (in volt) in a region is given by

2 2
V = 6x – 8xy – 8y + 6yz – 4x
Then electric force acting on a point charge of 2C placed at the origin will be
(A) 2 N (B) 6 N
(C) 8 N (D) 20 N

11. Two containers of equal volume contain identical gases at pressures P1 and P2 and absolute
temperatures T1 and T2 respectively. The vessels are joined and the gas reaches a common
pressure P and a common temperature T. Then
P P  1P P 
(A) P   1  2  T (B) P   1  2  T
 T1 T2  2  T1 T2 
 P T  P2 T1   P T  P2 T1 
(C) P   1 22 2  T (D) P   1 22 2  T
 T1  T2   T1  T2 

12. The half life period of a radioactive element X is same as the mean life time of another radioactive
element Y. Initially both of them have same number of atoms. Then,
(A) X and Y have the same decay rate initially.
(B) X and Y decay at the same rate always.
(C) Y will decay at a faster rate than X.
(D) X will decay at a faster rate than Y.

13. The temperature of source and sink of a Carnot Engine are 327°C and 27°C respectively. The
efficiency of the engine is
 27  27
(A) 1    (B)
 327  327
(C) 0.5 (D) 0.7

R 2
14. For a gas  . Possibly the gas is
CP 3
(A) diatomic (B) monatomic
(C) a mixture of diatomic and monatomic (D) polyatomic

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15. A wire has a non-uniform cross-section as shown in the

figure. A steady current is flowing through it. Then the
drift speed of the electrons
(A) is constant throughout the wire
(B) decreases from A to B
(C) increases from A to B A
(D) varies randomly
B

16. Which of the following is most suitable for the core of the electromagnets?
(A) Air (B) Steel
(C) Soft iron (D) Cu-Ni alloy

17. A black body radiates power P and maximum energy is radiated by it around a wavelength 0.
The temperature of the black body is now changed such that it radiates maximum energy around
3
the wavelength 0 . The power radiated by it now is
4
256 16
(A) P (B) P
81 9
64 4
(C) P (D) P
27 3

18. The plane of the dip circle is set in the geographical meridian and apparent dip is θ1. It is then set
in a vertical plane perpendicular to the geographical meridian, the apparent dip becomes θ2. The
angle of declination  at that place is given by
(A) tan   tan 1 tan 2 (B) tan   tan2 1  tan2 2
tan 1 tan 2
(C) tan   (D) tan  
tan 2 tan 1

19. According to Faraday’s law of electromagnetic induction

(A) the direction of induced current is such that it opposes the cause producing it.
(B) the magnitude of induced emf produced in a coil is directly proportional to the rate of change
of magnetic flux.
(C) The direction of induced emf is such that it opposes the cause producing it.
(D) None of the above.

Space for Rough work

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20. If  is phase difference between current and voltage, the wattless component of current is
(A) I cos (B) I sin
2
(C) I tan (D) I cos 

21. A particle is vibrating in S.H.M. Its velocities are v 1 and v2 when the displacements from the mean
position are y1 and y2, respectively, then its time period is
y12  y22 v12  v 22
(A) 2 (B) 2
v12  v 22 y12  y22
v 22  v12 y12  y 22
(C) 2 (D) 2
y12  y 22 v 22  v12

22. The first overtone of an open pipe has frequency n. The first overtone of a closed pipe of the
same length will have frequency
n
(A) (B) 2n
2
3n 4n
(C) (D)
4 3

23. Statement-1: The resolving power of a telescope is more if the diameter of the objective lens is
more.
Statement-2: Objective lens of large diameter collects more light.
(A) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for
Statement-1.
(B) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for
Statement-1.
(C) Statement-1 is true; Statement-1 is false.
(D) Statement-1 is false; Statement-1 is true.

24. Light of wavelength 6328 Å is incident normally on a slit having a width of 0.2 mm. the distance of
the screen from the slit is 0.9 m. The angular width of the central maximum is
(A) 0.09 degree (B) 0.72 degree
(C) 0.18 degree (D) 0.36 degree

25. A clear sheet of polaroid is placed on the top of similar sheet so that their axes make an angle
3
sin1   with each other. The ratio of intensity of the emergent light to that of unpolarised
5
incident light is
(A) 16 : 25 (B) 9 : 25
(C) 4 : 5 (D) 8 : 25

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26. The vernier constant of a travelling microscope is 0.001 cm. If 49 main scale divisions coincide
with 50 main scale divisions, then the value of 1 main scale division is
(A) 0.1 mm (B) 0.5 mm
(C) 0.4 mm (D) 1 mm

27. For circuit shown in figure IE = 4 mA, IB = 40μA. What are the E C
values of  and IC?
(A) 0.99, 3.96 mA
(B) 1.01, 4.04 mA
(C) 0.97, 4.04 mA
(D) 0.99, 4.04 Ma B

28. The average value of electric energy density in an electromagnetic wave is (E0 is peak value)
1 E2
(A) 0E02 (B) 0
2 2 0
1
(C) 0E02 (D) 0E02
4

29. A body is projected away from the earth with a speed 3v, where v, is the escape velocity. The
speed of the body at infinity will be
(A) ve (B) 2v e
(C) 2ve (D) 2 2v e

30. A vessel of cross-sectional area A contains a liquid to a height H1. If a hole having cross-sectional
area a is made at the bottom of the vessel, then the time taken by the liquid level to decrease
from H1 and H2 is
A g A 2
(A) H1  H2  (B) H1  H2 
a 2   a g 

a g a 2
(C) H1  H2  (D) H1  H2 
A 2  A g 

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Chemistry PART – II

SECTION – A

Single Correct Choice Type

This section contains 30 multiple choice questions. Each question has four choices (A), (B), (C) and (D)
out of which ONLY ONE is correct.

1. Which of the following tests you would perform to identify the functional group present in salicylic
acid?
(A) FeCl3 and NaHCO3 test (B) FeCl3 and NaOH test
(C) FeCl3 and 2, 4-dinitrophenylhydrazine test (D) FeCl3 and Schiff’s reagent test

2. A metal oxide is yellow when hot and white when cold. The metal oxide is
(A) CuO (B) ZnO
(C) PbO (D) All

3. A metal chloride solution on mixing with K2CrO4 solution gives a yellow ppt. insoluble in acetic
acid. The metal may be.
(A) Mercury (B) Zinc
(C) Silver (D) Lead

4. Enthalpy of dissociation of water in kJ mol–1 is

(A) +13.7 (B) +57.3
(C) +18.0 (D) +18  4.2

5. For the preparation of iodoform from acetone; we require

(A) KI (B) KI3
(C) KOI (D) KIO3

6. Cl2 gas passed through Lassaigne’s extract containing CCl4. If the extract contains both NaBr and
NaI then which colour will appear in the CCl4 layer?
(A) Violet colour (B) Brown colour
(C) Green colour (D) Yellow colour

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7. Which one of the following does not involve peptisation?

(A) Fe(OH)3 precipitate shaken with a small amount of dil. HCl
(B) Fe(OH)3 precipitate shaken with FeCl3 solution
(C) AuCl3 solution shaken with SnCl2 solution
(D) AgNO3 solution shaken with KI solution

8. The increasing order of stability of the following carbocation is

CH3 CH2 CH3 C CH2 H2C CH CH2

I II III IV
(A) IV < I < III < II (B) II < I < IV < III
(C) I < II < III < IV (D) III < I < IV < II

9. The product obtained when is oxidized with HIO4 is

CHO COOH
(A) (B)
CHO CHO

COOH CHO
(C) (D)
COOH CH2 OH

O
SeO2
10. CH3 C CH3     Se  H2O here, X is:

O O O

(A) CH3 C C H (B) CH3 C O CH3

(C) CH3 C CH2 OH (D) None of the above

11. The weight of a molecule of the compound C60H122 is

(A) 1.4  1021 g (B) 1.09  10 21 g
(C) 5.025  1023 g (D) 16.023  1023 g

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12. Benzoic acid undergoes dimerisation in benzene solution. The van’t Hoff factor (i) is
(A) i = (1 – x) (B) i = (1 + x)
(C) i = (1 – x/2) (D) i = (1 + x/2)

13. Which of the following has maximum dipole moment?

Cl Cl
Cl Cl
(A) (B)

Cl Cl
Cl Cl
Cl

Cl
Cl

14. The ONO bond angle is maximum in

(A) NO3– (B) NO2–
(C) NO2 (D) NO2+

15. Calculate the partial pressure of carbon monoxide from the following data

CaCO3 (s)   CaO(s)  CO2 (g); K p  8  102

CO2 (g)  C(s)  2CO(g); Kp  2

(A) 0.2 (B) 0.4
(C) 1.6 (D) 4

16. The diagonal partner of element B is

(A) Li (B) Al
(C) Si (D) Mg

17. Hybridization of the nitrogen atom and electron geometry around nitrogen atom in pyridine

is
N
(A) Sp2, planar triangular (B) Sp3, pyramidal
3 2
(C) Sp , tetrahedral (D) Sp , v-shape

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18. The energy of the second Bohr orbit of the hydrogen atom is –328 kJ mol–1; hence the energy of
fourth Bohr orbit would be
–1 –1
(A) –1312 kJ mol (B) –82 kJ mol
–1
(C) –41 kJ mol (D) –164 kJ mol–1

19. The ionic radii of isoelectronic species N3–, O2– and F– in Å are in the order
(A) 1.36, 1.40, 1.71 (B) 1.36, 1.71, 1.40
(C) 1.71, 1.40, 1.36 (D) 1.71, 1.36, 1.40

20. What will be the correct order of vapour pressure of water, acetone and ether at 30ºC. Given that
among these compounds, water has maximum boiling point and ether has minimum boiling
point?
(A) Water < ether < acetone (B) Water < acetone < ether
(C) Ether < acetone < water (D) Acetone < ether < water

21. Conjugate base of H2 is

(A) H3 (B) H3
(C) H+ (D) H–

22. How many ml of 0.05 M KMnO4 (acid) are required to oxidize 2.0 gram of FeSO4 in dilute
solution?
(A) 52.63 (B) 42.63
(C) 12.63 (D) 21.36

23. If the bond lengths of C  C, C  N are x and y Å respectively N  N bond length is given by
x  2y
(A) x + 2y (B)
2
(C) x + y (D) 2y – x

24. Which of the following complex species does not involve inner orbital hybridization?
(A) [CoF6]3– (B) [Co(NH3)6]3+
3–
(C) [Fe(CN)6] (D) [Cr(NH3)6]3+

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25. For the reaction,

A(g)  2B(g)  2C(g)  3D(g)
The value of H at 27ºC is 19.0 K cal. The value of E for the reaction would be
–1 –1
(given R = 2.0 cal K mol )
(A) 20.8 Kcal (B) 19.8 Kcal
(C) 18.8 Kcal (D) 17.8 Kcal

26. For the reaction


C 2H6 (g)  
 C2H4 (g)  H2 (g)
The kp = 0.05 atm. The value of Gº of the reaction at 627ºC would be
(A) 11.19 kJ mol–1 (B) 22.40 kJ mol–1
–1
(C) 33.57 kJ mol (D) 27.98 kJ mol–1

27. The following reaction described as:

Br
NaOH 
OH    
O CH3
(A) SN1 reaction with racemisation
(B) Intramolecular SN2 reaction with Walden inversion
(C) Intramolecular SN2 reaction with retention of configuration
(D) Intramolecular SN1 reaction with racemisation

28. 75% of the first order reaction was completed in 32 min. 50% of the reaction was completed in
(A) 24 min (B) 8 min
(C) 16 min (D) 4 min

29. pH of a solution produced when an aqueous solution of pH 6 is mixed with an equal volume of an
aqueous solution of pH 3 is about
(A) 4.5 (B) 4.0
(C) 4.3 (D) 3.3

O
C H ONa
2 CH3 2 5
COC 2H5  A
30. , which is true about A.
(A) A forms oxime (B) A shows tautomerism
(C) A shows idoform test (D) All of them

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Mathematics PART – III

SECTION – A

Straight Objective Type

This section contains 30 multiple choice questions numbered 1 to 30. Each question has 4 choices (A),
(B), (C) and (D), out of which ONLY ONE is correct

1 1
1  2 1 1  2 1
1. The value of  tan  x  2  dx   cot  x  2  dx is equal to (where [.] denotes greatest integer
1 1
function)
3  1  3  1 
(A) 1   (B) 1  
4  2 4  2
 1 
(C)  1   (D) none of these
4 2

2. The number of ordered pairs of positive integers (m, n) satisfying m  2n  60, n  2m  60 is

(A) 240 (B) 480
(C) 900 (D) none of these

3. Let ‘P’ be a point which does not lie outside the triangle ABC, A  (3, 2), B  (0, 0), C  (0, 4) and
satisfies d(P, A)  maximum {d(P, B), d (P, C)}, then maximum distance of P from side BC
where d(P, A) gives the distance between P and A, is
3 4
(A) (B)
4 3
(C) 3 (D) 0

4. The mth term of an arithmetic progression is x and the nth term is y. Then the sum of the first
(m + n) terms is
mn xy mn xy
(A) (x  y)  (B) (x  y) 
2  m  n  2  m  n 
1xy xy  1xy xy
(C)    (D)  
2 m  n m  n  2  m  n m  n 

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 1 x2 
5. Let the f(x)  sin1   is
 2x 
(A) Continuous but not differentiable at x = 1 (B) Differentiable at x = 1
(C) Neither continuous nor differentiable at x = 1 (D) Continuous everywhere

5n1  3n  22n
6. lim is equal to
n  5n  2n  32n  3

(A) 5 (B) 3
(C) 1 (D) zero

7. If the point P( x1  t(x2  x1 ), y1  t(y 2  y1 ) ) divides the join of A( x1,y1 ) and B (x2 , y2 ) internally
then
(A) t < 0 (B) 0 < t < 1
(C) t > 1 (D) t = 1

8. If x1, x 2 , x3 as well as y1, y 2 , y3 are in G. P with the same common ratio, then the points
(x1, y1 ), (x2 , y2 ) and (x 3 , y 3 )
(A) Lie on a straight line (B) Lie on an ellipse
(C) Lie on a circle (D) are vertices of a triangle

9. If in a ABC, cos A + 2 cos B + cos C = 2, then a, b, c are in

(A) AP (B) HP
(C) GP (D) none of these

1
x2  t 2
10. If f(x) =  2  t dt, then the curve y = f(x) represents a
0
(A) Straight line (B) Parabola
(C) Hyperbola (D) None of these

11. Let P and Q be points (4, –4) and (9, 6) of the parabola y2 = 4a(x – b). Let R be a point on the arc
of the parabola between P and Q. Then the area of PRQ is largest when
(A) PRQ = 90° (B) The point R is (4, 4)
1 
(C) The point R is  , 1 (D) none of these
4 

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12. For what values of a, m and b, Lagrange's mean value theorem is applicable to the function f(x)
 3 x0
 2
for x  [0, 2], f(x) =  x  a 0  x  1
 mx  b 1  x  2

(A) a = 3, m = – 2, b = 0 (B) a = 3, m = – 2, b = 4
(C) a = 3, m = 2, b = 0 (D) No such a, m, b exist

13. Eccentricity of the hyperbola whose asymptotes are given by 3x + 2y – 5 = 0 and 2x – 3y – 5 = 0

is
3
(A) 2 (B)
2
5
(C) 2 (D)
3

x2 y 2
14. Consider the ellipse   1 . Let P, Q, R, S be four points on this ellipse such that the
3 1
normals drawn from these points are concurrent at (2, 2) then the centre of the conic (apart from
the given ellipse) on which these 4 points lie is
(A) P, Q, R, S lie only on the given ellipse and on no other conic
(B) (3, –1)
(C) (3, 1)
(D) (0, 0)

x
 t 2  2t  1 
15. Let f(x) =  cos   dt, 0 < x < 2. Then f(x)
0  5 
(A) increases monotonically (B) decreasing monotonically
(C) has one point of local maximum (D) has one point of local minimum

x5 dx
16. If I   , then I is equal to
1  x3
5 3 3 1
2 2 2 2
(A) (1  x 3 ) 2  (1  x 3 ) 2  c (B) (1  x 3 ) 2  (1  x 3 ) 2  c
9 3 9 3
(C) log x  1  x 3  c (D) x 2 log(1  x3 )  c

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17. If w is an imaginary cube root of unity, then value of the expression

2 2 2
2(1 + w)(1 + w ) + 3(2 + w)(2 + w ) + ..... + (n + 1)(n + w)(n + w ) is
1 1
(A) n2 (n  1)2  n (B) n2 (n  1)2  n
4 4
1
(C) (n  1)2  n (D) none of these
4

18 If z be a complex number satisfying | z |2 2(z  z)  5  0 , then complex number z + 3 + 2i will lie

on
(A) Circle with centre 1 – 2i radius 4 (B) Circle with centre 1 + 2i radius 4
(C) Circle with centre 1 + 2i radius 3 (D) Circle with centre 1 – 2i radius 3

7 6 x2  13
2
19. If one of the roots of the equation 2 x  13 2 = 0 is x = 2, then sum of all other five
x2  13 3 7
roots is
(A) 2 15 (B) –2
(C) 20  15  2 (D) none of these

20. f: R  R, f(x) is differentiable such that f(f(x)) = k(x 5 + x), (k  0), then f(x) is always
(A) Increasing (B) Decreasing
(C) Either increasing or decreasing (D) Non-monotonic
  
21. Let a and b are two perpendicular unit vectors. If c is another unit vector equally inclined at
 
angle  to the vectors a and b , then the set of exhaustive values of  in [0, 2] is
   
(A)  0,  (B) 0, 
 2  4
  3    3 
(C)  ,  (D)  , 
2 4  4 4 

22. Let A, B, C be point with position vectors r1  2iˆ  ˆj  k,

ˆ r  ˆi  2ˆj  3kˆ and r  3iˆ  ˆj  2kˆ relative
2 3
to the origin ‘O’. The shortest distance between point B and plane OAC is
(A) 10 (B) 5
5 5
(C) (D) 2
7 7

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23. The maximum number of regions in which 10 circles can divide a plane is
(A) 512 (B) 1024
(C) 2048 (D) None of these

24. If [x] and {x} denotes the greatest integer and fractional part function, then the number of real x,
satisfying the equation (x – 2)[x] = {x} – 1, is
(A) 0 (B) 1
(C) 2 (D) infinite

26. Consider the 12 face diagonals of a rectangular block. How many pairs of them are skew lines?
(A) 24 (B) 30
(C) 48 (D) 60

27. Suppose M = {(x, y)| |xy| = 1, x > 0} and N = {(x, y)| tan–1 x + tan–1 y = } which of the following
statements is true?
(A) M  N = {(x, y)| |xy| = 1}
(B) M  N = M
(C) M  N = N
(D) M  N = {(x, y)| |xy| = 1, x, y are non negative simultaneously}

 1 x 
28. Let F(x) be such that F    x for all x  –1 which of the following statements is true?
 1 x 
 1 x 
(A) F(–2 – x) = –2 – F(x) (B) F   x   F  , x  1
 1 x 
 1
(C) F    F  x  , x  0 (D) F(F(x)) = –x
x

29. Which of the following is logically equivalent to p  (p  q)?

(A) ~ p  (p  q) (B) q  (p  q)
(C) ~ p  ~ q (D) none of these

30. The mean of the 200 observation is 48 and their standard deviation is 3. Then sum of the squares
of all the observations
(A) 46200 (B) 42600
(C) 62400 (D) none of these

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ANSWERS, HINTS & SOLUTIONS

CRT –II
(Main)
ALL INDIA TEST SERIES

S. No. PHYSICS CHEMISTRY MATHEMATICS

1. C A D
2. B B B
3. B D C
4. D B A
5. C C C
6. B A D
7. A C B
8. B B A
9. D B A
10. D A B
11. B A C
12. C C B
13. C A A
14. B D B
15. B B C
16. C C B
17. A A A
18. C B C
19. B C B
20. B B C
21. D D D
22. C A D
23. A D B
24. D A D
25. D D C
26. B B B
27. A B B
28. D C A
29. D D B
30. B D D

Physics PART – I
SECTION – A
1. Since the compartment (including passengers) is stationary, so the combined centre of mass of
compartment and passengers is fixed i.e. C2 is fixed. When the passengers move here and
there in the compartment then in an attempt to keep C2 fixed C1 has to move.

2. Since the balance has equal arms and it is a false

balance, so the pans must be having different weights.
Let w1 and w2 be the weights of left pan and right pan
respectively. Let w be the true weight of object. When
placed in left pan it weights x i.e. a total weight of w + w1
on left balances a total weight of w2 + x on the right pan.
 w + w1 = w + x ..............(1)
Similarly it weighs y when placed in right pan i.e.
y + w1 = w + w2 ..............(2)
From (1)
w1 – w2 = x – w ............................(3)
From (2)
w1 – w2 = w – y .............................(4)
x–w=w–y
xy
 w
2

h1 h
3. The center gravity of liquid in the first column is and that in the second column is 2 .
2 2
Let litial Potential Energy be Ui
h  h 
 Ui  (Ah1)g  1   (Ah2 )g  2 
2
   2 
1
 Ui = Ag(h12  h22 )
2
h1  h2
When the water column in the two equavalises then equivalent height is and the center of
2
h1  h2
gravity in both the columns is . If Uf be the final potential energy in both the columns then
4
  h  h2   h1  h2  
Uf  2  A  1   g
  2   4  
1
 Uf  Ag(h1  h2 )2
4
Since work done is equal to the decrease in potential energy,
So, w = Ui – Uf
2
h h 
 w  Ag  1 2 
 2 

5. From the concepts of projectile motion,

1
2r  gt 2
2
Since x = AB = (2v)t
4r
 t
g
r
 AB  4v
g

6. y = ax2
dy  dx  
  y  (2ax)x  2acx   x  c
dt  dt 
d2 y 
Further a   y  2acx  2ac 2
dt 2

7. (0)A  4v  (A  4)V
4v
 V
A4
Negative sign indicates the recoil speed.

8. Let the extension produced in the spring due to rotation be x.

Then, for the body to be in equilibrium
Fspring  Fcentrifugal
 kx  m(  x)2
m2
 x
k  m2
 
10. E  V
  U ˆ U ˆ U 
 E    ˆi j k 
 x y z 
U
 6  8y 2  8x
x
U
 16xy  8  6z
y
U
 6y
z
At (0, 0, 0)
U U U
6,  8 , 0
x y z

 | E |(0,0,0)  6iˆ  8 ˆj

 | E | 10NC1
 
 | F | q | E |

 | F | 20 N

11. Since P1V = n1RT 1

and P2V = n2RT2
After joining the vessels we have
P(2V) = (n1 + n2)RT
PV P V 
 P(2V)   1  2  RT
 RT1 RT2 
1  P1 P2 
 P   T
2  T1 T2 

0.693 1
12. 
X Y
 Y > X
 dN 
Since,      XN
 dt  X
 dN 
And      YN
 dt  Y
 dN   dN 
     
 dt  Y  dt  X
Decay rate of Y > Decay rate of X.

1
15. vd 
neA

16. Soft iron has highest susceptibility.

17. mT = constant

4
 T  T0
3
Since, P  T 4 (from Stefan’s Law)
256
 Pnew  P
81

tan 
18. tan 1 
cos 
tan  tan 
and tan 2  
cos(90   ) sin 
tan 
 cos   ..................(1)
tan 1
tan 
and sin   ..................(2)
tan 2
Dividing (2) by (1), we have
tan 1
tan   .
tan 2
19. A and C are statements of Lenz’s Law. B is the correct option.
20. Iv sin component of current gives no power consumption and hence is called wattless current.

21. v2 = 2(a2 – x2)

 v12  2 (a2  y12 ) ...................(1)
 v 22  2 (a2  y 22 ) ....................(2)
From (1) and (2), we get
y12  y22
T  2
v 22  v12
 v 
22. n  2 
 2 
 v  3n
n'  3   
 4  4
a
23. Resolving power of telescope =
1.22
where a is the diameter of objective lens and  is the wavelength of light used. On increasing a,
more light is collected by objective lens and so, the image formed is more bright. Thus,
resolving power of telescope increases.

D
24. 
d
So, angular with, 2θ is
2 2
2  
D d
6328  10 10

0.2  103
 θ = 3164 × 10–6 radian
180
   3164  106  degree

 θ = 0.18 degree
 2θ = 0.36 degree.
I
25. If I is the intensity of the incident unpolarised light, the intensity transmitted by the first is
. This
2
is the intensity of incident light on the second polaroid. Intensity transmitted by the second
I 2 3
polaroid is   cos θ, where θ is the angle between the axes. Here sinθ = , cos θ is therefore
2
  5
4
.
5
2
I I 4 8
cos2       I
2 2 5 25
8
is the required ratio.
25

26. 50 VSD = 49 MSD

49
1 VSD = MSD
50
 VC = 1 MSD – 1 VSD
49
 VC  1MSD  MSD
50

1
 VC  MSD
50
 1 MSD = 50 (VC) = 50 (0.001 cm)
 1 MSD = 0.05 cm = 0.5 mm
 1 MSD = 0.5 mm

27. IC = IE – IB = 4 mA – 40 μA = (4 – 0.04) = 3.96 mA

I 3.96
 C   0.99
IE 4

28. Electric energy density

1 2
u0  0Erms
2
E
Since, Erms  0
2
1
 u0  0E02
4

29. By the Law of conservation of energy

(U + K)surface = (U + K)∞
GMm 1 1
  m(3v1 )2  0  mv 2
R 2 2
GM 9v 2e 1 2
    v
R 2 2
2 2GM
Since, v e 
R
2 2
v 9v 1
  e  e  v2
2 2 2
 v 2  8v 2e
 v  2 2v e

30. Let h be the height of the water level at any instant. Then the rate of decrease of water level is
dh
 . Therefore
dt
dh
A  av  a 2gh
dt
dh a
   2gh
dt A
H2 1
dh  a 
  1/2
 2g   dt
H1 h A 0
H2 a 
 2 h1/ 2    2g  t
H1
A 
 a 
 2  H1  H2    2g  t
A 
A 2
 t H1  H2 
a g 

Chemistry PART – II
SECTION – A
1. Salicylic acid contains – OH group and – COOH group. Therefore FeCl3 and NaHCO3 tests are
used.

2. ZnO gives yellow colour on heating and white on cooling.

3. PbCl2  K 2CrO4  PbCrO4  2KCl

yellow ppt.


4. Dissociation of H2O(H2O    
 H  OH) is opposite to neutralization. Hence, enthalpy of
dissociation is equal to enthalpy of neutralization but has opposite sign.

5. For preparation of iodoform reagent which can supply I+ is require i.e. KOI or KOH/I2.

6. I– ion is more easily oxidized than Br– ion and hence violet colour due to I2 appears first.

7. AuCl3 is reduced by SnCl2 to form colloidal sol of gold.

There is no peptisation involved.

8. In accordance with Huckel rule, carbocation (III) is aromatic and hence most stable. Carbocation
(IV) is stabilized by resonance and hence more stable than carbocation (I and II). In carbocation
(II), +ve charge is present on a more electronegative sp-hybridized carbon while in carbocation
(I), it is present on a less electronegative sp2-hybridised carbon and hence carbocation (I) is more
stable than carbocation (II). Thus, the overall stability increases in the order : II < I < IV < III.

O
HIO4 COOH
9. CHO
OH

O
10. SeO2 oxidizes – CH2 – next to carbonyl group to another C group.
O O
SeO2
CH3 C CH3   CH3
 C CHO  Se  H2 O
x

11. 1 Mole of C60H122 = 60  12 g + 122  1 g = 842 g

= 6.022  1023 molecules
Weight of 1 molecule = (842/ 6.022  1023) gram.

12. 2C 6H5 COOH   (C6H5 COOH)2

Before asso. 1 mol
After asso. 1 – x x/2
Total = 1 – x + x/2 = 1 – x/2
x
1
2 x
i  1
1 2

Cl 2 Cl
13. 1 3 , resultant bond moment of C1 – Cl and C3 – Cl bonds adds up to the bond

moment of C2 – Cl. Hence, it has maximum value.

14. NO2+ has no unshared electron. It has only bond pairs in two directions. Hence, shape is linear
O N O with bond angle = 180º.

15. For reaction (i), K p  PCO2  8  10 2 (given)

2
PCO
For reaction (ii), K p   2 (given)
PCO
2
PCO
 2 or PCO  0.4
8  10 2

16. Boron is diagonally related to Si.

17. In , N is linked to two c-atoms and has one lone pair. Hence, electron geometry is
N
triangular planar and hybridization is Sp2.

k
18. E2   2
 328 kJ mol1
2
K = 1312 kJ mol–1
k
E4   2
4
1312
  82 kJ mol1
16

19. Among isoelectronic ions, size increases the –ve charge increases.
i.e. F– < O2– < N–3

20. Greater the boiling point, less is the vapour pressure. Hence, the correct order of vapour pressure
wil be
Water < acetone < ether

21. Conjugate base = conjugate acid – H+

= H 2 – H + = H –.

22. The redox changes are

Mn7   5e   Mn2
Fe2   Fe3   1e
152
Equivalent weight of FeSO4 =
1
Milli equivalent of KMnO4 = Meq. Of FeSO4

2
 0.05  5  V   1000
15 2 1
 V  52.63 ml

r r
23. C2  N2  rCN
2 2
X r r x
  N2  y  N2  y 
2 2 2 2
r 2Y  X
 N2 
2 2
 rN2  2y  x

7 2
24. Outer elect. Configuration of Co atom = 3d 4s
3+ 6
Outer elect. Configuration of Co ion = 3d
–
As F is a weak ligand, it cannot cause forcible pairing of electrons within the 3d level. Hence
3 2
sp d hybridization will result outer orbital complex.

25. n(g) = np – nr = 5 – 3 = 2
H = E + ng RT
E = H – ng RT
= 19–(2  2 × 10–3  300)
= 19 – 1.2 = 17.8 K cal.

26. Gº = – RT ln Kp = – (8.314 J k–1 mol–1) (900 K)

(2.303)log(0.05) = 22400 J mol–1.

27.
NaOH 
OH 
  
O
Br O CH3
Br

28. 75% of reaction is completed in two half-lives.

29. In the final solution, [H+] = (10–6 + 10–3)/ 2

 0.5005  10 3  5.005  104
pH   log[H ]   log(5.005  104 )
 4  0.6994  3.3

30. O O O
C H ONa
2 CH3 COC2H5 
2 5
 H3C C CH2 C OC2H5
A

(A) forms oxime due to keto group, shows tautomerism, also shows idoform test due to
O
CH3 C
group.

Mathematics PART – III

SECTION – A

1 1 3 1 1 1
1.  x2     x2  
2 2 2 2 2 2
 2 1  2 1
 x  2   0,1  x  2   1,0
   
 2 1   2 1 
x  2   1  x  2   1
   
1 1
1  x2   2 1  x 2   0
2 2
1 3 1 1
 x2    x2 
2 2 2 2
1 3 1
 | x | |x|
2 2 2
1 3 1

2 2 2
  3  1   3  1 3  1  1 
I  tan1 1dx  1 tan
1
1dx  1 cot
1
( 1)dx      
3
4  2  4 2 4  2 
 
2 2 2

   

4 2  2 3  2  6  
2 2
 32   2 2
 32 
2. m  2n  60  n  2m  60
1
For m = 1; 1  2n  60 and n  2  60   n  2  n  1, 2
2
For m = 2; 2  2n  60 and n  4  60  1  n  4  n = 1, 2, 3, 4
3
For m = 3; 3  2n  60 and n  6  60   n  6  n = 2, 3, 4, 5, 6
2
m=4 n = 2, 3, 4, 5, 6, 7, 8
m=5 n = 3, 4, …….. 10
m=6 n = 3, 4 ………12
Total order pair =
m 1 m 2 m 3 m 4 m 5 m 6 m 6 m 8 m 9 m 13 m 14 m 15 m 16
2
 4 
  57 
  8
 10
11  13
   14  .......  
20  22
 23
  23
 

22
  22
   21

 21  20

   20
   19
  19
   18
  18
   17
  17
   16
  16
 

6  12  18
 .............  42
   46  2(22  21  .......  16)
7 terms

7 7
  (6  42)  46  2   (22  16)
2 2  7  24  46  7  38  480

3. In the figure O is circumcentre and shaded region shows location of P

Distance of P from side BC is maximum if P is at A

mn
4. Let first term is a then x = a + (m – 1)d, y = a + (n – 1)d and s = [2a  (m  n  1)d]
2
(x + y) = 2a + (m + n – 2)d and (x – y) = (m – n)d
 xy   xy 
d=    (x + y) = 2a + (m + n – 1)d –  m  n 
 m  n   
m  n xy
S=   (x  y)  m  n 
 2  

1 1 
5. sin 1    x   check f(1–) = sin–11+, f(1) = sin–11, f(1+) = sin–11–
2  x 
Hence function is discontinuous so non differentiable

5.5n  3n  4n
6. lim  0 , divide numerator and denominator by highest power i.e. (9n)
n  5n  2n  27  9n

7. P(x1 + t (x2 – x1), y1 + t(y2 – y1)

t x  (1  t)x1 t y  (1  t)y1
x 2  y 2
t  1 t t  1 t
For internal division
t > 0 and (1 – t) > 0  0 < t < 1

8. x1 x/r y1 y/r
x2 x y2 y
x3 xr y3 yr
Slope of line joining (x1, y1) and (x2, y2)
yy / r y r  y r(y  y / r)
m  Slope of line joining (x2, y2) and (x3, y3)  m  
xx / r x r  x r(x  x / r)
Slope is same hence (x1, y1), (x2, y2) and (x3, y3) are in straight line

9. cos A + 2 cos B + cos C = 2

cos A + cos C = 2(1 – cos B)
AC A C B
2cos .cos  2  2sin2
2 2 2
B A C 2B A C B A C A C B
sin cos  2sin , cos     2sin , cos cos  sin sin  2sin
2 2 2  2 2  2 2 2 2 2 2
s(s  a) s(s  c) (s  b)(s  c) (s  a)(s  b) (s  a)(s  c)
 2
bc ab bc ab ac
a + c = 2b i.e. a, b, c, in AP

1 1
x2  t 2  x2 4 
10 f(x) = 
2t
dt =   2  t  t  2  2  t  dt
0 0
1
 t2  1
=  (x2  4) n| 2  t |   2t  = (x2 + 4) n2 – – 2
 2 0 2
5
i.e. y = x2 n2 + 4 n2 – which is a parabola
2
2
11. Since (4, –4) and (9, 6) lie on y = 4a(x – b)
 4 = a(4 – b) and 9 = a(9 – b)

 a = 1 and b = 0
 The parabola is y2 = 4x
Let the point R be (t2, 2t)
4 4 1 4 4 1
1 1 1 5 10
 Area of PRQ = 9 6 1 = 5 10 0 = 2
2 2 2 2 2 t  4 2t  4
t 2t 1 t  4 2t  4 0
2
1 1 5  1 125
= (10t + 20 – 10t2 + 40) = – 5t2 + 5t + 30 = –5(t2 – t + ) + 30 + = –5  t   +
2 4 4  2 4
1
Area in largest when t =
2
1
 Coordinates of the point R are ( , 1)
4

 3 x 0
 2
12. f(x) =  a  x 0  x 1
mx  b 1  x  2

For Lagrange’s mean value theorem f(x) must be continuous in [0, 2] and differentiable in (0, 2)
Hence, a = 3, m + b = 2
 2x 0  x  1
f(x) = 
 m 1 x  2
 m = – 2 and hence b = 4

13. It is rectangular hyperbola  e  2

14. Normal to ellipse is ax sec   by cos ec  a2  b2

 x 2  2x  1   (x  1)2 
15. f(x) = cos   = cos  
 5   5 
1 (x  1)2 9
Since 0  x  2,  
5 5 5
2
(x  1) (x  1)2 
cos = 0 only when =
5 5 2
5
f(x) > 0 if 0 < x < 1
2
5
f(x) < 0 if  1< x < 2
2
5
 x=  1 is a point of local maximum
2

x5 dx
16. I
1  x3
1 + x = t2
3

3x2 dx = 2t dt
2 (t 2  1)t dt 2 2 2 1 
I    (t  1)dt   t 3   t   c
3 t 3 3 3 
2 2 2 2 2
  t 3  3t   c  t 3  t  c  (1  x 3 )3/2  (1  x 3 )1/2  c
9 9 3 9 3

n n
1
17.  (r  1)(r  ) (r  2 )   (r 3  1)  4 n2 (n  1)2  n
r 1 r 1

18. zz   z  z    0
zz  2z  2z  5  0
z0  2 r  45
r=3
| z  2 | 3

19. 7 {7(x2  13)  6}  {14  2(x2  13}  (x 2  13) {6  (x 2  13)2 }  0

   2  0

20. f(f(x)) = k(x5 + x) (k  0)

f(f(x)) . f(x) = k(5x4 + 1)  Can be + or –
But it may not be –ve. Because if f(x)  ing then f(f(x)).f(x) > 0
and If f(x)  ing then f(f(x)).f(x) > 0 i.e. f(x) is either  ing or  ing (also k > 0)

21. Let c makes an angle  from â  bˆ then cos2  + cos2  + cos2  = 1
2 cos2  = 1– cos2  b̂
0  1  cos2  1
1 1 1
0  2cos2   1  0  cos2     cos  â
2 2 2
  3  â × b̂
  , 
4 4 

22. A = (2, –1, 1), B = (1, 2, 3), C = (3, 1, 2)


OA  2iˆ  ˆj  kˆ

OC  3iˆ  ˆj  2kˆ
ˆi ˆj kˆ
 
OA  OC  2 1 1  3iˆ  ˆj  5kˆ
3 1 2
 Equation of plane OAC: 3x + y – 5z = d
Put (0, 0, 0), d = 0
 Plane is 3x + y –5z = 0
| 3  1 2  5  3 | 10 5
Shortest distance =  2
9  1  25 35 7

23. Initially there is one plane

When C1 is introduced: 1 part is increased
When C2 is introduced: 2 part is increased
When C3 is introduced: 4 part is increased
When C4 is introduced: 8 part is increased
1(210  1)
 Total number of parts = 1 + 1 + 2 + 4 + 8 + 16 + .....  1  = 1024
2 1

24. for x  2, LHS is always non negative and RHS is always –ve
Hence, for x  2 no solution
If 1  x < 2 (x – 2) = (x – 1) – 1 = x – 2 which is an identity
For 0  x < 1, LHS is '0' and RHS is (–)ve  no solution
x < 0, LHS is (+)ve, RHS is (–)ve no solution

26. Consider a fixed face diagonal. It is parallel to corresponding one on the opposite face. It
intersects the other diagonal of the same face and one diagonal on each of the four neighbouring
faces. The other five face diagonals form five pairs of skew lines with the fixed one. By symmetry,
each of the twelve faces diagonals. The total count of pair of skew lines is 12  5 = 60 but this
must be halved since each pair is counted twice.

27. Clearly N = 
Now M consists of the curve xy = 1 and xy = –1 in the first and fourth quadrant
Hence, N  M

1 x 1 x
28. y , F y  x 
1 x 1 x
3x
F  2  x       2  F  x 
 1 x 
 1
F(F(x)) = –x, F    F  x 
x

29. p  (p  q) and q  (p  q) both are tautologies, while formulae

~ p  (p  q) and ~ p  ~ q are not

30. x  48 , n = 200 standard deviation = 3

Variance =  x 2    x 
n  n 
 

9
 x 2   48 2
200
2
x  46200

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Paper 1

Time Allotted: 3 Hours Maximum Marks: 180

 Pl ea s e r ea d t h e i n s t r u c t i o n s c a r ef u ll y . Yo u a r e a l l o t t ed 5 m i n u t es
s p ec i f i c a ll y f o r t h i s p u r p o s e.
ALL INDIA TEST SERIES

 Yo u a r e n o t a ll o wed t o l ea v e t h e E xa m i n at i o n Ha l l b ef o r e t h e en d o f
t h e t es t .

B. Filling of OMR Sheet

C. Marking Scheme For All Three Parts.

(i) Section-A (01 to 10) contains 10 multiple choice questions which have only one correct answer.
Each question carries +2 marks for correct answer. There is no negative marking.

Section-A (11 to 15) contains 5 multiple choice questions which have more than one correct
answer. Each question carries +4 marks for correct answer and – 1 mark for wrong answer.

(ii) Section-C (01 to 05) contains 5 Numerical based questions with answers as numerical value
and each question carries +4 marks for correct answer and – 1 mark for wrong answer.

Name of the Candidate

Enrolment No.

Useful Data

PHYSICS
2
Acceleration due to gravity g = 10 m/s

Planck constant h = 6.6 1034 J-s

Charge of electron e = 1.6  1019 C

Mass of electron me = 9.1  1031 kg

2 2
Permittivity of free space 0 = 8.85  1012 C /N-m

Density of water water = 103 kg/m3

Atmospheric pressure Pa = 105 N/m 2

Gas constant R = 8.314 J K1 mol1

CHEMISTRY

Gas Constant R = 8.314 J K1 mol1

Atomic No: H=1, He = 2, Li=3, Be=4, B=5, C=6, N=7, O=8,

Physics PART – I

SECTION – A
Single Correct Choice Type

This section contains 10 multiple choice questions. Each question has four choices (A), (B), (C) and
(D) out of which ONLY ONE option is correct.

1. In the figure shown, a parallel beam of light is incident on the l

plane of the slits of a Young’s double slit experiment. Light x

incident on the slit, S, passes through a medium of variable S1

refractive index   1  ax (where x is distance from the plane of

O
slits as shown) upto a distance 'l' before falling on S1. Rest of the
space is filled with air. If at ‘O’ a minima is formed, then the S2
minimum value of the positive constant a is: S1O  S2 O Screen

 
(A) (B) 2
l l
2
l l2
(C) (D)
 2

2. The fundamental frequency of a sonometer wire of length l is f0 . A bridge is now introduced at a

distance l from the centre of the wire  l  l . The number of beats heard if both sides of the
bridges are set into vibration in their fundamental modes are
8f l f l
(A) 0 (B) 0
l l
2f0 l 4f0 l
(C) (D)
l l

Space for Rough work

3. A light ray 1 is incident on a plane mirror M. The mirror is

rotated in the direction as shown in fig. by an arrow at frequency
9
rps . The light reflected by the mirror is received on the wall W 1
 37 0
at a distance 10 m from the axis of rotation. When the angle of
0
incidence is 37 , the speed of the spot on the wall is:
Wall
10m
(A) 10m / s (B) 1000m / s
(C) 500m / s (D) 100m / s

nd
4. The photon radiated from hydrogen corresponding to 2 line of Lyman series is absorbed by a
hydrogen like atom x in 2nd excited state. As a result the hydrogen like atom ‘x’ makes a transition
to nth orbit. Then
(A) x  He ,n  4 (B) x  Li ,n  6
(C) x  He ,n  6 (D) x  Li ,n  9

5. If a tuning fork of frequency  f0  340 Hz tolerance 1% is used in resonance column method .
The first and the second resonance are measured at l1  24.0 cm and l2  74.0 cm. The
maximum permissible error in speed of sound is:
(A) 1.2 % (B) 1.8 %
(C) 1 % (D) 0.8 %

6. Axis of a solid cylinder of infinite length and radius R lies along y – axis it carries a uniformly
R R
distributed current ‘I’ along + y direction. Magnetic field at a point  ,y,  is
2 2
I I

(A) 0 ˆi  kˆ
4R
 (B) 0 ˆj  kˆ
2R
 
 0I I
(C)
4R
ĵ (D) 0 ˆi  kˆ
4R
 
Space for Rough work

7. Find the potential difference across C2 : When steady state reached. C2=6µF
C
(A) 6 V (B) 4 V 2
(C) Zero (D) 2 V 4
C1=4µF
2
D 2 E

10V

8. A battery of emf 0  12V is connected across a 4m long uniform 0

R=8
wire having resistance 4/m. The cells of small emfs 1  2V and
N
2  4V having internal resistance 2 and 6 respectively, are A B
connected as shown in the figure. If galvanometer shows no 1 r1 G
deflection at the point N, the distance of point N from the point A is
equal to : 2 r2

1 1
(A) m (B) m
6 3
(C) 25cm (D) 50cm

9. A milliameter of range 10 mA and resistance 9 is joined in a –i

circuit as shown. The meter gives full-scale deflection for
current I when A and B are used as its terminals, i.e., current i –i
enters at A and leaves at B (C is left isolated). The value of I
is
(A) 100 mA (B) 900 mA
(C) 1 A (D) 1.1 A

10. In the circuit shown alongside, each battery is 5V and has an

internal resistance of 0.2 ohm. The reading in the ideal voltmeter V
V
is.
(A) 0 V (B) 1 V
(C) 2 V (D) 3 V

Space for Rough work

Multiple Correct Answer(s) Type

This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D)
out of which ONE or MORE THAN ONE are correct.

11. The instantaneous velocity v of a particle is related to its displacement x according to the relation
v = ax + b, where a > 0 and b  a/7. Which of the following statement(s) is (are) true if x = 0 at
t = 0.
b
(A)The displacement of the particle at time t is x = (eat  1)
a
(B)The particle will experience a retardation if b < 0
(C)The particle will be at rest at time t = 0
(D)The acceleration of the particle is constant

12. An moving body may have

(A)Varying speed without having varying velocity
(B)Varying velocity without having varying speed
(C)Nonzero acceleration without having varying velocity
(D)Nonzero acceleration without having varying speed

13. Balls A and B are thrown from two points lying on the same
horizontal plane separated by a distance 120 m. Which of the 50ms–1
following statement(s) is(are) correct.
(A) The balls can never meet 30ms–1
(B) The balls can meet if the ball B is thrown 1st later A 37º B
(C) The balls can meet if the balls are thrown at the same time
120m
(D) The two balls meet at a height of 45 m

14. Which of the following statements are correct

(A) Work by kinetic function on an object can be zero
(B) Work by static friction on an object can be non zero
(C) Work by normal force on an object can be non zero
(D) Work by force which is always perpendicular to its acceleration is zero

15. A 1.5kg box is initially at rest on a horizontal surface when at t  0, a horizontal force

F  1.8t ˆi N (with t in sec), is applied to the box. The acceleration of the box as a function of
time t is given by

a  0, for 0  t  2.8 sec

a  (1.2t  2.4)iˆ for t  2.8 sec
(A) The coefficient of static friction between the box and the surface is 0.24
(B) The friction force on the box at t  2 sec is 3.6 N
(C) The coefficient of kinetic friction between the box and surface is 0.24
(D) The friction force on the box at t  4 sec is 3.6 N

Space for Rough work

SECTION –C

Integer Answer Type

This section contains 5 questions. Each question, when worked out will result in one integer from 0 to 9
(both inclusive).

1. All the strings, springs and pulleys shown in figure are ideal. Initially the system is in
equilibrium and blocks are at rest. Now, the upper spring is cut. The acceleration of 1 m
2 2 m m 5
block 1 just after cutting the upper spring is found y×g m/s . Find y?
3 m

4 2m

2. Two identical blocks each having a mass of 20 kg is connected to each other by a light
inextensible string as shown and is placed over a rough surface. Pulleys are connected to the
blocks. The acceleration of block B after 1 sec is y/13 m/s2 , after the application of the time
varying force of 40t N, where t is in sec.

3. A ball is projected on a smooth inclined plane in a direction

perpendicular to line of greatest slope with velocity of 8m/s. It’s 8m/s
speed (in m/s) after 1 sec is 2K then value of K will be. 37°

4. A train of length l = 350 m starts moving rectilinearly with constant acceleration a = 3 × 10–2 m/s2. After
30 s the headlight of the train is switched on (event 1). After 60s from this event the tail signal
light is switched on (event 2). The distance (in m) between these events in the reference frames
fixed to the earth is 121K then what will be value of K.

5. A particle of mass m is released in a circular tube of radius R. If a spring m

of constant K is placed in the tube as shown in figure and angle made by
radius vector with vertical when particle will come at instant rest is 60°, 
 6mg  
then the value of K is then find the value of  :
2 R

Space for Rough work

Chemistry PART – II

SECTION – A

Straight Objective Type

This section contains 10 multiple choice questions. Each question has four choices (A), (B), (C) and (D)
out of which ONLY ONE is correct.

1. Which of the following property is not common in both VBT as well as MOT?
(A) LCAO (B) Losing of nature of atomic orbital
(C) Mixing of atomic orbital (D) Wave nature of electron

2. Which of the following method is not used to convert ROH  RBr ?

1o
HBr NaBr
(A) ROH  (B) ROH 

NaBr NaBr
(C) ROH 
 (D) ROH  

Conc.H2SO 4 H3PO4

3. The Molecularity of a reaction is :

(A) Always two
(B) Same as its order
(C) Different than the order
(D) May be same or different as compared to order

4. During micelle formation hydrocarbon part is directed towards

(A) centre (B) periphery
(C) randomly (D) Alternatively

5. Which of the following is having maximum hydration energy?

(A) V3+ (B) Fe2+
3+
(C) Cr (D) Cu2+

6. Which of the following is an application of Bohr’s theory in an orbit?

(A) frequency of an electron (B) nodal plane
(C) no. of waves (D) probability of finding of electron

Space for Rough work

7. In which of the following Gf is maximum in acidic medium?

(A) O (B)

O O
O O
(C) (D)

 X
8. RX
o

 RH
3
Reagent (X) cannot be used in the above conversion
(A) LiAlH4 (B) NaBH4
(C) Raney Ni (D) All

9. Which type of nuclear spin isomer of hydrogen exist at the temperature at which sulphonation of
phenol gives major para substituted product
(A) Ortho (B) Para
(C) both (D) cant be defined

10. VBT can’t explain the geometry of

3 3
(A)  Cr  NH3 6  (B)  Co  NH3 6 
2 2
(C)  Cu  NH3 4  (D)  Zn  NH3 4 

Space for Rough work

Multiple Correct Choice Type

This section contains 5 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for its
answer, out which ONE OR MORE is/are correct.

11. Which of the following ions can be separated by using NH4Cl and NH4OH?
3+ 3+ 3+ 2+
(A) Fe and Cr (B) Cr and Co
3+ 3+ 3+ 2+
(C) Cr and Al (D) Al and Ba

12. Which of the following has/have three isomers (including geometrical optical, ionisation, hydrated,
linkage isomerism)?
(A) [Zn(NH3)(H2O) (Cl) (Br)] (B) [Cr(NH3)4(H2O)2]Br
(C) [Cr(NH3)5Cl]NO2 (D) K3[Co(C2O4)3]

13. Consider the following reactions


I
(I) :   (X)

H I

I
(II) :   (Y)

H I

I
(III) :   (Z)

H I
Select correct statement (s)
(A) All the carbocations (X), (Y) and (Z) are secondary
(B) X is stabilised by conjugation with the double bond and is formed fastest
(C) Z is least stable and formed slowest
(D) Z is conjugated cyclic system with four  electrons and thus is anti aromatic

Space for Rough work

14. In the following reaction

O
MeNH2  H
  Me N O
excess
Intermediates can be
(A) OH (B) OH

Me N

Me NH
OH
(C) (D) O
OH2

OH Me N
Me N

15. Which of the folllowing possess two lone pairs at central atom and square planar shape?
(A) SF4 (B) XeO4
(C) XeF4 (D) ICl4

Space for Rough work

SECTION –C
Integer Answer Type

This section contains 5 questions. Each question, when worked out will result in one integer from 0 to 9
(both inclusive).

1. Conducting wall

Porous diaphragm

C2H4  Pt / H2
Insulating wall

Liquid A Liquid B

In how many cases mixing of liquid (A) and liquid (B) is more effective if liquid (A) is H2O and
liquid (B) can be HCl, CH3CH2OH, HNO3 and CH3OCH3

2. The electron in the ground state for a H-atom absorbs a photon of wavelength 97.20 nm and it
reaches to maximum level ‘n’ then when it returns from this level to ground state. Maximum
numbers of lines can be obtained in its spectrum?

3. The emf of the cell,

M(s)|MnXn(0.01M) || MnXn(0.1M)/M(s) is found to be 0.0295 V at 25oC. If MnXn is a soluble salt,
find the value of n.

4. How many complexes will show optical but not geometrical isomerism?
2
Pt  en 2 Cl2  , Pt Py  NH3ClBr 
4
Pt  en  ox  Cl2  , Pt  en 3 

5. 25 ml of 11.2 volume H2O2 solution is titrated with 25 ml KMnO4/H+. If 1L same KMnO4 is used to
oxidize 1 mole of an element X so that it is converted into a single species in which ‘X’ contains a
o
single electron. If this single species is irradiated with light radiation of 114 A so that electron is
excited into a particular energy level. Determine no. of transitions possible from this energy level
to ground state

Space for Rough work

Mathematics PART – III

SECTION – A

Straight Objective Type

This section contains 10 multiple choice questions numbered 1 to 10. Each question has 4 choices (A),
(B), (C) and (D), out of which ONLY ONE is correct

3 3
 1
1. The minimum value of the function f(x) = x 2  x 2  4  x   for all permissible real x, is
 x
(A) –10 (B) –6
(C) –7 (D) –8

 
e2010  
1 1  ln x 
2. The value of the definite integral  1 dx , equals
e
x  x 
 ln x ln   
  ln x  
(A) 2009 – ln(2010 – ln 2010) (B) 2010 – ln(2009 – ln 2009)
(C) 2009 – ln(2010 – ln 2009) (D) 2010 – ln(2010 – ln 2010)

3. Let z1, z2 and z3 be three points on |z| = 1 and z1 + z2 + z3 = 0. If 1, 2 and 3 be the arguments of
z1, z2, z3 respectively then cos(1 – 2) + cos(2 – 3) + cos(3 – 1) is equal to
3
(A) 0 (B) –
2
3
(C) –1 (D)
2

4. A point (x, y) where function f(x) = [sin [x]] in (0, 2  ) is not continuous, is (where [.] denotes
greatest integer  x)
(A) (3, 0) (B) (2, 0)
(C) (1, 0) (D) (4, –1)

Space for rough work

5. The solution of differential equation 2x 3y dy + (1 – y2)(x2y2 + y2 – 1) dx = 0

2 2 2 2 2 2
(A) x y = (cx + 1)(1 – y ) (B) x y = (cx + 1)(1 + y )
2 2 2
(C) x y = (cx – 1)(1 – y ) (D) None of these

6. Let S be the array of integral points (x, y, z) with x = 0, 1, 2; y = 0, 1, 2, 3 and z = 0, 1, 2, 3, 4. If

two points are chosen from S, then the probability that their mid point is in S is
4 2
(A) (B)
59 59
52
(C) (D) None of these
177

7. The number of ways in which 15 boys and 2 girls can sit in a row such that between the girls at
the most 2 boys sit is
(A) 17! – (12! × 3!) (B) 17! – (12C3 × 3!)
12
(C) 17! – ( C3 × 15!) (D) 17! – (91 × 2! × 15!)
   
8. If a , b and c are three mutually perpendicular unit vectors and d is a unit vector which makes
      
equal angle with a , b and c , then | a + b + c + d |2 is equal to
(A) 4 (B) 4 ± 3
(C) 4 ± 2 3 (D) None of these

x y z
9. The line   makes an isosceles triangle with the planes 2x + y + 3z – 1 = 0 and
k 2 12
x + 2y – 3z – 1 = 0 then value of k is
(A) 71 (B) –72
(C) 73 (D) 74

 Min f(t) : 0  t  x ; 0 x 1
10. If f(x) = 4x3 – x2 – 2x + 1 and g(x) =  then
3  x ; 1 x  2
 1 3 5
g   + g   + g   has the value equal to
4
  4
  4
7 9
(A) (B)
4 4
13 5
(C) (D)
4 2

Space for rough work

Multiple Correct Choice Type

This section contains 5 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for
its answer, out of which ONE OR MORE is/are correct
2 2 2 2
11. The circles x + y – 4x – 81 = 0, x + y + 24x – 81 = 0 intersect each other at points A and B. A
line through point A meet one circle at P and a parallel line through B meet the other circle at Q.
Then the locus of the mid point of PQ is
(A) (x + 5)2 + (y + 0)2 = 25 (B) (x – 5)2 + (y – 0)2 = 25
2 2
(C) x + y + 10x = 0 (D) x2 + y2 – 10x = 0

12. Let S be the set of real values of parameter  for which the equation
f(x) = 2x3 – 3(2 + )x2 + 12x has exactly one local maximum and exactly one local minimum.
Then S is a subset of
(A) (–4, ) (B) (–3, 3)
(C) (3, ) (D) (–, 0)

14. If x2 – 3x + 2 be a factor of the expression x 4 – px2 + q, then

(A) roots of equation x4 – px2 + q = 0 are ± 1, ± 2
(B) roots of equation x4 – px2 + q = 0 are 1, 2, a + ib, a – ib (a, b R b ≠ 0)
(C) value of p is 5
(D) value of q is 4

n 4
39
15. If f(x) = ae2x + bex + cx satisfies the condition f(0) = –1, f(log 2) = 31 and  (f(x)  cx) dx = ,
0
2
then
(A) a = 5 (B) b = –6
(C) c = 2 (D) a = 3

Space for rough work

SECTION – C

(One Integer Value Correct Type)

This section contains 5 questions. Each question, when worked out will result in one integer from 0 to 9
(both inclusive)

1. A straight line passing through origin O meets the lines 4x + 2y = 3 and 2x + y + 6 = 0 at the
points P and Q respectively, then the point O divides the segment PQ in the ratio 1 : k then k is
equal to _____

2. The number of real solutions of cos1 x  cos1 2x   is _____

a b c 
3. If trace (A) > 0 and abc = 1 where A =  b c a  and AA = I, then the value of a3 + b3 + c3 is
 c a b 
_____

x
4. Let f: [0, )  R be a continuous strictly increasing function such that f 3  x    t f 2  t  dt  x  0
0
then value of f(6) is _____

x2 y2
5. The hyperbola – = 1 passes through the point of intersection of the lines x – 3 5 y = 0
a2 b2
and 5 x – 2y = 13 and the length of its latus rectum is 4/3 units. Then the value of 9e2 – 1 where
e is the eccentricity of the hyperbola is _____

Space for rough work

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ANSWERS, HINTS & SOLUTIONS

CRT –II
(Paper-1)
Q. No. PHYSICS CHEMISTRY MATHEMATICS
B A A
1.
ALL INDIA TEST SERIES

A D A
2.
B D B
3.
D A D
4.
A C C
5.
A A C
6.
B A D
7.
C A C
8.
C A D
9.
A C D
10.
A, C
A,B B, D
11.

B,D B, C C, D
12.
C,D A, B, C, D A, D
13.
A, B, C A, C, D
14. A,B,C
A,B,D C, D
15. A, B
2 1 4
1.
6 3 0
2.
5 2 4
3.
2 1 6
4.
6 3 9
5.

Physics PART – I
SECTION – A

1. Optical path length of ray entering through S1

L
aL2
x1   dx  L 
0
2
aL2
Path difference of ray from S1 and S2  x 
2
For minima at O
aL2 

2 2

v
2. Fundamental frequency f0 
2L
Beat frequency = f1  f2
v v
 
L  L 
2   L  2   L 
 2   2 
v 2 2 
   
2   L  2L  L  2L  
 4L  8f0 L
 2f0L  2 2

 L  4L  l
3. When mirror rotates, by  , reflected ray
rotates by 2  36 rad/s. 
dH d 10cot 
speed of spot =  H  10cot 
dt dt  
 d 
  10 cos ec 2  
 dt 
 Wall
 10  10m
 2
 36 
  0.6  
= 1000 m/s

4. Energy of nth state in H-atom is same as energy if 3nth state in Li .

5. v  2f0  l2  l1 
 v  f l  l1
   0  2
 v max f0 l2  l1
1 0.1  0.1
 
100 74  24
 1.2%

x
R R 
6. Magnitude of field at Point P  , y,  is
2 2
0Ir 0I R 0I 
B  2
  B
2 2R 2 2 2R R P
ˆi  kˆ 2 0
Unit vector in the direction of field, B̂  45 z
2 y

10
7. i  1A ; V  iR  1 4  4V
10

 6   12 
8. 2   2    4x  x = 25 cm.
 8   8  16 

9. Ammeter and 0.9  is in series

10  10 3  9.9   i  0.1 = i  0.99
So current = 0.99 + 0.01 = 1 Amp.

Chemistry PART – II

SECTION – A
1. VBT and MOT both consider wave nature of e-.

2. H3PO4 can act as a dehydrating agent.

4. Hydrocarbon part is hydrophobic part

3
5. Cr H2O 6  has maximum CFSE

v
6.  and both v and r can be determined by Bohr’s theory .
2 r

7. OH is aromatic hence Gf of O is maximum in acidic medium.

LiAlH4
8. RX  Alkene

9. is major at high T, at high temperature para hydrogen convert into ortho hydrogen.

SO3H

2
10. Cu  NH3  4  has square planar shape

12. (C) : [Cr(NH3)5Cl]NO2

: [Cr(NH3)5NO2]Cl
: [Cr(NH3)5ONO]Cl

15.
F F Cl Cl

Xe I

F F Cl Cl

SECTION –C

1. In H2O and CH3CH2OH mixing shows +ve deviation while in other cases it shows –ve deviation.

2. Electron will reach n = 4, so the maximum possible transitions for single electron are 4 → 3,
3 → 2, 2 → 1.

3. nM  MnnA  ne  at anode 

Mnnc  ne  
 2M at cathode 
 
Net all reaction: Mnnc 
 MnnA

Mnn 
0.059
log   
C
 0.0295 
n Mn 
n
 A
n = 2

4
4. Pt  en 3  is optically active but does not show geometrical isomerism.

5. N1V1  N2 V2
H2O2  KMnO4 / H

2 x 25 = N x 25
N=2
Again NV1 1 = gm eq. of (X)
KMnO4 / H

 2 x 1 = gm eq. of (X)
and n factor of x = 2
o
Since X is converted into single e species Hence X is Li2+ and light radiation of 114 A
corresponds to 108.77 eV.
 e is excited into 3rd energy level and total 3 transitions are possible.

Mathematics PART – III

SECTION – A

3 3
 1 D
1. f(x) = x 2  x 2  4 x  
 x I
3 2
 1   1   1  
f(x) =  x   –3 x   – 4  x    2
 x  x  x  2 3

1
Let x  = t (x > 0)
x
3 2
Let g(t) = t – 3t – 4t + 8
3 2
Now g(t) = t – 4t – 3t + 8 where t  [2, )
g'(t) = 3t2 – 8t – 3 = (t – 3)(3t + 1); g'(t) = 0  t = 3 (t  –1/3)
g''(t) = 6t – 8
g''(3) = 10 > 0  g(3) is minimum
g(3) = 27 – 9 – 36 + 8 = –10

2010
2010  ln 2010
 1 t  1
2. Substituting lnx = t, we get   1  dt = 2009   du (u = t – ln t)
1  t  t  (ln t)   1
u
= 2009 – ln(2010 – ln 2010)

3. Since, |z1| = |z2| = |z3| = 1 and z1 + z2 + z3 = 0

 Triangle ABC is an equilateral triangle
2
 1 – 2 = 2 – 3 = 3 – 1 =
3
3
 cos(1 – 2) + cos(2 – 3) + cos(3 – 1) = –
2

4. In (0, 2) check for 1, 2, 3, 4, 5, 6 function is not continuous for 4

Hence, (x, y) is (4, –1)

5 2x3ydy + (1 – y2)(x2y2 + y2 – 1) dx = 0, Let 1 – y2 = t  2y dy = –dt

3 2 3 2 2 2 2
–x dt + t(x (1 – t) – t) dx = 0  –x dt + tx dx – t x dx – t dx = 0
tdx  xdt  x 2  1  x  1
x2(t dx – x dt) = t2(x2 + 1) dx  2
  2  dx  d     1  2  dx
t  x  t  x 
x 1 x2  1
Integrate  x c  x  (1  y 2 )  c(1  y 2 )
t x x
x2 = (x2 – 1)(1 – y2) + xc(1 – y2)  0 = –x2y2 – 1 + y2 + xc – xcy2
 x2y2 = (cx – 1)(1 – y2)

6. Total number of coordinates in S = 3 × 4 × 5 = 60

 Total no. of way of selecting two points = 60C2
Now the selection of (x1, y1,z1 ) and (x 2 ,y 2 ,z2 ) should be in such a way that
x1 + x2, y1 + y2 and z1 + z2 are even
Now x1, x2, may be
 Favourable points = 5 × 8 × 13 = 520
520 52
 Probability = 
30  59 177

7. Number of ways = total – those in which at least three boys between the girls = 17  a
Now, x G1 y G2 z
x + y + z = 15
Min x = 0 max x = 12
Min y = 3 max y = 15
Min z = 0 max z = 12
Total solution = coefficient of x15 in (x0 + ….. x12)(x3 + x4 + ….. + x15)(x0 + ….. + x12)
3
 1  x13 
= ….. x12 in (1 + x + ….. x 12)3 = ….. x12 in  12 13 3
 = ….. x in (1  x ) (1  x)
3

 1  x 
12 3 1
= C2 14 C2
 a 14 C2  2  15 ways = 17  91 2 15
   
8. Let d makes angle  from a, b and c then cos2  + cos2  + cos2  = 1
  
( a, b, c mutually perpendicular)
1
 cos  
3
                   
| a  b  c  d |2 | a |2  | b |2  | c |2  | d |2 2(a  b  a  c  a  d  b  c  b  d  c  d)
= 1 + 1 + 1 + 1 + 2 (0 + 0 + cos + 0 + cos + cos ) = 4 + 6 cos = 4  2 3

9. The line makes equal angle with planes hence with normal
   
a n an
i.e.  1   2
| a || n1 | | a || n2 |
 ˆ n  2iˆ  ˆj  3k,
ˆ n  ˆi  2ˆj  3kˆ
a  k ˆi  2ˆj  12k, 1 2
2k  2  36 k  4  36
  k  74
14 14

10. f(x) = 12x2 – 2x – 2 D I

= 2[6x2 – x – 1]
= 2(3x + 1)(2x – 1)

–1/3 0 1/2 1

11. S1: x2 + y2 – 4x – 81 = 0 S2: x2 + y2 + 24x – 81 = 0

S1 – S 2 = 0 28x = 0
x = 0 equation of AB Now let A  (0, a)  A is on x = 0
Now put A in S1 or S2 0 + a2 + 0 – 81 = 0
a=±9 A = (0, 9), B = (0, –9)
Now PA, y – 9 = mx  y = mx + 9  BQ, y + 9 = mx  y = mx – 9
Now let P(x1, y1) and Q(x2, y2)
18m  24 9m2  24m  9
Now solve S1 and PA then x1  and y1 
2
m 1 m2  1
18m  4 9m2  4m  9
Now solve S2 and BQ then x 2  and y 2 
m2  1 m2  1
Let mid pt of PQ is (h, k) then x 1 + x2 = 2h and y1 + y2 = 2k
20
x1  x 2  2  2h
m 1

h 1

k m
20m 20
y1  y 2  2
 2k  2
 2h
m 1 k 
  1
h
h(h2 + k2 + 10h) = 0

12. For one local max and min f(x) = 0 will have two real and unequal roots
2 2 2
f(x) = 6x – 6(2 + )x + 12  D = (2 + ) – 8 > 0  ( – 2) > 0    R – {2}

13 P lies on the line 2x + 3y + 1 = 0 ….. (1)

|PA – PB| is maximum if P, A, B are collinear.
Equation of the line AB is x + y = 2 ….. (2)
P is the point of intersection of (1) and (2) which is (7, – 5)
Equation of perpendicular bisector of AB is
y=x ….. (3)
 1 1
For minimum value of |PA – PB|, P is the point of intersection of (1) and (3) which is   ,  
 5 5

14. Since x2 – 3x + 2 is factors of x 4 – px2 + q, then roots of x 2 – 3x + 2 = 0

i.e. 2, 1 are also roots of x4 – px2 + q = 0
but if a is roots of x4 – px2 + q = 0, then –a is also
roots of x4 – px2 + q = 0 are ±1, ±2
and 1 – p + q = 0 and 16 – 4p + q = 0, p = 5 and q = 4

15. f(x) = ae2x + bex + cx

Since f(0) = a + b i.e. a + b = – 1 ….. (1)
f(x) = 2ae2x + bex + c
 f(log 2) = 2ae2 log 2 + belog 2 + c
= 8a + 2b + c = 8a + 2b + c = 31 ….. (2)
n 4 n 4

 (ae2x + ex + cx – cx) dx =  (ae2x + bex) dx

0 0
log 4
 ae 2x
 a e2 n4 a a 15a 39
=   bex  = + be n 4 – – b = 8a + 4b – –b= + 3b =
 2  0 2 2 2 2 2
i.e. 15a + 6b = 39 ….. (3)
From (1), (2) and (3) we get 9a = 45
 a = 5, b = – 6 and c = 3

SECTION – C

1. Straight line from Origin is y = mx

3 3m
4x + 2y = 3, y = mx , x1  , y1 
(2m  4) (2m  4)
6 6m
2x + y + 6 = 0, y = mx, x 2  , y2 
(m  2) (m  2)
6m 3mk 6 3k 3k
  6 
O
(m  2) 2(m  2)
 O
m  2 2(m  2)
 2  0  3 k  6 , k  12  4
k 1 k 1 k 1 2 3

2. cos–1 x + cos–1 2x = –, cos–1 x  [0, ]

cos–1 x + cos–1 2x = –  no solution

3. abc = 1, trace A = a + b + c > 0

a b c  a b c   a2  b2  c 2 ab  bc  ca ac  ab  bc 
 
I =  b c a   b c a  =  ab  bc  ca b2  c 2  a2 bc  ca  ab 
 c a b   c a b   2 2 2 
 ca  ab  bc bc  ca  ab c  a  b 
 a2 + b2 + c2 = 1, ab + bc + ca = 0
(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca) = 1
a+b+c=1
 a3 + b3 + c3 = 3abc + (a + b + c) (a2 + b2 + c2 – ab – bc – ca) = 3 + 1(1 – 0) = 4

x
4. f 3 (x)   t  f 2 (t)dt ….. (1)
0
Differentiate with respect to x
x
3f 2 (x)  f '(x)  x  f 2 (x)  f '(x) 
3
x2
 f(x)  C ….. (2)
6
But from (1) f 3(0)= 0  f(0) = 0
x2 62
 From (2) C = 0  f(x)   f(6)  6
6 6

5. The line x – 3 5 y = 0 and 5 x – 2y = 13 intersect at the point. Since the given hyperbola
passes through this point,
(3 5 )2 1 45 1
2 2
= 1 or 2 – 2 = 1 ….. (1)
a b a b
2b2 4
Also length of the latus rectum = = (given) ….. (2)
a 3
45 3
From (1) and (2) we get 2
– =1
a 2a
15
 a = 6 or –
2
 a > 0 therefore we have a = 6
 b2 = 4
2 10 2
e = hence 9e – 1 = 10 – 1 = 9
9

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Paper 2

Time Allotted: 3 Hours Maximum Marks: 180

 Pl ea s e r ea d t h e i n s t r u c t i o n s c a r ef u l l y. Yo u a r e a l l o t t ed 5 m i n ut es
s p ec i f i c a ll y f o r t h i s p u r p o s e.
ALL INDIA TEST SERIES

 Yo u a r e n o t a ll o wed t o l ea v e t h e E xa m i n at i o n Ha l l b ef o r e t h e en d o f
t h e t es t .

B. Filling of OMR Sheet

C. Marking Scheme For All Three Parts.

(i) Section-A (01 to 08) contains 8 multiple choice questions which have one or more than one
correct answer. Each question carries +3 marks for correct answer and – 1 mark for wrong
answer.

Section-A (09 to 16) contains 4 paragraphs with each having 2 questions. Each question carries
+3 marks for correct answer and – 1 mark for wrong answer.

Section-A (17 – 20) contains 4 Matching Lists Type questions: Each question has four
statements in LIST I & 4 statements in LIST II. The codes for lists have choices (A), (B), (C), (D)
out of which only one is correct. Each question carries +3 marks for correct answer and
– 1 mark for wrong answer.

Name of the Candidate

Enrolment No.

Useful Data

PHYSICS
Acceleration due to gravity g = 10 m/s2

Planck constant h = 6.6 1034 J-s

Charge of electron e = 1.6  1019 C

Mass of electron me = 9.1  1031 kg

Permittivity of free space 0 = 8.85  1012 C2/N-m2

Density of water water = 103 kg/m3

Atmospheric pressure Pa = 105 N/m2

Gas constant R = 8.314 J K1 mol1

CHEMISTRY

Gas Constant R = 8.314 J K1 mol1

Atomic No: H=1, He = 2, Li=3, Be=4, B=5, C=6, N=7, O=8,

Physics PART – I

SECTION – A

Multi Correct Choice Type

This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D)
out of which ONE or MORE THAN ONE are correct.

ke2
1. Suppose the potential energy between electron and proton at a distance r is given by .
3r 3
Application of Bohr’s theory to hydrogen atom in this case shows that
(A) Energy in the nth orbit is proportional to n6
(B) Energy is proportional to m3 .  m : mass of electron 
(C) Energy of the nth orbit is proportional to n2
(D) Energy is proportional to m3 (m = mass of electron)

P2
2. During an experiment an ideal gas is found to obey a condition = constant (  = density of

gas). The gas is initially at temperature T0, pressure P0 and density 0 . The gas expands such
0
that density changes to . Then which of the following is/are correct:
2
(A) The pressure of the gas changes to 2P0
(B) The temperature of the gas changes to 2T0
(C) The graph of the above process on the P – T diagram is parabola.
(D) The graph of the above process on the P – T diagram is hyperbola.

3. The galvanometer shown in the figure has resistance 10 .

It is shunted by a series combination of a resistance S = 1
and an ideal cell of emf 2V. A current 2A passes as shown.
Then choose the correct statement(s).
(A) The reading of the galvanometer is 1A
(B) The reading of the galvanometer is zero
(C) The potential difference across the resistance S is 1.5 V
(D) The potential difference across the resistance S is 2V

Space for Rough work

4. Figure shows two blocks A and B connected to an ideal 10 kg

pulley string system. In this system when bodies are B
released then:
(A) Acceleration of block A is 1 m/s2
(B) Acceleration of block B is 2 m/s2
(C) Tension in string connected to block B is 40 N
(D) Tension in string connected to block B is 80 N

60 kg A

5. In the circuit shown, each capacitor has a capacitance C. The

emf of the cell is E. If the switch S is closed
(A) positive charge will flow out of the positive terminal of the cell
(B) positive charge will enter the positive terminal of the cell
(C) The amount of charge flowing through the cell will be CE.
(D) The amount of charge flowing through the cell will be 4/3 CE.

6. A point-like body of mass m and of charge Q is initially at rest on the

horizontal tabletop as shown in the figure and is given a v 0 vertical,
upward speed. There is a horizontal electric field E. Acceleration due
to gravity = g. E
v0
v2
(A) Maximum height attained  0 m, Q
2g
2QEv 20
(B) Particle will collide with the table at horizontal distance
mg2
(C) At the highest point acceleration will be only in horizontal direction.
mg
(D) Particle will collide with the table at an angle tan   with horizontal.
2QE

Space for Rough work

7. X and Y are large parallel conducting plates close to each X Y

other. Each face has an area A. X is given a charge Q, Y is
without any charge. Point A, B and C are shown.
A B C

Q
(A) Electric field at B is
20 A
Q
(B) Electric field at B is
0 A
(C) The fields at A, B and C are of same magnitude
(D) The fields at A and C are in opposite direction.

8. In the given AC circuit, when switch S is at position 1, L= 3 mH 1


the source emf leads current by . Now, if the switch S
6
is at position 2, then 2
1000 F
C=
3
~
s V=V0sin 1000 t
 
(A) current leads source emf by (B) current leads source emf by
4 3
 
(C) source emf leads current by (D) source emf leads current by
4 3

Space for Rough work

Comprehension Type
This section contains 4 groups of questions. Each group has 2 multiple choice question based on a
paragraph. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which ONLY ONE is
correct

Paragraph for Questions 09 & 10

A block of mass M is kept on a smooth horizontal floor. A variable force t=0 F = kt
F = kt is acting on the block at an angle of  with the horizontal. At initial u=0 
moment the block is at rest (the situation shown in the figure below). M
The block leaves the contact with floor at time t = t0 Smooth

9. The value of t0 is
Mg Mg
(A) (B)
k cos  k sin
Mgtan  Mgcot 
(C) (D)
k k

10. The displacement of the block till time t = t0 is

M2 g3 cot  cosec 2  M2 g3 cot  cosec 2 
(A) (B)
6k 2 2k 2
M g tan  sec 2 
2 3
M g tan  sec 2 
2 3
(C) (D)
6k 2 2k 2

Paragraph for Questions 11 & 12

A force acting on a particle depends on the particles position in y a C
the x-y plane. This force is given by the expression B
F  x 2 y 2 ˆi  x 2 y 2 ˆj(N). The particle is moved in the x-y plane a
along various paths as shown in the figure.
O A x

11. Work done by the force along the path OAC is

a5 a4
(A) J (B) J
3 9
2a5
(C) J (D) zero
5

12. Work done by the force along the path OC is

a5 3a5
(A) J (B) J
3 8
2a5 a5
(C) J (D)  J
5 3

Space for Rough work

Paragraph for Questions 13 & 14

A pendulum bob has mass m. The length of pendulum is  . y
z
It is initially at rest. A particle P of mass m/2 moving g

horizontally along –ve x-direction with velocity 2g collides 

2 x

with the bob and comes to rest. When the bob comes to rest
momentarily, another particle Q of mass m moving
horizontally along z-direction collides with the bob and sticks
to it. It is observed that the bob now moves along a
horizontal circle. The floor is a horizontal surface at a
distance 2  below the point of suspension of the pendulum.

13. Tension in string immediately after the first collision is

(A)2mg (B)mg
3 5
(C) mg (D) mg
2 2

14. The height of circular path above the floor is

3 4
(A) (B)
2 3
5
(C) (D) data not sufficient
4

Space for Rough work

Paragraph for Questions 15 & 16

A uniform wire frame of linear mass density  having three A P B

sides each of length 2a is kept on a smooth horizontal surface. y
An impulse J is applied at one end as shown in the figure. P is
the midpoint of AB. Now answer the following question. x

15. The angular velocity of system just after the impulse

3J J
(A) 2
(B)
22a 22a2
2J 4J
(C) 2
(D)
22a 22a2

16. The velocity of point P just after the impulse is

J J
(A) ĵ (B) ĵ
a 6 a
J  2 ˆ 1 ˆ J  1 ˆ 1 ˆ
(C)  i  j (D)  i  j
a  11 6  a  11 6 

Space for Rough work

(Match list Type)

This section contains 4 multiple choice questions. Each question has matching lists. The codes for the
lists have choices (A), (B), (C) and (D) out of which ONLY ONE is correct

17. In L – C – R series circuit suppose r is the resonance frequency, then match the following table
List-I List-II
(P) If   r 1. Current will lead the voltage
(Q) If   r 2. Voltage will lead the current
If   2r 3. XL  2XC
(R)
(S) If   r 4. Current and voltage are in phase
Codes:
P Q R S
(A) 4 2 1 3
(B) 2 4 2 1
(C) 1 2 3 4
(D) 2 4 3 1

18. A ray of light falls normally on an equilateral prism of refractive index   3 . Match the following
table :
i1 = Angle of incidence
i2 = Angle of emergence from second surface of prism
r1 = Angle of refraction at the first surface of prism
r2 = Angle of incidence at second surface of prism
List-I List-II
(P) Angle i1 1. 0°
(Q) Angle r1 2. 90°
3.
(R) Angle i2 60°
(S) Angle r2 4. None
Codes:
P Q R S
(A) 4 2 1 3
(B) 2 3 4 1
(C) 1 2 3 4
(D) 1 1 4 3

Space for Rough work

19. In the arrangement shown in fig. match

the following:

List-I List-II
(P) Velocity of centre of mass 1. 2 SI unit
Velocity of combined mass when
(Q) compression in the spring is 2. 1 SI unit
maximum
3.
(R) Maximum compression in the spring 4 SI unit
Maximum potential energy stored in
(S) 4. 0.5 SI unit
the spring
Codes:
P Q R S
(A) 4 2 1 3
(B) 2 3 4 1
(C) 2 2 2 1
(D) 1 2 2 1

20. A tube is inverted in a mercury vessel P

as shown in fig. P is the pressure
inside the tube above the mercury level h
as shown.

List-I List-II
(P) If P is increased, then height h 1. Increase
(Q) If P is increased, then pressure at O 2. Decreases
If P is increased, then pressure at 1 3.
(R) Remains same
cm above O
If the vessel is moved upwards with Increase or decreases depending on
(S) 4.
acceleration (a<g), then height h the size of the tube
Codes:
P Q R S
(A) 4 2 1 3
(B) 2 3 3 2
(C) 1 2 3 4
(D) 2 4 3 1

Space for Rough work

Chemistry PART – II

SECTION – A

Straight Objective Type

This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D)
out of which ONE or MORE than is/are correct.

1. Select pair(s) in which both produces same coloured ppt. with CrO24 ion
(A) Ba2+, Sr2+ (B) Pb2+, Ba2+
+ 2+ +
(C) Ag , Ba (D) Ag , Hg22

2. Which of the following property (s) is/are related to solution of rosy red complex of Ni2+ and
dimethyl glyoximate ligand?
(A) Geometry around Ni is square planar and diamagnetic in nature
(B) Complex is stabilized by internal hydrogen bonding
(C) it is an organometallic complex
(D) Five member chelate ring is formed when each molecule of dimethyl glyoximate is bonded
with metal ion.

3. In the given reaction

O O
R C  X
OH  R C OCH3

X will be
(A) CH2N2 (B) CH3OH/H+
(C) MeCOOH (D) Me2SO4

4. In a cell, Zn(s) | Zn2+ ||H+ | H2 (pt)

the addition of H2SO4 to the cathode compartment will
(A) Decrease E (B) Increase E
(C) Shift equilibrium to left (D) Shift equilibrium to right

Space for Rough work

5. Select the correct staement (s) among the following

(A) Increase in concentration of reactant increases the rate of zero order reaction
(B) Rate constant K is equal to collision frequency A, if Ea = 0
(C) Rate constant K is equal to collision frequency A if Ea = 
(D) Graph of log10K vs 1/T is straightline

6. Self reduction is not used in the extraction of

(A) Lead (B) Manganese
(C) Copper (D) Iron

7. The Cl2 evolve at anode during electrolysis of fused MgCl2 which produces 13 g Mg. (Atomic
mass of Mg = 24.3) is
(A) 11.98 lit (B) 0.535 mole
(C) 37.98 gm (D) 1.068 mole

8. Which are aromatic compounds?

(A) (B) O

Space for Rough work

Paragraph for Question Nos. 9 to 10

An organic compound (X) is subjected to reductive ozonolysis to give a product (A) which on heating with
dil. NaOH gives another product (B). Compound (B) on heating with sodium hypoiodite followed by
acidification produces (C). Compound (C) on treatment with SOCl2 and then with diazomethane followed
by hydrolysis gives a product which on heating gives 1, 2-dimethyl cyclopentene.
Now answer the following questions.

9. Organic compound (X) in the above passage is:

(A) 1,2-dimethyl cyclopentene (B) 1,2-dimethyl hexene
(C) 1,2-dimethyl cyclohexene (D) 2,5-dimethyl-3-hexene.

10. Intermediate species formed in the reaction of (C) with SOCl2/CH2N2/H2O/ is:

(A) O (B)
C C H CH C O

(C) O (D) All of these

CH2 C OH

Space for Rough work

Paragraph for Question Nos. 11 to 12

N2 H2 O
A 
Burn
B  C  D  gas 
CO 2

Milkyness E 

AClX  green colour in flame.

11. What is “A”?

(A) Ca (B) Mg
(C) Ba (D) None

D
12. KI  excess   HgI2  F 
KOH
G
G is
(A) Hg (B) +
NH2 I
+
O NH2 I Hg
O
Hg Hg
(C) Hg (D) None
+
O NH3 I

Space for Rough work

Paragraph for Question Nos. 13 to 14

Redox reactions play a vital role in chemistry and biology. The values of Eored of two half cell reactions
decide which way the reaction is expected to proceed. A simple example is Daniel cell in which zinc goes
into solution and copper gets deposited. Given below are a set of half-cell reactions (acidic medium)
along with their Eo values.
I2  2e   2I  ; E o  0.54V
Cl2  2e   2Cl  ; E o  1.36V
Mn3   e  Mn2  ; E o  1.50V
Fe3   e  Fe2  ; E o  0.77V
O2  4H   4e   2H2 O; E o  1.23V

13. Fe3+ is stable but Mn3+ is not stable in acidic solution because:
(A) O2 oxidises Mn2+ to Mn3+
2+ 2+ 3+ 3+
(B) O2 oxidises both Mn and Fe into Mn and Fe respectively.
3+
(C) Fe oxidises H2O to O2
(D) Mn3+ oxidises H2O to O2.

14. Among the following identify the correct statement:

(A) Chloride ion is oxidized by O2 (B) Fe2+ ion is oxidized by iodine
(C) Iodide ion is oxidized by chlorine (D) Mn2+ is oxidized by chlorine.

Space for Rough work

Paragraph for Question Nos. 15 to 16

When 20.02 g of a white solid (X) is heated, 4.4 g of an acid gas (A) and 1.8 g of a neutral gas (B) are
evolved leaving behind a solid residue (Y) of weight 13.8 g (A) turns lime water milky and (B) condenses
into a liquid which changes anhydrous copper sulphate blue. The aqueous solution of Y is alkaline to
litmus and gives 19.7 g of white precipitate (Z) with barium chloride solution. (Z) gives carbon-dioxide with
an acid.

15. The white solid ‘X’ is:

(A) metal sulphate (B) metal chloride
(C) metal carbonate (D) metal bicarbonate

16. Atomic weight of the metal is:

(A) 23 g/mole (B) 39.0 g/mole
(C) 6 g/mole (D) 24 g/mole

Space for Rough work

(Match list Type)

This section contains 4 multiple choice questions. Each question has matching lists. The codes for the
lists have choices (A), (B), (C) and (D) out of which ONLY ONE is correct

17. Match the List – I with List – II.

List I List II
(Molecule) (Shape)
(P) N2O (1) Angular
(Q) F2O (2) Tetrahedral
(R) BF4 (3) Planar triangular
(S) CH3 (4) Linear

Codes:
P Q R S
(A) 1 2 3 4
(B) 2 1 3 4
(C) 2 1 4 3
(D) 4 1 2 3

18. Match List-I with List-II.

List – I List – II
Transformation Name
O
(P) O (1) Carbyl amine reaction
H 3C C Cl

AlCl
 C CH3
3

(Q) O O (2) Wittin Reaction

MCPBA
O


(R) O (3) Bayer-Villiger

R
oxidation
C  Ph3P  CH2  C CH2
R R
R
(S) CHCl3 /KOH
Ph  NH2  Ph  NC (4) Friedel Craft’s
Acylation
Codes:
P Q R S
(A) 1 2 3 4
(B) 2 1 3 4
(C) 2 1 4 3
(D) 4 3 2 1

Space for Rough work

19. Match List-I with List-II.

List-I List-II
(P) Fe(NO3)2 (aq) (1) Only cationic hydrogen
(Q) KClO4 (aq) (2) Only anionic hydrolysis
(R) HCOONa (aq) (3) Both cationic as anionic hydrolysis
(S) NH4CN (aq) (4) No hydrolysis
Codes:
P Q R S
(A) 1 2 3 4
(B) 1 4 2 3
(C) 2 1 4 3
(D) 4 3 2 1

20. Identify the correct matching from the codes given below
Reaction Product

(P) Borax  (1) H2B 4 O7

(Q) Borax  NH4 Cl   (2) NaBO2  B2O3
HCl
(R) Borax   (3) BN
(S) Borax  H2O (4) H3BO3

Codes:
P Q R S
(A) 1 2 3 4
(B) 2 1 3 4
(C) 2 3 4 1
(D) 4 3 2 1

Space for Rough work

Mathematics PART – III

SECTION – A

Straight Objective Type

This section contains 8 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D),
out of which ONE or MORE than is/are correct

1. Number of ways in which 200 people can be divided in 100 couples is

(200)!
(A) 100 (B) 1.3.5. ......199
2 (100)!
 101   102   200  (200)!
(C)    2  .......  2  (D)
 2     (100)!

2. Consider the quadratic equation (a + c – b) x2 + 2cx + (b + c – a) = 0 where a, b, c are distinct

real number and a + c – b  0. If both the roots of the equation are rational then the numbers
which must be rational are
c
(A) a, b, c (B)
ab
bc a b2
(C) (D)
ac b ca

3. If a, b, c are first three terms of a G.P. if the harmonic mean of a and b is 12 and arithmetic mean
of b and c is 3, then
(A) no term of this G.P. is square of an integer (B) arithmetic mean of a, b, c is 3
(C) b = ± 6 (D) common ratio of this G.P. is 2

Space for rough work

x2 y2
4. If the normal at any given point P on ellipse + = 1 meets its auxiliary circle at Q and R such
a2 b2
that QOR = 90º, where O is centre of ellipse, then
(A) 2(a2 – b2)2 = a4 sec2  + a2b2 cosec2  (B) a4 + 5a2 b2 + 2b4 = a4 tan2  + a2b2 cot2 
4 2 2 4 3 4 4 2 2 3
(C) a + 5b a + 2b ≥ 2a b (D) a + 2b ≥ 5a b + 2a b

x 2  5x  3 2x  5 3
2
5. If 3x  x  4 6x  1 9  ax 3  bx 2  cx  d then which of the following are correct?
7x 2  6x  9 14x  6 21
(A) a = 0 (B) b = 0
(C) c = 0 (D) d = 0

6. Let (1 + x2)2 (1 + x)n = A0 + A1 x + A2 x2 + ...... If A0, A1, A2 are in A.P. then the value of n is
(A) 2 (B) 3
(C) 5 (D) 7

7. A line passes through the points whose position vectors are î + ĵ – 2 k̂ and î – 3 ĵ + k̂ . The
position vector of a point on it at a unit distance from the first point is/are
1 ˆ ˆ ˆ 1 ˆ
(A) (5i  j  7k) (B) (5i  9ˆj  13k)
ˆ
5 5
1 ˆ ˆ ˆ 1 ˆ
(C) (6i  j  7k) (D) (4i  9 ˆj  13k)
ˆ
5 5

 
dx x 2 dx
8. Let u =  and v =  then
0 x 4  7x 2  1 0 x  7x 2  1
4

(A) v > u (B) 6 v = 

Space for rough work

Comprehension Type

This section contains 4 groups of questions. Each group has 2 multiple choice question based on a
paragraph. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which ONLY ONE is
correct

Paragraph for Question Nos. 9 to 10

Read the following write up carefully and answer the following questions:
Continuous probability distributions, a continuous distribution is one in which the variate may take any
value between certain limits a and b, a < b. Suppose that the probability of the variate X falling in the
1 1
infinitesimal interval x  dx to x + dx is expressible as f(x) dx, where f(x) is a continuous function of x
2 2
1 1
Symbolically, P(x – dx  X  x + dx) = f(x) dx where f(x) is called the probability density function
2 2
(abbreviated as p.d.f.) or simply density function. The continuous curve y = f(x) is called probability curve;
and when this is symmetrical, the distribution is said to be symmetrical. Clearly, the probability density
function possesses the following properties
(i) f(x)  0 for every x in the interval [a, b], a < b
b
(ii)  f(x) dx = 1, a, b > 0 since the total area under the curve is unity
a
(iii) Furthermore, we define for any [c, d], where c, d  [a, b], c < d;
d
P(c  X  d) =  f(x) dx ..... (1)
c
We define F(x), the cumulative distribution function (abbreviated as c.d.f.) of the random variate X
where F(x) = P(X  x)
x
or F(x) =  f(x) dx ..... (2)
a

2x ; 0  x  1 1
9. If f(x) =  then the probability that x  is
0 ; x 1 2
1 1
(A) (B)
2 4
3 1
(C) (D)
4 8

10. Suppose the life in hours (x) of a certain kind of radio tube has the probability density function
100
f(x) = 2 when x > 100 and zero when x < 100. Then the probability that none of three such
x
tubes in a given radio set will have to be replaced during the first 150 hours of operation, is
1 8
(A) (B)
27 27
1 26
(C) (D)
225 27

Space for rough work

Paragraph for Question Nos. 11 to 12

Read the following write up carefully and answer the following questions:
Let f be a function defined so that every element of the co-domain has at most two pre-images and there
is at least one element in the co-domain which has exactly two pre-images we shall call this function as
“two-one” function. A two-one function is definitely a many one function but vice-versa is not true. For
example, y = |ex – 1| is a “two-one” function. y = x 3 – x is a many one function but not a “two-one”
function. In the light of above definition answers the following questions:

11. In the following functions which one is a “two-one” function

(A) y = n x (B) y = x2 sin x
(C) y = x3 + 3x + 1 (D) y = x4 – x + 1

12. Let f(x) = {x} be the fractional part function. For what domain is the function “two-one”?
1 5  1 3
(A)  ,  (B)   , 
2 2  2 2
(C) [1, 2) (D) none of these

Paragraph for Question Nos. 13 to 14

Read the following write up carefully and answer the following questions:
v( x)
dy dy
If y =  f(t) dt , let us define in a different manner as = v(x) f 2(v(x)) – u(x) f 2(u(x)) and the
u( x)
dx dx
 dy 
equation of the tangent at (a, b) as y – b =   (x – a)
 dx (a, b)

x2
2
13. If y = t dt , then equation of tangent at x = 1 is
x
(A) y = x + 1 (B) x + y = 1
(C) y = x – 1 (D) y = x

x
2 d
14. If F(x) =  et /2
(1 – t2) dt, then F(x) at x = 1 is
1
dx
(A) 0 (B) –1
(C) 1 (D) 2

Space for rough work

Paragraph for Question Nos. 15 to 16

Read the following write up carefully and answer the following questions:
In reducing a given equation to the standard forms  sin x  sin  etc  we apply several trigonometric or
algebraic transformations. As a result of which the canonical form finally obtained may not be equivalent
to the original equation resulting either in loss of solutions or in the appearance of fake solutions

15. The solution set of the equation 5  2sin x  6 sin x  1 is

n  2 n  n 
(A) n   1 sin1    or x  n   1 (B) n   1
 9 6 6
n 2
(C) n   1 sin1   (D) Null set
9

16. The equation 2 cot 2x  3 cot 3x  tan 2x has

   
(A) two solutions in  0,  (B) one solution in  0, 
 3  3
(C) no solution in  ,   (D) none of these

(Match list Type)

This section contains 4 multiple choice questions. Each question has matching lists. The codes for the
lists have choices (A), (B), (C) and (D) out of which ONLY ONE is correct

17. Matching the following List-I with List-II.

List – I List – II
(P) The least positive integral solution of x2 – 4x > cot–1 x 1. 1
(Q) The least positive integral value of a for which
2. 2
f(x) = 2ex – ae–x + (2a + 1)x – 3 is increasing
(R) Let f: R  R be such that f(a) = 1, f(a) = 2,
1
 f 2 (a  x)  x k
3. 5
then lim   = e , then k is equal to
x0
 f(a) 
(S) Number of integral values of x which satisfy equation
4. 4
sin–1((3 – x)(x – 1)) + sin–1(2 – |x|) = π/2 is/are
Codes:
P Q R S
(A) 3 1 2 4
(B) 1 2 3 4
(C) 4 3 2 1
(D) 3 1 4 2

Space for rough work

18. Matching the following List-I with List-II.

List – I List – II
2
(P) If y = 2x + c – 4 is a normal to the parabola y = 4x, then value of
1. 12
‘–c’ is

x2 y 2
(Q) If the mid point of a chord of the ellipse + = 1 is (0, 3), then
16 25
2. 8
4k
length of the chord is , then k + 4 is equal to
5
(R) The product of the lengths of the perpendiculars from the two
x2 y2
focii on any tangent to the hyperbola – = 1 is k , then k 3. 3
25 3
is
2 2
(S) The line x + y = a touches the hyperbola x – 2y = 18, if a is
4. 9
equal to ±b, then value of |b| is
Codes:
P Q R S
(A) 1 2 3 4
(B) 2 1 3 4
(C) 2 1 4 3
(D) 4 3 2 1

19. Matching the following List-I with List-II.

List – I List – II
(P) If the sum of unit vectors is a unit vector, then magnitude of difference is 1
1.
x , then the value of x is 2

   
(Q) If f(x) = sin x,  x  0,  , f(x) + f( – x) = 2,  x   ,  and
 2  2  2. 2
f(x) = f(2 – x),  x  (, 2), then the area enclosed by y = f(x) and the
x-axis is k sq. units. Then k is

(R) Let f and g be differentiable function satisfying g(a) = 2, g(a) = b and

3. 3
g = f–1 then f(b) is equal to

 1  x2  1  2x 1  x 2  1
(S) The derivative of tan1   w.r.t tan1   at x = 0 is 4.
 x   1  2x 2  4
   

Codes:
P Q R S
(A) 1 2 3 4
(B) 3 2 1 4
(C) 3 1 4 2
(D) 4 3 2 1

Space for rough work

FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942
website: www.fiitjee.com
TG ~ @bohring_bot < 3 25
AITS-CRT-II-(Paper-2)-PCM-JEE(Advanced)/15

20. Matching the following List-I with List-II

List – I List – II

(P) If a, b, c  N and a, b, c  100, then probability that a, b, c are 5

1.
individually divisible by both 2 and 3 28
nd
(Q) Probability that in a year of 22 century chosen at random there will 31
2.
be 53 Sundays 32

(R) If a unbiased coin is tossed 5 times then probability of getting at 5

3.
least 1 head is 12
3 1 2 4
(S) If P(A  B)  ,P(A  B)  , P(A)  , then P(A  B) is 4.
4 4 3 1155

Codes:
P Q R S
(A) 4 1 3 2
(B) 1 2 3 4
(C) 3 4 2 1
(D) 4 1 2 3

Space for rough work

ANSWERS, HINTS & SOLUTIONS

CRT –II
(Paper-2)

Q. No. PHYSICS CHEMISTRY MATHEMATICS

A, B A, B, D A, B, C
1.
ALL INDIA TEST SERIES

B, D A, B, D B, C
2.
B, D A, B, D A, B
3.
B, D B, D A, D
4.
B, D A, B, C
5. A,D
A,B,D B, D A, B
6.
A,C,D A, B, C A, B
7.
A A, B, C, D B, C, D
8.
B C B
9.
A D B
10.
A C D
11.
C A B
12.
C D C
13.
C C A
14.

A D B
15.
D B C
16.
A D D
17.
C D C
18.
C B B
19.
B C D
20.

Physics PART – I
SECTION – A

dU ke2
1. F   4
dr r
mv 2
F
r
ke2 mv 2
 4  ... 1
r r
nh
and mvr  ...  2 
2
ke2 42 m Am
From (1) and (2) r   2
n2 n2 n
2
1 1 ke
T.E. = P.E   3
2 2  Am 
3 2 
 n 
6 3
T.E.  n , T.E.  m

P2
2.  constant

P2  nRT 
 = constant
MP
 PT  constant
3. In loop ABCDA 2A A i 10  B
10i  2   2  i   0
i0
Potential difference across S   2  i 1  2V 2 -i
D C
1 2V

8. L/R = tan/6  R = 3
tan = 1/CR=1
 =/4, leading P.D.

Chemistry PART – II

SECTION – A
1. CrO24
Ba 2
  BaCrO 4 

2 CrO24 
Pb   PbCrO 4  Yellow ppt.

CrO24 
Sr 2   SrCrO 4 


CrO24
2Ag   Ag2CrO 4 
 Re d ppt.
CrO24
Hg22   Hg2CrO 4 

2. O OH
H3C C N N C CH3

+
Ni 2

H3C C N N C CH3
HO O
[Ni(dmg)2]

4. 

Zn  s   2H  aq  2
 Zn  aq   H2 ....(1)
 Zn2  H2 
0.059
E E0cell
 log  
2
.....(2)
n H 
 
Equilibrium (1) will shift in forward direction and cell potential E will increase.

5. Rate of zero order reactions is independent of the concentration of reaction

In this equation K = AeEa / RT
K = A when Ea = 0
Ea
logK  log A 
2.303RT
Plot of log K against 1/T is straight line with negative slope.

7. At anode At cathode

2Cl  Cl2  2e 
Mg2  2e   Mg
No. of gm equivalent of Mg at cathode   No of gm equivalent of Cl2 at Anode 
13 WCl2

24.3 / 2 35.5

WCl2  37.98 gm
or nCl2  0.535
VCl2 = 11.98 lit at NTP

9.
O3 H2O Zn O OH O
O (Self Aldol Condensation)

(iodoform reaction in H+) NaIO / 

CH3 (i) SOCl2 CH3

(ii) CH2 N2

O O
(iii) H2O / H 
CO2
CH2 C OH (Arndt Estert Synthesis) C OH
(1,2-dimethyl
cyclopentene)

10.
O SOCl2 O
O
C OH C Cl
C Cl CH2 N N
H2C N N

O O O
N2 H 
C C H C CH N N C CH2 N N

(Carbene)

H2O 
CO2
CH C O CH2 COOH
(Unstable) (1,2-dimethyl cyclopentene)

N HO
11. Ba 2
Burning
 Ba3N2 
2
 Ba  OH 2  NH3 D 
A B 
BaCl2  green flame

Hg
+
KI
HgI2  K 2HgI4 3

NH
O NH2 I
12. KOH
Hg

13. Reaction of Mn3+ with H2O is spontaneous.

14. Reduction potential of I2 is less than Cl2.

Solution for the Q. No. 15 to 16.

(i) A turns lime water milky so ‘A’ may be CO2 or SO2
(ii) B condenses into liquid turning anhydrous CuSO4 to blue, so ‘B’ is H2O.
(iii) Y gives alkaline solution and its solution forms white ppt. Z with BaCl2 and Z on heating with acid
gives CO2 so, Z is BaCO3 and Y is metal carbonate.
(iv) Since, Y and A are formed from ‘X’ and thus X is metal bicarbonate.
X  A  B  Y
(v)
20.02 g 4.4 g 1.8 g 13.8 g

(vi) The above data reveal that,

2MHCO3   CO2  H2 O  M2 CO3
 X  A  B  Y
4.4 of CO2 is obtained by 20.02 g of MHCO3
44 g CO2 is obtained by 200.2 g of MHCO3
200.2
Molecular weight of MHCO3   100.1
2
 Atomic wt. of M  39.0 g/mole
Thus metal is potassium.

17. BF4 has sp3 hybridisation, CH3 is sp2, F2O is sp3 with two lone pairs on O. Also N2O is linear.

Mathematics PART – III

SECTION – A

(200)! (200)!
1. = = 1.3.5 ........ . 199
2! . 2!. .....2! 100 ! . 2100
   (100)!
100 times

(200)!  101   102   200 

100
=    .......  
100 ! . 2  2  2   2 

bc a
2. Since, f(–1) = 0 the other root is – Q
ac b
c
1
Which can be written as – a  b
c
1
a b
c
Hence, Q
ab

3. a, b, c are in GP
2ab bc
= 12 , =3
ab 2
2a . ar
= 12, ar + ar2 = 6
a  ar
ar = 6(1 + r), ar(1 + r) = 6
i.e. 6(1 + r)2 = 6, i.e. (1 + r)2 = 1
 1 + r = ± 1 i.e. r = 0 or r = –2
 r = –2
If r = –2, then a = 3
 The GP is 3, –6, 12, –24, 48
3  6  12
Clearly no term of the series is square of an integer AM of a, b, c is = =3
3

4. Homogenizing auxiliary circle with normal at P

2 2 a2 (ax sec   by cos ec)2
x +y =
(a2  b2 )2
a 4 sec 2  a2b2 cos ec 2 
QOR = 90º  –1 + –1 + =0
(a2  b2 )2 (a2  b2 )2
2 2 2 4 2 2 2 2
2(a – b ) = a sec  + a b cosec 
a4 – 5a2b2 + 2b4 = a4 tan2  + a2b2 cot2  ≥ 2a3b.

x 2  5x  3 2x  5 3
2
5. f(x) = 3x  x  4 6x  1 9
2
7x  6x  9 14x  6 21
f(x) = 0
 f(x) is a constant polynomial and f(0) ≠ 0
d≠0

6. (1 + x2)2 (1 + x)n = (1 + 2x2 + x4) (nC0 + nC1 x + nC2 x2 + ......)

= nC0 + nC1 x + (nC2 + 2·nC0) x2 + ......
Hence, A0 = 1; A1 = nC1; A2 = nC2 + 2 which are in A.P.

   AB P B
7. OP = r = OA ±  A 1
| AB | P
  
OB  OA r
ˆ ˆ ˆ
= i  j  2k ±  
| OB  OA |
4ˆj  3kˆ
= ˆi  ˆj  2kˆ ± O
| 4ˆj  3kˆ |
1
= ˆi  ˆj  2kˆ ± (4ˆj  3k)
ˆ
5
1 ˆ ˆ ˆ , 1 (5iˆ  9ˆj  13k)
ˆ
= (5i  j  7k)
5 5

8. Put x = 1/t in u or v  u = v
Now consider u + v

1/2 1/2
 1 1/2 1
9. Px   =  f(x) dx =  2x dx =  x2  =
 2  0 4
x 0 0

10. Probability that life of a radio tube is less than or equal to 150 hours
150 150 150
100 100  1 1  1
=  f(x) dx =  2
dx =  = –100    =
0 100 x x 100 150 100  3
2
 Probability that life is greater than 150 hrs =
3
2 2 2 8
 Probability that none of these have life less than on equal to 150 hrs = × × = h
3 3 3 27

11. y = n x has graph here an element in co-domain can have 4 pre-images

y = x2 sin x is zero for infinitely many values of x
y = x3 + 3x + 1 is M.I. and hence one-one function
y = x4 – x + 1 has graph
4
 y = x – x + 1 is two-one

12. From graph it is clear that any horizontal lines cuts the graph at either 1
 1 3
one or two points between   , 
 2 2 –1 1 O 1 3 2

2 2

13. At x = 1, y = 0
dy 4 2 2 2
= 2x·(x ) – (x ) = 2 – 1 = 1
dx
 Equation of tangent y – 0 = 1 (x – 1), y = x – 1

x
2
2
14. F(x) =  et /2
(1 – t ) dt
1
2
 x2 
F(x) =  e 2 (1  x2 ) 
 
 
 F(1) = e1/2.0 = 0

15. Let sin x = t; t  [–1, 1]

1
5  2t  6t  1  0 i.e. t
6
1 
So, t  ,1
6 
1 2 2  1 
Now squaring, ,  , t   not possible t   ,1 
5  2t  (6t  1) , t 
2 9 9  6 
1 1  
So, t  , sin x   sin , x  n  ( 1)n
2 2 6 6

cos 2x cos 3x sin 2x

16. 2 3 
sin2x sin3x cos2x
(2cos 2 x  1) 4 cos3 x  3 cos x 2sin x cos x /4 /4
2 3 
2sin x cos x 3 sin x  4 sin3 x 2cos2 x  1
1
Now Let cos2 x = t in above i.e. t   0,1   
2 
These positions can’t be the answer
4t  3 2t(1  t)
Now we have 2t  1  3t 
4t  1 2t  1
1
Solving t = ± 1 but t   0,1   
2 
So there is no t, so there is no x

17. (P) Form the graph it is clear that x = 5 is the least positive integral value
for which x2 – 4x > cot–1 x. y =
(Q) f(x) = 2ex + ae–x + 2a + 1 y = 
–x 2x x
= e (2e + (2a + 1)e + a)
–x 2x x
= 2e (e + (a + 1/2)e + a/2)
–x x x
= 2e (e + a)(e + 1/2) 0 (4, 0) (5, 0)
For f(x) to be increasing f(x)  0  x  R
x
or e + a  0 x  R
or a  0
1  f 2 (a  x)  f (a)  2
lim  f (a  x) (f (a))2
 f 2 (a  x)  x x  0 xf(a)
 lim
(R) lim   = e  
= e x 0 x = e4
x0
 f(a) 
k=4
–1 2 –1
(S) sin (–x + 4x – 3) = cos (2 – |x|)
2
0  –x + 4x – 3  1  x  [1, 3] ..... (1)
Also, 0  2 – |x|  1  x  [–2, –1]  [1, 2] ..... (2)
From (1) and (2) x  {1, 2}

18. (P) Equation of any normal to y2 = 4x is y = –tx + 2t + t3

Comparing it with y = 2x + c – 4
t = –2 and 2t + t3 = c – 4
 c = 4 – 4 – 8 = –8
 –c = 8
3y 9
(Q) Equation of the chord whose mid point is (0, 3) is –1= –1
25 25
x2 y2 x2 9 16
i.e. y = 3 it intersects the ellipse + = 1 at  1 
16 25 16 25 25
16
x=±
5
32
 Length of the chord =
5
4k 32
Thus  k=8
5 5
(R) The product of the lengths of the perpendiculars from the two focii on any tangent to the hyperbola
x2 y2
– = 1 is 3
25 3
 3 = k , hence k = 9
(S) Since, x + y = a touches the hyperbola
x2 – 2y2 = 18
 x2 – 2(a – x)2 = 18 has equal roots
i.e., x2 – 4ax + 18 + 2a2 = 0 has equal roots
 16a2 – 4(18 + 2a2) = 0
8a2 – 72 = 0  a =  3
 |b| = 3

  2 A
19. (P) From figure a  b 
3   
 2 ab b
2
and a  b  2  1 
  B
a 60º
C
ab  3  x 
b
x=3  
ab
(Q) f(x) = sin x, f(x) + f( – x) = 2
 
f(x) = 2 – f( – x) = 2 – sin ( – x) = 2 – sin x, where x   , 
2 
 3 
f(x) = f(2 – x) = 2 – sin(2 – x), where x   , 2
 2 
  
sin x , x  0, 
  2
  
2  sin x , x   ,  
 2 
f  x  
2  sin x , x   , 3 
  
 2

  3 
  sin x , x   , 2 
 2 


 /2  3  /2 2
Area =  sin xdx   2  sin xdx    2  sin x  dx     sin x  dx
0  /2  3  /2
 
= 1  2   1  2   1  1  2 sq. units
2 2
(R) Since g = f –1 so fog(x) = x
Therefore f(g(x)) g(x) = 1
f(g(a)) g(a) = 1
1
 f(b) 2 = 1  f(b) =
2
(S) Putting x = tan 
1  x 2  1 sec   1 
  tan
x tan  2
1 x2  1  1
This u  tan1   tan1 x
x 2 2
2x 1  x2
Putting x = sin , v  tan1  2sin1 x
1  2x 2
1 1
du 2 1  x 2 du  1
   
dv 2 dv  x 0 4
1 x2

20. (P) We have to select 3 no. from {1, 2, .....100}

Number division by 6 are 6, 12 ..... 96
16
C 4
Pr ob  100 3 
C3 1155
(Q) 22nd century 2200, 2201 ..... 2299
2200 leap year (L), 2201 Non leap year (N)
1 3
In 22nd century: 25 L and 75 N, P(L)  P(N) 
4 4
366 2
For L: 366 days;  52 
7 7
365 1
For N: 365 days;  52 
7 7
1 2 3 1 5
Now P(53 Sunday) = P(L) P(53S/L) + P(N)·P(53 Sun/N)     
4 7 4 7 28
5 0
 1  1 1 31
5
(R) P(at least one head) = 1 – P (no head) = 1  C5      1  
2  2 32 32
3 9 1 3 A B S(12)
(S) P(A  B)   , P(A  B)  
4 12 4 12
8 4 5 1 3 5
P(A)  ; P(A)  , P(A  B) 
12 12 12
3

Time Allotted: 3 Hours Maximum Marks: 360

t h e t es t .

INSTRUCTIONS

A. General Instructions
1. Attempt ALL the questions. Answers have to be marked on the OMR sheets.
2. This question paper contains Three Parts.
3. Part-I is Physics, Part-II is Chemistry and Part-III is Mathematics.
4. Each part has only one section: Section-A.
5. Rough spaces are provided for rough work inside the question paper. No additional sheets will be
provided for rough work.
6. Blank Papers, clip boards, log tables, slide rule, calculator, cellular phones, pagers and electronic
devices, in any form, are not allowed.

B. Filling of OMR Sheet

C. Marking Scheme For All Three Parts.

(i) Section-A (01 to 30) contains 30 multiple choice questions which have only one correct answer.
Each question carries +4 marks for correct answer and – 1 mark for wrong answer.

Name of the Candidate

Enrolment No.

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Useful Data

PHYSICS
Acceleration due to gravity g = 10 m/s2

Planck constant h = 6.6 1034 J-s

Charge of electron e = 1.6  1019 C

Mass of electron me = 9.1  1031 kg

2 2
Permittivity of free space 0 = 8.85  1012 C /N-m

Density of water water = 103 kg/m3

Atmospheric pressure Pa = 105 N/m2

Gas constant R = 8.314 J K1 mol1

CHEMISTRY

Gas Constant R = 8.314 J K1 mol1

Atomic No: H=1, He = 2, Li=3, Be=4, B=5, C=6, N=7, O=8,

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Physics PART – I

SECTION – A

Straight Objective Type

This section contains 30 multiple choice questions numbered 1 to 30. Each question has 4 choices (A),
(B), (C) and (D), out of which ONLY ONE is correct

1. In the figure shown, pan C is massless. All strings and pulleys are
ideal. Block B is dropped from a height l on pan C. Collision between l A
B
block B and pan C is perfectly inelastic. Just after collision tension in M
string A differ from tension in string A before collision by a magnitude M
of l
Mg
(A) Mg (B)
2
C
Mg
(C) (D) 2Mg
4

2. An infinite grid is made using resistor R1 and R 2 R1 R1 R1

C E
as shown. The potential difference between A A
V
and B is V, between C and D is , between E
3 R2 R2 R2 
V
and F is i.e. potential difference becomes B
9 D F
R
one third after each pair. The value of 1 is
R2
4
(A) 1 (B)
3
2
(C) (D) 2
3

Space for rough work

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3. A uniform disc of mass m and radius R is rotating with angular velocity   on 

a smooth horizontal surface. Another identical disc is moving translationally
with velocity v as shown. v
When they touch each other, they stick together. The angular velocity of centre of mass of the
system after contact will be
2v  R
(A) zero (B)
6R
v 
(C) (D)
R 6

4. A capacitor of capacitance 2C is charged to a potential 2V and +– R

then connected to an uncharged capacitor of capacitance C with
a resistance R to a battery of emf 4V as shown. 2C, 2V C
The key is closed at t = 0. The charge on uncharged capacitor
as a function of time will be
K
4V
t t
  4CV  
(A) 4CV 1  e 2RC  (B) 1  e 2RC 
  3 
  
3t 3t
  4CV  
(C) 4CV 1  e 2RC  (D) 1  e 2RC 
  3 
  

5. A series circuit consisting of a capacitor and a coil with active resistance is connected to a source
of harmonic voltage whose frequency can be varied, keeping the voltage amplitude constant. At
frequencies 1 and 2 , the current amplitudes are n times less than the resonance amplitude.
Then resonance frequency will be
  2
(A) 1 (B) 1  2
2
(C) 1  2 (D) 12

6. If the tension in the string changes by 21%, the fundamental frequency of string changes by 15
Hz. Which of the following statement is correct?
(A) original fundamental frequency is nearly 150 Hz
(B) velocity of propagation changes by 4.5%
(C) velocity of propagation changes by 10%
(D) fundamental wavelength changes by 15%

Space for rough work

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7. Two moles of an ideal monatomic gas are taken from state A to state B B
through a process as shown in figure. The work done by the gas in the P T2
process is A
(A) R(T2  T1 ) (B) 2R(T2  T1 ) T1
1 1
(C) R(T2  T1 ) (D) (T2  T1 ) V
2 4R

8. When an ideal gas undergoes an isothermal expansion, the pressure of the gas decreases due to
(A) decrease in change in momentum per collision
(B) decrease in frequency of collision with the container
(C) both decrease in frequency of collision and change in momentum per collision
(D) none of these

 4
9. A vessel is filled with water     upto a height 20 cm. A concave
 3 
mirror of radius of curvature 40 cm is placed 15 cm above the water level. 15cm
The position of final image of an object placed 20 cm below the water
surface will be
(A) 45 cm below the water surface 20cm
(B) 60 cm above the water surface
(C) 45 cm above the water surface
(D) 60 cm below the water surface

10. An object is placed in front of convex lens of focal length f=10cm

(f = 10 cm). The distance d for which image of the object will
concide with the object (irrespective of focal length of the mirror)
(A) 30
(B) 20 O
(C) 0
(D) 15 15cm d
11. The probability that a radioactive nucleus will survive 10 times longer than its half life, is
(A) 2–20 (B) 2–10
5
 1
(C)   (D) data insufficient
 2

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12. In a screw gauge, there are 100 divisions in circular scale and each main scale division is of
th
1 mm. When there is no gap between the jaws, 97 divisions coincides with the main scale zero
and zero of main scale is not visible. While measuring the diameter of a ball, the circular scale is
between 3 mm mark and 4 mm mark such that the 76th division of circular scale coincides with
the reference line. Select the correct alternative
(A) the least count of the micrometer is 0.01 cm
(B) the zero error is – 0.04 mm
(C) the diameter of the ball is 3.79 cm
(D) the main scale reading is 4 mm

13. A uniform solid square plate ABCD of mass m and side a is

moving in x – y horizontal smooth plane. The velocity of centre of
vcm  v0 (2iˆ  4 ˆj )
mass is v0 (2iˆ  4ˆj) m/s. The end A of square plate is suddenly y
fixed by a pin, find the new velocity of centre of mass of square B C
(A) v0 (2iˆ  4ˆj)
3v 0 ˆ 3v 0 ˆ
(B) (  i)  j
4 4
3v 0 ˆ 3v 0 ˆ A D
(C) i j
2 2 x
(D) 3v 0 ˆi  3v 0 ˆj

14. Three moles of an ideal gas goes to a cyclic process shown in P

figure. The work done by the gas during the process is 2P0 A
(A) 2.36RT0
(B) 0.58RT0
P0
(C) 1.16RT0 C B
(D) 0.58RT0 T
O T0 2T0

15. A uniform ring of mass ‘m’ and radius ‘R’ is projected horizontally with velocity v0 on a rough
horizontal floor, so that it starts off with a purely sliding motion and it acquires a purely rolling
motion after moving a distance d. If the coefficient of friction between the ground and ring is ,
then work done by the friction in the process is
1
(A) – mgd (B)  mv 02
4
1
(C) mgd (D)  mv 20
8

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16. In the circuit shown in figure, the switch ‘S’ is closed at t = 0. L1 R1

After closing the switch, in steady state, the value of current in
the resistor R3 R2
(A) increases E R3
(B) decreases L2
(C) remains constant S
(D) becomes zero

17. A particle is moving in x – y plane with a constant speed v0 such that its y displacement is given
2v x
3v 0
by y  e . Where v x is component of velocity along x axis. y displacement of the particle
1
when slope of tangent on its path is  , is
3
(A) e1 (B) e 2
(C) zero (D)  2 e

18. A uniform string of large length L of uniform cross section, mass per unit length  and free at both
ends is revolving round the earth at equator in vertical position in such a way that its position
relative to the earth is always the same and its lower end is slightly above the earth’s surface. If
the string is pulled down by a small distance x, then work done by gravity is
 R  GMxL
(A) GMx   (B)
RL R(R  L)
GMx 2 GML2
(C) (D)
R(R  L) R(R  L)

19. A particle on a spring executes SHM. If the mass of the particle and the amplitude are both
doubled then the magnitude of the maximum acceleration of the particle will change by a factor of
(A) 4 (B) 8
(C) 2 (D) 1

20. Two waves travel down the same string. The waves have the same velocity, frequency (f0), and
wavelength but different phase constants ( 1 > 2 ) and amplitudes (A1 > A2)
According to the principle of superposition, the resultant wave has an amplitude A such that
(A) A = A1 + A2 (B) A = A1 – A2
(C) A1  A  A2 (D) A1 – A2  A  A1 + A2

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22. Light traveling through three transparent substances follows

the path shown in figure. Arrange the indices of refraction in n1
order from smallest to largest. Note that total internal
reflection does occur on the bottom surface of medium 2
(A) n1 < n2 < n3 n2
(B) n < n < n
2 1 3
(C) n1 < n3 < n2
n3
(D) n3 < n1 < n2

23. Light shines through a spherical air bubble underwater. What type of optical device does the
bubble act like?
(A) Converging (B) Diverging
(C) Planar (D) none of these

24. For fun, a student constructs a three-slit interference experiment. If the light from each of the
three slits arrives in phase at the central maximum, then how will the intensity I compare to the
intensity I0 from a single slit?
(A) I = I0 (B) I = 3I0
(C) I = 6I0 (D) I = 9I0

25. An ideal gas has an initial pressure of 3 units and an initial volume of 4
units. The table gives the final pressure and volume of the gas (in same a b c d e
unit) in five processes. Which processes start and end on the same
isotherm?
P 12 6 5 4 2
(A) a, b, d (B) a, b, d, e V 1 2 7 3 12
(C) only c (D) d, e

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26. If two moles of diatomic gas and one mole of mono atomic gas are mixed then the ratio of specific
heats is
7 5
(A) (B)
3 4
19 15
(C) (D)
13 19

27. The K wavelength of an element with atomic number Z = 17 is given by .

What is the atomic number of an element with a K wavelength of 4?
(A) 4 (B) 5
(C) 7 (D) 9

28. For a small angled prism, angle of prism A, the angle of minimum
deviation    varies with the refractive index of the prism as shown in
Q

the graph
(A)point P corresponds to  > 1
(B)slope of the line PQ = A/2
(C)slope = A
(D)none of the above statements is true O P M

29. A light ray from air is incident (as shown in figure) at one
end of a glass fiber (refractive index  = 1.5) making an Air
incidence angle of 60º on the lateral surface, so that it Air
undergoes a total internal reflection. How much time would
it take to traverse the straight fiber of length 1 km 60º
(A) 3.33  sec Glass
(B) 6.67  sec
(C) 5.77  sec
(D) none of these

30. Two ideal slits S1 and S2 are at a distance d apart, and illuminated
by light of wavelength  passing through an ideal source slit S
placed on the line through S2 as shown. The distance between the
planes of slits and the source slit is D. A screen is held at a S1
distance D from the plane of the slits. The minimum value of d for
which there is darkness at O is S O
S2
3D
(A) (B) D
2
D D D
(C) (D) 3D
2

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10

Chemistry PART – II

SECTION – A

Straight Objective Type

This section contains 30 multiple choice questions numbered 1 to 30. Each question has 4 choices (A),
(B), (C) and (D), out of which ONLY ONE is correct

1. The basicity of Aliphatic Amines in gas phase is

(A) 1° amine > 2° amine > 3° amine (B) 3° amine > 2° amine > 1° amine
(C) 2° amine > 1° amine > 3° amine (D) 2° amine > 3° amine > 1° amine

2. Br
O

CH3
How many stereo isomers are possible for above compound?
(A) 14 (B) 16
(C) 12 (D) 18

3. The major product of the given reaction is


AlCl
?
3

(A) (B)

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4. O
O
H3 PO4
 P 

(A) O (B) OH

O O

5. H CH3
H3C C C CH3  ?

H3C H

(A) NH2  NH2 | OH (B) H2 | Ni
(C) H2 | Pd  BaSO 4 (D) H2 | LI. liq NH2

6. Which of the following halide readily undergo SN1 reaction?

(A) (B)
Cl

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7. OH OH
CHO
 i  aq. NaOH

 ii CHCl3 , H2O

The intermediate involved is:

(A) : CH2 (B) : CCl2
 
(C) CH3 (D) CCl2

8. When accetaldehyde is heated with Fehling’s solution. It gives a precipitate of

(A) Cu (B) CuO
(C) Cu2O (D) Cu + Cu2O + CuO

9.

D3 O

D O
 P 
2

The product (P) formed finally is?

(A) D (B) D
D D

D D
D D

(C) D (D) None of these

D D

10. N OH


H Polymerisation
 C   D
n
What is D?
(A) Nylon-66 (B) Nylon – 6
(C) Bekalite (D) Buna – 9

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11. The electronic configuration of four elements are

1
(1) [Xe]6s
14 1 2
(2) [Xe]4f , 5d , 6s
2 5
(3) [Ar]4s 4p
(4) [Ar]3d7, 4s2
Which one of the following statements about these elements is not correct?
(A) 1 is strong reducing agent.
(B) 2 is a d-block element.
(C) 3 has high electron affinity.
(D) The compound formed between 1 and 3 is ionic.

12. The IP1, IP2, IP3, IP4 and IP5 of an element are 7.1, 14.3, 34.5, 46.8, 162.2 eV respectively. The
element is likely to be
(A) Na (B) Si
(C) F (D) Ca

13. H2S is bubbled into 0.2 M NaCN solution which is 0.02 M in each [Cd(CN)4]2– and [Ag(CN)2]–. The
H2S produces 1 × 10-9 M sulphide ion in the solution.
Given: K sp Ag2S  1 10 50 M3 ;K sp CdS  7.1 10 28 M2
1 2
K instability  Ag  CN2   1 10 20 M2 , K instability Cd  CN 4   7.8  10 18 M
Identify the correct statement:
(A) Ag2S precipitates first from the solution.
(B) CdS precipitates first from the solution.
(C) None of them precipitates under the given conditions.
(D) Ag2S precipitates at a sulphide concentration 1 × 10+15 M.

14. Extraction of Ag from commercial lead is done by

(A) Parke’s Process (B) Clarke’s Process
(C) Kroll’s Process (D) Electrolytic Process

15. Most stable polymeric hydride among the following is:

(A) CaH2 (B) MgH2
(C) BaH2 (D) SrH2

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16. NaOH  O3  X  products

X is
(A) Na2O (B) NaO3
(C) NaO2 (D) Na2O2

17. Arrange the bond angle order for Cl2O, ClO2, F2O, Br2O:
(A) OF2 > OCl2 > OBr2 > ClO2 (B) OF2 < OCl2 < OBr2 < ClO2
(C) ClO2 < OF2 < OCl2 < OBr2 (D) OF2 < OCl2 < ClO2 < OBr2

18. A d-Block element forms octahedral complex but its magnetic moment remains same either in
strong field or in weak field ligand. Which of the following is/are correct?
(A) Element always forms colourless compound.
(B) Number of electrons in t2g orbitals are higher than in eg orbitals.
(C) It can have either d3 or d8 configuration.
(D) It can have either d7 or d8 configuration.

19. Among the following complexes which will be attracted by magnetic field?
2 2 2
Ni  CN  
 4
, NiCl4  , Ni  CO  4  , Ni H2 O 6 
(I) (II) (III) (IV)
(A) (I) only (B) (I) and (IV)
(C) (II), (III) and (IV) (D) (II) and (IV)

20. H2S gas on passing through CdSO4 solution will give

(A) Black ppt (B) Yellow ppt
(C) Organge ppt (D) No ppt

21. The vapour density of a mixture containing NO2 and N2O4 is 27.6. The mole fraction of N2O4 in
the mixture is
(A) 0.1 (B) 0.2
(C) 0.5 (D) 0.8

22. Calculate the maximum number of electrons which may have magnetic quantum number m = + 1
and spin quantum number s = +1/2 in chromium.
(A) 0 (B) 1
(C) 4 (D) 3

23. 36.5% HCl has density equal to 1.2 g ml–1. The molarity (M) and molality (m), respectively are
(A) 15.7, 15.7 (B) 12, 12
(C) 15.7, 12 (D) 12, 15.7

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24. 3KClO3  3H2 SO4  3KHSO4  HClO 4  2ClO2  H2O

Equivalent weight of KClO3 is
M M
(A) (B)
4 2
 M M M
(C)  M   (D)   
 2 4 2

25. An ideal gas obeying kinetic theory of gases can be liquefied, if

(A) It’s temperature is more than critical temperature Tc.
(B) Its pressure is more than critical pressure Pc.
(C) It’s pressure is more than Pc at a temperature less than TC.
(D) It cannot be liquefied at any value of P and T

26. A sample of Argon gas at 1 atm pressure and 27°C expands reversibly and adiabatically from
1.25 dm3 to 2.5 dm3. Calculate the enthalpy. Change in this process. Cvm for Argon is 12.48
–1 –1
Jk mol
(A) –114.52 J (B) –110.32 J
(C) –100.42 J (D) –126.42 J

27. In reduction


 CH3 CH3  CO  g , if the initial pressure of CH3COCH3(g) is 150 mm and at
CH3 COCH3 
 g  g
1
equilibrium the mole fraction of CO(g) is , then the value Kp is
3
(A) 50 mm (B) 100 mm
(C) 33.3 mm (D) 75 mm

28. Calculate pH of saturated solution of

Mg(OH)2, Ksp for Mg(OH)2 is 8.9  10–12
(A) 9.4 (B) 11.2
(C) 10.4 (D) 9.8

29. For a hypothetical elementary reaction:

2B
K1 K1 1
Where 
A K2 2
K2 2C
Initially only 2 moles of A are present. The total no. of moles of A, B and C at the end of 75%
reaction are
(A) 2 (B) 3
(C) 4 (D) 3.5

30. A crystal is made of particles X and Y. X forms f cc packing and Y occupies all the octahedral
voids. If all the particles along one body diagonal are removed then the formula of the crystal
would be:
(A) X4Y3 (B) X5Y4
(C) X4Y5 (D) None of these

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16

Mathematics PART – III

SECTION – A

Straight Objective Type

This section contains 30 multiple choice questions numbered 1 to 30. Each question has 4 choices (A),
(B), (C) and (D), out of which ONLY ONE is correct

2 2n n n n
1. If ,  are the roots of the equation x + px + q = 0 and also of the equation x + p x + q = 0 and
 
if , are the roots of the equation x n + 1 + (x + 1)n = 0 then n must be an
 
(A) even integer (B) odd integer
(C) irrational number (D) none of these

2. A pole 50m high stands on a building 250 m high to an observer at a height of 300 m, the building
and the pole subtend equal angles. The distance of the observer from the top of the pole is
(A) 25 m (B) 50 m
(C) 25 3m (D) 25 6m

1
  n 1

3. Let P    10 2  then log0.01(P) is

n 1 


(A) 0 (B) –1
(C) 1 (D) none of these

x
4. The greatest value of the function y  2
(a, b > 0)
ax  b
1 1
(A) (B)
ab 2 ab
(C) ab (D) 2 ab

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n n n
5. If a variable takes the values 0, 1, 2, ….., n with frequencies 1, C1, C2, …, Cn, then the A.M. is
2n
(A) n (B)
n
n
(C) n + 1 (D)
2
15
6. The greatest coefficient in the expansion of (a + b + c + d) is
15! 15!
(A) 2 3
(B)
 3!  4! 3! 4!3
15!
(C) (D) none of these
 3!3  4!2
7. Let S be the set of all triangles and R+ be the set of positive real numbers, then the function,
+
f: S  R , f() = area of , where   S is
(A) injective but not surjective (B) surjective but not injective
(C) injective as well as surjective (D) neither injective nor surjective

8. Tangents are drawn to x2 + y2 = 16 from P(0, h). These tangents meet x-axis at A and B. If the
area of PAB is minimum, then value of h is
(A) 12 2 (B) 8 2
(C) 4 2 (D) 2

x2 y 2
9. If  +  = 3 then the chord joining the points  and  for the hyperbola   1 passes
a2 b2
through
(A) focus (B) centre
(C) one of the end points of the transverse axis (D) one of the end points of the conjugates axes

x y x y
10. For a given non-zero value of m each of the lines   m and   m meets the hyperbola
a b a b
x2 y 2
  1 at a point, sum of the ordinates of these points, is
a2 b2

(A)

a 1  m2  (B)
b 1  m2  
m m
ab
(C) 0 (D)
2m

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11. Two points P and Q are given in the rectangular cartesian co-ordinates on the curve y  2x 2 ,
 
such that OP  ˆi  1 and OQ  ˆi  2 where î is a unit vector along the x-axis. Then the
 
magnitude of OQ  4OP is
(A) 8 (B) 10
(C) 12 (D) 16

12. The image of A(p, q) in the line y = –x is B and the image of B in the line y = x is C then the mid
point of AC is
 p  q p  q
(A) (p + q, p + q) (B)  ,
 2 2 
(C) (p – q, q – p) (D) (0, 0)

13. Let PQR be a triangle let P be the point (1, 2), y = x is the perpendicular bisector of PQ and
x – 2y + 1 = 0 is the angel bisector of R. If equation of QR is given by x + y – 5 = 0, then the
value of  +  is
(A) 1 (B) 2
(C) 3 (D) 4

14. If the tangents are drawn from any point on the line x + y = 3 to the circle x2 + y2 = 9, then the
chord of contact passes through the point
(A) (3, 5) (B) (3, 3)
(C) (5, 3) (D) none of these

15. A is a point (a, b) in the first quadrant. If the two circles which passes through A and touches the
coordinate axes cut at right angles then
(A) a2 – 6ab + b2 = 0 (B) a2 + 2ab – b2 = 0
2 2
(C) a – 4ab + b = 0 (D) a2 – 8ab + b2 = 0

16. If P be a point on the parabola y2 = 3(2x – 3) and M is the foot of perpendicular drawn from P on
the directrix of the parabola, then length of each sides of an equilateral triangle SMP (where S is
the focus of the parabola), is
(A) 2 (B) 4
(C) 6 (D) 8

17. If the chord of contact of tangents from a point P to the parabola y2 = 4ax touches the parabola
x2 = 4by, then locus of P is
(A) circle (B) parabola
(C) ellipse (D) hyperbola

Space for rough work

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AITS-CRT-III-PCM-JEE(Main)/15

x 2 y2
18. A tangent to the ellipse   1 at any points meet the line x = 0 at a point Q. Let R be the
25 16
image of Q in the line y = x, then circle whose extremities of a diameter are Q and R passes
through a fixed point, the fixed point is
(A) (3, 0) (B) (5, 0)
(C) (0, 0) (D) (4, 0)

x2 y2
19. Number of distinct normal lines that can be drawn to ellipse   1 from the point P(0, 6) is
169 25
(A) 1 (B) 2
(C) 3 (D) 4

20. The conditional (p ^ q)  p is

(A) a tautology (B) a contradiction
(C) neither a tautology nor a contradiction (D) none of these

21. If f(x) = (x – 1)(x – 2)(x – 3)(x – 4), then out of the three roots of f(x) = 0
(A) three are positive (B) three are negative
(C) two are complex (D) none of these

22. Let f(x) g(y) = g(y) – f(x) g(y)  x, y  R and g(0) = 1, g(0) = 1, f(0) = –5, then
(A) f(0) = 8 (B) f(1) = e
(C) g(1) = e (D) g(–1) = e–1

23. At an election, a voter may vote for any number of candidates not greater than the number to be
elected, there are 10 candidates and 4 are to be elected. The number of ways in which a voter
can vote for atleast one candidate is
(A) 385 (B) 1110
(C) 5040 (D) 4050

24. The number of 5 digit numbers that can be made using the digits 1 and 2 and in which at least
one digit in different, is
(A) 30 (B) 31
(C) 32 (D) 35

Space for rough work

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20

25. If two numbers a and b are chosen from the set of integers 1, 2, 3, 4, ….., 39 then, the probability
that the equation 7a – 9b = 0 is satisfied, is
1 2
(A) (B)
247 247
4 5
(C) (D)
741 741

26. Two numbers are selected randomly from set S = {1, 2, 3, 4, 5, 6} without replacement one by
one the probability that minimum of the two number is less than 4, is
1 14
(A) (B)
15 15
1 4
(C) (D)
5 5

dx
27. If   g  x   c , then g(x) is equal to
3
sin x cos x
2 2
(A)  (B) 
cos x tan x
2 2
(C) (D)
cot x tan x

28. The number of real solution of 6x + 1 = 8x – 27x – 1 is/are

(A) exactly one (B) exactly two
(C) more than two (D) none of these

29. The value of   (0, 2) for which the equation is 2 sin2  – 5 sin  + 2 > 0, is
    5    5 
(A)  0,    , 2  (B)  , 
 6  6  8 6 
     5   41 
(C)  0,    ,  (D)  , 
 8   6 6   48 

30. If f: R  R and g: R  R are two function such that f(x) + f(x) = –xg(x)f(x) and g(x) > 0  x  R
then f 2(x) + (f(x))2 has
(A) maxima at x = 0 (B) min at x = 0
(C) point of inflexion at x = 0 (D) none of these

Space for rough work

ANSWERS, HINTS & SOLUTIONS

CRT – III
(Main)
PHYSICS CHEMISTRY MATHEMATICS
ALL INDIA TEST SERIES

1. B B A
2. B B D
3. B B B
4. D D B
5. D D D
6. A A B
7. A B B
8. B C C
9. A A B
10. A B C
11. B B B
12. C B D
13. B B B
14. C A B
15. B B C
16. B B C
17. A B D
18. B B C
19. D D C
20. A B A
21. C B A
22. A D C
23. A D A
24. D C A
25. A D C
26. C A D
27. D D B
28. C C B
29. D D A
30. C B A

Physics PART – I
SECTION – A

1.   Td t  Mv  Mu and  T d t  Mv TA
u
v  Tdt
2 M
 Tdt
v2 u2 3Mg v
TA new  Mg  M  Mg  M 
l 4l 2
TA old  Mg u
v

2. Let equivalent resistance between A and B is R. R1 R1

V i A i
iR  V, iR   i  i i – i
3 3
V 2i R2 R
 R2
3 3
B
R  2R2
2V
iR1 
3
R1 4

R2 3

3. Conserving angular momentum

3mR2 
mvR   3mR 2 
2
2v  3R
  
6R

4. Let at any time a charge q flows through the circuit. The + – R + –

circuit is as shown q
2RC 2C C
Time constant ( ) 
3

4V
3t
4CV  
 q 1  e 2RC 
3  

VM
5. IM1 
2
2  1 
R   1L  
 1C 
VM
Similarly, IM2 
2
2  1 
R   2L  
 2 C 
Since, both IM1 and IM2 are n times less than resonance current amplitude,

 IM1  IM2
VM VM
 
2 2
 1   1 
R2   1L   R2   2L  
  C
1   2C 

1 1
or 1L    2L
1C 2 C
12LC  1 1  22LC
or 
1C 2 C
or 12LC(1  2 )  (1  2 )  0
or 1  12LC  0
1
 12   02
LC
 0  12

1 T
6. f0 
2l 
df0 1 dT

f0 2 T
T
f0  2  15  100  150 Hz.
21

7. W   PdV and P  V

9. The object distance for the mirror

 20 
   15   30 cm
 4 / 3 
For mirror,
1 1 1
 
v u f
1 1 1
    v  60 cm
v 30 20
The distance from water surface equals 60 – 15 = 45 cm

Pitch
12. Least count =
Number of divisions on circular scale
1mm
=  0.01mm  0.001 cm
100
Zero error = – 0.03 mm
Measurement = 3.76 – (– 0.03) = 3.79 mm

 a a
13. rcm  ˆi  ˆj (considering A to be origin AD to x-axis and AB to be y-axis)
2 2
   a a 
Li  mrcm  v cm  m  ˆi  ˆj   v 0 (2iˆ  4 ˆj)
 2 2 
= mav (2kˆ  k)
0
ˆ  mav kˆ
0

m a2   ma2 ma2 
L f   (a2  a2 )  m      
12 2  6 2 
From conservation of angular momentum about point A
L f  Li
2
ma2   mav 0
3
3v
  0
2a
 a 3v a 3v0
v cm    0 
2 2a 2 2 2

14. If V0 be volume of the gas at C, P0 V0  3RT0

W = 3R(2T0 )ln 2  P0 V0
= 6RT0 ln 2  3RT0
= 3RT0 [2  0.693  1]
= 3RT0 [1.386  1]
= 1.16RT0

15. mv 0R = I + mvR
v
= mR2    mvR
R
v0
v =
2
1
Ki = mv 02
2
2
1 1 1 v  1 v2
Kf = mv 2  I2  m  0   mR2 02
2 2 2  2  2 4R
1
= mv 20
4
1
So Wf =  mv 02
4

E
16. Initial current in R3 , ii 
R1  R 3
E R2
Final current if  
R2R3 R2  R3
R1 
R2  R3
R2E
=
R1 (R2  R3 )  R2R3
E
if 
R1R3
 R1  R 3
R2
As if  ii , so current in the resistor R3 will decrease.

dy 1 vy
17.  
dx 3 vx
So v x   v y 3
v 2x  v y2  v 02
3v 2y  v 2y  v 02
v0
vy  
2
3v 0
vx  
2
2 3v 0

3v 0 2
So y  e
y  e1

18. If the string is pulled down by distance x, it is equivalent that a mass of x is removed from the
upper end of the string and is connected to the lower end
So Wg  U  Ui  Uf
GM(x)  GMx 
Wg   
RL  R 
1 1 
= GMx  
 R R  L 
K
19. amax = 2A = A
m
If mass and amplitude both are doubled maximum acceleration will not change.

20. Resultant wave amplitude is given by

A = A12  A 22  2A1A 2 cos(1  2 )
Value of cos(1 – 2) lies between 1 and – 1
Hence A1 – A2  A  A, + A2

24. If three waves meet in same phase then the resultant amplitude will be algebraic addition of all
three amplitudes
R = 3A
A  Amplitude of superimposed wave
R  Amplitude of resultant wave
than I = 9I0 as [I  R2]

25. If process starts and ends on same isotherm then product of P and V must be constant hence
process a, b, d will be on same isotherm.

n1Cp1  n2Cp2
26.  mix 
n1Cv1  n2Cv 2
7 5
2  R  1 R
= 2 2  19
5 3 13
2 R 1 R
2 2

l 1 1
27.  R(Z  1)2  2  2  .....(i)
K 1 2 
l 1 1
 R(Z  1)2  2  2  ....(ii)
K  1 2 
Solving (i) and (ii) Z = 9

28.  = A( – 1) Slope of line will be A.

 
30. x = (SS1 + S1O) – (SO) =
2 2

2 D2  d2  2D 
2
2
 d  
2D 1  2
 1 
 2D  2
D
d =
2

Chemistry PART – II
SECTION – A
1. In gas phase inductive effect play a key role.
n
2. Geometrical isomers = 8 = 2
3
Optical isomer = 8 = 2 = 8
= 16
3. Due to bulky effect.

4. O OH

O O
H


OH OH

O O

OH O
O O


Tautomerise



5. Birch reduction.

6. 3o carbocation is more stable.

7. : CCl2 intermediate.

8. CH3 CHO  2CuO  CH3 COOH  Cu2O 

Re d ppt.

9. H H

 D   D D

D H

D D D
D O
  
3

D D
D

10. H
N OH
N O
C O
H Polymerisation

  
NH (CH2)5 C
n
Caprolactum Nylon-6

11. It is belong to f-block – Lu element.

12. IP4 – IP5 required more energy, So IVA group elements.

15. H H
Mg Mg
H H

1
16. 3NaOH + 2O3  2NaO3 + NaOH.H2O(s) + O2
2

17. Cl2O, ClO2 , F2 O, Br2O

B.A 110 118 105 112

19. (II) and (IV) have unpaired electron.

20. H2S  CdSO4  CdS

Yellow ppt

21. Let 1 mole of mixture has x mole N2O4

2  27.6 = x(92) + (1 – x) 46
x = 0.2
2p 3p 3d
22. m 1 1 1
s 1/ 2 1/ 2 1/ 2

% by weight  10  d 36.5  10  1.2

23. M   12 M
M.w 36.5
36.5  1000 1000
m   15.7 m
36.5  100  36.5  63.5

24. ClO3  ClO4  2e  x  2  oxidation

x = 5, x=7
e  ClO3  ClO2  x  1 reduction
x = 5, x = 4
M
EW  M 
2

25. Ideal gas has no force of attraction and has negligible volume. Hence it can’t be liquefied at any T
and P.

26. H = nxCp  T Cp = Cv + K
PV –1
n  0.05 Cp = 20.794 JK
RT
r 1
For reversible adiable  TV   cons tan t
T2 V2r 1
 T1V1r 1 r = 1.66 for argon
T2 = 189.85 k  T = – 110.15
Thus H = 0.05  20.794  –110.15) = – 114.52 J

27. 

 CH3CH3  g   CO  g
CH3 COCH3 
At t  0 150 0 0
At t  t 150  x x x
x 1

150  x 3
x = 75 mm.

28. 

Mg  OH2  2
 Mg  2OH

s 2s

4s3 = 8.9  10–12,

s = 1.3  10–4 m
 [OH] = 2  1.305  10–4
pOH = 3.58832
pH = 10.4168

29. 1 mole of A will form 2 moles of B and C after completion of reaction when 75%. A converted into
B and C then total no. of moles
0.5   2  1.5   3.5 moles

30. Along one body diagonal 2X atoms from 2 corners, one Y particle (at the centre of cube) will be
removed.
1 15
So effective no. of X particle in a unit cell = 4  2  
8 4
And effective no. of Y particle in a unit cell = 4 – 1 = 3
15
X:Y  :3
4
5:4

Mathematics PART – III

SECTION – A
n n n
1.  +  = –p ….. (1)

Again since is a root of equation

xn + 1 + (x + 1)n = 0
n n
  
   1    1  0
  
n n n
 +  + ( + ) = 0
n + n = –( + )n = –(–p)n ….. (2)
From equation (1) and (2), we get
n n n n
–p = –(–p)  p = (–p)
n n n n
 p = (–1) p  (–1) = 1
 n is an even integer

2. From the figure, OM is bisector of angle BOP of triangle OBP

250 OB O (observer)
  P
50 OP 
 50 m
2 2
300  x
 5 M
x
 25x2  x2 = (300)2 300 m
 24x2 = (300)2
300  300 250 m
 x2   3250  25  25  6
24
 x  25 6. x m
A B

1 1 1
P  101  10 2  10 2  10 2 …..
2 3
3.
1
1 1 1 1
1  2  3 ..... 1
P  10 2 2 2
=  102  100 10 2
log0.01P = log0.01(100) = –1

ax 2  b
4.  ax 2b  x ab
2
Consequently, for all x > 0
x x 1
y 2  
ax  b 2x ab 2 ab
n n 1
n
n n
0.1  1 C1  2 C2  3 C3    n Cn n n n r r Cr  1
5. x =
1 n C1 n C2   n Cn
=  r n Cr = r 1
n
r 0
 n Cr
r 0
n
n n 1Cr 1
r 1 n2n 1 n
= = =
n
2n 2
 n Cr
r 0

n! 15!
6. Greatest coefficient is m r r
 3
 q!  q  1!  3! 4!
ln = 15 and m = 4, q = 3 and r = 37
 15 = 4  3 – 13

7. Two triangles may have equal areas

 f is not one-one
Since each positive real number can represent area of a triangle
 f is onto

32 P(0, h)
8. Area of PAB =
sin 2
 C
(PAB)min when 2 
2


  B O A
4
 h = OC cosec  = 4 2

x   y    

9. Equation of chord joining  and  is cos    sin    cos  
a  2  b  2   2 
  +  = 3
x  y
cos   0
a  2  b
It passes through the centre (0, 0)

x y
10. Ordinate of the point of intersection of the line   m and the hyperbola, given by
a b
 x y  x y 2y   2y 
 a  b  a  b  b   1 i.e. m  m  b   1
    

i.e. y 

b 1  m2 
2m
x y
Similarly ordinate of the point of intersection of the line   m and the hyperbola is given by
a b

y

b m2  1 
2m
Sum of the ordinates is zero

11. Let P(x1,y1 ) , Q(x 2 , y 2 ) be the given points on y  2x 2

 y1  2x1  2 and y 2  2x2 2
 
OP  î = projection of OP on x-axis = x1 = 1  y1 = 2
 
and OQ  î = projection of OQ on x-axis = x2 = 2  y2 = 16
 
 OP   î  2ˆj , OQ  2iˆ  16ˆj
 
 OQ  4OP  6iˆ  8iˆ
 
 OQ  4OP  6iˆ  8ˆj  10

12. The point B is (–q, –p) and the point C is (–p, –q) mid-point of AC is (0, 0)

13. The point Q is (2, 1)

x 1 y  2 4
Image of P(1, 2) in the line x – 2y + 1 = 0 is given by  
1 2 5
9 2
 Ordinate of the point are  , 
5 5
Since this point lies on QR
 Equation of QR is 3x – y – 5 = 0
+=2

14. Let (, 3 – ) be any point on x + y = 3

 Equation of chord of contact is x + (3 – )y = 9
i.e. (x – y) + 3y – 9 = 0
The chord passes through the point (3, 3) for all values of 

15. Equation of the two circles be (x – r)2 + (y – r)2 = r2 i.e. x2 + y2 – 2rx – 2ry + r2 = 0 where r = r1 and
r2 condition of orthogonally gives 2r1r2 + 2r1r2 = r12  r22
 4r1r2 = r12  r22 , circle passes through (a, b)
 a2 + b2 – 2ra – 2rb + r2 = 0
i.e. r2 – 2r(a + b) + a2 + b2 = 0
 4(a2 + b2) = 4(a + b)2 – 2(a2 + b2)
 a2 – 4ab + b2 = 0

 3
16. y2  6  x  
 2
Equation of directrix x = 0
3 3 
Let coordinate P be   t 2 , 3t 
 2 2 
 Coordinate of M are (0, 3t)
 MS  9  9t 2
3 3
MP   t 2
2 2
2
3 3  9 2
2
 9 + 9t =   t 2   1  t 2
 2 2  4
 
 4 = 1 + t2
 Length of side = 6

17. yy1 = 2a(x + x1)

 2a  
x2 = 4by = 4b    x  x1  
 y1  
2
 y1x – 8ab x – 8ab x1 = 0
D = 0 gives xy = –2ab

x cos  y sin 
18. Equation of the tangent to the ellipse at P(5 cos , 4 sin ) is  1
5 4
It meets the line x = 0 at Q(0, 4 cosec )
Image of in the line y = x is R(4 cosec , 0)
Equation of the circle is x(x – 4 cosec ) + y(y – 4 cosec ) = 0
2 2
i.e. x + y – 4(x + y) cosec  = 0
 Each member of the family passes through the intersection of x2 + y2 = 0 and x + y = 0
i.e. The point (0, 0)

x2 y2
19. Equation of normal at (13 cos , 5 sin ) of  1
169 25
13 sin 
 y  5 sin     x  13cos   it passes through (0, 6) then cos (15 + 72 sin ) = 0
5 cos 
5
 cos  = 0, sin   
24
 5 
   sin1  
 24 

21. Given, f(x) = (x – 1)(x – 2)(x – 3)(x – 4)

Consider the interval [1, 2]
Since, f(x) is a polynomial function, it is continuous in [1, 2] and differentiable in (1, 2)
Also, f(1) = 0 = f(2)
By Rolle’s theorem these exists a point x = c in (1, 2) such that f(c) = 0 i.e. these exists a root of
f(x) = 0 in the interval (1, 2) which is positive
Similarly, two positive roots of f(x) = 0 exist in the open interval (2, 3) and (3, 4)
Hence, the three roots of f(x) are positive

22. f(x) g(y) + f(x) g(y) = g(y)

 (f(x) + f(x)) g(y) = g(y)
g'  y 
 f(x) + f(x) = = constant ….. (1)
g y
g'  0  1
Put y = 0  
g0 1
g'  y 
Now from equation (1), we get f(x) + f(x) = 1 and 1
g y
 g(y) = ey

10 10 10 10
23. C1  C2  C3  C4  10  45  120  210  385

24. Total number of numbers without restriction two numbers have all the digits equal
Hence, the required number of numbers = 25 – 2 = 30

7
25. Given 7a – 9b = 0  b = a
9
So, the number of pairs (a, b) can be (9, 7), (18, 14) (27, 21) and (36, 28)
4 4
Required probability = 39 
C2 741

26. Two numbers are selected from {1, 2, 3, 4, 5, 6}

 n(S) = 6  5 (as one by one with out replacement) favourable events = minimum of the two
numbers is less than 4
 n(E) = 6  4
As for the minimum of the two is less then 4, we can select one from {1, 2, 3, 4} and other from
{1, 2, 3, 4, 5, 6}
n  E  24 4
 Required probability =  
n  S  30 5

dx dx cosec 2 x
27.  = =
 sin2 x cot x  cot x dx
sin4 x cot x
Put cot x = t2
cosec2 x dx = –2t dt
2t 2
=   dt = –2t + c = 2 cot x  c =  c
t tan x

28. Let a = 2x, b = –3x – 1, c = –1

 a3 + b3 + c3 = 3abc
1 2
Now, a3 + b3 + c3 – 3abc =  a  b  c    a  b  factorizing and get a + b + c = 0 as a  b  c
2
Hence, the given equation is; 2x – 3x – 1 – 1 = 0
By trial x = 1 and 2 are solution of this
Now, 2x = 3x – 1 + 1 has no other solution
Number of real solution exactly two

29. 2 sin2  – 5 sin  + 2 > 0

1
sin   ( –1  sin   1)
2
    5 
   0,    , 2 
 6   6 

30. Let h(x) = f2(x) + (f(x))2

 h(x) = –2x(f(x))2 g(x)
 x = 0 is the point of maximum

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Paper 1

Time Allotted: 3 Hours Maximum Marks: 234

 Pl ea s e r ea d t h e i n s t r u c t i o n s c a r ef u ll y . Yo u a r e a l l o t t ed 5 m i n u t es
s p ec i f i c a ll y f o r t h i s p u r p o s e.
ALL INDIA TEST SERIES

 Yo u a r e n o t a l l o wed t o l ea v e t h e E xa m i n at i o n Ha l l b ef o r e t h e en d o f
t h e t es t .

INSTRUCTIONS
Caution: Question Paper CODE as given above MUST be correctly marked in the answer
OMR sheet before attempting the paper. Wrong CODE or no CODE will give wrong
results.

B. Filling of OMR Sheet

C. Marking Scheme For All Three Parts.

(i) Section-A (01 – 08) contains 8 multiple choice questions which have only one correct answer.
Each question carries +3 marks for correct answer and – 1 mark for wrong answer.
Section-A (09 – 12) contains 4 multiple choice questions which have more than one correct
answer. Each question carries +4 marks for correct answer. There is no negative marking.
Section-A (13 to 18) contains 3 paragraphs with each having 2 questions. Each question carries
+3 marks for correct answer and – 1 mark for wrong answer.
(ii) Section-C (01 – 05) contains 5 Numerical based questions with answers as numerical value and
each question carries +4 marks for correct answer. There is no negative marking.

Name of the Candidate

Enrolment No.

Useful Data

PHYSICS
2
Acceleration due to gravity g = 10 m/s

Planck constant h = 6.6 1034 J-s

Charge of electron e = 1.6  1019 C

Mass of electron me = 9.1  1031 kg

Permittivity of free space 0 = 8.85  1012 C2/N-m2

Density of water water = 103 kg/m3

Atmospheric pressure Pa = 105 N/m2

Gas constant R = 8.314 J K1 mol1

CHEMISTRY

Gas Constant R = 8.314 J K1 mol1

Atomic No: H=1, He = 2, Li=3, Be=4, B=5, C=6, N=7, O=8,

Physics PART – I

SECTION – A

Straight Objective Type

This section contains 8 multiple choice questions numbered 1 to 8. Each question has 4 choices (A),
(B), (C) and (D), out of which ONLY ONE is correct

1. A spaceship is sent to investigate a planet of mass M and radius R. While hanging motionless in
space at a distance 5R from the centre of the planet, the ship fires an instrument package of
mass m (m < < mass of spaceship) with speed v 0 at an angle  w.r.t. radial line between the
centre of planet and spaceship. For what value of , the package will just graze the surface of the
GM
planet. Given v 0 
5R
(A) 30o (B) 60o
(C) 37o (D) none of these

 a r
2. A charge distribution generates a radial electric field E  2 e b rˆ
r
Where a and b are constants. The total charge giving rise to this electric field is
(A) 40 a (B) 40 b
(C) 40 ab (D) 0

3. Suppose an electrostatic potential has a maximum at point A and a minimum at point B. Choose
the correct alternative:
(A) point A is stable equilibrium point for both positive and negative charge
(B) point B is stable equilibrium point for both positive and negative charge
(C) point A is stable equilibrium point for positive and point B is stable equilibrium point for
negative charge
(D) point B is stable equilibrium point for positive and point A is stable equilibrium point for
negative charge

Space for rough work

4. Choose the correct alternatives for the circuit shown in figure. C

(A) The total energy is stored only in the electric field and is changing
(B) The total energy is stored only in the magnetic field and is
changing
(C) The total energy is stored in both electric and magnetic field and i
is changing L
(D) The total energy is stored in both electric and magnetic field and
is constant

5. A glass rod of rectangular cross section is bent into the shape as

shown in figure. The inner and outer contours of the curved portion
of the rod have fixed radii R and R + a respectively. The refractive
index of the glass is 1.5. For a given R  , what is the maximum
value of ‘a’ for which all the light entering at A will merge at B?
(A) R (B) 0.5 R
R B A
(C) (D) 2 R
3

6. A wire segment is bent into the shape of an y

Archimedes spiral as shown in figure. The i
b
equation of the curve is, r     a   for

0    . The magnetic field at P is r

x
– (a + P a

 0i ˆ 0 i
(A) k (B) kˆ
2r  2a  b 
 0i  b  ˆ  0i  b  ˆ
(C) ln  1   k (D) ln  1   k
4b  a  2b  a 

Space for rough work

7. The graph shows current and voltage y

vs time in a driven RLC circuit at
particular driving frequency. At this
I V
frequency the circuit is dominated by
(A) Inductance
(B) Capacitance
t
(C) Both (A) & (B)
(D) None of these

8. The volume of one mole of an ideal gas with adiabatic exponent  is varied according to the law
a
V , where a is a constant. Find the amount of heat obtained by the gas in this process if the
T
gas temperature increased by T.
2RT  2    R T  2   
(A) (B)
 1  1
RT  2    3RT  2   
(C) (D)
2    1 2    1

Multiple Correct Choice Type

This section contains 4 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for
its answer, out of which ONE OR MORE is/are correct

9. A drum of mass m1 and radius r1 rotates freely with initial angular velocity 0 . A second drum
with mass m2 and radius r2  r2  r1  is mounted on same axle and is at rest although it is free to
rotate. A thin layer of sand with mass m is distributed on inner surface of smaller drum. At t = 0,
small perforations in the inner drum are opened. The sand starts to fly out at a constant rate
 kg/sec and sticks to the outer drum. Ignore the transit time of the sand. Choose the correct
alternatives.
2
t0  r1 
(A) Angular speed of outer drum at time t is  
m2  t  r2 
(B) Difference in final angular speeds of two drums is 0.
 m r22  r12  m2r22 
(C) Difference in final angular speeds of two drums is 



  m  m2  r2 2  0
 
(D) Angular speed of inner drum remains constant.

Space for rough work

10. A Yo-Yo of mass m has an axle of radius b and a spool of

radius R. The Yo-Yo is placed upright on a table and string R
 b
is pulled with a horizontal force F to the right as shown in F
figure.
The static friction coefficient between Yo-Yo and table is  s . Now there are two cases.
case (a): The string is pulled very gently?
case (b): The string is jerked hard?
The Yo – Yo will rotate:
(A) Clockwise in case (a),
(B) Anticlockwise in case (a)
(C) Clockwise in case (b)
(D) Anticlockwise in case (b)

11. Holding the upper plate at + 5V relative to the ground of the A

lower plate, what is true about the electric potential at the
following locations. C
B
(A) V(A) > V(B)
(B) V(B)  V(C)
(C) V(D) > V(A)
(D) V(C) > V(D) D

12. The equation of a transverse wave travelling along a string is given by

y  0.4 sin    0.5x  200t   where y and x are measured in cm and t in seconds. Suppose that
you clamp the string at two points L cm apart and you observe a standing wave of the same
wavelength as above transverse wave. For what values of L less than 10 cm is this possible?
(A) 2 cm (B) 4 cm
(C) 9 cm (D) 8 cm

Space for rough work

Comprehension Type

This section contains 3 groups of questions. Each group has 2 multiple choice question based on a
paragraph. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which ONLY ONE is
correct

Paragraph for Questions 13 & 14

A flutes can be regarded as a tube open at both ends. It will emit a musical Valve
note if the flutist excites a standing wave in the air the column inside the tube. B
Speed of sound in air = 344 m/s. C D E
BC  CD  DE  0.035m.

13. The lowest musical note that can be played on flutes is C (261.7 H2). What must be length of the
tube? Assume that air column is vibrating in its fundamental mode.
(A) 0.251 m (B) 0.355 m
(C) 0.657 m (D) 0.87 m

14. In order to produce higher musical notes, the flutist, opens valves along the tube. Since the holes
in these valves are large, an open valve has same effect as shorting tube. The flutist opens one
valve to play C, two valves to play D, etc. find frequency of flutes when valve C is opened keeping
D and E are closed. Assume flute vibrates in its fundamental mode.
(A) 276.5 hz (B) 305.5 hz
(C) 150.5 hz (D) 450.5 hz

Paragraph for Questions 15 & 16

A Rydberg hydrogenic atom is one in which the electron possess a very large quantum number e.g.
n = 100. Take the electron charge to be – e (e > 0). The binding energy of Rydberg electron may be taken
as Eb  103 eV, e  1.6  10 19 C

15. What angular frequency of a photon can ionize such an atom?

(A) 1016 rad/s (B) 1010 rad/s
8
(C) 10 rad/s (D) 109 rad/s

16. Now consider the electron in the Rydberg hydrogenic atom to be unbounded and force for all
pratical purpose. Supposing such Rydberg atoms are infected uniformly into an oscillatory
ˆ Provided by an electromagnetic wave, let the speed of the electron at
electric field F0 cos t k.
time of infection (t = 0), be v = 0. Find maximum speed of electron.
eF0 eF0
(A) (B)
2m 3m
eF0 eF0
(C) (D)
5m m

Space for rough work

Paragraph for Questions 17 & 18

When two slabs of N-type and P-type semiconductor are put in contact, the relative affinities of the
materials cause electrons to migrate out of N-type material across the function to P-type material. This
leaves behind a volume in the N-type material that is positively charged and orates a negatively charged
volume in P-type material.
Let us model this as two infinite slabs of charge, both of thickness a with the function lying on the plane z
= 0.
The N-type material lies in the range 0 < z < a and has uniform charge density 0 . Thus:
0 0za

  x,y,z     z    0 a  z  0
 0 z a

The give answer of questions

17. Find electric field in religion –a < z < 0

 a  z  20  a  z 
(A) 0 k̂ (B) k̂
0 0
0  a  z 
(C) 0 (D) k̂
0

18. Magnitude of potential difference between points z = –a to z = a

 a2 30 a2
(A) 0 (B)
40 20
0 a2 0 a 2
(C) (D)
0 60

Space for rough work

SECTION – C

(One Integer Value Correct Type)

This section contains 5 questions. Each question, when worked out will result in one integer from 0 to 9
(both inclusive)

1. A parallel plate capacitor of capacitance C is connected to a battery of emf  until fully charged.
It is then disconnected from battery and the plates are moved by a distance x so that potential
difference between the plates is increased by four times. What is the volume of the dielectric (in
m3) necessary to fill the region between the plates.
Given C = 8.85  F, x = 3 mm

2. Find maximum charge store on capacitor in  C. R1 = 6 

C = 2 F
+ R2 = 2 
 =16 V

3. In the circuit shown in the figure the emf of battery is 9v and its V1 V2
internal resistance is 15. The two identical voltmeters can be
considered ideal. Let V1 and V11 reading of 1st voltmeter when
10Ω 20Ω
switch is open and closed respectively. Similarly, V2 and V21 be
the reading of 2nd voltmeter when switch is open and closed
V1  V2 9V
respectively. Then find value of 2 ? 15Ω
V1  V11

4. A cube with sides of length 60  

2  1 cm and mass M is moving with an initial speed v 0 along a
frictionless table. When it reaches the end of the table, it is caught abruptly by a short lip. What is
the minimum speed v 0 in m/s such that it falls of the table?

5. Two beetles A and B hold the ends of a slightly stretched rubber band laid out on a horizontal
table top. The initial positions of A and B are (0, 16 m) and (0, –4 m). A knot tied on the band is
initially located at the origin. The beetle A starts running in +Y direction at constant acceleration
and B moves in +X direction at constant speed of 2 5 m/s. If the knot passes through point (8,
4), find the acceleration of A.

Space for rough work

Chemistry PART – II

SECTION – A

Straight Objective Type

This section contains 8 multiple choice questions numbered 1 to 8. Each question has 4 choices (A),
(B), (C) and (D), out of which ONLY ONE is correct

O
1.
H C OH , CH3COOH can be separated by
(A) NaHCO3 (B) Tollen’s reagent
(C) Neutral FeCl3 (D) All of these

2. Write the basicity order for following compounds

H3C CH3
NH2 N NH2
O 2N NO2 O 2N NO 2

(I) (II) (III)

(A) III > II > I (B) II > I > III
(C) II > III > I (D) III > I > II

3. Which one of the following gives positive test with Fehling solution?
(A) (B)

O OCH3 O CH3
(C) (D)
CH3
O
O OH

211
4. The radioactive decay 83 Bi  Tl207 , takes place in 100 L closed vessel at 27°C. Starting
81
211
with 2 moles of 83 Bi  t1/ 2  130 sec  . The pressure development in the vessel after 520 sec
will be
(A) 1.875 atm (B) 0.2155 atm
(C) 0.4618 atm (D) 4.618 atm

Space for rough work

5. The forward rate constant for the elementary reversible gaseous reaction

C 2H6  –3 –1
 2CH3 is 1.57  10 s at 100 K
What is the rate constant for the backward reaction at this temperature if 10–4 moles of CH3 and
10 moles of C2H6 are present in a 10 litre vessel at equilibrium
(A) 1.57  106 L mol–1s–1 (B) 1.57  1010 L mol–1s–1
11 –1 –1
(C) 1.57  10 L mol s (D) 1.57  107 L mol–1 s–1

6. A mixture of nitrogen and water vapours is admitted to a flask at 760 torr which contains sufficient
solid drying agent after long time the pressure reached a steady value of 722 torr. If the
experiment is done at 27°C and drying agent increase in weight by 0.9 gm, what is the volume of
flask? Neglect any possible vapour pressure of drying agent and volume occupied by drying
agent.
(A) 443.34 L (B) 246.3 L
(C) 12.315 L (D) 24.63 L

7. Among KO2, AlO2, BaO2 and NO2 , unpaired electron is present in

(A) NO2 and BaO2 (B) KO2 and AlO2
(C) KO2 only (D) BaO2 only

8. Which of the following sentence is wrong about H2O and HF

(A) Both the compounds resemble each other in solubility.
(B) Heat of formation of HF and H2O are close to each other.
(C) Like H2O, H2F2 is a weak acid and both are ionising solvents.
(D) The boiling points of HF and H2O are high due to hydrogen bonding

Space for rough work

Multiple Correct Choice Type

This section contains 4 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for
its answer, out which ONE OR MORE is/are correct

9. Which of the following order is wrong with respect to alkali metals?

(A) In gaseous state the order of reactivity is Li < Na < K < Rb < Cs.
(B) conducting power Cs+ > Li+ > K+ > Na+ > Rb+ (in aqueous Solution).
(C) density order Li > Na > K > Rb > Cs.
(D) In gaseous sate the order of reactivity is Li > Na > K > Rb > Cs.

10. Among the following statements, the correct statements are:

(A) Calamine and cerrusite are carbonate ores
(B) Rutile and cuprite are oxide ores
(C) Zinc blende and pyrites are sulphides ores
(D) malachite and azurite are sulphate ores of Cu

11. Which of the following is correct for an ideal solution?

(A) Hmix  0 and Vmix  0 (B) Vmix  0 and Smix  0
(C) Hmix  0 and Smix  0 (D) Gmix  0 and Smix  0

12. Which of the following statements are correct

(A) Benzene is more reactive than pyridine towards electrophilic aromatic substitution
(B) Benzene is less reactive than pyrrole towards electrophilic aromatic substitution
(C) Pyridine is more reactive than benzene towards nucleophilic aromatic substitution
(D) Pyridine is more reactive than Benzene towards electrophilic aromatic substitution

Space for rough work

Comprehension Type

Paragraph for Question Nos. 13 to 14

The cell potential (Ecell) of a reaction is related as G = –nFEcell
For reversible reaction
at constant pressure d(G) = –(S).dT
G = H = T.S
 d  G  
G  H  T  
 dT P
 dEcell 
 dT  is known as temperature coefficient of the e.m.f of the cell.
 P

 dE 
13. The temperature coefficient of the e.m.f of cell   is given by
 dT P
nF S
(A) (B)
S nF
S
(C) (D) –nFE
nFT

14. At 300 K, H for the reaction

Zn  s   AgCl s   ZnCl2 aq  2Ag  s  is –218 kJ/mol. While the e.m.f. of the cell was
 dE 
1.015 V. Then the value of   for the cell is:
 dT P
(A) –4.2  10–4 VK–1 (B) –3.81  10–4 VK–1
–1
(C) 0.11 VK (D) 7.62  10–4 VK–1

Space for rough work

Paragraph for Question Nos. 15 to 16

An unknown mixture contains one or two of the following: CaCO3, BaCl2, AgNO3, Na2SO4, ZnSO4 and
NaOH. The mixture is completely soluble in water and solution gives pink colour with phenolphthalein.
When dilute hydrochloride acid is gradually added to the solution, a precipitate is formed which dissolves
with further addition of the acid.

15. Which of the following combination of compounds is soluble in water?

(A) BaCl2 and AgNO3 (B) AgNO3 and NaOH
(C) BaCl2 and Na2SO4 (D) ZnSO4 and excess NaOH

16. The aqueous solution of mixture gives white precipitate with dil HCl which dissolves in excess of
dil. HCl. It confirms
(A) BaCl2 + NaOH (B) Na2SO4 + NaOH
(C) ZnSO4 + NaOH (D) AgNO3 + NaOH

Space for rough work

Paragraph for Question Nos. 17 to 18

Cl O
OH H 2 SO 4
H3C CH CH2 CH2 C Br
 i (i)
 A  
 B 

ii Zn Hg / HCl  ii  KMnO4 / 
 iii  AlCl3
H3CO

17. In the above sequence of reactions compound A is:

(A) (B)

H3CO
H3CO
(C) (D)

H3CO H3CO

18. In the above sequence of reactions compound B is:

(A) COOH (B)
COOH

H3CO COOH
COOH H3CO COOH
(C) COOH (D) HOOC COOH

H3CO COOH H3CO COOH

Space for rough work

SECTION – C

(One Integer Value Correct Type)

This section contains 5 questions. Each question, when worked out will result in one integer from 0 to 9
(both inclusive)

–8
1. If the lowest energy X-rays having  = 3.055  10 m, estimate the minimum difference in energy
between two Bohr’s orbits such that an electronic transition would correspond to the emission of
an X-ray. Assuming that the electrons in other shells exert no influence, at what Z (minimum)
would transition from the second energy level to the first result in the emission of an X-ray?

2. A vessel of 250 litre was filled with 0.01 mole of Sb2S3 and 0.01 mole of H2 to attain the
equilibrium at 440°C as


Sb2S3  s   3H2  g 
 2Sb  s   3H2S  g  . In a separate experiment this H2S is sufficient to
2+
precipitate 1.19 g PbS from Pb solution. Then calculate Kc of the above reaction.

3. In metaborate structural arrangement, the number of B – O – B bonds is/are:

4. O O

dil OH
H3C C H  H3C CH2 C H  ?
The number of possible products is/are:

Hydrolysis Na CH CH  Cl H SO
5. A 
 B 
 C 
3 2
 D 
2
HgSO
4
E
4

E is 2-butanone then ‘A’ has how many  bonds?

Space for rough work

Mathematics PART – III

SECTION – A

Straight Objective Type

This section contains 8 multiple choice questions numbered 1 to 8. Each question has 4 choices (A),
(B), (C) and (D), out of which ONLY ONE is correct

 1 
1. If a1 is the greatest value of f(x), where f  x    (where [.] denotes the greatest integer
 2   sin x  
 
n 2
 1
function) and an 1   an , then lim an is
n  1 n

(A) 1 (B) e2
(C) ln 2 (D) none of these

      n
2. If n be a positive integer such that sin    cos    , then
 2n   2n  2
(A) 6  n  8 (B) 4 < n  8
(C) 4  n  8 (D) 4 < n < 8

 
3. Let f(x) = tan x and g(f(x)) = f  x   , where f(x) and g(x) are real values functions for all
 4
possible value of x, then f(g(x)) is
 x  1
(A) tan   (B) tan (x – 1) – tan (x + 1)
 x  1

f  x  1 x
(C) (D) 4
f x  1 
x
4

Space for rough work

4. For which of the hyperbola, we can have more than one pair of perpendicular tangents?
x2 y 2 x2 y 2
(A)  1 (B)   1
4 9 4 9
(C) x2 – y2 = 4 (D) xy = 4

5. A ray of light passing through the point P(1, 2) is reflected at a point Q on the x-axis and then
passes through (5, 3), then the equation of reflected ray is
(A) 5x – 4y = 13 (B) 5x – 4y = –3
(C) 4x + 5y = 14 (D) 4x – 5y = –6

6. On the line segment joining (1, 0) and (3, 0) an equilateral triangle is drawn having its vertex in
the fourth quadrant, then radical centre of the circles described on its sides as diameter is
 1 
(A)  3, 
 3
 (B) 3,  3  
 1 
(C)  2, 


3
(D) 2,  3  
7. Let P and Q be points (4, –4) and (9, 6) of the parabola y2 = 4a(x – b). Let R be a point on the arc
of the parabola between P and Q, then the area of PRQ is largest when
(A) PRQ = 90º (B) the point R is (4, 4)
1   1
(C) the point R is  , 1 (D) the point R is  1, 
4   4

8. If f(x) is a decreasing function then the set of values of ‘k’ for which the major axis of the ellipse
x2 y2
  1 is the x-axis, is
 
f k 2  2k  5 f  k  11
(A) k  (–2, 3) (B) k  (–3, 2)
(C) k  (–, –3)  (2, ) (D) k  (–, –2)  (3, )

Space for rough work

Multiple Correct Choice Type

This section contains 4 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for
its answer, out which ONE OR MORE is/are correct

9. If
x 7/6
 x5/6   Z
dx  P 
3

3zQ 3z 2


 log z   R where
 1/2 1/3
 3 1 T

x1/3 x 2  x  1   x1/2 x 2
 x 1  
1/6
 1 
z   x   1  1 , then
 x 
(A) P = 6 (B) Q – P = 4
(C) T + Q = 3 (D) Q – P + T = 5

tan2 x 2
10. If e sin 4xdx  k cos4 xetan x
 c , then k is less than
(A) 3 (B) 4
(C) 5 (D) none of these

t
sin3 x
11. Let f  t    3 cos2 x dx , then

2

 18
(A) f    0 (B) f  0   
 2 7
 18
(C) f     (D) none of these
2
  7

12. If f(x) = ax2 + bx + c = 0 has real roots and its coefficients are odd positive integers then
(A) f(x) = 0 always has irrational roots
(B) discriminant is a perfect square
(C) if ac = 1, then equation must have exactly one root ‘’ such that [] = –1 (where [.] G.I.F)
(D) Equation f(x) = 0 has rational roots

Space for rough work

Comprehension Type

Paragraph for Question Nos. 13 to 14

Read the following write up carefully and answer the following questions:
Let S be the set of points {(x, y): 0  x  3, 0  y  4, where x, y  I}. Two distinct points are randomly
chosen from S. It is given that mid-point of the line segment joining A(x1, y1) and B(x2, y2) is
 x1  x 2 y1  y 2 
 , 
 2 2 

13. The probability that the mid-point of the line segment of the two points chosen belongs to S, is
21 42
(A) (B)
95 95
36
(C) (D) none of these
95

14. If two points chosen are (x1, y1) and (x2, y2), then the probability that x1  x2 and y1 = y2 with
 x1  x 2 y1  y 2 
 ,   S , is
 2 2 
8 12
(A) (B)
95 95
14
(C) (D) none of these
95

Space for rough work

Paragraph for Question Nos. 15 to 16

Read the following write up carefully and answer the following questions:
Equation of the form P(sin x  cos x, sin x cos x) = 0 where P(y, z) is a polynomial, can be valued by the
2  t2  1 
change cos x  sin x = t; 1  2 sin x cos x = t . Reduce the given equation into P  t, 0
 2 

15. If sin x + cos x = 1 + sin x cos x, then x is equal to

 
(A)  2n or 2n, n  I (B)  2n or (2n + 1), n  I
2 4
(C) 2n, n  I (D) none of these

16. If sin4 x + cos4 x = sin x cos x, then x is equal to


(A)  6n  1 (B) n
6

(C)  4n  1 (D) none of these
4

Paragraph for Question Nos. 17 to 18

Read the following write up carefully and answer the following questions:
 x    x   
Consider f  x   where 0 <  <  <  < 
 x    x   
17. Number of local maxima of f(x) will be
(A) 0 (B) 1
(C) 2 (D) can’t be determined

18. Function f: D  R, (where D is the domain and R is the set of real numbers), will be
(A) one-one (B) many one
(C) bijective (D) none of these

Space for rough work

SECTION – C

(One Integer Value Correct Type)

This section contains 5 questions. Each question, when worked out will result in one integer from 0 to 9
(both inclusive)

1
1. Let p(x) be a polynomial of degree 11 such that p  x   for x = 0, 1, 2, ….., 11 then p(12) is
1 x
_____

2. If the coefficient of the second, third and fourth terms in the expansion of (1 + x)n are in A.P. then
n is _____

8
3. A tangent is drawn to the curve y  in xy-plane at the point A(x0 , y0 ) where x0  2 , and the
x2
 
tangent cuts the x-axis at a point B. Then AB  OB (where O is origin) is equal to _____

4. Let f(x) = 30 – 2x – x3, then find the number of positive integral values of x which satisfies
f(f(f(x))) > f(f(–x)) _____

5. If x + sin y = 2010 and x + 2010 cos y = 2009, 0 < y 


then find the value of
 x  y  (where [.]
2 1005
denotes the greatest integer function) _____

Space for rough work

ANSWERS, HINTS & SOLUTIONS

CRT–III
(Paper-1)

Q. No. PHYSICS CHEMISTRY MATHEMATICS

1. C B C
ALL INDIA TEST SERIES

2. D C D

3. D C A

4. D C B

5. B D A

6. C D C

7. B C C

8. B A B

9. A, C, D B, C, D A, C

10. A, D A, B, C A, B, C

11. A, B, D A, B, D A, B

12. A, B, D A, B, C A, C

13. C B A

14. A B D

15. A D A

16. D C C

17. A A B

18. C B B

1. 4 2 0

2. 8 1 7

3. 1 3 3

4. 4 4 2

5. 8 2 2

Physics PART – I

SECTION – A

1. Angular momentum conservation

v0
about the centre of the planet, v1
mv0  5 R sin   mv1R  R
v1  5 v 0 sin  -------(1) M
Energy conservation gives,
1 GMm 1 GMm
mv 0 2   mv12  ------(2)
2 5R 2 R
Solve equation (1) and (2),
  37

2. Assume a Gaussian surface to be a sphere of radius r and take limit as r  . The flux,
  q
 E.dA  0
q
 E.dA  0
a r /b q
lim 2
e  4r 2 
r  r 0
q
4a  lim e r /b 
r  0
q=0

5. The smallest angle of incoming beam should be greater than critical angle
of the medium.
The smallest angle incoming beam should be greater than critical angle of Ra 
the medium.
sin   sin c R
R 1

Ra 
a     1 R
R
amax 
2

 0i d   rˆ i  i
6. dB 
4 r 2
4r
   
 0 2 drˆ r  rdˆ   ˆ r  0 2 r d kˆ
4r
ˆ
Where ̂ r and   are unit vectors along radial and azimuthal directions
  i d
dB  0 kˆ
4 r
  i d
B   dB   0 kˆ
0 4  b 
a  
  
 i  b
 0 ln  1   kˆ
4b  a 

7. Since current leads the voltage

Hence circuit is capacitive.

8. For 1 mole gas,

PV = RT
PV 2  aR  constant
Compare with PV x  constant
x=2
R R
molar heat capacity of gas C  
 1 x 1
R
C 2   
 1
RT
Q  nCT  2   
 1

9. As the sand leaves the inner drum through the open holes, it does not 2
exert any force of the drum, angular momentum remains conserved.
At time t,
m1r12 0   m1  t  r12 0  m2  t  r22 2
1
tr12 0
2  r1
m2  t  r22
r2
The speed of inner drum does not change.

10. In first case, we can write torque equation about bottom point because it coincides with
instantaneous axis of rotation while in second case, you will write torque equation of centre of
mass.

11. V  A   5V
V  B   V  C   2.5V
V D   0

2
12. K  0.5     4cm


L   2cm
2

 L    4cm

3
L   6cm
2

 L  2  8cm

 v
13.     0.657m
2 2f

nv
14. f
2
n 1
V = 344
  0.657  0.035
f = 276.5 hz

15. Hf > Eb

6.6  10 34  103  1.6  1019
2
  1.6  1012 rad/s
F
16. a  0 cos t
m
eF0
v sin t
m
eF
v max  0
m

17. Here z is co-ordinate a+z z

  a  z  A
EA  0
0 E
S
0  a  z 
E
0

   0 0  a  z  a 0  a  z  
18.  dv     dr    a 0 dz  0 0 dz 
E

0 a2

0

SECTION – C

1. Initially,
0 A
C -----(1)
d
V0  Ed ------(2)
and Vf  Edf
4V0  E  d  x 
4Ed  E  d  x 
x
d
3
Volume = A(d + x)
Cd
=  d  x 
0
4x 2 C

90


2. V
R
1 1
R2
C
Q  CV 
R
1 1
R2

3. When switch is open

V1  V2  6V
V1  V2  3V
When switch is closed
V11  2V; V21  4V
V21  V2 43
 1
V1  V11 32

4. If length of side = 2a = 60  
2  1 cm
angular momentum conservation,
Mav 0  I 
Mav 0
 .....(1)
I
1 2
2

I  Mga 2  1 ......(2) 
2
I  Icm  M a 2  
8Ma2
 ....(3)
3
Solve these equation,
v 0  4 m/s

Chemistry PART – II

SECTION – A

O
1. Tollen’s reagent gives +ve test for H C OH not CH3COOH
.

2. Due to steric hindrance in II the loan pair e are not involved in resonance in I electron
withdrawing effect of NO2 is high.

3. 



O OH OH O
H

211
4. 83 Bi  Tl207  2 He 4 ;
81
total time = n  half life
moles of substance left after n halve.
initial moles 2
=  4  0.125
2n 2
moles of He produced
= 2 – 0.125 = 1.875
Pressure developed due to
1.875  0.0821 300
He =
100
= 0.4618 atm

2
K f  CH3 
5. K eq   ,  CH3   10 5 m
Kb C2H6 
5 2
1.57  103 10 
 , K b  1.57  107 L mol1s1
Kb 1

38
6. PH2 O  760  722  atm;
760
0.9
nH2O 
18
0.9 0.0821 300
V   760;
18 38
V = 24.63 L

7. KO2 has only unpaired e  remaining all are diamagnetic.

8. Both compounds resemble each other in solubility.

12. Order of reactivity towards E. A. substitution

 
N
N
H

d   G 
13. From equation above S =
dT
d  nFE  dE dE S
S  = nF  or    
dT dT  dT P nF

 dE 
14. H  nF   T  nFEcell
 dT P
 dE 
–218  1000 = 2  96500  300    2  96500  1.015
 dT P
 dE  4 1
   3.81 10 VK
 dT P

16. Mixture contains NaOH + ZnSO4

NaOH  ZnSO 4  Na 2 ZnO2  Na 2SO 4  2H2 O
so lub le

Na2 ZnO2  2HCl  2NaCl  Zn  OH2 

Zn  OH 2  2HCl  ZnCl2  2H2O
so lub le
18. O

Cl O C
H3C CH CH2 CH2 C Br
Zn Hg / HCl

H3CO H3CO
Cl
CH3

OH
AlCl

3

H3CO H3CO
Cl (A) H3CO
CH3

COOH
KMnO / 

4


H3CO COOH

SECTION –C

1. E 
hc

 
6.63  1034 J.s 3  108 m / s 
 3.055  10 8 m
–18
= 6.52  10 J.

3
EH 
4

2.176  1018 J 
–18
= 1.63  10
E = EH(Z)2

2
Z 
E


6.52  1018
4

EH 
1.63  10 18 
Z=2

2. 

Sb2S3  s   3H2  g 
 2Sb  s   3H2S  g
0.01  x 0.01  3x 2x 3x
3x = 0.005
H2 S  Pb2  PbS  2H
1.19
no. of moles of PbS formed =  0.005 moles
238
 0.005 
at equilibrium H2    
 250 
3
 0.005 
Kc    1
 0.005 

3.
O
B
O O
Metaborate BO2

B B
O O O

4. OH O OH
H3C C CH C H H3C CH2 C CH2 CHO
H CH3 H

OH OH

H3C C CH2 CHO H3C CH2 C CH CHO

H H CH3

5. C
A  CaC2 Ca
C
B = HC CH
C = HC CNa
D = HC C CH2 CH3
O

E=
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AITS-CRT-III-(Paper-1)-PCM(Sol)-JEE(Advanced)/15

Mathematics PART – III

SECTION – A
1. a1 = 1
1
a2 = 1 
2
1 1
a3 = 1  
2 3


1 1 1
a = 1     .....  ln2
2 3 4

      n
2. Given, sin    cos   
 2n   2n  2
            n
 sin2    cos2    2sin   cos   
2n
  2n
  2n
   2n  4
 n  n4
 1  sin     sin   
n 4 n 4

For n = 2, the given equation is not satisfied considering n > 1 and n  2, 0  sin    1
n
n4
 0 1  4 < n < 8
4

   tan x  1
3. g(f(x)) = tan  x   
 4  tan x  1
x 1
 g x 
x 1
 x  1
f(g(x)) = tan  
 x  1

x2 y 2
4. Locus of point of intersection of perpendicular tangents is director circle for   1 equation
a2 b2
of director circle is x2 + y2 = a2 – b2 which is real if a > b

5. Image of P in x-axis is (1, –2) y

(5, 3)
5 P(1, 2)
 Equation of reflected ray is (y – 3) =  x  5 
4
  
 5x – 4y = 13 
x
O Q(a, 0)

6. Radical centre of the circles described on the sides of a triangle y

Q(1, 0) S R(3, 0)
as diameters is the orthocentre of the triangle x x
 OS = 2 T
 1
ST  QS tan  
6 3
 1 
 Quadrants of T is  2,   P
 3 y
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10

2
7.  Since, (4, –4) and (9, 6) lie on y = 4a(x – b)
 4 = a(4 – b) and 9 = a(9 – b)
 a = 1 and b = 0
 The parabola is y2 = 4x
2
Let the points R be (t , 2t)
4 4 1 2
1  1  125
 Area of PRQ = 9 6 1 = 5  t   
2 2  2 4
t 2t 1
1
 Area in largest when t 
2
1 
 Coordinate of the point R is  , 1
4 

8. f(x) is a decreasing function and for major axis to be x-axis

f(k2 + 2k + 5) > f(k + 11)
 k2 + 2k + 5 < k + 11
 k  (–3, 2)

x 7/6  x 5/6 x 7 /6  x5/6

9. 1/2 1/3
 1/2 1/3

x1/3 x 2  x  1  
 x1/2 x 2  x  1   1 
x 5 /6  x   1
 1 
 x 5/6  x   1
 x   x 
1
1 1/6
x2  1 
= 1/2 1/3
put Z   x   1 1
 1   1   x 
 x   1   x   1
 x   x 
After solving we get, P = 6, T = 2 and Q = 1

tan2 x
10. e sin 4xdx
Applying integration by parts
 sin 4xdx    e 
2
tan2 x
etan x
 2 tan x sec 2 x  sin 4xdx dx
1 tan2 x 1 2
 cos 4x   etan x tan x sec 2 x 2 2cos2 x  1  1 dx continue solving like this
2
= 
4
e
2 
   
11. Put z = cos x, dz = –sin x dx
t t 2
sin3 x 
f t    3 cos2 x 
dx    1  z2 z  3 dz

 
2 2

12. If a, b and c are odd integers then discriminant = b2 – 4ac  perfect square
 Roots are always irrational
If ac = 1  a = 1, c = 1 and b  3
 a, b and c = odd
 c(a – b + c) < 0  f(0) f(–1) < 0
   (–1, 0)

13.-14. Total points generated = 20

Two points can be selected in 20 C2  190 ways
(i) Number of points with all coordinates odd = 4
(ii) Number of points with x odd, y even = 6

(iii) Number of points with x even, y odd = 4

(iv) Number of points with both even = 6
 Required number of points = 4 C2  6 C2  4 C2  6 C2  42
42 21
(1) Required probability = 
190 95
(2) (x1  x2 and y1 = y2) = 2  5 = 10
10 1
Required probability = 
190 19

15. sin x + cos x = 1 + sin x cos x

t2  1
 t  1 (Let sin x + cos x = t)
2
2t = 2 + t2 – 1  t2 – 2t + 1 = 0
2
(t – 1) = 0  t = 1
 sin x + cos x = 1
  1
cos  x   
 4 2
 
x   2n 
4 4

x  2n 
4

 x = 2n or x = 2n + , n  I
2

16. sin4 x + cos4 x = sin x cos x

(sin2 x + cos2 x)2 – 2 sin2 x cos2 x = sin x cos x
1 – 2p2 = p ( sin x cos x = p)
(p + 1)(2p – 1) = 0
sin x cos x = –1 2 sin x cos x –1 = 0
sin x cos x = –1 sin 2x = 1
 
sin 2x = –2 (not possible) 2x = 2n +  x = (4n + 1)
2 4

 x    x   
17.-18. Graph of f  x   x-axis
 x    x   
y=1

y-axis
O    

SECTION – C

1. Let g(x) = (x + 1) p(x) – 1 ….. (1)

g(0) = g(1) = g(2) = ….. = g(11) = 0
 g(x) = kx(x – 1)(x – 2) ….. (x – 11) ….. (2)
Also, g(–1) = –1
1
 k
12

Now, from equation (1) and (2), we get

1
 x  1 p  x   1   x  x  1 x  2  .....  x  11
12
 p(12) = 0

n
2. C1, n C2 , nC3 an in A.P.
2n  n  1 n  n  1n  2 
n
2 6
 n = 2 or 7
Since n C3 demands that n should be  3 then n can not be 2
n=7

8
3. x0  2  y0  2
4
 Equation of the tangent is y  2x  6 which cuts x-axis at (3, 0)
Hence
 B  (3, 0) 
 AB  (3  2)iˆ  (0  2)jˆ and OB  3iˆ
 
 AB  OB  3

4. f(x) = –2 – 3x2 < 0  f(x) is decreasing

 f(f(f(x))) > f(f(–x))
 f(f(x)) < f(–x)  f(x) > –x
 30 – 2x + x3 > – x  x3 + x – 30 < 0
 (x – 3)(x2 + 3x + 10) < 0  x < 3
Positive values of integer x = 1, 2

x  sin y  2010
x  2010cos y  2009
5. substracting
sin y  2010cos y  1
 sin y = 1 + 2010 cos y
 
Possible if y   0  y 
2 2
 x = 2010 – 1 = 2009
 
2009  
 x  y   2
  2
1005 1005

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CONCEPT RECAPITULATION TEST - III
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Paper 2

Time Allotted: 3 Hours Maximum Marks: 234

 Pl ea s e r ea d t h e i n s t r u c t i o n s c a r ef u ll y . Yo u a r e a l l o t t ed 5 m i n u t es
s p ec i f i c a ll y f o r t h i s p u r p o s e.
ALL INDIA TEST SERIES

 Yo u a r e n o t a l l o wed t o l ea v e t h e E xa m i n at i o n Ha l l b ef o r e t h e en d o f
t h e t es t .

INSTRUCTIONS
Caution: Question Paper CODE as given above MUST be correctly marked in the answer
OMR sheet before attempting the paper. Wrong CODE or no CODE will give wrong
results.

B. Filling of OMR Sheet

C. Marking Scheme For All Three Parts.

Name of the Candidate

Enrolment No.

Useful Data

PHYSICS
2
Acceleration due to gravity g = 10 m/s

Planck constant h = 6.6 1034 J-s

Charge of electron e = 1.6  1019 C

Mass of electron me = 9.1  1031 kg

Permittivity of free space 0 = 8.85  1012 C2/N-m2

Density of water water = 103 kg/m3

Atmospheric pressure Pa = 105 N/m2

Gas constant R = 8.314 J K1 mol1

CHEMISTRY

Gas Constant R = 8.314 J K1 mol1

Atomic No: H=1, He = 2, Li=3, Be=4, B=5, C=6, N=7, O=8,

Physics PART – I

SECTION – A

Straight Objective Type

This section contains 8 multiple choice questions numbered 1 to 8. Each question has 4 choices (A),
(B), (C) and (D), out of which ONLY ONE is correct

1. Two coherent sources S1 and S2 are situated as shown in I0 I0

figure. Here S1 is fixed but S2 is performing SHM with
amplitude 2 cm. Initially S2 is at right extreme. At time S1 S2 P
4 cm

t (where  = angular frequency of SHM):  4
2

(A) Phase difference between waves produce by S1 and S2 at point P is .
2
(B) Intensity at point P is 7I0.
(C) Intensity at point P is 4I0
(D) Intensity at point P is zero

2. A disc of mass m and radius R is projected with initially velocity V0 and initial angular velocity zero
on rough surface. Then, which of the following is INCORRECT.
m

V0
R

rough surface
P

(A) Angular momentum of disc about point P is conserved

(B) Friction force will be zero after pure rolling start
V
(C) Final angular velocity of disc will be 0
R
2V0
(D) Final velocity of centre of mass will be
3

Space for rough work

3. There is a conductor with cavity of radius R as shown in

figure. A point charge Q is placed at R/2 distance from
centre of cavity. A point P is at r distance from centre of Q
conductor. Then, which of the following is INCORRECT. R C r P
(A) charge density at inner surface of cavity will be non 2 R
uniform 2R
(B) charge density at surface of conductor will be uniform
Q
(C) electric field at point P is
40r 2
(D) force on charge Q is zero.

4. A metal rod PQ (carrying current from P to Q) is placed perpendicular to the

infinite long wire carrying current i0. If this arrangement lies in a horizontal
plane; then the rod PQ will rotate:
(A) clockwise (B) anticlockwise i0 P Q
(C) along PQ (D) does not rotate

Space for rough work

5. Current growth in two L – R circuits (2) and (3) is shown in figure (1) after switch S closed. Then
L1 R1 L2 R2
i

(b)
(c)

t
(1) S S
V V
(2) (3)
(A) R1 > R2 (B) R1 = R2
(C) L1 > L2 (D) L1 = L2

6. A uniform disc of mass m and diameter 2 R moves

M
forward towards another uniform disc of mass 2 m V0
M
& diameter 2 R on a frictionless surface as shown
in figure 2M
2M

When the first disc contacts the second, they stick to each other and move as a single object.
Then, find the incorrect option.
V
(A) the velocity of combined disc after the collision is 0 .
3

(B) the angular velocity of combined disc after the collision is 0 .
3
V
8 0
(C) If 0  , the combined disc will not rotate
3 R
19
(D) If combined disc does not rotate, the energy loss is MV0 2 .
9

Space for rough work

7. Two strings with linear mass densities 1  0.1 kg/m and  2  0.3 kg/m are joined seamlessly.
They are under tension of 20 N. A travelling wave of triangular shape is moving from lighter to
heavier string.
(A) The reflection coefficient at interface is zero
(B) The reflection coefficient at interface is 2  3
(C) The transmission coefficient at interface is 1
(D) The transmission coefficient at interface is 3  5

8. Consider a piston – cylinder arrangement with spring in its natural

length. Due to heat transfer, the gas expands until the piston hits the k = 10000 N/ m
2
stop. The mass of piston is 10 kg and area is 78 cm . Initial and final
specific internal energies are 214 and 337 k J / kg. Then, find the
incorrect option.
(A) Initial pressure of gas is 112 k Pa 5 cm
(B) Work done by gas on the piston is 56.5 J
(C) Work done by gas on the piston is 100 J
(D) Heat transfer for 0.5 gas is 118 J

Multiple Correct Choice Type

This section contains 4 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for
its answer, out of which ONE OR MORE is/are correct

9. In the circuit shown, the switch S has 50 k

been closed for a long time. At time
t = 0, the switch is opened. It remains
open for a long time T, after which it is 10 V
closed again. S 10 F

100 k

(A) voltage drop across 100 k  resister is 10e t /1.5 V for t < T
 t /1.5
20 e
(B) voltage drop across 100 k  resister is V for t < T
3

(C) voltage drop across 100 k  resister is 100 eT  t  e
T  3t  /3
 for t > T
(D) The time constant of the circuit is 1.5 sec for t > T

Space for rough work

10. Which type of electrostatic field E is not possible?

(A) E  yiˆ (B) E  xjˆ
(C) E  yiˆ  xjˆ 
(D) x 2 3yiˆ  xjˆ 
11. The threshold wavelength for photoelectric emission from a material is 5200Å, photoelectrons
will be emitted when this material is illuminated with monochromatic radiation from a
(A) 50 W infrared lamp (B) 100 W infrared lamp
(C) 50 W ultraviolet lamp (D) 100 W ultraviolet lamp

12. A very large number of fine smoke particles each of mass m1 are trapped inside a small glass
container, which contains an ideal gas of molecular mass m 2 at constant pressure P and density
d. On viewing the particles through a powerful microscope, the ratio of the rms speed of the
smoke particles to that of the gas molecules is K.
(A) smoke particles are in random translatory motion
(B) m1 = K2m2
(C) m2 = K2m1
(D) rms speed of the smoke particles will change if (p : d) changes.

Space for rough work

Comprehension Type

Paragraph for Questions 13 & 14

A person is sitting on a stool that is initially not rotating and is holding

a spinning wheel. The moment of inertia of the person and stool
about a vertical axis passing through the centre of stool is Isp. The
moment of inertia of the wheel about an axis, perpendicular to the
plane of the wheel, passing through the centre of mass of the wheel
 1
is I    ISP . The mass of wheels m . Suppose that the person
4
holds the wheel as shown in the sketch such that the distance of an
axis passing through the centre of mass of wheel to the axis of
1
rotation of stool is d and md2  I. Suppose the wheel is spinning
3
initially at an angular velocity s . The person then turns the spinning
wheel upside down. You may ignore any frictional torque in the
bearing of the stool.

13. What is angular speed of person and stool after the spinning wheel is turned upside down?
3 3
(A) ws (B) ws
4 2
3 3
(C) ws (D) ws
8 7

14. Linear impulse given by person to wheel.

3 3
(A) mdw s (B) m dw s
4 2
3 3
(C) m dw s (D) m dw s
8 7

Space for rough work

Paragraph for Questions 15 & 16

A rectangular plank of sides 300 m x 400 m is placed over a

S
horizontal plane at an inclination of 30o as shown in the figure.
(Take g = 10 m/s2) 400 m
R
Q
300 m
30o

15. A particle is projected from one base corner at an angle of 30o with the plank but in a vertical
plane which carries line of maximum possible length of plank with speed 10 m/s. Then, if time
after which particle strikes the plank is T sec, find T 21 .
(A) 5 sec (B) 2 7 sec
2
(C) sec (D) None
3

16. If a particle was projected with speed 10 m/s from one base corner at an angle 30o with the plank,
but in a vertical plane which carries a line over the plank with least possible slope, then time after
which particle lands over the plank is
2
(A) 1 sec (B) sec
3
(C) 2 sec (D) Never

Space for rough work

Paragraph for Questions 17 & 18

An insect which can crawl only, but can not fly, has to reach from one
corner to diagonally opposite corner inside a room ABCDEFGH.
E 3 F
5 5
H G
4 3 4
4 B 4
A 3 5
5
D 3 C
17. Find the minimum path length which the insect may select to reach the
target point [AD = 5 m, DH = 4 m, DC = 3 m]
(A) 10 m (B) 90 m
(C) 5 2 m (D) 74 m

18. If while crawling over wall velocity of insect is 2 m/hr and over the floor it is 4 m/hr, then find the
minimum possible time insect may take to reach the target point.
(A) More than 3.37 hours (B) 3.37 hours
(C) Less then 3.37 hours (D) Less than 1 hour

Space for rough work

SECTION – C

(One Integer Value Correct Type)

This section contains 5 questions. Each question, when worked out will result in one integer from 0 to 9
(both inclusive)

1. If two spherical bubbles of same radius 2 m. and same liquid coalesece to form a common plane
surface, then find the gap between the centres of the bubbles.

2. If the no. of times the electron goes round the first Bohr orbit of hydrogen atom in 1S is
2P
P  1015 per second then find approximate value of .
13

3. A uniformly charged non conducting disc of radius 2 m, mass 1 kg and charges Q = 4C is lying
over smooth horizontal plane. The space carries a uniform vertical magnetic field of 1 Tesla find
the angular velocity of disc just after reversal in direction of magnetic field.

4. Sound wave travels through air between two points A & B located at a gap of 30 m. if temperature
between them varies linearly from 1730 C to 127oC. Time taken by sound signal to move from A
M
to B is K where  is adiabatic exponent of air and M its molecular weight. Then find value
R
of K.

5. A biconvex thin lens of focal length 30 cm and aperture diameter 10

cm is broken into four identical pieces. Two of the pieces are placed
in a coordinate system as shown in the diagram. Find the magnitude
of the y coordinate of the image if an object is placed at origin. O 20 cm

Space for rough work

Chemistry PART – II

SECTION – A

Straight Objective Type

This section contains 8 multiple choice questions numbered 1 to 8. Each question has 4 choices (A),
(B), (C) and (D), out of which ONLY ONE is correct

1. Write the order of reactivity of halogens towards electrophilic addition reactions of alkenes
through cyclohalonium intermediate:
(A) F2 > Cl2 > Br2 > I2 (B) I2 > Br2 > Cl2 > F2
(C) I2 > Br2 > F2 > Cl2 (D) Br2 > Cl2 > I2 > F2

2. CH3
KOH
X 
 

CH3

Find out the structure of ‘X’

(A) O O (B) O

(C) O O (D) None of these

3. Which is the correct order of following compounds towards nucleophilc substitution via 1st order
mechanism?
Br Br
Br

(I) (II) (III) (IV)

(A) I > IV > II > III (B) I > II > III > IV
(C) I > IV > III > II (D) II > I > III > IV

Space for rough work

4. Determine which reagent is in excess and by how much if 100 g P4O6 is treated with 100g KMnO4
in HCl solution to form H3PO4 and MnCl2?
(A) P4O6, 12.05g (B) KMnO4, 12.05g
(C) P4O6, 13.01g (D) KMnO4, 13.01g

5. 

The equilibrium pressure of NH4CN  s  
 NH3  g   HCN  g  is 0.298 atm. Calculate KP. If
NH4 CN  s  is allowed to decompose in presence of NH3 at 0.25 atm. Calculate partial pressure of
HCN at equilibrium.
(A) KP = 0.022 atm 2, 0.0694 atm (B) KP = 0.022 atm 2, 0.0492 atm
2 2
(C) KP = 0.022 atm , 0.0542 atm (D) KP = 0.022 atm , 0.0601 atm

6. 

AgBr  s   2S2 O32   aq. 
3
 Ag S2 O3 2  aq   Br  aq.


Given K sp AgBr   5  1013 , K  5  1013

 3
f Ag S2 O3 2 
What is the molar solubility of AgBr in 0.1 M Na2S2O3?
(A) 0.5 M (B) 0.45 M
(C) 0.045 M (D) None of these
7. Energy of d-orbitals in square pyramidal geometry
(A) dxz  dyz  dz 2  dxy  dx 2  y 2 (B) dxz  dyz  dxy  dz2  dx 2  y 2
(C) dxz  dyz  dxy  dx 2  y 2  dz 2 (D) dxz  dyz  dxy  dx 2  y 2  dz 2

8. When alkali metals dissolved in liq. NH3 it turns blue colour solution and after some time it
disappears. Then
(A) The solution acts as good conductor.
(B) The solution shows diamagnetic nature.
(C) The solution acts as strong oxidising agent.
(D) The solution liberates H2 gas.

Space for rough work

Multiple Correct Choice Type

This section contains 4 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for
its answer, out which ONE OR MORE is/are correct

9. O
?
H3C C OH   H3C CH2 OH
For above reaction, the reagent used is:
(A) LiAlH4 (B) NaBH4
(C) B2H6 (D) Ni/H2

10. Which of the following possess two lone pairs at central atom and has square planar shape?
(A) SF4 (B) XeO4
(C) XeF4 (D) ICl4

11. In co-ordination compounds, the valence bond theory does not explain:
(A) Magnetic moment (B) Colour property
(C) Square planar geometry of Cu2+ complex (D) Square planar geometry of Pt2+, Pd2+

12. 1 atm pressure is equal to:

(A) 760 torr (B) 1.01325  105 pascal
(C) 1.01325  105 N/m2 (D) 1.01325 bar

Space for rough work

Comprehension Type

Paragraph for Question Nos. 13 to 14

hydrolysis i  NaI
Cu / 
Metalcarbide   A   B   Mesitylene
ii CH Cl 3

Hot KMnO4

C D
‘C’ gives effervescence with NaHCO3

13. ‘B’ is
(A) HC CH (B) H3C CH2 C C H
(C) H3C C C H (D) H3C C C CH3

14. Mesitylene and ‘D’ respectively are:

(A) CH3 (B) CH3

, CH3COOH , CO 2

CH3 H3C CH3

CH3
(C) CH3 (D) CH3

, CO 2 , HCOOH

CH3 H3C CH3

CH3

Space for rough work

Paragraph for Question Nos. 15 to 16

A triatomic linear (neglect vibration degree of freedom) ideal gas of two mole is taken through a reversible
process starting from ‘A’ as shown in figure
(C)

Vol (L)
(B)

(A)

Temperature (K)

VB
The volume ratio  4 and if the temperature at A is –73°C
VA

15. Work done by the gas in AB process is

(A) 3.32 kJ (B) - 3.32 kJ
(C) 9.97 kJ (D) - 9.97 kJ

16. Total enthalpy change in the process (A →C) is:

(A) 3000 R (B) 4200 R
(C) 2100 R (D) 0

Space for rough work

Paragraph for Question Nos. 17 to 18

Many bond angles can be explained by either electronegativity (or) size arguments. Molecules with larger
difference in electronegativity values between central and outer atoms have smaller bond angles.

17. Which has the smallest bond angle (X—O—X) in the given molecules?
(A) OSF2 (B) OSCl2
(C) OSBr2 (D) OSI2

18. Consider the following iodides:

PI3 AsI3 SbI3
102o 100.2o 99 o
The bond angle is maximum in PI3 which is
(A) due to small size of phosphorus (B) due to more bp–bp repulsion in PI3
(C) due to less electronegativity of ‘P’ (D) None of the above

Space for rough work

SECTION – C

(One Integer Value Correct Type)

This section contains 5 questions. Each question, when worked out will result in one integer from 0 to 9
(both inclusive)

1. H3C C C H  A  B  Marsh gas

H3C C N / Excess of H2O

o
Cu / 300 C CH MgBr / H O
F   E 
3 2
D
No. of  bonds present in F

2. C6H14O is an 3o alcohol (A) which on Cu/300°C gives an alkene (having four hyperconjugative
structures) which reacts with NBS in presence of light gives Bromo substituted compound (B). B
again reacts with Benzene in presence of AlCl3 gives final product C. ‘C’ reacts with strong
oxidising agent such as KMnO4 to give CO2 as one of the products. What is the number of 1o
carbons present in (A).

3. Ammonium nitrate on strong heating, is decomposed to give a gas (A) and water. It is neutral
gas. (A) on heating at 1300°C gives two gases (B) and (C). (B) reacts with FeSO4 solution to form
brown coloured compound (D). Number of unpaired electrons present in (D):

4. In Goldschmidt aluminothermic process, thermite mixture contains how many parts of Fe2O3
along with one part of aluminium?

5. At 48°C, the vapour pressure of pure CS2 is 850 torr. A solution of 2.0 g of sulphur in 100 g of
CS2 has a vapour pressure 844.9 torr. Determine the atomicity of sulphur molecule.

Space for rough work

Mathematics PART – III

SECTION – A

Straight Objective Type

This section contains 8 multiple choice questions numbered 1 to 8. Each question has 4 choices (A),
(B), (C) and (D), out of which ONLY ONE is correct

1. Let a, b and c be non-zero real numbers, such that

1 2

 1  cos x  ax   1  cos x  ax  bx  c  dx
8 2 8 2 2
 bx  c dx = then, the equation ax + bx + c = 0
0 0
has
(A) no roots in (0, 2) (B) atleast one root in (1, 2)
(C) a double root in (0, 2) (D) two imaginary roots

bcos x b  sin x
2. The value of b, such that the equation  posses solutions,
 
2cos 2x  1 cos x  3 sin2 x tan x
2

belong to the set

 1 1 
(A)  ,  (B)  ,  
 2  2 
 1
(C) (–, ) (D)  ,   1,  
 2

3. Two forces whose magnitudes are 2 gm wt and 3 gm wt act on a particle in the directions of the
vectors 2iˆ  4ˆj  4kˆ and 4iˆ  4ˆj  2kˆ respectively. If the particle is displaced from the origin to the
point (1, 2, 2), then work done is
(A) 6 gm-cm (B) 4 gm-cm
(C) 5 gm-cm (D) none of these

Space for rough work

4. If a, b, c, d be in G.P. then (a2 + b2 + c2)(b2 + c2 + d2) is

2
(A) ab + bc + cd (B) (ab + bc + cd)
2
(C) (ab – bc – cd) (D) 0

5. The value of ‘a’ for which ax2 + sin–1 (x2 – 2x + 2) + cos–1 (x2 – 2x + 2) = 0 has a real solution, is
 
(A) (B) 
2 2
2 2
(C) (D) 
 

 p  1
 x sin   , x  0
6. Let f  x    x , then f(x) is continuous but not differentiable at x = 0 if
 0 , x 0

(A) p < 0 (B) p = 0
(C) 0 < p  1 (D) p  1

n
7. If A1, A3, ….. A2n – 1 are n skew–symmetric matrices of same order then B    2r  1  A 2r 1 2r 1
r 1
will be
(A) symmetric (B) skew–symmetric
(C) neither symmetric nor skew–symmetric (D) data is inadequate

8. In an acute angled triangle ABC, vertex A is (1, 2). Let H is the orthocentre and M is the mid point
of side BC. On the line HM, a point T is taken such that HM = MT and equation of lines BT and
CT are x – 2y + 3 = 0 and x + y + 1 = 0 respectively. Find equation of circumcircle of ABC.
 5  2  5  2
(A)  x  1  x     y  2   y    0 (B)  x  1  x     y  2   y    0
 8  3  8  3
 5   2   5   2
(C)  x  1  x     y  2   y    0 (D)  x  1  x     y  2   y    0
 8  3  8  3

Space for rough work

Multiple Correct Choice Type

This section contains 4 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for
its answer, out which ONE OR MORE is/are correct

 5   2 
9. Let tan1  tan    , tan1   tan    . Then
 4   3 
(A)  >  (B) 4 – 3 = 0
7
(C)     (D)  +  = 
12

 1
10. If f  x    x    x  , x [2, 2] (where [.] denotes the G.I.F.), then f(x) is discontinuous at x is
 2 
equal to
(A) 2 (B) 2
(C) 0 (D) 1

11. If D1 and D2 are diagonal matrices of order 3, then

(A) D1D2 is a diagonal matrix (B) D1D2 = D2D1
2 2
(C) D1  D2 is a diagonal matrix (D) none of these

/2
n
12. Let un   cos
0
x cosnx dx it turns out that u1, u2, u3, ..... form a G.P. then

1 1
(A) the common ratio of G.P. is (B) the common ratio of G.P. is
2 4
 
(C) then un  2n
(D) then un  n 1
2 2

Space for rough work

Comprehension Type

Paragraph for Question Nos. 13 to 14

Read the following write up carefully and answer the following questions:
If the three equation x2 + ax + 12 = 0, x 2 + bx + 15 = 0 and x 2 + (a + b)x + 36 = 0 have a common possible
root

13. Then value of a and b are

(A) a = 7, b = 8 (B) a = 8, b = 7
(C) a = 0, b = 0 (D) none of these

14. Then the common possible roots are

(A) 1 (B) 2
(C) 3 (D) 4

Paragraph for Question Nos. 15 to 16

Read the following write up carefully and answer the following questions:
x2 y 2
P: y2 = 8x and E:   1 are equation of parabola and ellipse respectively
4 15

15. Equation of a tangent common to both the parabola P and ellipse E is

(A) x – 2y + 8 = 0 (B) x + y + 9 = 0
(C) x + 2y – 8 = 0 (D) x – 2y – 8 = 0

16. Equation of the normal at the point of contact of the common tangent, which (i.e. tangent) makes
an acute angle with positive direction of x-axis to the parabola P, is
(A) 2x + y = 24 (B) 2x + y + 24 = 0
(C) 2x + y = 48 (D) 2x + y + 48 = 0

Space for rough work

Paragraph for Question Nos. 17 to 18

Read the following write up carefully and answer the following questions:
2 2
Let f(x) = |ln|{x} – 1||  x  [2, 3) and g(x) = f(x). If h(x) = f (x) + g (x) such that h(2) = –3, then
(where {x} denotes fractional part of x)

17. h(x) + g(x) = 0 has

(A) at least one root in [2, 3) (B) exactly one root in [2, 3)
(C) non real root in [2, 3) (D) can’t say

18. h(x) + g(x) = 0 has

(A) at least one root in [2, 3) (B) exactly one root in [2, 3)
(C) no real root in [2, 3) (D) can’t say

SECTION – C

(One Integer Value Correct Type)

This section contains 5 questions. Each question, when worked out will result in one integer from 0 to 9
(both inclusive)

cos x 1  4cos 2x  1 1  

1. If  sin x  4 sin x cos2 x dx  2 ln h  x   2 ln f  x   c then, f    h   is equal to _____
2 2

x2 f x
f ' 4
2. If  f  t  dt  x cos x and  t 2 dt  x cos x then is equal to _____
0 0
16

3. Let f: R  R be a twice differentiable function satisfying f(x) – 5f(x) + 6f(x)  0  x  0 if f(0) = 1,

f(0) = 0. If f(x) satisfies f(x)  ah(bx) – bh(ax);  x  0, then the value of (a + b) h(0) is equal to
_____

4. Let f(x) be a differentiable function such that f(x) + f(x) = 4xe–x · sin 2x and f(0) = 0 if
n
pe
lim  f  k   2
then p is equal to _____
n 
k 1

e  1 
 
5. If the range of the function cos 1 1  x 2 is a,  then find the value of b _____
 b

Space for rough work

ANSWERS, HINTS & SOLUTIONS

CRT–III
(Paper-2)

Q. No. PHYSICS CHEMISTRY MATHEMATICS

1. A D B
ALL INDIA TEST SERIES

2. C B A

3. D A A

4. A C B

5. B A B

6. B C C

7. B A B

8. C A D

9. B, C A, C, D B, C

10. A, B C, D B, C, D

11. C, D B, C, D A, B, C

12. A, C, D A, B, C, D A, D

13. C C A

14. C B C

15. A C A

16. A B A

17. D A B

18. C B C

1. 2 3 1

2. 1 4 1

3. 4 3 5

4. 2 3 2

5. 5 8 2

Physics PART – I
SECTION – A

2
1.   X

2 
  4 
16 2

I  4I0 cos2  2I0
4

3
2. mV0R  mR 2 
2
2V0

3R
2V0
Vcm  R 
3

3. Electric field lines kill off on surface of conductor

4. Torque will be in clockwise sense.

5. Since maximum current passing through two circuits is same and equal to V/R1 in (b) and V/R2 in
(c) then R1 = R2
Current growth equation of L – R circuit is :

i  i0 1  e t /  
 tR
 
i  i0 1  eL

 
di i0R  tR/L
Now,  e
dt L
At t = 0 slope of I vs t curve:
di i0R

dt L
From graph (a) : slope of curve (b) > slope of curve (c)
R1 R2

L1 L 2
Since, R1 = R2
L1 < L2

6. Momentum conservation.
MV0  2MV
V
V 0
3
Angular momentum conservation about centre of mass,
2
1 2 4  MR 2 4  1  2 
MR  0  MV0R    M  R    2M R 2  2M  R 2   
2 3  2 3  2  3  
3 8 V0
 0 
25 25 R

For   0
8V0
0 
3R
The energy loss,
E  Ei  Ef
1 1  MR2  2 1 19
 MV0 2   2
     3M V  MV0 2
2 2 2  2 9

T
7. V1   10 2
1
T 10 2
V2  
2 3
Reflection coefficient
V  V1
R 2  2 3
V1  V2
Transmission coefficient,
2V2
T  3 1
V1  V2

8. Initially,
Pgas  P0 A  mg  112.5 kPa
x
w   Pgas A dx
0
x
  P0 A  mg  kx dx  56.5 J
0
Q  m u  w  118 J

9. When the switch ‘S’ is opened,

  RC  1.5 sec
Voltage across the capacitor,

Vc  10 1  e t /1.5 for t < T 
dQ dVc
The current, i  C
dt dt
200
3
 106 e t /1.5 for t < T  
Voltage drop across 100 k  resister,
20  t /1.5
V  iR  e for t < T
3
After t = T,
  100 k  10 F  1 sec
Voltage across capacitor,
Vc  Vo t
  t  T  / 

 10 1  e 
 T /1.5 
,e 
   t  T  /1

The charge on capacitor,

Q  Vc  C
The current through 100 k  resistor,
dV
i  C c  103 1  e T /1.5 1  e  
dt
  t  T /1
 
FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942
website: www.fiitjee.com
TG ~ @bohring_bot < 3
AITS-CRT-III-(Paper-2)-PCM(Sol)-JEE(Advanced)/15
4

  V    V 
10. If     then E exists.
y  x  x  y 

o
11. For emission of photoelectrons, incident  5200 A .

12. m1V12  m2 V22  3KT

3RT 3PM
 
NA NA d

13. Angular momentum conservation about axel of stool.

I w s  Isp w f  I  w f  w s   md2 wf
3
wf  ws
8

14. Linear impulse  p  mv cm  0

3
 mdw s
8

15. Slope will be now different such that

200 2
sin   
500 5
21
So, cos  
25
2u sin30o
Hence, T =
gcos 

16. For minimum slope, line will be PQ, i.e. a horizontal line
2u sin30o
Hence, time = = 1 sec.
g

17. 72  52  74

18. Snell’s law

For minimum time it will follow a path which could have been
followed by a light ray under refraction, but not the previous 5
one. Definitely time taken would be less than what could have
been over straight path.
3 4

SECTION – C

1. 4T . cos  . dl  2T . dl
1 
cos  
2

  60o c1 d c2
d R 2
So,  R cos 60 
2 2
c 1c 2 = d = R = 2 m

1 v
2. fn   n
2rn 2rn
vn
2.18  106
f1 = sec 1
2  3.14  0.529  1010
= 6.56  1015 sec 1
P = 6.56
2P
 1 approximately.
13

Q  x 2B
3. dQ  2xdx. 2
, Ex 
R 2 t
R x 2B Q
L    dt  IW =  . 2xdx dt
0 2 t R2
MR2 QBR 2 QB
  4
2 2 M

v1  v 2 t M
4. Vavg = Since variation is linear, t = 2
2 Vavg R

R  T1  T2 
=  
M  2 


1 1
5.   30 , f = 60
f f
1 1 1 1 1 1
   v1  30 ,  
v1 20 60 v 30 60
I v 60
m  2
0 v 1 30
I
  2  I  10
5
y coordinate = 5

Chemistry PART – II

SECTION – A

1. Normally not possible to F2 because it is highly exothermic with I2 reversible reaction takes place
due to more E.N of Cl2 compare to Br2. It does not give e  pair to carbocation.

2.
H O KOH
 
H 

CH3

O O

3. II, III are unstable because of their rigid frame work, to stabilize themselves by collapsing to the
stable intermediate planer state. II is more stable than III because the greater rigidity about bridge
head carbon with a one-carbon than with a two – carbon bridge.

5e  Mn7   Mn2

4.
P 3
4
 4P5   8e

100  5  1000
Meq. Of KMnO4 =  3164.56
158
100  8  1000
Meq. Of P4O6 =  3638.02
219.9
Meq. Of P4O6 in excess = 473.46
w  8  1000
  473.46 WP4O6 in excess  13.01g
219.9



NH4CN  s  
 NH3  g   HCN  g 
5.
P P
At eq.
 Total pressure = 2P = 0.298 atm
p = 0.149
1 1
Also KP = PNH3
 PHCN  0.149  0.149  0.022 atm 2
Now PNH3  0.25 atm


NH4 CN  s  
 NH3  g   HCN  g 
Initial 0.25 0
1
At eq 0.25  P P1
1 1
KP = P (0.25 + P )
1
P = 0.0694 atm

6. 

AgBr  s  
 Ag  aq.  Br  aq.
 
K1  K sp


Ag  aq   2S2 O32   aq. 
3
 Ag  S2O3  2  aq. K2  Kf



AgBr  2S2 O32  aq. 
3
 Ag  S 2O3  2  Br

K  K sp  K f  25

x2 x2
K 2
 2
 25
 0.1  2x   0.1  2x 
 x  0.045 M .

8. M  M  e 

M  e   x  y  NH3  M NH3  x   e NH3  y 

 
solvated metal solvated e

on s tand

M  NH2

11. V. B.T does not explain colour, Cu2+, Pt2+, Pd2+ square planar geometries with proper reason.

14. CH3

hydrolysis NaI Cu / 
CaC2 
 CH  CH 
CH Cl
 H3C C C H  
3

H3C CH3
Hot KMnO4

CH3  COOH  CO2

15. W = –P.V = –nRT = – 2  8.314  600 = – 9.97 kJ

16. HTotal  HAB  HBC

 nCp,m T  0
7
 2  R   800  200 
2
= 4200 R

17. Due to high electronegativity of ‘F’ bp–bp repulsion is minimum in OSF2. Steric effect may also be
the explanation.

18. Among P, As and Sb, P is most electronegative and so it exerts the strongest pull on shared
electrons causing more bp–bp repulsion.
SECTION – C

1. H3C C C H  CH3MgBr  H3C C C MgBr  CH4

(A ) (C)
(B)
H3C C N
/ Excess of H2O
OH
o CH3MgBr / H2O
Cu / 300 C H3C C C C O
 H3C C C C CH3 
CH3 CH3
D 
H3C C C C CH2
CH3

o NBS / h
Cu / 300 C
    AlCl
3

KMnO
4
  No reaction
OH Br

Due to absence of Benzoic hydrogens.


3. NH4NO3  N2O 2H2 O
A
o
1300 C
N2 O   2NO N2
B   C
NO  FeSO4  FeSO4 .NO
 D
Brown Compound

Po  P n W
5.  
P N W N
850  844.9 2  76
= 
844.9 M  100
M  252
252
N= 8
32

Mathematics PART – III

SECTION – A

x
1.  
Consider f  x    1  cos8 x ax2  bx  c dx 
0
Obviously, f(x) is continuous in [1, 2] and differentiable in (1, 2). Also f(1) = f(2)
By Rolle’s theorem, there exist one point
k  (1, 2) such that f(k) = 0 ….. (1)
Now, f(k) = (1 + cos8 k)(ak2 + bk + c) = 0
 ak2 + bk + c = 0 (1 + cos8 k  0)
2
 x = k is root of ax + bx + c = 0 where k  (1, 2)
Here, atleast one root  (1, 2)

2.  2 cos 2x – 1 = 2(cos2 x – sin2 x) – (sin2 x + cos2 x) = cos2 x – 3 sin2 x

The given equation reduces to b sin x = b + sin x, then
b
sin x 
b 1
 1  sin x  1
b
 1  1
b 1
b
  1
b 1
b
and 1
b 1
1
 b
2
 1
 b   , 
 2 

3. The vectors representing forces are 2 × unit vector along (2iˆ  4ˆj  4k)
ˆ
and 3 × unit vector along (4iˆ  4 ˆj  2k)
ˆ
1 ˆ
i.e., ˆ and 1 (4iˆ  4ˆj  2k)
(2i  4ˆj  4k) ˆ
3 2
1
 The resultant is (8iˆ  2ˆj  7k)
ˆ
3
The displacement is given by the vector ˆi  2jˆ  kˆ
1 ˆ
Hence the work done  (8i  2ˆj  7k)
ˆ  (iˆ  2ˆj  2k)
ˆ  6 gm-cm
3

4. Let r be the common ration of the given G.P., then b = ar, c = ar2, and d = ar3
Then (a2 + b2 + c2)(b2 + c2 + d2)
= (a2 + a2r2 + a2r4)(a2r2 + a2r4) + a2r6
2 2 4 2 2 2 4
= a (1 + r + r ) a r (1 + r + r )
4 2 2 4 2
= a r (1 + r + r )
= (a2r + a2r3 + a2r5)2
= (a·ar + ar·ar2 + ar2·ar3)2
= (ab + bc + cd)2

8. M is mid point of BC and HT A(1, 2)

 BTCH is a parallelogram
 CH is perpendicular BA and CH is parallel BT
 BT is perpendicular AB
Similarly CT is perpendicular AC H
Now, a circle drawn taking AT as diameter will pass M
through B and C B C
i.e. points A, B, T, C are concyclic points x – 2y + 3 = 0 x+y+1=0
 5  2
 Required circle  x  1  x     y  2   y    0 T  5 2
 8  3  , 
 3 3

12. First use integrating by parts to show that

/2
n 1
un   cos
0
x sin x sinnx dx

/2 / 2
=  cosn1 x cos(n  1)x   cos
n
x cosnx dx (Use sin nx sin x = cos(n – 1)x – cos nx cos x)
0 0
= un – 1 – un
u 1
 n 
un1 2
 1  
We have, u1  , un  n1   n1
4 2 4 2

13.-14. Let  be the common root of the three equations are their other roots be , ,  respectively
  +  = –a,  = 12
 +  = –b,  = 15
 +  = –(a + b),  = 36
 ( + ) + ( + ) = –(a + b) =  + 
++= ….. (1)
Again ( +  + ) = 12 + 15 + 36 = 63
or (2 – ) = 63 by (1)
or 2 = 2 = 63 or 72 – 2 = 63
  = 36
 2 = 9;  = 3, –3
 = 3   = 4,  = 5,  = 12
 = –3   = –4,  = –5,  = –12
 a = –( + ) = 7, –7; b = –( + ) = 8, –8

2
15. Equation of any tangent to the parabola y2 = 8x is y  mx 
m
x2 y 2
Since, it touches  1 ( c2 = a2m2 + b2)
4 15
2
2 2 1
   4m  15  m  
m
  2
Hence, equations of tangent are x  2y + 8 = 0

1
16. When m  then slope of normal is –2 and equation of the normal to the parabola is
2
y = –2x – 2(2)(–2) – 2(–2)3 ( y = mx – 2am – am3)
 y = –2x + 8 + 16  2x + y = 24

SECTION – C

1.
cos x 1  4 cos2x 
  
cos x 1  4 1  2 sin2 x 
 =

cos x 5  8 sin2 x

cos x

4 sin x 
 2
sin x 1  4cos x  2
sin x 1  4  4 sin x  2
sin x 5  4 sin x  sin x 5  4 sin2 x   
Put t = sin x; dt = cos x dx

x2
2.  f  t  dt  x cos x
0

 
2x  f x2  cos x  x sin x
Put x = 2
4f(4) = 1
1
f 4 
4
f x

And f(x),  t 2 dt  x cos x

0
{f(x)2·f(x) = cos x – x sin x
Put x = 4

3. f(x) – 5f(x) + 6f(x)  0  f(x) – 2f(x)  3(f(x) – 2f(x))

 g(x)  3g(x); where g(x) = f(x) – 2f(x)
Multiply by e–3x and then we can write it as
d

dx
  –3x
g  x  e 3x  0  y = g(x)e is non decreasing

 x  0 (Given)
 g(x)e–3x  g(0) e–0  g(x) e–3x  – 2  g(x)  –2e3x
 g(x) = f(x) – 2f(x), g(0) = f(0) – 2f(0) = 0 – 2
 f(x) – 2f(x)  –2e3x
Multiply it by e–2x and than we can write it as
d

dx
  –2x x
f  x  e 2x  2e x  0  f(x) e + 2e  3

 x  0, ex  1 ….. (1)
 We have e–2x f(x) + 2ex  3
or f(x)  3e2x – 2e3x ….. (2)
f(x)  ah(bx) – bh(ax) ….. (3)
From equation (2) and (3), we get a = 3, b = 2
 (a + b) h(0) = (3 + 2) e0 = 5

dy
4.  y  4xe x sin 2x
dx
x
Integrating factor = e
 Equation becomes ye x  4  x sin 2xdx
 yex = (sin 2x – 2x cos 2x) + c
 f(0) = 0  c = 0
 y = f(x) = (sin 2x – 2x cos 2x)e–x
f(k) = e–k(0 – 2k(1))
f(k) = –2k e–k
n
2e
lim  f k   2 1 e  2e 2   3e3   ..... upto   = 2
n 
k 1
e 

1

Time Allotted: 3 Hours Maximum Marks: 360

t h e t es t .

INSTRUCTIONS

A. General Instructions
1. Attempt ALL the questions. Answers have to be marked on the OMR sheets.
2. This question paper contains Three Parts.
3. Part-I is Physics, Part-II is Chemistry and Part-III is Mathematics.
4. Each part has only one section: Section-A.
5. Rough spaces are provided for rough work inside the question paper. No additional sheets will be
provided for rough work.
6. Blank Papers, clip boards, log tables, slide rule, calculator, cellular phones, pagers and electronic
devices, in any form, are not allowed.

B. Filling of OMR Sheet

C. Marking Scheme For All Three Parts.

(i) Section-A (01 to 30) contains 30 multiple choice questions which have only one correct answer.
Each question carries +4 marks for correct answer and – 1 mark for wrong answer.

Name of the Candidate

Enrolment No.

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Useful Data

PHYSICS
Acceleration due to gravity g = 10 m/s2

Planck constant h = 6.6 1034 J-s

Charge of electron e = 1.6  1019 C

Mass of electron me = 9.1  1031 kg

Permittivity of free space 0 = 8.85  1012 C2/N-m2

3 3
Density of water water = 10 kg/m

Atmospheric pressure Pa = 105 N/m2

Gas constant R = 8.314 J K1 mol1

CHEMISTRY

Gas Constant R = 8.314 J K1 mol1

Atomic No: H=1, He = 2, Li=3, Be=4, B=5, C=6, N=7, O=8,

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Physics PART – I

SECTION – A
Single Correct Choice Type

This section contains 30 multiple choice questions. Each question has four choices (A), (B), (C)
and (D) out of which ONLY ONE is correct.

1. An object of mass m is moving in a uniform circular motion in the xy plane. The circle has
radius R and the object is moving around the circle with speed v. The motion is projected
onto the x axis where it appears as simple harmonic motion according to x(t) = R cos (  t
+  ), here  is equal to
2
(A) v/R (B) m R
(C) R/v (D) v/(R sin  t)

2. In a standing wave on a string, the spacing between nodes is x. If the tension in the
string is doubled but the frequency of the standing waves is fixed, then the spacing
between the nodes will change to
(A) 2 x (B) x
(C) x/2 (D) x/ 2

3. A 10 cm long rubber band obeys Hooke’s law. When the rubber band is stretched to a
total length of 12 cm the lowest resonant frequency is f0. The rubber band is then
stretched to a length of 13 cm. The lowest resonant frequency will now be
(A) higher than f0
(B) the same as f0
(C) lower than f0
(D) changed, but the direction of the change depends on the elastic constant and the
original tension

4. Objects A and B are initially in thermal equilibrium. Objects A and C are originally not in
thermal equilibrium, but the two are placed in thermal contact and quickly reach thermal
equilibrium. After doing this
(A) B and C will also be in thermal equilibrium
(B) B and C could be in thermal equilibrium, but might not be
(C) B and C cannot be in thermal equilibrium
(D) none of these

Space for Rough work

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5. Four different containers each hold 0.5 moles of one of the following gases. Which gas is
at the highest temperature?
(A) 8.0 L of helium gas at 120 kPa (B) 6.0 L of neon gas at 160 kPa
(C) 4.0 L of argon gas at 250 kPa (D) 3.0 L of krypton gas at 300 kPa

6. A ray of light strikes one of two mirrors that meet at a 60º

angle as shown in figure. As the angle of incidence  is
increased, the angle between that final reflected ray and
the incident ray
(A) increases
(B) decreases 
(C) remains the same 60º
(D) the question cannot be answered without solving each
incident angle individually

7. Which of these statements is most correct?

(A) diffraction occurs only for transverse waves
(B) diffraction is proof that light can behave like a wave
(C) diffraction explains rainbows
(D) red sunsets are a diffraction phenomena

8. A gas mixture consists of molecules of type 1, 2 and 3 with molecular masses m 1 > m 2 >
m 3. Which gas has maximum average translational kinetic energy?
(A) 1 (B) 2
(C) 3 (D) all has same average KE

9. The figure here shows four paths traversed by a gas P

on a P – V diagram. Which path has max change in
internal energy of gas?
(A)1 1•
(B)2 • 2
(C)3 • T3
• •
(D)4 • T2
3 4
• T1 V

10. A carnot engine operates between the temperature TH = 850 K and TL = 300 K. The
engine performs 1200 J of work each cycle which takes 0.25 sec. How much energy is
delivered as heat to the low temperature reservoir each cycle?
(A) 655 J (B) 1855 J
(C) 1200 J (D) 600 J

Space for Rough work

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11. Of the following quantities, which one has dimensions different from the remaining three
(A) energy per unit volume
(B) force per unit area
(C) product of voltage and charge per unit volume
(D) angular momentum per unit mass

12. The resistance R = V/i where V = 100  5 volts and i = 10  0.2 amperes. What is the
total error in R
(A) 5% (B) 7%
(C) 5.2% (D) 5/2%

13. A ray of light travels from an optically denser to rarer medium. The critical angle for the
two media is C. The maximum possible deviation of the ray will be
 
(A)   C  (B) 2C
 2 
(C)  – 2C (D)  – C

14. 150, 225, 300, 375, is a series of resonant frequency. Which frequency is missing lower
than 400Hz
(A) 250 (B) 350
(C) 75 (D) 200

15. The following four waves are sent along four wires with the same linear mass densities.
Which string has maximum tension
(A)y1 = 3 sin (9x – 3t) (B) y2 = 6 sin (2x – t)
(C) y3 = 1 sin (4x – t) (D) y4 = 2 sin (x – 2t)

16. The figure shows the face and

interface temperature of a
composite slab consisting of four
materials, of identical thickness, 25ºC 15ºC 10ºC –5ºC –10ºC
through which the heat transfer
is steady. Which material has
maximum thermal conductivity
(A) a a b c d
(B) b
(C) c
(D) d

Space for Rough work

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17. The wavelength of radiation emitted is  0 , when an electron jumps from the third to the
second orbit of hydrogen atom. For the electron jump from the fourth to second orbit of
the hydrogen atom, the wavelength of radiation emitted will be
16 20
(A) 0 (B) 0
25 27
27 25
(C) 0 (D) 0
20 16

18. A radioactive substance disintegrates as follows 92 X234  87 Y 222 . The number of  and -
particles that are emitted in this process are
(A) 3, 5 (B) 5, 3
(C) 3, 3 (D) 3, 1

19. Suppose the daughter nucleus in a nuclear decay is itself radioactive. If d and  m
denote the decay constants of daughter and mother nuclei and Nd and Nm the numbers of
daughter and mother nuclei present at a time, then the number of daughter nuclei
becomes constant when
(A) mNm = dN d (B) mNd = dN m
(C) Nm – Nd = m –d (D) Nm + N d = m + d

20. Consider the fission reaction, 92 U236  X117  Y117  n  n two nuclei of same mass
number 117 are found plus two neutrons. The binding energy per nucleon of X and Y is
8.5 MeV whereas of U236 is 7.6 MeV. The total energy liberated will be about
(A) 2 MeV (B) 20 MeV
(C) 200 MeV (D) 2000 MeV

235
21. Energy released in the fission of a single 92 U nucleus is 200 MeV. The fission rate of
235
92 U fuelled reactor operating at a power level of 5 watt is
10 11
(A) 1.56 × 10 /sec (B) 1.56 × 10 /sec
20
(C) 1.56 × 10 /sec (D) 1.56 × 10–17/sec

22. In the circuit shown, the current through the ideal diode is 80
(A) 75 mA
(B) 20 mA
(C) 100 mA
2V 20
(D) 25 mA

Space for Rough work

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23. The forward biased diode is

(A) – 2V + 2V (B) – 4V – 3V
(C) 3V 5V (D) 0V – 2V

24. At t  0 , a particle is at rest at x  0 where x is the position of the particle with respect to
3x 2
origin. The acceleration of the particle depends on its position as a  . The velocity of
2
the particle at t = 1s will be
(A) 8 m/s (B) 4 m/s
(C) 2 m/s (D) 0

25. All the strings, springs and pulleys shown in figure are ideal. Initially
the system is in equilibrium and blocks are at rest. Now, the upper
spring is cut. The acceleration of block 1 just after cutting the upper 1 m
spring will be 2 m m 5
g g
(A) (B)
3 2 3 m
2g
(C) 2g (D)
3
4 2m

26. A water tank having a uniform cross-sectional area A has an orifice of area of cross-
section a at the bottom through which water is discharging. There is constant inflow of Q
into the tank through another pipe. If a 2g  K , the time taken for the water level to rise
from h1 to h2 will be

A   Q  K h2  
(A) t 
K2
K

 h1  h2   Qln  Q  K 
h1  

3A   Q  K h2  
(B) t  K
K 2 
 h1  h2   Qln  Q  K 
h1  


2A   Q  K h2  
(C) t  K
K 2 
 h1  h2   Qln  Q  K 
h1  

A   Q  K h2  
(D) t 
2K 2
K

 h1  h2   Qln  Q  K 
h1  


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27. A block of mass m is performing SHM with an amplitude A, on a extreme

frictionless surface. When it is at extreme position, a bullet of position mean
mass m moving with a velocity v 0 collides and gets embedded t = 0 position
into it at a time t = 0 (mv 02  2kA 2 ) then the displacement x at v0 m m
any time t will be x=0
A
 k  k
(A) x   2A sin  t   / 4 (B) x   2A sin t
 2m  2m
 
 k  k
(C) x   2A sin  t (D) x   A sin t
 2m 3  m
 
+
28. An electron of KE 100eV collides with stationary helium ion He in its ground state and
excites it to a higher energy level. After collision He + ion emits two photons in succession
with wavelength 1085Å and 304Å. Then the KE of electron after collision will be (Neglect
+
the recoil speed of He )
(A) 47.7 eV (B) 94.4 eV
(C) 20.2 eV (D) 10.1 eV

29. Two parallel conducting rails, separated by a distance l,

are placed horizontally in a region of uniform magnetic
 l m B m V0
field B , perpendicular to the plane of the rails as shown l
in the figure. Two conducting wires, each of length l are
placed so as to slide on parallel conducting rails without
friction. One of the wires is given a velocity v0 parallel to
the rails. Loss in kinetic energy of the system till the
steady state is achieved will be
3
(A) Zero (B) mv 02
4
3 1
(C) mv 02 (D) mv 02
8 4

30. A round metal hoop is suspended on the edge by a hook. The hoop can oscillate side to
side in the plane of the hoop, or it can oscillate back and forth in a direction perpendicular
to the plane of the hoop. For which mode will the frequency of oscillation be larger?
(A) oscillations in the plane of the hoop
(B) oscillations perpendicular to the plane of the hoop
(C) the frequency of oscillation will be the same in either mode
(D) none of these

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Chemistry PART – II

SECTION – A
Single Correct Choice Type

This section contains 30 multiple choice questions. Each question has four choices (A), (B), (C)
and (D) out of which ONLY ONE is correct.

1. Minerals kinoite is a silicate anion having chain of three SiO4 tetrahedra. If in kinoite the
three SiO4 tetrahedra shares corners with adjacent tetrahedra then the overall charge on
silicate anion is:
(A) – 4 (B) – 8
(C) – 6 (D) – 2

2. When compounds
Cu  NO3 2 Hg NO3 2 NaNO3 AgNO3
I II III IV
are heated (upto 500°C) seperately then a brown gas is formed in case of
(A) I and II (B) II and III
(C) II, III and IV (D) I, II and IV

3. Consider the following order:

(I) Thermal stability: BeSO4 < MgSO4 < CaSO4 < SrSO4 < BaSO4.
(II) Bond angle: OCl2 > SF2 > ASH3 > H2Se
(III) Bond order of O2 , NO and N2 is 2.5.
(IV) NCl3 on hydrolysis gives HOCl but BCl3 gives HCl.
Identify which of the above four statements are correct.
(A) I, II, III and IV (B) II, III and IV
(C) I, II and IV (D) I and II

4. Identify the following statements as True (indicated by T) or False (indicated by F)

(I) Dipole moment order of NH3 > SbH3 > AsH3 > PH3
(II) F2 on reaction with water produces HF, O2 and O3
(III) Number of S—S bonds in H2SnO6 are n–2
(IV) PH3 can be produced by hydrolysis of Ca3P2
(A) TFTT (B) TTFF
(C) TFFT (D) TTFT

5. Identify the incorrect statement for XeF6

(A) XeF6 reacts with excess H2O to form XeO3.
(B) XeF6 may act as a fluoride donor during reaction with group I element.
(C) XeF6 on reaction with small amount of H2O form XeOF4
(D) None of the above

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6. For complex [Co(NH3)5CO3]ClO3, what is the spin only magnetic moment, oxidation
number coordination number and d-electrons on transition metal.
(A) 3 BM, 0, 7, 2 (B) 15 BM, 3, 6, 0
(C) 0 BM, 3, 6, 6 (D) 15 MB, 2, 7, 3

7. Identify the incorrect statement

(A) HgCl2 is prepared by heating mercury in chlorine.
(B) HgO decomposes on heating whereas Al2O3 has high thermal stability
(C) Mac-Aurthor’s process is used to extract platinum
(D) NO2 undergoes disproportionation reaction under alkaline medium.

8. Identify the incorrect:

(A) Ni2  aq   dim ethy lglyoxime  aq   NH3  aq   Re d ppt / colour
(B) Hg2   aq   SCN  aq   CO2   aq  Blue ppt / colour
(C) When Mn(NO3)2 (aq) is treated with sodium bismuthate in presence of concentration
HNO3 a purple coloured solution is obtained.
(D) None of these

9. Which of the following have same number of p-electrons as the 4d-electrons of

Palladium?
(A) Si (B) S
(C) P (D) None of these

10. If the longest wavelength in Balmer series of Li2+ is ‘x’ then the value of shortest
+
wavelength of He Paschen series will be
36x 20x
(A) (B)
5 36
45x 36x
(C) (D)
16 20

11. The ratio of rms velocity of N2, average velocity of H2 and most probable velocity of CO at
27°C is
3 4 1
(A) : : (B) 1.224 : 1.128 : 1
28  14
(C) 0.33 : 1.113 : 0.27 (D) Both (A) and (C)

12. Element ‘X’ having atomic mass equal to 109.75 g crystallises in face centred cubic (fcc)
structure. If the density of crystal is 9.0 g/cm3 then the edge length of the unit is
(A) 450 pm (B) 750 pm
(C) 900 pm (D) 600 pm

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13. In solubility of hydroxide A(OH)2 in a buffer of pH = 11 is found to be 0.0416 g/L. Molar

solubility of A(OH)2 in pure water would be (in mol/lit). Use molecular weight of A(OH)2 as
104 g/mol
(A) 10–3 (B) 2.16  10–3
–4 –4
(C) 10 (D) 2.16  10

14. Consider following equilibrium at 400 K and 50 atm


3A  2B  2C  D
If the equilibrium pressure of A is 5 atm then Kc will be:
(A) 38.32 (B) 19.16
(C) 9.58 (D) 28.74

15. For adiabatic reversible expansion, initial pressure and temperature are respectively 1
atm and 27°C. If final pressure is 0.25 atm then final temperature will be: (Assume the
gas is monoatomic)
(A) –100°C (B) –50°C
(C) –25°C (D) 0°C

16. For the combustion of C6H10 with required amount of oxygen. Which of the following is
correct? {consider the state of each reactant and product same as at room temperature}.
(A) H > E (B) H = E
(C) H < E (D) None of these

17. 10 g sample of compound ‘X’ undergoes a first order reaction. If t1/2 for this reaction is 20
min then the time needed to consume Tg of sample is
(A) 40 min (B) 32 min
(C) 34.75 min (D) 38 min

18. Identify the incorrect statement regarding adsorption of gas on solid surface.
(A) Adsorption can be endothermic
(B) Physisorption may get transformed into a Chemisorption at high temperature.
(C) Chemisorption is very slow as compared to physisorption because chemisorption
have higher activation energy.
(D) Chemisorption is more exothermic than physisorption

19. Elevation in boiling point of aqueous glucose solution is 0.104 K. Osmotic pressure for
same aqueous glucose solution was found to be 7.38 atm at 300 K. The density of
aqueous glucose solution is:
Take R = 0.082 L. atm. Mol–1 K–1 and Kb = 0.52 K Kg mol–1
–1 –1
(A) 1.036 g mL (B) 1.054 g mL
–1
(C) 1.00 g mL (D) None of these

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20. Identify the total number or aromatic compound from following:

N S S
O
(A) Six (B) Five
(C) Four (D) Three

21. The number of monobromination product(s) in the following reaction is:

Br2 /h
  monobromination

(A) 5 (B) 7
(C) 6 (D) 8

22. Identify the major produt (Z) in the following reaction sequence.
NBS(1 equiv) Mg (i) CO
  X 
THF
 Y  2
(ii) H O
Z
3

(A) (B) COOH

COOH

(C) COOH (D)

23. Arrange the following compounds in decreasing order of their basicity:

NH2
CH3 H
N O N O N

NO 2
I II III IV
(A) I > IV > II > III (B) I > III > IV > II
(C) I > IV > III > II (D) I > III > II > IV

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24. In the following reaction sequence:

OH
Al O 2 i  O
(i)  Na  

2 3

 A 
(ii) Me S
B 
(ii) NH Cl H O
 ' X'
2 4 2

The major product ‘X’ is

(A) (B) OH

H H
(C) (D) Both B and C
O

25. For the following reaction sequence

NH2 NH2
NO 2

X Y Z 3  i H O , 
  A   B   C 
 
 ii OH

Identify the correct sequence of reagents X, Y and Z

X Y Z
(A) Conc. HNO3 CH3COCl/base Conc. H2SO4
(B) CH3COCl, Base Conc. HNO3 Conc. H2SO4
(C) Conc. H2SO4 CH3COCl, base Conc. HNO3
(D) CH3COCl, base Conc. H2SO4 Conc. HNO3

26. In the following reaction identify the major product ‘X’:

O O
C OH OH
 i LiAlH
H OCH3 

4
 A 
 Y
H (ii) H3O

(A) (B) O
HO OH

HO OCH3

(C) O (D) Both (B) and (C)

H OH

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27. Identify incorrect statement regarding Hunsdieker reaction:

(A) It can be given by silver salt of carboxylic acid
(B) Major product of reaction is halide of corresponding carboxylic acid
(C) It follows radical mechanism
(D) None of the above

28. What will be the net change on following amino acid at pH = 3

O NH2
H2 N C CH CH COOH
NH2
(A) +2 (B) +3
(C) –1 (D) Zero

29. When C4H10 undergoes chlorination by reaction with Cl2(g) in presence of ultraviolet light
then how many stereoisomers are possible for dichloroderivatives.
(A) Two (B) Three
(C) Four (D) Five

30. The expression of Kp for the following reaction will be:



 MnO  s   CO2  g  H2O     KCl(aq)
KMnO4  aq   H2C2O4  aq   HCl  aq  
PCO2 is partial pressure of carbon dioxide
2
(A) Kp  PCO2  
(B) K p  PCO2
5 10
(C) Kp  PCO2   (D) KP   PCO 2

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Mathematics PART – III

SECTION – A
Single Correct Choice Type

This section contains 30 multiple choice questions. Each question has four choices (A), (B), (C)
and (D) out of which ONLY ONE is correct.

tan x cos  sin x   sin  sin x 

1. lim is
x  0 cos  tan x  tan x  sin  tan x 
1
(A)  (B) –1
2
1
(C) (D) 1
2

2.  3  x   1x
If f is true for exhaustive domain of ‘f’, then domain of f(x) is

(A)   3, 3  (B) 0, 3 

(C)  0, 3  (D) 0, 3 

 

3. If  is a root of z 5  z3  z  3  0 then
(A) |  |  1 (B) |  |  1
1
(C)  lies on or outside the circle z  (D)  lies inside the unit circle | z |  1
2

a 2  b2  c 2
4. In a quadrilateral ABCD with sides a, b, c, d the range of is
d2
1  2 
(A)  ,   (B)  ,  
3  3 
(C) 1,   (D)  0,  

5. If ai is negative or positive according as i is odd or even and equation

x 4  a1x3  a2 x 2  a3 x  5  0 has four real roots then minimum values of a1a3 is equal to
(A) 20 (B) 40
(C) 80 (D) 160

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f x  5
6. Let a function f(x) has the property f  x  2   , then which of the following must
f x  3
equal to f(2014)
(A) f(2004) (B) f(2010)
(C) f(2006) (D) f(2018)

e   x  1
x
7. lim (where n  I), is
x n
x2
1
(A) e (B) e 
2
(C) e – 2 (D) none of these
2 3
8. From (1, 2), tangents are drawn to the curve y – 2x – 4y + 8 = 0. Then
(A) sum of x–coordinates of points of contact is zero
(B) sum of x–coordinates of points of contact is 6
(C) sum of y–coordinates of points of contact is zero
(D) sum of y–coordinates of points of contact is 4

b b x
9. If  f  x dx   f  x  dx  a  b  also f(x)  0 for any x  (a, b) and g  x    f  x dx, then
a a 0
b

 f  x g  x  dx when a and b are positive

a
(A) cannot be positive (B) cannot be negative
(C) can be less than –1 (D) nothing can be said

10. Three straight lines are drawn through a point ‘P’ lying in the interior of the ABC and
parallel to its sides. The areas of the three resulting triangles with P as the vertex are s1,
s2 and s3 then area of ABC is equal to
2
(A)  s1  s2  s3  (B) (s1 + s2 + s3)

(C) s1s2  s2 s3  s3 s1 (D) s1 + s2 + s3 + 2  s1s2  s2s3 

11. Let a real valued differentiable function f(x) be strictly monotonic on [a, b], then
b f (b)
2 2
(A) 
  f(x)   f b    dx   2x b  f  x  dx
1

a f (a)
b f (b)
2 2
(B)   f(x)   f  a   dx    
2x b  f 1  x  dx
a f (a)
b f (b)
2 2
 x b  f  x 
1
(C)  f(x)  f a  dx  dx
a f (a)
(D) none of these

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1 i
    must be
2 n
12. If z0 =
2

, then the value of the product 1  z0  1  z02 1  z02 .......... 1  z02

n 1  1 
(A) 22 (B) 1  i   1  n  , if n  1
 22 
5
(C) 1  i , if n  1 (D) 0
4

N
13. Let N be any four digit number say x 1 x2 x3 x 4. Then maximum value of
x1  x 2  x 3  x 4
is equal to
1111
(A) 1000 (B)
4
(C) 800 (D) none of these

ecot x
14.  sin2 x [2 ln cosec x + sin 2x] dx is equal to
cot x cot x
(A) 2e ln (cosec x) + c (B) e ln x + c
cotx
(C) e (ln (cosec x) + c (D) none of these
2 2
15. Let P be any moving point on the circle x + y  2x = 1. AB be the chord of contact of this
2 2
point w.r.t the circle x + y  2x = 0. The locus of the circumcentre of the triangle CAB, (C
being centre of the circles) is
2 2 2 2
(A) 2x + 2y  4x + 1 = 0 (B) x + y  4x + 2 = 0
2 2 2 2
(C) x + y  4x + 1 = 0 (D) 2x + 2y  4x + 3 = 0

16. ABCD is a quadrilateral with side lengths AB = 4, BC = 10, CD = 6 and AD = 6, and

diagonal BD = 8 units. If the incircles of triangles ABD and BCD touch BD at P and Q
respectively, then area of quadrilateral C1PC2Q (where C1 and C2 are incentres of triangle
ABD and BCD respectively), is
15
(A) 3  sq. units (B) 3 sq. units
2
15
(C) sq. units (D) none of these
6

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17. Let S = {1, 2, 3, ….. n} be a set and two numbers x and y are drawn with replacement
2 2
from S. Let Pk be the probability that x – y is divisible by k and [x] denotes greatest
integer less than or equal to x. The value of P2 is equal to
2
 n  
1  2 
(A) (B)    
2  n 
 
 
2 2
n  n  n   n 
 2   2  2   2 
2
(C) 1  2.  2    (D) 1    2   
n  n  n  n 
   
   

18. A eccentric person starts writing numbers from 1 to n in a row such that ith number is
written i2 times. The 500th digit from the starting in such a number is
(A) 0 (B) 1
(C) 2 (D) 3

19. The value of a for which 4sin x  a2sin x  1  0  x  R is

3 3
(A) a   (B) a  
2 2
(C) a  2 (D) No such value exists

20. The number of tangents which can be drawn to the ellipse 16x 2  25y 2  400, such that
sum of perpendicular distance from the foci to any tangent is 8, is
(A) 1 (B) 2
(C) 4 (D) none of these

21. Let a, b  R where ab  0 , then the graph of the straight line ax  y  b  0 and the conic
section bx 2  ay 2  ab can be
y y

(A) x (B) x

y
y

Space for rough work

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website: www.fiitjee.com
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 
22. The differential equation of a curve, passing through  0,   and (t, 0) is
 4
 dy   dy   x  t
cos y   e  x   sin y  e  x    e e , the value of t.e e is
 dx   dx 
(A) –1 (B) 1
(C) 2 (D) –2

23. If (1, 2, p) , (2, 2p, -6) and ( 2  2, 1, 1) are the ordered triplets of form (x, y, z) which
x y z x y z x y z
satisfies all the equations.    1,    1 and    1 then  can be
a b c b c a c a b
(A) 2 (B) –2
(C) –4 (D) 6

24. If f(x) is a continuous and differentiable function such that f(tk )  1 for k  1, 2, .....n,
k
1
where tk   then
r 1 r(r  1)(r  2)
 1  1
(A) f    1 (B) f    1
4  2
 1  1
(C) f     0 (D) f     1
 2 4

x 2  3x  2
25. The sum of the real values that the function f  x   can not take is
x2  x  6
(A) 2 (B) – 3
6 4
(C) (D)
5 5

26. Let z1, z2 , z3 be the coordinates of vertices A, B, C of the triangle ABC. If W1  z1  z 2 ,

W2  z3  z2 and A  90 , then
W 
(A) Re  2   2 (B) Re (W1W2 )  | W1 |2
 W1 
W 
(C) Re (W1W2 )  | W2 |2 (D) Re  1   1
 W2 

Space for rough work

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website: www.fiitjee.com
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27. Let cos is a root of the equation 169x 2  26x  35  0 , 1  x  0 , then sin 2 is
144 144
(A) (B) 
169 169
120 131
(C) (D) 
169 169

28. Let (1  x  x 2 )n  a0  a1x  a 2 x 2 .........a2n x 2n , where n is odd integer

S1  a0  a4  a8  ..............
S2  a1  a5  a9  ..............
S3  a 2  a6  a10  ..............
S4  a3  a7  a11  ..............
(A) S1  2S3 (B) S2  S 4
(C) S2  S4  0 (D) S1  S2  S3 or S1  S3  S4

29. Let f(n) be the sum of first n terms of sequence 0, 1, 1, 2, 2, 3, 3, 4, 4 ….., then
n2
(A) f(n)  {n is even number)
6
n2  1
(B) f(n)  {n is odd number)
6
(C) f(n  m)  f(n  m)  nm , where n, m  I (n  m)
4nm  1
(D) f(n  m)  f(n  m)  , where n, m  I (n  m)
1
   
30. If A, B, C and D are four non-zero vectors in the same plane no two of which are
collinear then which of the following hold(s) good?
       
(A) (A  B).(C  D)  0 (B) (A  C).(B  D)  0
         
(C) (A  B)  (C  D)  0 (D) (A  C)  (B  D)  0

Space for rough work

S. No. PHYSICS CHEMISTRY MATHEMATICS

ALL INDIA TEST SERIES

1. A B A
2. B D D
3. D A A
4. B D A
5. C B C
6. C C C
7. B C C
8. D D D
9. D B B
10. A C A
11. D D B
12. B C C
13. A B A
14. C B A
15. C A A
16. C C A
17. B C C
18. D A B
19. A B B
20. C B B
21. B A B
22. B C B
23. D C B
24. D B A
25. C D C
26. C C B
27. A D C
28. A A D
29. D C C
30. A D C

Physics PART – I
SECTION – A

1. Angular frequency of SHM is equal to angular velocity of particle  = V/R.

T
2. V = n = n(2x)  T  x
M
If tension is twice then x will increase to 2 x.

1 T
3. f0  [M = mass/length]
2l M
for elastic wire T = Kx
x  elongation from natural length
Hence frequency depends on K and original tension.

5. Temperature is maximum for which product of P and V is maximum as no. of moles is same for
each containers
PV = nRT
T  PV
Hence option (c) is true.

6. S = [ – 2] + [ – 2 (60 – )]

= 2 – 2 – 120 + 2 = constant
Angle of incidence and final reflected ray does not depend on .
60º – 

 60º

8. It only depends on temperature. In a mixture all gases has same temperature.

9. U  nCV T From graph

Maximum change in internal [T3 > T2 > T1] energy is in process 4.

 300  11
10. n = 1  
 850  17
Work done in each cycle is
W 1200
n= Qin =  17
Qin 11
1200  17
Qrej = Qin – W =  1200
11
1200  6
= = 654.54
11

11. Quantities in option a, b, c has dimensions of [M L–1T2] but angular momentum per unit mass has
2 –1
dimension [Mº L T ].

R V I
12.  100 =  100   100
R V I
5 0.2
=  100   100
100 10
Total percentage error = 7%

 
13. S = 2 –   C 
C 2 

= C
2

14. Series of resonant frequency must be integral multiple of fundamental frequency. This series is
integral multiple of 75 H2. Hence missing frequency is 75 H2.

T
15. V V T

String with maximum tension will have maximum velocity.
V1 = 3 m/sec, V2 = 2m/s, V3 = 4m/s
1
V4 = m/s. Hence third string will have maximum tension.
2

dQ 
16.  KA
dt x
Rate of heat transfer, cross section area and thickness is same for all slab therefore
1
K

Slab which has minimum difference of temperature across interface will have maximum thermal
conductivity.
Hence (c) option is true.

19. The number of mother nuclei decaying in a short time interval dt is = Nmmdt. But death of a
mother nucleus implies the birth of a daughter nucleus.
The number of daughter nuclei decaying in the same time interval is = Ndddt.
The number of daughter nuclei will remain constant when Nmmdt = Ndddt
or Nmm = Ndd

21. Let fission rate be n per second; then

n × 200 × 106 × 1.6 × 10–19 = 5 × 1
5
 n  1.56  1011 / sec
3.2  10 11

2
22. I =  25 mA
80

24. The acceleration of particle at t  0 is 0.

25. After the upper spring is cut, the block 1 and 2 will move with same acceleration.
 a  2g

dV
26.  (Q  av)
dt
Adh
 (Q  av)
dt
h2 t
Adh
 Q K   dt
h1 h 0

2A   Q  K h2  
t K
K 2 
 h1  h2   Qln  Q  K 
h1  


2
1 1 V 1
27. KA 2  2m 0  KA '2 (by energy and momentum considerations)
2 2 4 2
2 1 2 2
KA  mV0  KA '  A '  2A
2
5
A '  2A   
4
x  A ' sin   ' t   
K
' 
2m
 K 
x   2A sin  t 
 2m 4 
 

28. 1  1085Å &  2  304Å

hc hc
E1   11.5eV E 2   40.8eV
1 2
KE of electron after collision
= 100 – 52.3 = 47.7 eV

29. Net force on the system is zero hence momentum is conserved so

mv 0  2mv
Both wires will have same velocity when steady state will be achieved.
1
K  mv 02
4

I 1 mgl
30. T = 2 f=
mgl 2 I
I is moment of inertia about axis passing from suspension point.
1
f , I about axis side to side in plane of loop is more hence f is greater in first case.
I

Chemistry PART – II
SECTION – A

1. O
O
Si
O O Si
O O O
O O
Si
O
8
Structure of silicate anion Si3 O10  
in mineral kinoite.

2. NO2 is a brown coloured gas:

 1
Ca NO3  2   CuO  2NO2   O2 
2

Hg  NO3 2   Hg  2NO2  O2 
 1
NaNO3   NaNO2   O2 
below 500o C 2
 1
AgNO3   Ag  NO2   O2 
2

3. Bond order of O2 is 2.5

Bond order of NO is 2.5
Bond order of N2 is 2.5
Bond order can be calculated using molecular orbital diagrams.

N Net dipole moment

H H
H
Bond angle around central atom decrease down the group. Also electronegativity of N > P > As >
Sb and electronegativity of H > P > As > Sb. Electronegative of P is very close that of H.

P Net dipole moment

H H
H
Ca3P2 + 6H2O  3Ca(OH)2 + 2PH3
2F2 + 2H2O  4HF + O2

3F2 + 3H2O  6HF + O3

5. XeF6 may act as fluoride donor when it reacts with non metal fluorides.


XeF6  SbF3   
 SbF6  XeF5

XeF6  excess H2O  XeO3  6HF

XeF6  H2O  XeOF4  2HF

6. Co
 NH3 5 CO3  ClO3
   
x 0 2 1
x + 0  5 + (–2)  1 = + 1
x=+3

Co3

3d 4s 4p

in Co3
complex

d2 sp3
Coordination number = 6
No. of unpaired e– (n) = 0
   n n  2   0
Number of ‘d’ electron = 6

7. Hg + X2  HgX2 (where X = F, Cl, Br or I)

Mac-Aurthor’s process is used in extraction of gold and silver
4M  8NaCN  2H2O  O2  4Na M  CN2   4NaOH Where, M = Ag, Au.

8. H3C C N OH H3C C N O
Ni2  NH3  Ni
H3C C N OH H3C C N OH

Red precipitate
2 2
Hg  Co(SCN)2  aq   Hg  SCN 2  CO
Blue precipitate

3Mn  NO3 2  aq  5NaBiO3  aq   9HNO3  aq  3HMnO4  aq  3H2O     5Bi NO3 3
Purple 

10 0
9. Pd  4d 5s  10  electron
2 2 6 2 4
S  1s 2s 2p 3s 3p  10 p electrons

10. Longest wavelength of Li2+, Balmer series  n1 = 2, n2 = 3

1  1 1  1 1 1 
 RH  32  2  2    RHZ 2  2  2  
x 2 3     n1 n2  

1 9  4
 RH  9   ....(1)
x  36 
+
Shortest wavelength of He Paschen series 
n1 = 3, n2 = 
1 1 1 
 RH  22  2  2  .....(2)
  3  
From equation (1) and (2)
5
RH  9 
 36 5  9 45
  
x 1 4  4 16
RH  4 
9
45x

16

11.
3RT 8RT 2RT
Vrms  Vavg  Vm.p 
M M M
For N2 at 27°C For H2 at 27°C For CO at 27°C
3RT 8RT 2RT
Vrms  Vavg  Vmp 
28 2 28
3 4 1
Vrms : Vavg : Vmp  : :
28  14
= 0.33 : 1.128 : 0.267

Atomic wt. 109.75

12. Mass of each atom =  23
 1.82  1022
NA 6.022  10
In FCC number of atom/unit cell = 4
Mass of unit cell = 4  1.82  10–22 = 7.29  10–22 g
mass 7.29  10 22 g
9g / cm3  density  
volume a3
7.29  1022 g
a3   81 10 24 cm3
9g / cm3
a = 900 pm

13. 

A  OH 2  2
 A  2OH


0.0416g / L
104g/ mol
 A 2    4  10 4 g / mol
 
OH   10 pOH  103
 
2
Ksp =  A 2  OH 
2

= 4  104 103  = 4 × 10-10
3
Ksp = 4s for A(OH)2

Ksp 3 4  10 10
S3   2.16  10 3
4 4

14. 

3A 
 2B  2C  D

Initial Po 0 0 0
o
Equilibrium P  2x 2x 2x x
o
Equilibrium pressure = P – 2x + 2x + 2x + x
50 = Po + 3x ………..(1)
From question 5 = Po – 2x ………..(2)
From equation (1) and (2)
55 = 5x
x = 11

Kp 
PB2 .PC2 .PD

 22 2  22 2 11
PA3 53
= 20614.5
K p  K c  RT  n
Kp 20614.5
Kc  n
 2
RT   0.082  400 
= 19.16

15. Adiabatic reversible expansion

T1P11  T2P21
1
P 
T2  T1  1 
 P2 
For mono atomic gas  = 1.66
T2 = 172.88
T2 = –100.27°C

1
16. C6H10     8 O2  g  6CO2  g   5H2 O   
2
H = E + ng RT
H = E + (6 – 8.5) RT
H = E – 2.5 RT

2.303 a
17. t log o
k at
0.693 0.693
k 
t1/ 2 20
2.303 10
t  log 7
0.693 10
20
= 34.75 min

18. Adsorption of gas on solid is always exothermic.

Tb 0.104K
19. m   0.2 mol Kg1
Kb 0.52K Kg Mol

  MRT  M 
RT
7.38 atm


0.082 L atm mol1 K 1  300 K 
–1
= 0.3 mol L
Assuming volume of water does not change significantly on addition of glucose
In 1 L solution  0.3 mol glucose and 1 kg water
mass of solution
density 
volume of solution

=
 
0.3 mol  180g mol1  1000g
1000 mL
= 1.054 g mL

20. 8e–. satisfying (4n + 2)e– rule 

only 6e– are in one plane and in
cyclic conjugate. Therefore, aromatic.

6e– satisfying (4n+2)e– rule 

aromatic
N
Due to repulsion between internal
H hydrogen atom  non aromatic
H

10e– satisfies (4n + 2)e– rule 

aromatic

S S
O
Non aromatic because cyclic
delocalisation of e– is abscent.

21. Br
Br
Br2 /h

 Br

Br
Br
Five products

22. Br MgBr COOH

NBS Mg (i) CO
20   
THF
 
2

10 Allylic bromination
(ii) H3O

23. CH3 Aliphatic tertiary amine, nitrogen electron pair is not in conjugation.
N

2
Nitrogen is sp hybridised. Nitrogen electron pair is not in
N
conjugation.

Nitrogen electron lone pair is in conjugation with phenyl ring.

NH2

H Nitrogen e– lone pair is in conjugate with two C=O groups leading to

two equivalent resonative structures.
O N O

24. OH OH
Al O
2 3 (i) O
3 (i)  Na 
    
 (ii) Me S 2 O (ii) NH4Cl, H2O

25. NH2 NHCOCH3 NHCOCH3 NHCOCH3

NO 2
CH3 COCl Conc. H2 SO4 Conc. HNO3

Base
 
 


NH2 SO 3H SO 3H

NO 2
(i) H3 O
 
(ii) OH

26. LiAlH4 can reduce both aldehyde as well as ester group into. But use OH OH protects aldehyde
group and hence only ester group is reduced.
O O O O
O O
H OCH3 OH OH H OCH3 (i) LiAlH
H OH

  4
 
H  ii H3O

27. Hunsdieker reaction

R  COOAg  Br2  R  Br  CO2   AgBr 
Free radical mechanism is followed.
O

28. At pH = 3, condition is highly acidic so all NH2 group will get protonated. Protonation of C NH2
group is very difficult. So at pH = 3 structure will be

O NH3
H2N C CH CH COOH
NH 3

29. Cl Cl
H3C CH 2 CH2 CHCl 2 H 2C CH2 CH2 CH2

Cl Cl Cl
H3C CH CH2 CH H3C CH2 C CH3
(R & S)
Cl
CH3
CH3 Cl
H3C CH CHCl 2
H3C CH2 CH CH2
(R & S)
Cl CH3 Cl
CH3 CH3
H 2C CH CH 2
H3C C CH2 Cl
Cl
Indicate chiral carbon

30. Balanced chemical equation is



 2MnO  s   10CO2  g   6H2O     2KCl(aq)
2KMnO4  aq  5H2C2 O4 aq  2HCl  aq  
Kp depends only on partial pressure of gaseous species
10

 K p  PCO2 

Mathematics PART – III

SECTION – A

sin  x  sin x  x  sin x 1

1. lim  lim 
x 0 sin  x  tan x  x 0 x  tan x 2

2. f  
3  x exists for 3x  0
i.e. Domain of ‘f’ is 0,  
But, for x  0  0  3  x  3

3. Suppose |  | 1 , we have
| 3 | |  5   3   ||  |5  |  |3  |  |  1  1  1
 3  3 A contradietion
Thus |  | 1   lies either on or outside the unit circle

4. Here a  0, b  0, c  0, d  0
 a2  b2  c 2  ab  bc  ca
 3(a2  b2  c 2 )  (a  b  c)2  d2
a2  b2  c 2 1
 
d2 3

5. Clearly from Descarte’s rule of sign equation has 4 positive real roots lets say , , , and 
    a1;    a2 ;    a3 and   5
         ... 4
using AM  GM  .   .4 33  3 3    5
4 4
a1 a3
– .  5  a1a3  80
4 4

8. Curve is (y – 2)2 = 2x3 – 4 ..... (1)

dy 3x 2

dx y  2

 Equation of tangent at (x1, y1)

3x12
y  y1   x  x1 
y1  2

As it passes through (1, 2)

3x12
 2  y1  1  x1 
y1  2
2
or  y1  2   3x12  x1  1

or 2x13  4  3x13  3x12 [As (x1, y1) lies on the curve]

or x13  3x12  4  0
or (x1 + 1) (x1 – 2)2 = 0
 x1 = –1 or 2
But x1  = –1 (it is clear from equation (1))
 x1 = 2
 y1 = 2  2 3
Points of contact are (2, 2 + 2 3 ) and (2, 2 – 2 3 )
 Sum of x–coordinates of points of contact is 4
 Sum of y–coordinates of points of contact is 4

b b
9. If  f  x dx   f  x  dx
a a

 f(x) is either positive in (a, b)

or f(x) is negative in (a, b) and also b > a
Also g(x) = f(x)
When f(x) > 0  g (x) > 0  g(x) is increasing function
When f(x) < 0  g (x) < 0  g(x) is decreasing function
Also g(0) = 0
b b

 f  x  g  x  dx   g  x  g'  x  dx
a a
2 2

=
 g b     g  a   which will be positive for 0 < a < b
2

A
s1 x12 s1 x1
10.  or 
 a2  a
s2 x 2 s3 x 3
Similarly  and  s1 s3
 a  a P

s2
B C
x1 x2 x3

-1
11. Let f (x) = t  f(t) = x
f (b) b b

  
2x b  f 1(x) dx   2f(t).(b  t).f '(t)dt = b [(f(t)) ]a –
2 b
 t .2f(t) .f '(t) dt
f(a) a a
b 2 2
= a  f  x     f  a    dx


1  z 0  1  z 0 2
1  z0 2
n 1
 
1  z0 2  1 
12.
1

1  z  . 1  z 
0
2 2

0
2n
=
1  z0

1  z0
 1  i   1  n  , if n > 1;
 22 
5
If n = 1, 1  z0  1  z02    4
1  i 

17. If one number is drawn from the set {1, 2, 3…….n} the probability  that it is divisible by 2 is given
n
2
by    
n

 P (the number is not divisible by 2) = 1 – 

2 2
Now, x – y = (x – y) (x + y)
2
 P2 = . + (1 – )(1 – ) = 1 – 2 + 2 

g
18. To exhaust all single digit numbers he must have written  i2  285 digits, to exhaust 10 he
i 1

must write 2  102 more digits i.e. 485 digits. So 500th digit will occur when he is writing 11

sin x 1
19. Taking 2 =t t2
2
1  t2  1 t2  3
a  a  min    a
t  t  2

20. If p1 and p2 are length of perpendiculars from foci to any tangent then p1p2  b2  16 and
p1  p2  8  p1  p2  4
Hence only two tangents are possible

21. Option (B) is correct because in this case a  0, b  0 (Let b  c )

x2 y 2
 cx 2  ay 2  ac   1
a c

dy x
22. (cos y  sin y)  (cos y  sin y)e  x  e e
dx
cos y  sin y  u
du x x
  e x u  ee  ue e  x
dx
t
 e e  t (Putting (t, 0))
t
 t.e e  1

23. Fro more than one triplet satisfying all three equations we have
1 1 1
a b c
1 1 1
0
b c a
1 1 1
c a b
 abc
 (x  y  z)  a satisfying the triplets
We get a  10, p  7   2  2  8  0
  2 and 4

 1 1
24.
k 

k 

If f(x) is continuous then lim f(t k )  f lim (tk )  f    1 (Where lim tk  )
 4 k  4
 1
Also as graph is smooth f     0
4

x 1
25. f x  x  2  f(x) can not take the value 1 and f(2).
x3

W2 W C z3
26.  1 e  i
W2 W1
W2 W
  2 e i  1  itan 
W1 W1
W1 90°
W1 W 2 
W2
W2
2

 W1 W2 ei  W1  i W1
2
W2 sin   
A z1 B z2

27. (13x  1)2  36

13x  1  6
7
x
13
5
x
13
5
 cos   
13
12
 sin   
13
120
 sin 2  
169

28. (1  x  x 2 )n  a0  a1x  a2 x2    a2n x2n ….. (1)

n
3 1
 a0  a2  a 4  
2
3n  1
a1  a3  a5  
2
Put x  i in equation (1) where i  1
If n  4m  1
 a1  a3  a5   1
a0  a2  a 4   0
3n  1
So, a1  a5  a9    S2
4
3n  3
a3  a7  a11    S4
4
Similarly if n  4m  3
3n  3 3n  1
Then, S 2  and S 4 
4 4

29. If n is even
 n  2  n n2
f(n)  2  1  2    
 2  2 4
n  1 (n  1)2 n  1 n2  1
If n is odd then f(n)  f(n  1)    
2 4 2 4

(n  m)2  (n  m)2
f(n  m)  f(n  m)  if both even
4
(n  m)2  1  (n  m)2  1
 it both odd = nm
4
   
30. Obviously A  C and B  D will be non-zero collinear vectors
       
 (A  C).(B  D)  0 and similarly (A  B), (C  D) will be non-zero collinear vectors
   
Hence (A  B)  (C  D)  0

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Paper 1

Time Allotted: 3 Hours Maximum Marks: 180

 Pl ea s e r ea d t h e i n s t r u c t i o n s c a r ef u ll y . Yo u a r e a l l o t t ed 5 m i n u t es
s p ec i f i c a ll y f o r t h i s p u r p o s e.
ALL INDIA TEST SERIES

 Yo u a r e n o t a l l o wed t o l ea v e t h e E xa m i n at i o n Ha l l b ef o r e t h e en d o f
t h e t es t .

INSTRUCTIONS

C. Marking Scheme For All Three Parts.

(i) Section-A (01 – 10) contains 10 multiple choice questions which have one or more than one
correct answer. Each question carries +3 marks for correct answer. There is no negative
marking.

(ii) Section-C (01 – 10) contains 10 Numerical based questions with answers as numerical value
from 0 to 9 and each question carries +3 marks for correct answer. There is no negative
marking.

Name of the Candidate

Enrolment No.

Useful Data

PHYSICS
Acceleration due to gravity g = 10 m/s2

Planck constant h = 6.6 1034 J-s

Charge of electron e = 1.6  1019 C

Mass of electron me = 9.1  1031 kg

Permittivity of free space 0 = 8.85  1012 C2/N-m2

Density of water water = 103 kg/m3

Atmospheric pressure Pa = 105 N/m2

Gas constant R = 8.314 J K1 mol1

CHEMISTRY

Gas Constant R = 8.314 J K1 mol1

Atomic No: H=1, He = 2, Li=3, Be=4, B=5, C=6, N=7, O=8,

Physics PART – I

  2 x2 
1. A particle is released at x = 1 in a force field F  x    2   xˆ , x  0. Which of the
x 2 
following statements is/are true?

(A) F  x  is conservative
(B) Angular momentum of the particle about the origin is const.
(C) The particle moves towards x  2
(D) The particle moves towards the origin

2. Light from a monochromatic source is incident normally on a small photo sensitive

surface S having work function . If power of the source is W and a is the distance
between the source and S, then
 wS 
(A) the number of photons striking the surface per unit time will be  2 
 4hca 
1
(B) the maximum energy of the emitted electrons will be (hc  )

1
(C) the stopping potential needed to stop the most energetic photons will be (hc  )
e
hc
(D) photo emission occurs only if 0   


3. In a one-dimensional collision between two identical particles A and B, B is stationary

and A has momentum P before impact. During impact, B gives impulse J to A.
(A) The total momentum of ‘A plus B’ system is P before and after the impact and (P – J)
during the impact
(B) During the impact, A gives impulse J to B
2J
(C) The coefficient of restitution is 1
P
J
(D) The coefficient of restitution is  1
P

Space for Rough work

4. The electric potential decreases uniformly from 120V to 80V as one moves from y-axis
from y = –2 cm to y = +2 cm. Then at origin
V V
(A) Electric field must be equal 20 (B) Electric field may be equal to 20
cm cm
(C) Electric potential must be 100V (D) Electric potential may be 60V

5. A conducting ring of radius ‘a’ is rotated about a point O

on its periphery as shown in the figure in a plane

perpendicular to uniform magnetic field B which exists
everywhere. The rotation velocity is w, then:

(A) v P – v 0 > 0 and vR – v 0 < 0 (B) v p = vR > v 0

2L  x 
6. A reflecting surface is represented by equation Y  sin   . A
  L 
ray travelling horizontally becomes vertical after refection. The
coordinates of the point(s). Where this ray is incident :

 L 2L  L 3L 
(A)  , (B)  , 
 4   3  
   
 3L 2 L  2L 3 L
(C)  ,  (D) ,
 4   3 


7. A V-shaped conducting wire is moved inside a

magnetic field as shown in figure. Magnetic
field is perpendicular to paper inwards. Then :
(A) Va = Vc
(B) Va > Vc
(C) Va > Vb
(D) Vc > Vb

Space for Rough work

8. The variation of gravitational field

intensity with distance from the centre
of a body is shown in the graph from
which one can conclude that

(A) Variation of gravitational field intensity is due to the spherical mass body of radius R
(B) Eg  r for r < R
E 9R
(C) the separation of two points having field g may be
4 4
(D) the separation of two points R1 is R is R/4

9. For an LCR circuit

(A) A and B represent R and Z respectively
(B) A and B represent Z and R respectively
(C) A, B, C and D represent Z, XL, R and Xc respectively
(D) for  = ’, the phase difference between current and
voltage becomes zero

10. An ideal gas undergoes at thermodynamics cycle as shown in

figure. Which of the following graphs represents the same cycle?

(A) (B)

Space for Rough work

SECTION – C
(One Integer Value Correct Type)

This section contains 10 questions. Each question, when worked out will result in one integer
from 0 to 9 (both inclusive).

 2abxt  a 2 x 2  b 2 t 2 
1. A travelling pulse is given by f(x, t) = A exp   , Where A, a, b, c are
 c2 
positive constants of appropriate dimensions, the speed of the pulse is nb/2a, n =

2. A free particle of mass ‘m’ is confined

to a region of length L. The de-Broglie
wave associated with the particle is
sinusoidal in nature as given in figure
n2 h2
the energy of the particle is ,n=
8 mL2

3. A concentric spherical volume of inner

radius a and outer radius ‘b’ is filled
with a material of finite conductivity
A
specified by   r   2 where A is a
r
positive constant of appropriate
dimensions.
The outer surface is grounded and the inner surface is maintained at potential V0. The
resistance of this configuration is (b–a)/nA, n = _________.

4. A rod of length 5m is placed at the edge of a smooth table as

shown in figure. It is hit horizontally at point B. If the
displacement of centre of mass in 1S is 5 2 m, then the
angular velocity of the rod is 5n rad/s, n = : (Given : g = 10
m/s2)

5. Two consequent plane waves of light of equal amplitude and each of wavelength

20×10–8m propagating at an angle of rad with respect to each other, fall almost
1080
normally on screen the fringe width (in mm) on the screen is 108/100n, n =
____________.

Space for Rough work

6. A directed beam from a small but

powerful searchlight placed on the
ground tracks a small plane flying
horizontally at a fixed height ‘h’ above
the ground with a uniform velocity V as
shown in figure below.

If the searchlight starts rotating with an instantaneous velocity 0 at t = 0, when the plane
was directly overhead, then at a large time t, instantaneous angular velocity w(f) is given
w0
by n , n = _____________
1  w 20 t 2

7. An electric dipole is constructed by fixing two circular charged rings,

each of radius a, with an insulating contact, one of these rings has a
total charge +Q and the other has total charge –Q. If the charge is
distributed uniformly along each ring, the dipole moment about the
point of contact will be n QaZˆ /4, n = ___________.

8. A dumbbell is placed in water of density . It is

observed that by attaching a mass m to the rod,
the dumbbell floats with rod horizontal on the
surface of water and each sphere exactly half
submerged as shown in figure.

The volume of the mass m is negligible the value of length L is nd

 V  3M ,
 V  2M
n = ________________.

9. A small sphere of mass m and radius r rolls without slipping on the inside a large
hemisphere of radius R. The axis of symmetry of the hemisphere is vertical. It starts at
the top from rest. The small sphere will exert a normal force on the hemisphere at its
17
bottom equal to n mg , n = _________.
14

10. An electron of mass ‘m’ and charge ‘e’ initially at rest gets accelerated by a constant
electric field E. The rate of change of De Broglie wavelength of this electron at time ‘t’ is
 h n 
  eE t  ; where n equals to_______.
 

Space for Rough work

Chemistry PART – II

1. The product of the following reaction,

CH3 
H / H2O
 ?
OOH
would be
CH3 CH3
(A) (B)
O OH
CH2
(C) (D)
O O
OH

(i) CH3 CHO

2.  Product
Mg
  
(ii) aq. NH4 Cl
Br Et 2O

The product of above reaction is

O = C  CH3 Cl

(A) (B)
Br COCH3
Br CH(OH)CH3

3. When 150 ml of ozonized oxygen was passed through red hot tube, the volume
increased by
10 ml then the volume percentage of ozone in the sample is
(A) 25% (B) 51%
(C) 13.33% (D) 17.84%

Space for Rough work

2
4. The volume of water needed to dissolve 1 mg of PbSO4 (Ksp = 1.44  108 M ) at 25°C is
approximately (molar mass of Pb = 207)
(A) 10 ml (B) 27 ml
(C) 43 ml (D) 80 ml

Cl
T D
Fe / 
5. Br
2 /  X

SiMe3
The product of the reaction (X) will be
Cl Br
T D T D

(A) (B)

Br SiMe3
Cl Cl
T D Br D

6. Which of the following is a salt of sulphurous acid?

(A) NaHSO4 (B) Na2S2O3
(C) Na2S4O6 (D) Na2SO 3

7. Which of the following is/are true for 3Methyl butanone

(A) This compound is an isomer of 4penten1ol
(B) It may be prepared by acidic Hg2+ catalyzed hydration of 3methyl1butyne
(C) It may prepared by CrO3 oxidation of 2methyl2butanol
(D) It can not give positive iodoform test.

8. In blast furnace a charge of oxide ore, limestone and coke is feed for metallurgical
extraction of iron. The correct statements regarding the process is/are
(A) at the top of furnace the reducing agent is CO
(B) at the top of furnace the reducing agent is C
(C) at the top of furnace the reducing agent is CO2
(D) at the bottom of furnace the reducing agent is C

Space for Rough work

3
9. The correct facts regarding shape and hybridization of complexes [CuCl5] and
3
[Ni(CN)5] will be
(A) both have similarly hybridized states on central metal ion
(B) both have different shapes
(C) both have different hybridized states on central metal ion
(D) both have similar shapes

10. On the basis of following graphs between volumes of gas adsorbed and pressure of gas

V(cc/g) V(cc/g)
I II

P(torr) P(torr)
which of the following is/are correct ?
(A) graph I is of chemisorption (B) graph II is of physiosorption
(C) graph II is of chemisorption (D) graph I is of physiosorption

Space for Rough work

SECTION – C
(One Integer Value Correct Type)

This section contains 10 questions. Each question, when worked out will result in one integer
from 0 to 9 (both inclusive).


NaCl NaOH 1. H
1. A   B   yellow solution   Blue solution.
H2SO4 2. H2 O2

(red vapours)
The blue colour obtained at the end of reaction is due to salt formation. The oxidation
number of the metal in the salt is

2. Find the quantum numbers of the excited state of electrons in He+ ion which on transition
to ground state and first excited state emit two photons of wavelengths, 30.4 nm and
7
108.5 nm respectively. (RH = 1.09678  10 m1)

3. Electron is now a day considered as probability wave function with quantized angular
momentum and energy. The angular momentum of e in 1s orbital is.

4. The number of atomic orbitals involved in the hybridization at central metal ion of
K3[Fe(CN)6] are

5. In an acidic buffer the ratio [acid]/[salt] is 0.5. If the concentration of acid make twice of its
initial value then what is value of 10pH. [log 2 = 0.3]

Space for Rough work

6. How many stereoisomers are obtained on complete hydrogenation of CH2 ?

CH2

7. 20 ml solution containing mixture of NaOH and Na2CO3 require 60 ml of 0.5 N H2SO4 for
the end point using phenolphthalein as indicator. After this when same solution was
titrated using methyl orange indicator. A further 20 ml of 0.5 N H2SO4 was needed for the
second end point. Calculate the molarity of NaOH.

8. A solution containing 2.68  103 mol of A n ions requires 1.61 103 mol of MnO 4 for the
complete oxidation of A n  to AO3 in acidic medium. What is the value of n?

9. 0.15 mole of pyridinium chloride has been added into 500cm3 of 0.2M pyridine solution.
What is the pH of the resulting solution assuming no change in volume. (Kb for
pyridine  1.5  10 9 )

10. A solution of 6.2 g ethylene glycol in 55 g H2O is cooled to 3.720 C . What is the weight
(in grams) of the ice separated from the solution?
(Kf H2O  1.86k mole 1kg )

Space for Rough work

Mathematics PART – III

1. If curve A  z : z  3  z  3  8 , where B  z : z  3  k, k  R  touches A internally


and C  z : z  3  z  3  4 , then 
(A) B also touches C internally (B) A and C intersect orthogonally
(C) k  1 (D) 1  k  2

2. If f(x) is a polynomial of degree three with leading coefficient as unity such that
f(1) = 1; f(3) = 9 and f’(1) = 2, then
(A) f(x) = 2x has at least one root between (1, 3)
(B) f(x) = 2 has at least one root between (1, 7/3)
(C) product of roots of f(x) = 0 is 4
(D) sum of root of f(x) = 0 is 3

3. The inequality sin | z | 0 may represent, where z is complex number

(A) an interior region of circle where centre is origin and whose radius is 
(B) a parabola whose vertex in (0, 0)
(C) a region between two concentric circles cantered at (0, 0) and having radii 2 and
3
(D) an ellipse of semi axes are  and 2

4. A straight line touches the rectangular hyperbola 9x 2  9y 2  8 and parabola y 2  32x

then equation of line is (are)
(A) 9x  3y  8  0 (B) 9x  3y  8  0
(C) 9x  3y  8  0 (D) 9x  3y  8  0

5. Which of the following is true about matrix?

(A) if A is a non–singular square matrix of order n  n and k is scalar (k, n)  (0, 1) then
2
| adj(kA) | is equal to kn n | A |n 1
(B) let A and B be 3  3 order matrices. If | A | 0, AB  O and AC  O then B  C
(C) A and B are two square matrixes of order n  n such that AB  O and A is non-
singular then B must be null matrix
(D) if AB is non–singular and A, B are square matrixes. Then A and B both must be
non–singular matrices

Space for rough work

6 For a given parabola y2 = 4ax, two variable chords PQ and RS at right angles are drawn
through the fixed point A(x1, y1) inside the parabola, making variable angles  and  with
1 1
x-axis. If r1, r2, r3, r4 are distances of P, Q, R and S from A, then the value of 
r1r2 r3r4
(A) independent of  (B) independent of 
(C) depends upon both  and  (D) is a constant

7. Consider the equation tan (5 cos ) = cot (5 sin ) for   (0, 2), then
(A) number of real solution is 28 (B) number of real solution is 14
2n  1 2n  1
(C) sin  + cos  = (D) sin  – cos  =
10 10

n
8. A is non singular square matrix of  n  n such that A 2   A , then det (adj (A )) is
(A) 1, if n is even (B) –1, if n is odd
(C) 1 if n is odd (D) –1 if n is even

9. Which of the following when simplified reduces to unity?

(A) log3 log27 log4 64 (B) 2 log18  2 8 
 2 
(C) log2 10  log2 
 5
 (D)  log 2 1  2  1
dx 1 1
10. If  p2 sin2 x  r 2 cos2 x  12 tan  3 tan x   c , then the value of p sin x + r cos x can be
6
(A) (B) 5
5
(C) 6 5 (D) –4

SECTION – C
(One Integer Value Correct Type)

This section contains 10 questions. Each question, when worked out will result in one integer
from 0 to 9 (both inclusive).

1. If  x  a   x  a2   0 +
has exactly one integral solution of x for a  I , then a possible
integral value of x is _______

2. If | z1  1| 1 and | z 2  4i | 1 , then the difference between maximum and minimum values
of | z1  z2 | is _____

Space for rough work

 
3. Two points A and B lie on the curve y  x 2 such that OA. ˆi  1, OB . ˆj  4 , where O is
 
origin and A and B lies in first and second quadrant respectively. Then OA.OB equals
_____

1  ab
4. If log245 175  a, log1715 875  b , then the value of is _____
ab

5. A circular section is formed by a thread of fixed length then

maximum area of section formed by thread
is equal to _____
maximum area of rectange formed by thread

6. For all positive real number x and y, such that x + y = 1, the maximum value of
 
36 x 4 y  xy 4 is _____

7. The number of solutions (x,y,z) to the system of equations

x  2y  4z  9
4yz  2xz  xy  13
xyz  3
such that at least two of x, y, z are integers is _____

8. If number of points common between | x |  | y |  1 and 2a x y  1  2 x  a y are 8

then value of 3a (where a  1) is _____

9. The value of the expression

1/7 A
lim  8y  3   y  1  y  2   y  3   2y  3   4y  5   2y  7    2y is of the form .
y   B
A
Then the value   (where [.] denotes the greatest integer function), is _____
B

x  x  n
10. If the number of the positive integral solutions of the equation      is n, then is
 9  11 3
_____ (where [.] denotes the greatest integer function)

Space for rough work

ANSWERS, HINTS & SOLUTIONS

CRT – IV
(Paper-1)
ALL INDIA TEST SERIES

Q. No. PHYSICS CHEMISTRY MATHEMATICS

1. A, B, C C A, B, C

2. A, B, C, D D A, B

3. B, C C A, C

4. B, C B B, C

5. B, D A A, C, D

6. B, D D A, B, D

7. A, C, D A, B B, C

8. A, B, C A, D A, C

9. C, D B, C B, D

10. A, B A, B A, B, D

1. 2 6 3

2. 3 5 4

3. 4 0 2

4. 6 6 5

5. 5 3 1

6. 1 3 3

7. 8 1 5

8. 2 2 2

9. 2 5 3

10. 2 5 8

Physics PART – I

SECTION – A

1.  F.dx is independent of path.

2. A, B, C, D
WS
Energy received at photo sensitive surface per unit time =
4a 2
 number of photons = energy per unit time + energy of a photon.

3. Conservation of momentum and impulse momentum theorem  Fdt  P

  v v ˆ v ˆ 
4. E    ˆi  j  k
  x y z 

dy  x 
6.  2cos    1
dx  L 
x  2
 ,
L 3 3
 L  2L
x   ,
3 3

 
7. 
ind   v  B .d
Gme
8. g  r  Re
inside R3
Gme
g 
out r2
R  Re

9. For RLC circuit

2
Z= R 2   xc  xL 

Xz = L
1
xC =
C
Ab =  = ’, x L = xC  z = R
Hence phase difference between current and voltage = 0

SECTION – C
2
 ax bt 
 
 c 
1. f(x,t)  e
b
v
A

2
h
P2  

2. K  
2m 2m
h2
K  2 .9
8L m

r 2 dr 1
3. R  A ar 2 

b  a 
4A

4. ucmx  5 m / s
So impulse = 5m,
 m 2
5m.  
2 12
30

L

D
5. 
d

6. X = h tan
dx d
 h sec 2 
dt dt
d v


dt h 1  tan2  
v vh
 
 v t 
h 1  2 
2 2
h 2
 v 2 t2 
 h 
v
Put 
h
get n = 1


7. b   q.2a  zˆ

 v
8. F3  d.v  d.  dv   3m  m  g
 2
txy = 0
L L
3mg  mgd  dv
3 2
 dv 
L  mg    mgd
 2 
mgd
 2d.
 dv  3mg 
L
 dv 
 mg 
 dv  2mg 

 2 

h h d h 1
10.     t
p eBt dt eE

Chemistry PART – II

SECTION – A

CH3 CH3 CH3


1. O  OH2 H2O
H2 O
  O+


H O OH O
OH
Hemiketal
2. Aryl chlorides do not form Grignard reagents in ether and THF is necessary as solvent, while aryl
bromide and iodide readily react in ether solution.
The aqueous NH4Cl is used to prevent the possibility of dehydration of alcohol which might occur
if acid is used.
Cl Cl Cl
Cl O
CH3  C
H
Mg NH Cl
 
CH  CH3 
4

Et 2O CH  CH3
Br MgBr
OMgBr OH
3. Let x ml ozone be present in 150 ml of ozonized oxygen.
heated
2O3   3O2
tube
x ml 3/2 x ml
Increasement in volume due to decomposition of ozone = 0.5 x = 10 ml
 x = 20 ml
20
Therefore, percentage of ozone in ozonized oxygen =  100 = 13.33%
150
4. Solubility of PbSO4 = K sp = 1.44  10 8 = 1.2  104 M
Solubility of PbSOor4 = 1.2  10–4  303  103 = 36.36 mg litre–1
1000
 Volume of water needed to dissolve 1 mg of PbSO4 = = 27.5 ml
36.36

5. The reaction is known as bromodesilylation.

6. Sulphurous acid is actually aqueous solution of SO2

2NaOH + SO2  Na2SO3 + H2O

8. Both C and CO act as reducing agent. Being gas CO reduction is achieved at the top of furnace
at lower temperature of 250oC.
3
9. [CuCl5]3 is dz sp hybridized and trigonal bipyramidal in shape while [Ni(CN)5]3 is d x2  y2 sp 3
hybridized and square pyramidal in shape.

10. As chemisorption is monolayer phenomenon while physiosorption is multilayered adsorption.

SECTION – C

 H
1.  CrO2Cl2 NaOH
K2Cr2O7 + H2SO4 + Cl   Na2CrO4  Na2Cr2O7
(A) (B) (yellow) H2O2

blue colour of CrO5

1 1 1
2.  RH Z 2  2  2 
  n1 n2 
For n1 = 1, n2 = ?
1 1 1
9
 1.09678  107  4  2  2 
30.4  10  1 n2 
 n2 = 2
For n1 = 2 (first excited state), n2 =?
1 1 1
9
 1.09678  107  4  2  2 
108.5  10 2 n2 
 n2 = 5

h
3. L      1
2
as for 1 s orbital  = 0  L = 0

4. This octahedral complex involve d2sp3 hybridization which is hybrid of 6 atomic orbitals.

5. pH = pKa+ log
S 
A 
pH = pKa + log 2

now acid added so that

 S =1
 A
pH = pKa + log 1
pH = pKa
 pH = pKa + log 2 pKa
= log 2 = 0.3
 10 pH = 3.

CH3 CH3 H3C

6. 3 H 2 / Ni
CH2   CH3 + H + H
H H H
CH2 H CH3 CH3
meso enantiomers

x m mole NaOH
7. Let the 20 ml alkali solution
y m mole Na2CO3
Using phenolphthalein indicator
nf of NaOH = 1
nf of Na2CO3 = 1
No. of meq. of base = 1  x +1  y = x + y
Meq. of acid = 60  0.5 = 30
Meq. of acid = Meq. base
x + y = 30
using methylorange indicator
No. of meq. of base Na2CO3 = 1  y = 4
Meq. of acid = 20  0.5 = 10
y = 10
x = 20
20
Molarity of NaOH = 1
20

8. MnO 4  Mn2  n- factor = 5

Equivalents of MnO 
4  1.61 10 3  5  8.05  10 3
Equivalents of A n   8.05  103
N –factor of AO3  5  n   5  n   2.68  103  8.05  103
5n  3 n2

0.15  1000
9. Pyridinium Chloride   0.3M Pyridine  0.2M
500
A mixture of pyridine and its salt pyridinium chloride forms a basic buffer and therefore

pOH   logK b  log

Salt 
Base
0.30
 
pOH   log 1.5  10 9  log
0.20
pOH   log1.5  log10 9  log1.5

pOH  9 pH  14  9 pH  5

6.2
10. No of mole of solute  nB    0.1 Tf  Kf  m
62
0.1 1000 1.86  0.1 1000
3.72  1.86  WA 
wA 3.72
WA  50
Weight of ice separated from solution  55  50  5 grams

Mathematics PART – III

SECTION – A
1. Con-focal conics; therefore orthogonal intersection and circle touches at vertices

2.  f(x)  x 2  (x  1)2 (x  3)

3. sin | z | 0
| z | (0, ) or (2, 3) or (4, 5)
 0 | z | , 2 | z | 3 so on

8 2
4. Let equation be y  mx  c since it touches the hyperbola c 2  (m  1)
9
8
Also it touches the parabola y 2  32x, we get c 
m
8 2 64 8
 (m  1)  2  (m2  9) (m2  8)  0  m  3 c  
9 m 3
Requires equation are 9x  3y  8  0, 9x  3y  8  0

2
5. | adj(kA) || kA |n1  k n n | A |n 1 (A) is correct
(B) Order of C is unknown. Order of C not necessarily 3  3 hence B, C need not be equal
 (B) is incorrect
Since AB = O and A is non-singular
 A 1(AB)  A 1O  IB  O  B  O
(C) is correct
Now | AB | 0 | A | 0 and | B | 0
 A an B both are non-singular
 (D) is correct

6. Let two perpendicular chords through A(x1, y1) be PQ P 2

and RS y = 4ax
x  x1 y  y1
Equation of PQ is  r
cos  sin  r1
Where tan  = slope of PQ
R
Any point on this line may be taken as r3
(x1 + r cos , y1 + r sin )
2
As the point lies on y = 4ax, A(x1, y1)
2
  y1  r sin   4a  x1  r cos   r2 r4
Q
 r 2 sin2   2y1 sin .r  4acos r  y12  4ax1  0
S

 r1r2 
 y12  4ax1 
2
sin 

Similarly r3r4 
y12  4ax1 
where tan  = slope of RS
sin2 
Since PQ is perpendicular to RS,

  = 90 +  or  – 90
In either case; sin2  = cos2 
1 1 sin2   cos2  1
Now    2 which is constant
r1r2 r3r4 y12  4ax1 y1  4ax1   
7. Since tan (5 cos ) = cot (5 sin )
2n  1
Or sin  + cos  = ,nI
10
2n  1  10 2  1   10 2  1 
 –1 < < 1      n   
10 2  2   2 
Hence total number of solution is 14

2
8. A 2   A  A   A if n is odd
 A  1 , A n adj A n  An I

 An Adj An  A nI

 Adj An  1
Similarly when n is even A  1
 Adj An  1

9. log3 log27 log4 64 = –1


2log18 2  8  log 3   2  
2  2 2 = log3 2 3 2   1
 2   2  3
log2 10  log2 
 5
 = log2  10 

 = log2 2 2 =
5 2
 
 log 2 1  2 1  1 
dx 1 1
10.  p2 sin2 x  r 2 cos2 x  12 tan  3 tan x   c
sec 2 x 2
 p2 tan2 x  r 2 dx (p tan x = t, p sec x dx = dt)

1 dt 1 t 1  p tan x 
 2 2
 tan1    c = tan1  c
p t r pr r  pr  r 
p
 pr = 12,  3  p = 6, r = 2
r
 –6 < 6 sin x + 2 cos x < 6 when sin x = 1

SECTION – C
1. When a = 2, x = 3

2. Centres are (0, 4), (1, 0) and radii are each unity
 The shortest distance = distance b/w centres – (sum of radii)  17  2
Maximum distance = distance b/w centres + (sum of radii)  17  2
 Diff. = 4

 
3. Let OA  x1i  x12 j ,OB  x2i  x 22 j

OA.i  1  x1i . i  x12 j . i  1
 x1  1  y1  x12  1

OB . j  4  (x 2i  x 22 j) .j  4
 x 22  4  x 2  2 ( B lies in II quad.)
 
OA.OB  (i  j).( 2i  4j)  2  4  2

4. 175  52.7, 245  5.72 , 875  53.7, 1715  5.73

Let   log5,   log7
log175 2  
a 
log245   2
log875 3  
b 
log1715   3
1  ab (  2) (  3)  (2  ) (3  )

a  b (2  ) (  3)  (  2) (3  )
5(2  2 )
 5
2   2

5. Let length   radius of sector  r , angle  

  2r

r
1 1    2r  1
A  r 2 .  r 2 
2 2  r  2
 r  2r 2   

dA   4r  d2 A
 0 r ,  2  0
dr 2 4 dr 2
r
A max   4r  2r   r 2  2 / 16
2
2
 2
Max. area of rectangle    
4 16

8. Shown in the figure

x  x 
10.  9   11  I
   
x x
 I   I  1, I   I  1, Solution is possible only when 11I  9I  9
9 11
9
 0I
2
4
Total number of solution  (9  2I)  1  24
I 0

Paper 2

Time Allotted: 3 Hours Maximum Marks: 180

 Pl ea s e r ea d t h e i n s t r u c t i o n s c a r ef u l l y. Yo u a r e a l l o t t ed 5 m i n ut es
s p ec i f i c a ll y f o r t h i s p u r p o s e.

ALL INDIA TEST SERIES

Yo u a r e n o t a l l o wed t o l ea v e t h e E xa m i n at i o n Ha l l b ef o r e t h e en d o f
t h e t es t .

Name of the Candidate

Enrolment No.

Useful Data

PHYSICS
Acceleration due to gravity g = 10 m/s2

Planck constant h = 6.6 1034 J-s

Charge of electron e = 1.6  1019 C

Mass of electron me = 9.1  1031 kg

Permittivity of free space 0 = 8.85  1012 C2/N-m2

Density of water water = 103 kg/m3

Atmospheric pressure Pa = 105 N/m2

Gas constant R = 8.314 J K1 mol1

CHEMISTRY

Gas Constant R = 8.314 J K1 mol1

Atomic No: H=1, He = 2, Li=3, Be=4, B=5, C=6, N=7, O=8,

Physics PART – I

SECTION – A
(Only One Option Correct Type)
This section contains 10 multiple choice questions. Each question has four choices (A), (B), (C)
and (D) out of which ONLY ONE option is correct.

1. Electrons in a sample of gas containing hydrogen like atom (Z = 3) are in fourth excited
state. When photons emitted only due to transition from third excited state to second
excited state are incident on a metal plate photoelectrons are ejected. The stopping
potential for these photoelectrons is 3.95 eV. Now, if only photons emitted due to
transition from fourth excited state to third excited state are incident on the same metal
plate, the stopping potential for the emitted photoelectrons will be approximately equal to:
(A) 0.85 eV (B) 0.75 eV
(C) 0.65 eV (D) None of these

2. The coefficient of friction between block A of mass m1 = 5 A m1 F

kg and block B of mass m2 = 10 kg is  = 0.5. There is no B m2 300N
friction force between block B and fixed horizontal
surface. A force of 300 N acts on block B in horizontal Fixed horizontal surface
direction and a horizontal force of magnitude F acts on
block A as shown, both towards right. Initially there is no
relative motion between the blocks. The minimum value of
F such that relative motion starts between A and B is:
(A) 200N (B) 187.5N
(C) 150 N (D) 0

3. A proton moves in the positive z-direction after being accelerated from rest through a
potential difference V. The proton then passes through a region with a uniform electric
field E in the positive x-direction and a uniform magnetic field B in the positive y-direction,
but the proton's trajectory is not affected. If the experiment were repeated using a
potential difference of 2V, the proton would then be:
(A) deflected in positive x-direction (B) deflected in negative x-direction
(C) deflected in positive y-direction (D) deflected in negative y-direction

Space for Rough work

4. In the circuit shown initially both keys

are open and capacitance of both 1 k2
–10 µC +10µC
capacitors is 1 F. Charges on each
capacitor are as shown. Now both the R2 =10
+10µ C –10µC
keys k1 and k2 are simultaneously R1 =10 R3 =10
closed. Just after closure of both keys,
the current through resistance R2 is:
(A) 1 ampere (B) 0.5 ampere
(C) 1.5 ampere (D) zero

5. A negative point charge is brought near –– +

– +
an uncharged solid dielectric sphere. – +
Positive induced charges appear on the – +
– +
side of the sphere nearest to the point
charge, negative induced charges on the
–
– A B C + –q
– +
opposite side. A, B and C are three points – + fixed point
– + charge
on diameter of the sphere that also – +
passes through the point charge. fixed dielectric sphere
Comparing only at points A, B and C on
the dielectric sphere, the potential is:
(A) Largest at C (B) Largest at A
(C) Largest at B (D) The same every where

6. A uniform thin, rod AB of length L and mass m is undergoing fixed axis rotation about end
A, such that end A remains stationary as shown. The kinetic energy of section AP of rod
is equal to kinetic energy of section BP of rod at an instant. Then the ratio of length AP
AP
and length AB, that is, :
AB
1 1
(A) (B) 1/3
2 2
1 1
(C) (D)
2 2 2

Space for Rough work

7. In displacement method, the distance between object and screen is 96 cm. The ratio of
length of two images formed by a convex lens placed between them is 4:
(A) Ratio of the length of object to the length of shorter image is 2.
(B) Distance between the two positions of the lens is 40 cm.
(C) Focal length of the lens is 64 cm.
(D) When the shorter image is formed, distance of the lens from the shorter image is
64 cm.

8. An unstable radioactive element A decays into a unstable daughter element B which in

turn decays into stable grand-daughter element C. At time t = 0, the element A consists
of N0 atoms while number of atoms of element B and element C are zero. The decay
constant  of A and B are same. Based on this situation choose the incorrect statement
from the following:
(A) At the instant activity of element B is zero, the number of atoms of element B are
maximum.
(B) The maximum number of atoms of element B present at any moment is less than
N0
.
2
(C) At the instant activity of element B is maximum, the activity of element A and element
B are equal.
(D) The activity of stable element C increases with time.

9. Which of the following statements is/are incorrect:

(A) A positively charged particle having non-zero initial velocity may move from region of
lower electric potential towards region of higher electric potential.
(B) Whenever a charged solid conductor is grounded, the net charge on conductor must
become zero.
(C) Even though the plates of a parallel plate capacitor are oppositely charged, the net
electrostatic potential energy stored inside the capacitor is positive.
(D) As the inter-plate distance of a parallel plate capacitor is slowly reduced to half its
initial value, the electric field inside the capacitor may remain same.

Space for Rough work

10. Which of the following statements is/are correct:

(A) If distance x between two particles at any time t is increasing then the distance
dx
between particles will be least if and only if = 0.
dt
(B) For a particle moving with constant acceleration, the rate of change of speed with
time must be a constant.
(C) The magnitude of average velocity of a particle moving along a straight line between
time t = 0 to t = T may be greater than speed of particle at time t = T.
(D) Several stone are projected from top of a tower with same speeds but at different
angles of projection. The velocities of all these stone just before reaching the ground
are equal.

Comprehension type (Only One Option Correct)

Paragraph for Questions 11 & 12

A variable force F = t is acting on block of mass 'm' F = t
at an angle  with horizontal as shown.

1 1 kg
(given m = 1 kg,  = 3 N/sec, µs = 0.75, µk = , sin
3
4 Rough
53 = ). Give answer of following.
5

11. Minimum time after which block starts moving:

(A) 1 sec
(B) 2 sec
(C) 3 sec
(D) Data insufficient as  is not mentioned

12. Speed of block when it leaves the horizontal surface for  = 37º approximately:
(A) 20 m/s (B) 25 m/s
(C) 28 m/s (D) 26 m/s

Space for Rough work

Paragraph for Questions 13 & 14

In the shown AC, L-C-R series circuit reading V2

of ideal voltmeter V1 and V3 are 300 volt each
and reading of hot wire ammeter A is 10 amp
then give the answer of followings. R
V1 V3

~ A
(200 volt, 50 hz)

13. Reading of V2 will be:

(A) 600 volt (B) 300 volt
(C) 400 volt (D) 200 volt

14. Value of resistance 'R' will be:

(A) 10  (B) 20 
(C) 30  (D) data insufficient

Paragraph for Questions 15 & 16

A particle of mass 'm' is moving in horizontal

circle inside a smooth inverted vertical cone
above height 'h' from apex. Half angle of cone
is ‘’ then give the answer of following.

15. Normal force by surface of cone is :

(A) mg cos  (B) mg sec 
(C) mg sin  (D) mg cosec 

16. Time period of one revolution if  increases keeping 'h' constant :

(A) will remain same (B) will decrease
(C) will increase (D) increase then decrease
Space for Rough work

(Matching List Type)

This section contains 4 multiple choice questions. Each question has matching lists. The
codes for the lists have choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

17. Match the statements from List I with those in List II and select the correct answer using
the code given below the lists.
List-I List-II
(P) Floating of a block in liquid (1) Work energy theorem
(Q) The expression for buoyant force (2) Mass conservation
(R) Equation of continuity (3) Equilibrium of a body
(S) Bernoulli’s theorem (4) Archimedes’ principle

P Q R S
(A) 1 2 3 4
(B) 2 1 4 3
(C) 4 1 3 2
(D) 3 4 2 1

Space for Rough work

18. Consider an arrangement shown in the figure.

The rod of mass 3m is constrained to move in
the vertical direction only. The wedge of 4m
moves in the horizontal direction. The slider S
of mass 2m moves on a fixed horizontal rod. At
a particular instant system is released and the
string connecting the slider S to the making
0
angle 60 with the rod. (Neglect friction
everywhere) Match the statements from List I
with those in List II and select the correct
answer using the code given below the lists.
List– I List - II
(P) 5amg (1) 3
If net force on the block of m at this instant is ,
37
then a =
(Q) 5bmg (2) 1
If net force on the wedge is , then b =
37
(R) 5 2 cmg (3) 2
If contact force between wedge and rod is ,
37
then c =
(S) If tension in the string between blocks having mass m (4) 8
21dmg
and 2m is , then d =
37

P Q R S
(A) 1 2 3 4
(B) 2 4 4 3
(C) 4 1 3 2
(D) 3 4 1 2

Space for Rough work

19. A source of sound, emitting frequency of 3000 Hz, moving towards a stationary reflecting
wall with speed 50 m/s. There are four observers A, B, C and D as shown in the figure.
(Detectors A & C are at rest and detectors B & D are moving with speed 10 m/s towards
right. Speed of sound is 350 m/s) Match the statements from List I with those in List II
and select the correct answer using the code given below the lists.
10 m/s V0 = 10 m/s
50 m/s
C B
D Source of A Stationary
sound wall
Car

[Rest] [Rest]
List-I contains detectors and List-II contains the beat frequency observed by the
detectors. Match the List-I with List-II.
List-I List-II
(P) Detector A (1) 200 Hz
(Q) Detector B (2) Zero
(R) Detector C (3) 900 Hz
(S) Detector D (4) 875 Hz

P Q R S
(A) 1 2 3 4
(B) 2 1 4 3
(C) 4 1 3 2
(D) 3 4 1 2

Space for Rough work

20. A semi circular disc of mass M and

radius R with linear charge density 
on its curved circumference is hinged
at its centre and placed in a uniform
electric field as shown in the figure
Match the statements from List I with those in List II and select the correct answer using
the code given below the lists.
List – I List – II
(P) The net force acting on the ring is (1) M

E
(Q) If the ring is slightly rotated about O and (2) RE
released, find its time period of oscillation
(R) The work done by an external agency to rotate (3) 2K
it through an angle  is E
R
Magnitude of electric field at ‘O’ will be (4) 
(S) 4ER2 sin2  
2

P Q R S
(A) 1 2 3 4
(B) 2 4 1 3
(C) 2 1 4 3
(D) 3 4 1 2

Space for Rough work

Chemistry PART – II

SECTION – A
(Only One Option Correct Type)
This section contains 10 multiple choice questions. Each question has four choices (A), (B), (C)
and (D) out of which ONLY ONE option is correct.

1. The solution of weak acid H2N2O2 decomposes spontaneously at room temperature. The
gaseous product obtained is
(A) NO (B) N2O
(C) N2 (D) NO2

2. The correct statement for different solvents at constant temperature is

(A) The higher is the molar mass of solvent larger is Kb.
(B) The higher is the molar mass of solvent smaller is Kb.
(C) The higher is the molar mass of solvent larger is Kf.
(D) The higher is the molar mass of solvent smaller is Kf.

3. The incorrect statement amongst the following is?

(A) Black phosphorous is thermodynamically most stable allotrope of phosphorous
(B) Fe(III) is thermodynamically more stable than Fe(II)
(C) Graphite is thermodynamically more stable than diamond
(D) White phosphorous is kinetically least stable allotrope of phosphorous and so is
graphite in comparison to diamond
(Note: All comparisons are in their respective standard states)

4. A certain amount of N2O4(g) is enclosed in a closed container at 127°C when following

equilibrium got set up at a total pressure of 10 atm.

N2O4(g)  
 2NO2(g)
If the concentration of NO2(g) in the equilibrium mixture be 8  105 ppm, the Kc
–1
(in mol L ) for the above reaction at 127°C is equal to
(A) 3.189 (B) 2.051
(C) 0.974 (D) 1.842

5. A metal M and its compounds can give the following observable changes in sequence of
reactions.

M  dil  colourless 

dil   white  
excess   colourless  
H2S  white 

HNO3  solution  NaOH ppt  NaOH  solution  ppt 

The metal M is
(A) Mg (B) Pb
(C) Zn (D) Sn

Space for Rough work

6. What is not correct about E1CB reaction?

(A) Mechanism is followed when substrate is having excessive acidic proton while
leaving group is a bad leaving group
(B) the rds is unimolecular and the intermediate is carbanionic
(C) the overall order of reaction is one
(D) the rate law of the reaction is as follows:
[S]2
Rate   , where S is the substrate and B– is the base
[B ]

7. Fe2O3(s) + 3CO(g, 1 atm)  2Fe(s) + 3CO2 (g, 1 atm), G25° = – 10 kJ

The G for the reaction
2Fe(s) + 3CO2 (g, 2 atm)  Fe2O3(s) + 3CO (g, 20 atm)
at the temperature 25°C is equal to
(A) +17.11 kJ (B) –7.11 kJ
(C) +27.11 kJ (D) – 5.70 kJ

8. Consider the following alcohols:

CH 2OH CH 2 OH CH 2OH CH 2OH

O2 N OCH 3 Br

I II III IV

The order of decreasing reactivities of these alcohols towards substitution with HBr is:
(A) III  I  IV  II (B) III  I  II  IV
(C) I  III  IV  II (D) I  III  II  IV

9. Which of the following is a pyrosilicate.

(A) Sc 2Si2O7 (B) Zn2SiO4
(C) Ca2Si3 O9 (D) Be3 Al2 Si6 O18

10. Boric acid is a weak acid but it behaves as a strong acid in presence of
(A) Glycerol (B) Borazole
(C) FeCl3 (D) AlCl3

Space for Rough work

Comprehension Type

Paragraph for Question Nos. 11 to 12

Br 2 alc. KOH Br 2 / CCl4

C5H10 (X) A B C
light

S
NB NaNH2(excess)
E
/
Zn
D
F Soluble in H 2O
l4
CC
r 2/
B

G
NaNH2 (excess)
Na2C10H8 (H)
X is a saturated hydrocarbon which gives only one monobromination product:

11. What is correct about D?

(A) D is an alkadiene
(B) D is an alkyne
(C) D is an aromatic ionic compound
(D) D can exist as two stereo isomeric forms

12. ‘F’ is:

(A) (B)

Space for Rough work

Paragraph for Question Nos. 13 to 14

In coordination chemistry there are a variety of methods applied to find out the structures of
complexes. One method involves treating the complex with known reagent and from the nature of
reaction, the formula of the complex can be predicted. An isomer of the complex
Co(en)2(H2O)Cl2Br, on reaction with concentrated H2SO4 (dehydrating agent) suffers loss in
weight and on reaction with AgNO3 solution it gives a white precipitate which is soluble in
NH3(aq).

13. The correct formula of the complex is

(A) [CoClBr(en)2]H2O (B) [CoCl(en)2(H2O)]BrCl
(C) [CoBr(en)2(H2O)]Cl2 (D) [CoBrCl(en)2]Cl.H2O

14. If one mole of original complex is treated with excess Pb(NO3)2 solution, then the number
of moles of white precipitate formed under cold conditions will be
(A) 0.5 (B) 1.0
(C) 0.0 (D) 3.0

Space for Rough work

Paragraph for Question Nos. 15 to 16

Two identical bulbs contain an excess of solid A (volatile) and solid B (non-volatile) respectively.

A( g )  P( g )  H 2 S( g ) Q( g )  H 2 S ( g )

A( s ) B( s )
I II
o
At 25 C equilibrium gets established as shown above. At equilibrium; total pressure inside both
bulbs is 10 mm whereas partial pressure of H2S inside bulb I & II is 4 and 5 mm of Hg
respectively.

Equilibria shown for bulb ─I is a mix of Equilibrium shown for bulb ─II

Decomosition
A (s)  


 P(g)  H2S(g) 
Decomosition
B(s)  


 Q(g)  H2S(g)

Sublim ation
A (s )  
 A (g)

15. If some A(g) is removed from bulb─I at equilibrium:

(A) decomposition equilibrium will shift forward
(B) sublimation equilibrium will shift forward
(C) both decomposition and sublimation equilibrium will shift forward
(D) no change will happen in the state of equilibrium
o
16. If a mixture of excess solids A & B is kept at 25 C in a sealed container, the pressure
developed inside the container at equilibrium will be:
(A) 2.5 mm of Hg (B) 3.9 mm of Hg
(C) 14.8 mm of Hg (D) 8.4 mm of Hg

Space for Rough work

(Matching List Type)

This section contains 4 multiple choice questions. Each question has matching lists. The
codes for the lists have choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

17. Match the following:-

List-I List -II
(Binary liquid mixture in (method used for separation)
ether)
(P) Aniline + N-methyl aniline (1) (i) addition of [Ag(NH3)2]+
(ii) filtration
(iii) treating precipitate with dil HCl
(Q) Phthalimide + benzyl amine (2) (i) addition of C6H5SO2Cl & aq NaOH
(ii) Treating aqueous layer with dil HCl, con.
HCl and Distilled over NaOH

(R) 1-propanol + propanone (3) (i) addition of NaOH

(ii) treating aqueous layer with HCl
(S) 1-Butyne + 2-butyne (4) (i) addition of saturated NaHSO3
(ii) filtration
(iii) heating precipitate with dil HCl

P Q R S
(A) 3 2 4 1
(B) 4 3 2 1
(C) 2 3 4 1
(D) 3 4 2 1

18. Match the following:

List – I List – II
(Nature of oxide product)
(P) gently
NH4NO3 
 (1) Acidic & diamagnetic

(Q) Pt
NH3  O2   (2) Acidic & paramagnetic


(R) Cu  HNO3  (3) Neutral & paramagnetic

 Conc 
(S) P4O10
HNO3   (4) Neutral & diamagnetic


P Q R S
(A) 1 2 3 4
(B) 3 4 1 2
(C) 4 2 1 3
(D) 4 3 2 1

Space for Rough work

19. Match the column.

List  I List  II
(P) CH3COCH2COOC3H7 (1) Gives positive iodoform test.

(Q) CH3CH2CH2CHO (2) Gives 2, 4DNP derivative.

(R) CH3CH(OH)CH=CH2 (3) Decolourises Baeyer’s reagent.

(S) C2H5OC(CH3)=CH2 (4) Do not give positive iodoform test.
(5) Can not show tautomerism.

P Q R S
(A) 2, 4 2, 4 3, 5 4, 5
(B) 2, 4 4 1,5 3, 4
(C) 2, 4 2, 4 1, 3, 5 3, 4, 5
(D) 2, 4 2 1, 3, 5 4, 5

20. Match the following:

List – I List - II
(P) Plot of log t1/2 vs log [A0] is a straight line with a slope (1) Zero order
of ‘-1’
nd
(Q) Plot of log t1/2 vs log [A0] is a straight line with a slope (2) 2 order
of ‘1’
0
(R) Plot of log Kp vs 1/T is a straight line with a negative (3) ∆H is negative
slope
(S) Plot of log Kp vs 1/T is a straight line with a positive (4) ∆H0 is positive
slope

P Q R S
(A) 1 2 3 4
(B) 2 1 3 4
(C) 1 2 4 3
(D) 2 1 4 3

Space for Rough work

Mathematics PART – III

SECTION – A
(Only One Option Correct Type)
This section contains 10 multiple choice questions. Each question has four choices (A), (B), (C)
and (D) out of which ONLY ONE option is correct.

 x2 , x0
1. If f  x    2 and x  0 is point of maxima, then
   2

 a  2a  b  2 x, x  0
(A) a   0, 2  for b   2, 2  (B) a  0, 2 for b  R   2,

2

(C) a  0, 2 for b  R   2, 2  (D) a  R   0, 2  for b  R   2,

2


2. If Pn  1  cos 2n1   sin2n 1 and Qn  1  tan 2n 1  such that

P1Q0   P2Q1   ......... PnQn1   1, then  may be
 
(A) (B)
64 256
 
(C) (D)
33 31

3. If h  x   f  x   g 1  x  is an increasing function from 0,   to 0,   such that

g' 1  x   f ''  x  and f '  0   0 , then
(A) g' 1  x   0 for x  0 (B) g' 1  x   0 for x  0
(C) g' 1  x   0 for x  0 (D) g' 1  x   0 for x  0

h1  x   h1 1

4. If f  x   x  1 , g1  x   x 3  x  1 and g  f  x    h  x  , then lim is equal to
x 1 x 1
(A) 2 (B) 4
(C) 1 (D) 8

r n
5.  
If f  x   tan x  tan2 x tan 2x and g  n    f 2r , then g  2013   tan 22014 is equal to  
r 0
(A) 0 (B) tan1
(C)  tan1 (D) 1

Space for rough work

 
6. If XY-plane is rotated about origin by an angle  in a.c.w. manner  0     , then the
 2
x y
equation of line   1 is changed by x cos   y sin   p . Then, correct statement is
a b
(A) a  p cos  , b  p sin  (B) a  p sec  , b  pcosec 
1 2 1 1 2 1 1
(C)    (D)  
p2 ab a b p2 a 2 b 2

7. If y 
x
and
 y  x  dx  f  y   c , then f(y) for
2

x  1 is equal
x   x  y  x  y 
2 2
x2 
x
x2  2
x  ........
to
5 1
(A) 1 (B)
2
5 1 2 5
(C) (D)
2 2

8. Suppose f and g are functions having second derivatives f" and g" everywhere, if f(x) g(x)
f "(x) g"(x)
= 1 for all x and f ' (x) and g' are never zero, then  equals
f '(x) g'(x)
f '(x) f '(x)
(A) (B) 2
f(x) f(x)
2g'(x) 2f '(x)
(C)  (D) 
g(x) f(x)

 1
9. Let A be the point of intersection of the curve y  log 1  x    log3 9x 2  6x  1 and the
3 
3
2 2
circle x + y = 17. B is also a point on the curve but lies inside the given circle such that
its abscissa is an integer, then max {AB} is
(A) 2 units (B) 3 units
(C) 4 units (D) none of these

2n
S3n a
10. If Sn denotes the sum of first n terms of an AP and an  , then Lt  r is equal
S2n  Sn n 
r 1 n
to
(A) 2 (B) 4
(C) 6 (D) 8

Space for rough work

Comprehension type (Only One Option Correct)

Paragraph for Question Nos. 11 to 12

Read the following write up carefully and answer the following questions:
b b

 f(x)d  (x)   (x).d  f(x)  (b)f(b)  (a).f(a)

a a

6
2
11.  (x  [x])d | 3  x | (where [.] denotes greatest integer function) is equal to
0
(A) 126 (B) 63
(C) 72 (D) 36

3
12.  [| x |]d | x | , (where [.] denote greatest integer function) is equal to
2
(A) 1 (B) 2
(C) 4 (D) none of these

Paragraph for Question Nos. 13 to 14

Read the following write up carefully and answer the following questions:
Let f(x, y) be defined  x, y ϵ N such that
1
(i) f(x, x  1)  ,  x ϵ N
3
(ii) f(x, y) = f(x, z) + f(z, y) – 2f(x, z) f(z, y), for all z such that x < z < y; x, y, z ϵ N

13. lim f(1, n) is equal to

n 

1
(A) 0 (B)
2
2 1
(C) (D)
3 3

14. f(2, 5) is equal to

13 8
(A) (B)
27 27
1 1
(C) (D)
27 3

Space for rough work

Paragraph for Question Nos. 15 to 16

Read the following write up carefully and answer the following questions:
7
Locus of the centre of a circle touching the circle x2 + y2 – 4y – 2x = internally and tangents on
4
which from (1, 2) is making an angle 60o with each other is director circle (C) of a variable ellipse
E. On this basis given answers to the following questions

15. The equation of the director circle (C) is

2 2 2 2
(A)  x  2    y  1  3 (B)  x  1   y  2   3
2 2 2 2
(C)  x  1   y  2   3 (D)  x  1   y  2   3

16. Locus of focus of variable ellipse (E) when major axis is parallel to x-axis, is
 3 
(A)  x, y  /
2
| x  1| 3 and y  2  
(B)  x, y  / 0 | x | 3 and y  2 
 
 3 
(C)  x, y  / 0 | x  1| 3 and y  2  (D)  x, y  /
 2
| x  1| 3 and y  2 


(Match List Type)

This section contains 4 multiple choice questions. Each question has matching lists. The codes
for the lists have choices (A), (B), (C) and (D) out of which ONLY ONE is correct

17. Match the following List-I with List-II

List – I List – II
x 2
sin t    2
(P) If f  x    dt for  x   , then f     f  x  dx 1.
3
t 2 2 2 3 3

(Q) If x  y  z  4 and x 2  y 2  z2  6 , then least value of x is 2. 2

(R) If all the two digit numbers from 19 to 92 are written 1

consecutively to form the number N = 19202122……9192, 3.
the largest power of 3 that divides N is 2

(S) If z  C such that z  1  z  3  8 , then minimum value of

1 4. 1
z  1 is equal to
3

Codes:
P Q R S
(A) 3 1 4 2
(B) 1 2 3 4
(C) 4 3 2 1
(D) 3 1 2 4

Space for rough work

18. Match the following List-I with List-II

List – I List – II
(P) If the expression
2
sin 2 () + sin2 (120° + ) + sin2 (120° – ) remains constant, 1.
  R, then the value of the constant is 3
2x x
(Q) If the sum of the solutions of the equation 2e – 5e + 4 = 0 is ln k
2. 1
then k equals
tan  1 cot  3
(R) If  , find the value of 3.
tan   tan3 3 cot   cot 3 2
(S) If 3 tan(  – 30°) = tan(  + 120°) then the value of cos 2  equals 4. 2

Codes:
P Q R S
(A) 4 3 2 1
(B) 3 4 1 2
(C) 3 4 2 1
(D) 1 2 3 4

19. Match the following List-I with List-II

List – I List – II
(P) The distance of the point of intersection of the planes
  
r . (2iˆ  ˆj)  0, r . (4ˆj  3k)
ˆ  24, and r . (12iˆ  7k)
ˆ  24 from 1. 256
the origin
(Q) The value of  such that the planes
x – 2y – z = 10, 3x + 3y +  z  8 and 2x – y + 2z = k are not 2. 36
able to make faces of a tetrahedron
(R) The product of distances of any vertex from the remaining
vertices of a nine-sided regular polygon inscribed in a circle or 3 81
radius 2 units

(S) If on the complex plane the numbers 72  56i 3 , a + b i 3 (a,

b  R, b  0) and 0 represent vertices of an equilateral triangle 4. 144
then the value of (a + b)2
Codes:
P Q R S
(A) 2 3 1 4
(B) 1 2 3 4
(C) 4 3 2 1
(D) 2 3 4 1

Space for rough work

FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942
website: www.fiitjee.com
TG ~ @bohring_bot < 3
24
AITS-CRT-IV-(Paper-2)-PCM-JEE(Advanced)/15

20. Match the following List-I with List-II

List – I List – II
(P) If angles A, B of ABC are roots of equation
   2  
3 tan2  3 tan  ( 3   )  0 , then C can     3  be 1.
2 2  3  6
equal to
sin A 2 
(Q) If  where A is a angle of ABC then A cannot be 2.
A  3


(R) In a ABC if sin A cosB  cos A sinB sinC  1 , then angle C can be 3.
2
k 2
(S) If 16cos4 x  k 2  2k  8 cos2 x  2 then possible values of is 4.
6 3

Codes:
P Q R S
(A) 1 2 3 4
(B) 2 4 3 1
(C) 2 4 1 3
(D) 4 3 2 1

Space for rough work

ANSWERS, HINTS & SOLUTIONS

CRT – IV
(Paper-2)
ALL INDIA TEST SERIES

Q. No. PHYSICS CHEMISTRY MATHEMATICS

1. B B B

2. D A C

3. B B A

4. D C B

5. B C C

6. B D C

7. A C C

8. D A B

9. B A B

10. C A C

11. B C B

12. C D B

13. D D B

14. A A A

15. D B B

16. C C D

17. D C A

18. B D B

19. B C D

20. C D B

Physics PART – I

SECTION – A

 1 1
1. hv = 13.6 (3)2  2  2  = 2.75 eV
4 5 
for n= 4 to n = 3
1 1
hv = (13.6) × (3)2  2  2  = 5.95 eV
3 4 
for shorter wavelength
3.95 = 5.95 –    = 2eV
for longer wavelength eVs = 2.75 – 2 = 0.75 eV.

2. If F = 0
300
Then assuming no relative motion acceleration of A + B = = 20 m/s2
15
20 m/s2 > g
where  = 0.5 and g = 10 m/s2
 relative motion shall exist. Hence F = 0 N.

3. The electrostatic force on proton is along positive x-axis and the magnetic force is along negative
x-axis. Initially net force on charge q is zero. Since the velocity of proton is increased in repeat
experiment, the magnetic force on proton shall increase and the proton would then be deflected
in negative x-direction.

4. The reference potential at point A and B be o and x respectively.

Applying Kirchoff laws
20  x 10  x 0  x
  0
R1 R2 R3
or x = 10 V thus i = 0.

6. The electric field of the negatively charged sphere is not entirely canceled inside the dielectric.
This field points toward the sphere and so a positive particle placed inside the dielectric would
tend to move toward the sphere. This means that the potential on the negative side of the
dielectric is higher than that on the positive side.

7. The KE of given section AP and PB will x

be equal if MI of each section AP and A P B
section PB about A is same. L
x2 L2 x2
IAP= (x) IPB= IAB – IAP = (L) – (x)
3 3 3
x3 L3 x 3
 IAP= IPB  = 
3 3 3
3
L L
or x3= or x = 1/3
2 2

 y 1  x  2
8. For first & second position = , =
x O y O
y2 
 2
= 1 =4
x  2

y
 = 2 and y + x = 96
x
 y = 64, x = 32

O y Position 2
= = 2  A is True
2 x y x
Distance between two position of lens = y – O
x = 32 cm  B is True Principle axis I2
xy 64  32 x y
Focal length of lens f =  I2
x  y 64  32
Position 1
64
=  C is True
3
Distance of lens from shorter image = x = 32
cm  D is True

9. NA = N0e–t
dNB
= NA – NB
dt
 dN  –t t
et  B  NB  = N0e · e
 dt 
e t · NB = N0 · t
NB = N0te–t
dNB
= N0 e t  et t 
dt
dNB  1
If = 0 then (NB)maximum i.e. at t =  
dt 
dNA
– = N0e–t
dt
d2NB dNB
2
= N0  et  et   2 e t  t  = 0 for to be maximum
dt dt
2
That is t = activity of A and B will be same


12. t cos   µs N for no relative motion

3
t cos   (10 –t sin )
4
3
t (cos  + sin ) 0
4
10
t1 = to start motion
4cos   3 sin 
10
tmin = as (4 cos  + 3 sin )max = 5 for  = 37°
5

a = t cos  – µk (10 – t sin ) = t (cos  + µs sin ) – µk 10

t2
t 2
v = (cos   k sin )   k 10t
2 t1

 (t 2  t 2 ) 
v =  2 1 (cos   k sin )  k 10(t 2  t1 )
 2 
(t 2  t1 )
= [ (t1 + t2) (cos  + µk sin ) – 2µk10]
2
 50 
  2
= 
9  3   50  2  4  1  3   20 
    
2   9  5 3 5  3 
16  68 20 
=   1
9 3 3 
16 48 928
= × =
9 3 27
256
V= m/s
9
t2 sin  = 10
10  50 
t2 = =  
 sin   9 
t1= 2

14. Irms (XL2  R2 )1/ 2 = 300 = Irms (X2C  R2 )1/ 2

give XC = XL
 200
so Irms = rms 10 =  R = 10 
2R 2R
XC2 + 102 = 302
XC = 20 2 
so VCmax = Imax · XC
= 10 2 × 20 2
= 400 volt
N sin 
16. N sin  = mg
2 N
mv
 N cos  = 
F.B.D. of particle
R N cos 

gR
tan  = = v = gR cot  
v2 mg

v = gh as h increases v increases.
2R 2h tan  h
T = = = 2 tan  as  increases T
v gh g
increases.

17. 3y defination

18. By NLM
N sin = 4 ma4 ...(1)
3mg – Ncos – T1 = 3 ma2 ...(2)
T1 – T2 cos 60 = 2 ma2 ...(3)
T2 – mg = ma1 ...(4)
a3 = a4 = a2
a2cos60 = a1
5mg 42mg
b  mg  
37 37
mg 21ma
3mg  Ncos   
2 4
5mg 37ma

2 4
 10g 
a 
 37 
mg mg 9ma2
So, T2   2mg  
2 2 4

19. f A = 0
2v 0 2  10
f B = f0   3000  200 Hz
c  vs 300
2cv s 2   350  50
f C = f0   3000  875 Hz
c 2  v 2s 400  300
360  2  50
f D =   3000  900Hz
400  300

20.  df   ERd cos 

R  RE   sin 
 2RE
t   ERd .R sin 


 ER2   cos 2 
  
2 

      
 ER 2   cos        cos      
 2   2  
t net  ER2  2 sin  
MR2
  –2E 2 
2
 2E 2 
  – 2 

 MR 
4EA
w
m

w  2ER2  sin d  td
0
2
w  2ER 1  cos  

Chemistry PART – II

SECTION – A
1. Hyponitrous acid is hydrate of nitrous oxide.

2. As most liquids follow Trouton’s rule which provides constant value of Sovap  90 J/K.
RMT02 RMT0
So K b  
1000  H0V 1000  S0V
as R and  S 0V are constant so Kb is proportional to MT0 (T0 increases along with M) while,
S 0f may vary with a good deal particularly when M is very high.

3. 2Fe3+ + Fe  3Fe2+, G° = –ve  Eocell   ve

8  105
4. pNO2   10  8 atm,  pN2O4  2 atm
106
2
pNO 82
Kp  2
  32 atm
pN2O4 2
32
Kc   0.974 mol L1
0.0821 400
(ng = 1)

6. Cl F Cl
I Cl C C 
k1
F  : B  
 Cl C CF 3  HB (fast)
k 1

H F carbanion
(S)

k2
II Cl2C CF2   Cl2 C  CF2  : F (slow)
F
Rate = k2 [Carbanion]
Also,
k [Carbonion][HB]
Kc  1 
k 1 [S] [: B  ]
k c [S][B  ]
 [Carbanion] = , so
[HB]
[S][B ] – –1
Rate = k2kc Rate  [S] [B ] [HB]
[HB]

7. Fe2O3(s) + 3CO(1 atm)  2Fe(s) + 3CO2 (1 atm),

G25o C given for reaction is Go25o C as all the species are in their standard states.
So, for the reaction
2Fe(s) + 3CO2 (2 atm)  Fe2O3(s) + 3CO (20 atm),
Go25o C will be equal to + 10 kJ

3
PCO (20)3
Q 3
 3
 103
PCO 2
2
G = G° + 2.303 RT log Q
–3 3
= 10 + 2.303  8.314  10  298 log 10
= 10 + 17.11
= 27.11 kJ

8. Substitution of —OH by Br atom takes place by carbocationic intermediate formation in rds. The
stability of carbocation involved is as follows:

CH2 CH2 CH2 CH2

> > >

OCH3 Br O2N

9. Sc 2 Si2 O7 is a pyrosilicate
Silicates containing Si2 O67  ios are pyrosilicate
Zn2 SiO4 orthosilicate
Ca3 Si3 O9 and Be3 Al2Si6 O18 are cyclic silicates


B  OH 3  H 2O B  OH 4  H 
10.

CH2 OH OH OH CH2 O OH
+
B B
CH2 OH 2H2
OH CH2 O OH
OH

CH2 OH

CH2 OH CH2 O O CH2

B
CH2 O O CH2

11. Br Br

Br2 /CCl4
Br
Br2 /light Alc. KOH/

Monobro min ation
    
(B) (C)
(X)

+
2 
NaNH excess H2O Na
   

(D)
Br
NBS Zn/

  

(B) (E) (F)

13. As it suffers loss in weight with H2SO4, ionization sphere contains H2O molecule. As complex also
–
gives white ppt. with AgNO3, the ionization sphere should contain Cl ion. So formula of the
complex may be [CoBrCl(en)2]Cl.H2O.

14. Only one Cl– ion is ionisable from the complex. So, 1 mole of complex will generate half mole of
PbCl2 precipitate.

15. Equilibria shown for bulb ─I is a mix of


Decomosition
A (s)  
 P(g)  H2 S(g)
 K p  4 2  16

Sub lim ation
A (s)  
 A (g) Kp  2


Decomosition
A (s)  
 P(g)  H2S(g) K p  16
16.
x x

Decomosition
B(s)  Q(g)  H2S(g) K p  25
y y
16  x(x  y), 25  y(x  y)
Totalpressure  2x  2y  V.P.of A (g)  14.8mm

19. 2, 4DNP test is given by aldehydes and simple ketones, Baeyer’s test is used to identify CC
unsaturation and iodoform test is given by methyl ketones or alcohols containing CH3CH(OH)
group. No condition for tautomerism in (D) and (D) but (A) can not give idoform test due to active
methylene group.

Mathematics PART – III

SECTION – A

1. If f  a   f a  h  and f  a   f a  h  for h  0 and h  0 , then x  a is a point of maxima

x x x sin x
2. cos cos 2 ......... cos n 
2 2 2 x
2n sin
2n

3.  h'  x   f '  x   g' 1  x   0

 f '  x   f ''  x   0
d x

dx

e f 'x  0 
 e x f '  x  is decreasing function
 e x f '  x   e0 f '  0  for x  0
 f '  x   0 x  0
 g' 1  x   0 x  0

4. f 1  x   x  1
 h1  x   x 3  x
d 1

dx
 
h  x   3x 2  1

5.  f  x   tan2x  tan x

 
 f 2n  tan 2n 1  tan2n
 g  n   tan2n1  tan1
g  2013   tan22014   tan1

6. Since distance from origin is unchanged

1 1 1 1
So, p   2  2  2
1 1 p a b

a2 b2
2
1 2  1 1
    
p2 ab  a b 

x
7. y 2
 y2  x2 y  x  0
x y
 x 
  2x  2  1

dy   2xy  1
   x y  
 2x 2  x2  y  
dx x 2  2y 
x2  y  y x

x2  y   y 


 y 

1 dy

 x2  y

d  ln y  
 y  x2
y dx 
x 2  y x  y2 dx  
x2  y x  y 2   
 ln y  c  
 y  x  dx 2

 x  y  x  y 
2 2

1 1  5
 f 'y  and for x  1 , y 2  y  1  0  y 
y 2

d
f "(x) g"(x)  f '(x) / g'(x) g' d  f '  2f '  x 
dx
8.  =
f '  x  .g'(x)
= .  
f x
 f(x).g(x)  1
f '  x  g'  x  f ' dx  g' 
2
 g'  x  
 1  1
9. y = log1/3  x    log3 9x 2  6x  1 = log1/3  x    log3 | 3x  1| = 1
 3   3 
1
Since, x  , maximum {AB} = 3
3

3n
2a   3n  1 d
10. an = 2   3;
2n n
2a   2n  1 d   2a   n  1 d
2 2
2n
1
 Lt  .3  6
n 
r 1 n

6 3 6 3 6
11.  
x2  [x] d | 3  x |      
x 2  [x] d(3  x)   x 2  [x] d(x  3)    x 2  [x] dx   x 2  [x]  
0 0 3 0 3
6 3 6 3 6

 x 
2
  [x] d | 3  x |   x 2 dx   x 2 dx   [x]dx   [x]dx
0 0 3 0 3
3 3 3 6 3 3 1
x3  x3 
=      [x]dx  3  [x  3]dx  9  (72  9)   [x]dx   [x]dx  3  3dx
3  0 3 0 0 0 0 0 0
= 72 – 18 + 9 = 72 – 9 = 63.

3 0 3
12.  | x | d | x |  [ x]d(  x)   [x]dx
2 2 0
0 3 0 3 3
=   [ x]dx   [x]dx   [x]dx   [x]dx   [x]dx  2(3  2)  2
2 0 2 0 2

n 1
1 1 1 1 1  1 
13.-14. f(1,n)  f(1,2)    ...   1    
3n  2 3n  2 3n  3 3 2   3  

15. Let r and R be radius of required & given circle respectively &
let centre is (h, k) A
By given condition 2
(h  1)  (k  2) 2
=Rr r
C
r 30°
Now, = sin 30°
AB B (1, 2)
r = AB/2 = (R  r)/2 (AB = R  r)
R
 r
3
R 2R
 (h  1)2  (k  2)2 = R  =
3 3
3 3
Now, R =
2
2 3 3
 (h  1)2  (k  2)2 =   3
3 2
 Locus is (x  1)2 + (y  2)2 = 3 which is C

16. When major axis of ellipse (E) is parallel to x- (1, 2+3)

axis then equation of E will be
 x  12  y  2 2
 1 …..(1)
a2 b2
F (1+3, 2)
Such that a2 + b2 = 3 A B (1, 2) C D
(1–3, 2) (1+(3/2), 2)
In this case one focus will lie in between F (1–(3/2), 2)
and D and other in between A and F.
i.e., 0 | x  1| 3 and y = 2 (1, 2–3)

sin x
17. (P)  xf  x    f  x  dx    xdx
x
(Q)  y  z  4  x
  y  z 2 
2 2
and y  z   
 2 
 
2
 4  x 
 6  x2   
 2 
 
2
 x2
3
(R) Sum of digits in N = 705 is divisible by 31 but not 32
(S) Locus of z is an ellipse where a = 4 and b  2 3
i.e. z  1 min  2 3
1
 z 1  2
3

18. (P) LHS = sin2  + 1 – [cos2 (120° + ) – sin2 (120° – )]

1 1 3
= sin2  + 1 – [(cos 240°)(cos 2)] = sin2  + 1 + cos2  sin2   1  1  2 sin2  =
2 2

2

(Q) 2e2x – 5ex + 4 = 0
Let roots be x1 and x2, product of the roots be
e x1  e x2  2 , e x1  x2  2
x1 + x2 = ln = 2
(R) a, b  R p1  p2  4 3 tan p1  p2  8 = tan p1p2  b2  16 – tan 3 p2
tan3
p1 2 tan 16x 2  25y 2  400, + tan 3  = 0  = –2
tan 
cot  1 1 1 2
Now    
cot   cot 3 1  tan   cot 3 tan  1 3
1 1
tan3 2
tan  cot 
Alternatively: Prove that  = 1 now proceed
tan   tan3 cot   cot 3
2 sin(   300 ) cos(   120 o ) 1
(S) . 
2 sin(   120o ) cos(   30o ) 3
1 1
cos2  cos2 
2 sin(2  900 )  sin150o 1 2 1 2 1
o o
 or  or 
sin(2  90 )  150 3 1 3 1 3
cos 2  cos2 
2 2
3 1
 3 cos 2  – = cos 2  +
2 2
2 cos 2   cos 2  = 1

19. (P) 2x + y = 0, 4y + 3z = 24 and 12x – 7z = 24. Solving we get the point intersection as
(–12, 24, –24) whose distance from origin is 36
1 2 1
(Q) For no unique solution  = 0  3 3  0
2 1 2

 1(9)   (3)  2 (9)  0    9    81

(R) Let the centre be at origin and the vertex at ( 2, 0) . The remaining vertices will be
 2 2 
corresponding to the complex numbers 2, 2 2 , 23 , ...., 28    cos  i sin  .
 9 9 
The product of distances
= | 2  2 || 2  2 2 | ...  ( 2)8 | 1   | | 1   2 | ... | |  8 |  16  9  144
(S) a  b 3i  (72  56 3i) . ei /3  (36  28 3i) ( 1  3i)
 a = 36 – 84 = –48, b = 36 + 28 = 64. (a + b)2 = 256

 
20. (P) 3 tan2  3 tan  ( 3   )  0
2 2
A B
tan  tan
c 2 2  
cot   3
2 A B   
1  tan tan 1  1  
2 2  3

 C
3

sin
sin A 2 if  A    
(Q) Clearly from figure, tan   tan ,  
A   2  
2
(R) sin A cosB  cos A sinB sin C  sin A cosB  cos A sinB

 1  sin(A  B)  sin(A  B)  1 or C 
2
2
 1 
(S) 16  cos2 x    (k  1)2  0 , x  n 
 4 3

Time Allotted: 3 Hours Maximum Marks: 432

 Pl ea s e r ea d t h e i n s t r u c t i o n s c a r ef u l l y. Yo u a r e a l l o t t ed 5 m i n ut es
s p ec i f i c a ll y f o r t h i s p u r p o s e.

ALL INDIA TEST SERIES

Yo u a r e n o t a ll o wed t o l ea v e t h e E xa m i n at i o n Ha ll b ef o r e t h e en d o f
t h e t es t .

B. Filling of OMR Sheet

C. Marking Scheme For All Three Parts.

(i) Section-A (01 to 03 and 10 to 12) contains 6 multiple choice questions which have only one
correct answer. Each question carries +8 marks for correct answer and – 2 mark for wrong
answer.

Section-A (04 to 09 and 13 to 30) contains 24 multiple choice questions which have only one
correct answer. Each question carries +4 marks for correct answer and – 1 mark for wrong
answer.

Name of the Candidate

Enrolment No.

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2

Useful Data

PHYSICS
Acceleration due to gravity g = 10 m/s2

Planck constant h = 6.6 1034 J-s

Charge of electron e = 1.6  1019 C

Mass of electron me = 9.1  1031 kg

Permittivity of free space 0 = 8.85  1012 C2/N-m2

3 3
Density of water water = 10 kg/m

Atmospheric pressure Pa = 105 N/m2

Gas constant R = 8.314 J K1 mol1

CHEMISTRY

Gas Constant R = 8.314 J K1 mol1

Atomic No: H=1, He = 2, Li=3, Be=4, B=5, C=6, N=7, O=8,

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Physics PART – I

SECTION – A
Single Correct Choice Type
This section contains 30 multiple choice questions. Each question has four choices (A), (B), (C)
and (D) out of which ONLY ONE is correct.

1. A projectile is fired with velocity v at angle of  with horizontal. What would be the radius
of curvature of path at instant when rate of change of speed of projectile is minimum.
v 2 cos2  v2
(A) (B)
g gcos
v 2 cos  v2
(C) (D)
g gcos2 

3 2
2. Position of particle is given by x = t 4t + 5t + 9. What would be the distance travelled
by particle from instant t = 0 to instant when particle changes its direction of
velocity for last time.
(A) 1.85 m (B) 2.15 m
(C) 2m (D) 2.85 m

3. If a particle have velocity and acceleration given

  
 1   
 2
 3 i  4 j  ms ,  4 i  3 j  ms respectively.
   
What would be the rate of change of speed at that instant?
(A) 4.8 ms2 (B) 5 ms2
2
(C) 3 ms (D) 4 ms2

4. A ball of mass 1 kg attached with string of length 10 m is free Y

to rotate in horizontal plane is subjected to constant force of
 
 2  i  j  N. What should be the velocity of projection at X
  10 m
o
45
position shown in figure. Such that it complete full revolution.
(A) 10 ms1 (B) 2 5 ms1
(C) 4 5 ms1 (D) 6 ms1 V

Space for Rough work

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3 3
5. Fluid of density 10 kg/m is kept in container. Container is
Y
moving with acceleration of 10 ms2 find the magnitude of
pressure difference between A and B as shown in figure.
4
(A) 10 (B) 0
(C) 2  104 (D) 2 × 104 a = 10ms1

X
A(1,0) B(2,0)

6. Coefficient of friction between all the surfaces is 0.5. f=t

1 kg
Which diagram would describe the friction between
ground and 2 kg block when f = t. N is applied on 1 kg
2 kg
block. It is given that string is not taut initially

(A) (B) f

f
5

5
t
0 5
15 20 25
0 t
5 10

-15
(C) f (D) f

5 5

15
t t
5 10 5 10

-5

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7. If vA = 10 m/sec what would be the velocity of

C. 1 = 30 0, 2 = 60o. If B is very heavy and
remains on ground.
10
(A) m/sec 2 1
3 10 m/sec
C A
(B) 10 3 ms1 B
(C) 10 ms1
(D) 5 3 ms1

8. A ball of mass m kept at top slide freely on frictionless

hemispherical surface. At what minimum angle  net force acting
on ball would be equal to mg 
(A) cos1 (2/3) (B) cos1 (1/2)
o
(C) 0 (D) 30 o

9. From the given velocity acceleration diagram identify in which case object is slowing and
turning to the left.
(A) (B) a

V a

10. A light ray is incident on a plane mirror M. The mirror is rotated

in direction as shown in figure by an arrows at frequency (9/)
revolution/sec. The light reflected by the mirror is received on the
wall W at a distance of 10 m from axis of rotation. speed of the
spot on wall when angle of incident = 37o is
(A) 10 m/sec (B) 1000 m/sec M 10 m
(C) 360 m/sec (D) 500 m/sec

Space for Rough work

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  
11. A ray of light moving along the vector  i  2 j  undergoes refraction at an interference of
 
5
two media which is x-z plane  for y > 0 is 2 while Y < 0 it is . The unit vector along
2
which the refracted ray moves is
  
 3 i  5 j  
4 i  3 j


(A)   (B)
34 5
 
    

 3 i 4 j   4i3 j 
(C)  (D)
 5   5 
   

12. A dipole is placed at origin of co-ordinate system as shown in figure. Y

(0, y)
Electric field at point P (0, y) is given as
K     K   
 X
(A) (B) 3   i  2 j  45o
 i  j
3y3   y  
K     K   

(C)  i 2 j (D) 3 
 i2 j
y3   2y  

13. Figure shows a large conducting sheet having + +

charge density  on both sheet. The surface potential + +
of sheet is V0. A sphere of radius r is placed at a + + K
+ +
distance  ( >> r) from the sheet. When the key is
+ + 
closed charge q appears on the surface. The values
+ +
 on sheet is given by

 q  40rV0   4 0rV0  q 

(A)   (B)  
 4 r   2r 
 40rV0  q 
(C)   (D) Zero
 2r 

Space for Rough work

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14. The resistance between A and B of given circuit at t = 0 C

A B
and t =  will be (it the switch is closed at t = 0).
(A)  at both t
(B) O at both t
(C) O and R respectively
(D) R at both t

15. In series LCR circuit voltmeter and ammeter reading

V 200 V 200 V
are
(A) V = 250 V, I = 4A
(B) V = 150 V, I = 2A 50 
(C) V = 1000V, I = 5A
(D) V = 100 V, I = 2A A

100V, 50 Hz

16. A radioactive substance X decays into another substance Y. Initially angle X was present.
x and y are distingration constant of x and y. Nx and Ny are the number of nuclei of x
and y at any time t. Number of nuclei Ny will be maximum when
Ny y Nx x
(A)  (B) 
Nx  Ny  x   y Nx  Ny  x   y
(C)  y Ny  Nx  x (D)  x Nx   xNy

17. A capillary tube is immersed vertically in water and the height of the water column is x.
When this arrangement is taken into a mine of depth d, the height of the water column is
x
y. If R is the radius of the earth, the ratio is
y
 d  d
(A)  1   (B)  1  
 R  R
R d R d
(C)   (D)  
R d R d

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18. The ends of a rod of length  and mass m are

attached to two identical springs as shown in
figure. The rod is free to rotate about its centre k
/2
O. The rod is depressed slightly at end A and
A
released. The time period of the resulting O
B
oscillation is
k /2

m 2m
(A) 2 (B) 2
2k k
2m 3m
(C)  (D) 
3k 2k

19. The period of oscillation of a simple pendulum of length L suspended from the roof of a
vehicle which moves without friction down an inclined plane of inclination , is given by
L L
(A) 2 (B) 2
gcos  gsin 
L L
(C) 2 (D) 2
g gtan 

20. In the circuit shown in figure, A

R R L
and B are two cells of the same
emf E and of internal resistances E E
rA and rB respectively. L is an ideal 2R
inductor and C is an ideal A B X Y
capacitor. The key K is closed. rA rB
When the current in the circuit R R
C
becomes steady, what should be
the value of R so that the potential
K
difference across the terminals of
cell A is zero
(A) R  rA  rB if rA  rB
(B) R  rA rB
1
(C) R   rA  rB 
2
(D) For no value of R will the potential difference between the terminals of cell A be equal
to zero

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21. A uniform metal rod is moving with a uniform velocity v parallel to a long straight wire
carrying a current I. The rod is perpendicular to the wire with its ends at distance r1 and r2
(with r2 > r1) from it. The emf induced in the rod is
 Iv r 
(A) Zero (B) 0 loge  2 
2  r1 
 0Iv r  0Iv  r1 
(C) loge  1  (D) 1  
2  r2  4  r2 

22. When a centimeter thick surface is illuminated with light of wavelength , the stopping
potential is V. When the same surface is illuminated by light of wavelength 2, the
stopping potential is V/3. The threshold wavelength for the surface is
4
(A) (B) 4
3
8
(C) 6 (D)
3
23. The radiations emitted from a radioactive Magnetic field
material separate into three distinct groups A, B (into the page)
and C when a magnetic field is directed into the A B
plane of the paper. The names of radiations A, × × × × ×
C
×
B and C, respectively, are (see figure) × × × × × ×
(A) ,  and  × × × ×
× ×
(B) ,  and  × × × ×
(C) ,  and  × ×
× × Radioactive
(D) ,  and  sample
× ×

24. The radioactivity of a sample is X at a time t1 and Y at a time t2. If the mean life of the
specimen is , the number of atoms that have disintegrated in the time interval (t2 t1) is
(A) X t1  Y t2 (B) X  Y
(C) (X  Y)/ (D) (X  Y) 

25. A block of silver of mass 4 kg hanging from a string is immersed in a liquid of relative
density 0.72. If relative density of silver is 10, then tension in the string will be:
[Take g = 10 m/s2]
(A) 37.12 N (B) 42 N
(C) 73 N (D) 21 N

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26. An electron is accelerated from rest through potential difference V. Then electron
experience force f in uniform magnetic field on increasing the p.d. to v force became 2f.
(V’/V) is equal to
(A) 1/4 (B) 2
(C) 1/2 (D) 4

27. Magnetic flux linked with a stationary loop of resistance R varies with time T as  = at (T
t) amount of heat generated in loop during that time is:
aT a 2T 2
(A) (B)
3R 3R
2 2
a T a 2T 3
(C) (D)
R 3R

28. A large tank filled with water to height h is to be emptied through a hole at bottom ratio of
 n n 
time taken to value of water level from  n to  &  to 0  is
 2  2 
1
(A) 2 (B)
2
1
(C) 2  1 (D)
2 1

29. A conductivity rod of length = 1.0 m is moving with uniform × × × × × × × × × × ×

A × × × × × × × × × × ×
speed v = 20m/sec in field of 4.0T. C = 10 t is connected as
shown in figure B × × × × × × × × × × ×
× × × × × × × × × × ×
(A) qA = +80 C, q B = 80 C
(B) qA = 80 C, q B = +80 C
(C) qA = 40 C, q B = + 40 C
(D) qA = +40 C, q B = 40 C

30. What resistor should be connected in parallel with 20 B 10

20 resistor in ADC branch so that potential
difference between B & D may be zero. A C
(A) 20  20 5

(B) 10 D
(C) 5
(D) 15 14V

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Chemistry PART – II

SECTION – A
Single Correct Choice Type

This section contains 30 multiple choice questions. Each question has four choices (A), (B), (C)
and (D) out of which ONLY ONE is correct.

1. The Vander Waal’s equation of corresponding states for 1 mole of gas is:
[Where Pc = critical pressure, Tc = critical temperature, Vc = critical volume]

 P V T
  ,   and   
 Pc Vc Tc 
 3  3
(A)      3  1  8R (B)      3  1  8R
   
 3  
(C)      3  1  8R (D)      3  1  8R
   3

0 0 +2 +3
2. Given that EFe 2 |Fe  0.44V; E
Fe 3 |Fe 2
 0.77. If Fe , Fe and Fe solid are kept together
then
(A) Fe3  increase (B) Fe3  decrease
   
2 
(C) Fe2 / Fe 3 remain unchanged 
(D) Fe  decrease

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3
conc.

H SO

2 4

SO3H
Mark out the incorrect statement:
(A) Hexa deutarated benzene gets sulphonated slower than simple benzene
(B) Electrophile in the reaction is SO3
(C) Dilution of acid during reaction promotes reversal (desulphonating)
(D) Polysulphonation is hard to proceed

4. To maintain the pH of 7.4 for blood at normal condition which is 2M in H2CO3 (at
equilibrium). What volume of 5M NaHCO3 solution is required to mix with 10ml of blood?
K(H2CO3) = 7.8  10–7
(A) 78.36 ml (B) 102 ml
(C) 52.71 ml (D) 89.01 ml

5. For fairly concentrated solution of a weak electrolyte AxBy, having concentration ‘C’ the
degree of dissociation is given by:
(A)   k eq xy / c (B)   K eqc / xy
1
(C)   K eq / c  x  y  (D)   K eq / c x  y 1.x x .y y  x y
 

6. XeF6 dissolves in anhydrous HF to give a good conductivity solution which contains:

(A) H and XeF7 ion (B) HF2 and XeF5
(C) HXeF6 and F  ion (D) None

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7. When the concentration of A is 0.1 M, it decomposes to give ‘x’ by a first order process
–2 –1
with a rate constant of 6.93  10 min . The reactant ‘A’ in the presence of catalyst gives
‘Y’ by a second order mechanism with a rate constant of 0.2 min–1M–1. In order that half
life of both the process is 10 min. One should start the initial concentration of A as:
(A) 0.01M (B) 5.0 M
(C) 10.0 M (D) 0.5 M

8. According to Arrhenius equation, the rate constant (K) and energy of activation (E) of a
reaction are related by:
(A) A  KeE / RT (B) K  AeE / RT
(C) K  Ae  E / RT
(D) K   AeE / RT

9. HOOC COOH
C C BD  THF H O , DO , D O

3
 
2 2 2
 Pr oduct
H H

(A) COOH (B) COOD

H D H D
H OD H OD
COOH COOD

(C) COOH (D) COOH

H D H D
H OH DO H
COOH COOH

10. In alkaline medium, H2O2 acts as an oxidising agent in its reaction with:
(A) Cr2(SO4)3 (B) Ag2O
(C) K3[Fe(CN)6] (D) K2Cr2O7

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11. (a) KNO2 (p)

1
 pka  pka2 
pH =
2
(b)  (q) 1
PhNH2Cl pH =  pkw  pka  pkb 
2
(c) NaHCO3 (r) 1
pH =  pkw  pkb  logc 
2
(d) NH4CN (s) 1
pH =  pkw  pka  logc 
2
For abcd correct order of matching is
(A) pqrs (B) qprs
(C) rspq (D) srpq

13. Mark out the correct order of nucleophilicity in DMF:

(A) CH3  OH  NH2  F  I (B) CH3  OH  NH2  I F

I  F  OH  NH2  CH3
(C) (D) CH3  NH2  OH F  I

14. Mark out the right combination of cell and conditious for spontenety
(A) Pt H2  | HCl | Pt H2 ; P1  P2
P1 1M P2
2
(B) Zn | Zn  C1  || Zn2  C1  | Zn; C1  C2
(C) Pt  Cl2 , 1 atm  | Cl  C1  || Cl  C2  | Pt  Cl2 , 1 atm  ; C2  C1
(D) Pt H2 , 1 atm  | HCl  C1  || HCl  C2  | Pt H2  , 1 atm; C1  C2

15. The dissolution of Al(OH)3 by a solution of NaOH results in the formation of

(A) [Al(H2O)4(OH)4]+ (B) [Al(H2O)3(OH)3]
–
(C) [Al(H2O)2(OH)4] (D) [Al(H2O)6(OH)3]

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16. Which of the following is polar and aromatic?

(A) (B)
O

(C) (D) None

17. The species that undergo disproportion in alkaline medium:

(A) Cl2 (B) MnO4
(C) HNO3 (D) ClO 4

18. The compound, whose stereo chemical formula is written below, exhibits are geometrical
isomers & y optical isomers

H
C C OH

H CH2 CH2 C CH3

H
The value of x & Y are:
(A) 4 & 4 (B) 2 & 2
(C) 2 & 4 (D) 4 & 2

HBr
19. Z  CH  CH2   Z  CH2  CH2  Br
Which of the following is Z?
(A) —Cl (B) —SO3H
(C) —OCH3 (D) —CH3

20. Rank the following in the order of increasing value of the equilibrium constant for
hydration:
O O
O (H3C)3C C C(CH3) 3

(I) (II) (III)

(A) I < II < III (B) III < I < II
(C) II < I < III (D) II < III < I

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21. To a one liter solution of 0.1 N HCl, 0.025 mol of NH4Cl are added. Assuming 80%
dissociation of the solutes, the freezing point of the solution is (Kf = 1.85 deg/molal)
(A) –0.33°C (B) –0.85°C
(C) –0.23°C (D) –0.416°C

22. The suggested mechanism for reaction:

CHCl3  Cl2  CCl4  HCl


Cl2 
1
 2Cl
K
K2

K
CHCl3  Cl 
3
 HCl  CCl3  slow 
 
K4
CCl3  Cl   CCl4
The experimental mechanism for reaction:
1/2
(A) R = K3[CHCl3][Cl2] (B) R = K3K2[CHCl3][Cl2]
(C) R = K3K1[CHCl3][Cl] (D) R = K3(Keq)1/2[CHCl3][Cl2]1/2

23. A catalyst increases the rate of reaction at 27°C by 10 times. By what amount it is
decreasing Ea at the same temperature.
(A) 2 kJ/mol (B) 1.3 kJ/mol
(C) 5.7 kJ /mol (D) 21 kJ/mol

24. CH3
Li / NH O
3
   3
Zn, CH , COOH
 Pr oduct
3

(A) O (B) O O
H3C C (CH2) 4CHO H3C C (CH2) 3 C CH3

(C) OHC CH CHO (D) O

CH3 H3C C CH2 CHO

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25. CH3
1. BH  THF
CH3 
3
  Pr oduct  major 
2. HO O H, OH
18 18

CH3

(A) 18 (B) 18
HO CH3 HO CH3

CH3 CH3
H H
H CH3 H3C H
(C) H CH3 (D) HO CH3

CH3 CH3
H H
HO CH3 H3C H

26. Which of the following is the best synthesis of 2-chloro-4-nitro benzoic acid?
Cl

O2N COOH

(A) (i) Heat benzoic acid with HNO3 + H2SO4

(i) Cl2, FeCl3, 
(B) (i) Toluene + (HNO3 + H2SO4)
(ii) K2Cr2O7, H2O, H2SO4, 
(iii) Cl2, FeCl3
(C) (i) Heat chlorobenzene with (HNCO3 + H2SO4)
(ii) CH3Cl, AlCl3
(ii) K2Cr2O7, H2O, H2SO4, Heat
(D) (i) Toluene + (HNO3 + H2SO4)
(ii) Cl2, FeCl3, Heat
(iii) K2Cr2O7, H2O, H2SO4, heat

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27. When a certain volume, V1 ml of a gaseous hydrocarbon is exploded with excess of

oxygen and the product coded to room temperature, the observed concentration in
volume was 1.5 V1 ml. A further contraction of 2V1 ml was observed on exposure of
product to alkali. Formulae of hydrocarbon is
(A) C2H4 (B) C2H2
(C) C2H6 (D) C3H4

28. The de-broglie wavelength of an electron that has been accelerated through a potential
difference of 100V is:
(A) 0.12264nm (B) 1.226m
(C) 1.226nm (D) 0.012264nm

29. The correct order of increasing C  O bond length of CO, CO32 , CO2 is:
(A) CO23  CO2  CO (B) CO2  CO32  CO
(C) CO  CO3  CO2
2
(D) CO  CO2  CO32

30. Ph

CH2 CH2 CH CH CH2

H+
A

Product A is:
(A) (B) Ph

Ph CH2 CH3
(C) (D)
CH2 CH3

Ph Ph

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Mathematics PART – III

SECTION – A
Single Correct Choice Type

This section contains 30 multiple choice questions. Each question has four choices (A), (B), (C)
and (D) out of which ONLY ONE is correct.

1. For the system of equations

 
log10 x 3  x 2  log5 y 2

log10 y 3
 y2   log
5 z2

log10 z 3
 z2   log
5 x2
Which of following is/are true?
(A) there are infinite number of solution
(B) there is unique solution with x, y, z  Q
(C) there are exactly two solution with x, y, z  Q
(D) there is no solution

2. Three points A, B and C are considered on a parabola. The tangents to the parabola at
these points form triangle MNP (NP being tangent at A, PM at B and MN at C). If the line
through B and parallel to axis of parabola intersect AC at L then quadrilateral LMNP
(A) is always a parallelogram
(B) can never be parallelogram
(C) is parallelogram only when ordinates of A, B, C are in AP
(D) has exactly two sides parallel to each other, always

e tan x  e x  ln  sec x  tan x   x

3. Let f  x   be a continuous function at x = 0. The value of
tan x  x
f(0) equals
1 2
(A) (B)
2 3
3
(C) (D) 2
2

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3
4. Let , ,  are the roots of the equation , 8x + 1001x + 2008 = 0. The value of
3 3 3
( + ) + ( + ) + ( + ) is
(A) 251 (B) 751
(C) 735 (D) 753

5. Number of four digit numbers of the form N = abcd which satisfy following three condition
(i) 4000 < N < 6000 (ii) N is a multiple of 5 (iii) 3  b < c  6 is equal to
(A) 12 (B) 18
(C) 24 (D) 48

6. A coin that comes up head with probability p > 0 and tails with probability 1 – p > 0
independently on each flip, is flipped eight times. Suppose the probability of three heads
1
and five tails is equal to of the probability of five heads and three tails. The value of p
25
is
5 2
(A) (B)
6 3
1 3
(C) (D)
6 4

10
 x
7. The power of ‘x’ which has the greatest coefficient in the expansion of  1   is
 2
(A) 2 (B) 3
(C) 4 (D) 5

8. The first two terms of a G.P. add up to 12. The sum of the 3rd and 4th term is 48. If the
term of the G.P. are alternatively positive and negative, then the first term is
(A) –4 (B) –12
(C) 12 (D) 14
3 4
9. If a curve is represented parametrically by the equations x = 4t + 3 and y = 4 + 3t and
 d2 x 
 2
 dy  is a constant then the value of n, is
n
 dx 
 
 dy 
(A) 3 (B) 4
(C) 5 (D) 6

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 1
10. The minimum value of the function f  x   x 3/2  x 3/2  4  x   for all permissible real x,
 x
is
(A) –10 (B) –6
(C) –7 (D) –8

11. The complete set of value of ‘a’ for which there exists at least one line that is tangent to
the graph of the curve y = x 3 – ax at one point and normal to the graph at another point is
given by
 4  4 
(A) a   ,   (B) a    ,  
 3  3 
 4   4
(C) a   ,   (D) a   , 
3   3

sin A sinB sinC c b a

12. In ABC, if      , then the value of A is (All symbols have
c sinB c b ab ac bc
there usual meaning)
(A) 120º (B) 90º
(C) 60º (D) 30º

0 2b c 
13. If A  a b c  is orthogonal matrix, then |abc| is equal to
a b c 
1 1
(A) (B)
2 3
1
(C) (D) 1
6

 n 
  aii 
14. If A  aij  , where aij  i100  j100 , then lim  i1101  equals
nn n   n 
 
 
1 1
(A) (B)
50 101
2 3
(C) (D)
101 101

Space for rough work

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2 2 2
15. If a + b + c + ab + bc + ca  0  a, b, c  R, then value of the determinant
 a  b  2 2 a2  b 2 1
2
1 b  c  2  b  c2
2
is equal to
2 2 2
c a 1 c  a  2
2 2 2
(A) 65 (B) a + b + c + 31
2 2 2
(C) 4(a + b + c ) (D) 0

2
16. If z is a complex number satisfying the equation |z – (1 + i)|2 = 2 and w  , then the
z
locus traced by ‘w’ in the complex plane is
(A) x – y – 1 = 0 (B) x + y – 1 = 0
(C) x – y + 1 = 0 (D) x + y + 1 = 0
2 2
17. The angle between pair of tangents drawn to the curve 7x – 12y = 84 from M(1, 2) is
1 –1
(A) 2 tan1 (B) 2 tan 2
2
 1 1 –1
(C) 2  tan1  tan1  (D) 2 tan 3
 3 2 

18. Tangents PA and PB are drawn to the circle x2 + y2 = 8 from any arbitrary point p on the
line
x + y = 4. The locus of the midpoint of chord of contact AB is
(A) x 2 + y2 + 2x + 2y = 0 (B) x 2 + y2 – 2x – 2y = 0
2 2
(C) x + y – 2x + 2y = 0 (D) x 2 + y2 + 2x – 2y = 0

19. Let A(5, 12), B(–13 cos , 13 sin ) and C(13 sin , –13 cos ) are vertices of ABC
where   R. The locus of the orthocentre of ABC is
(A) x – y + 7 = 0 (B) x – y – 7 = 0
(C) x + y – 7 = 0 (D) x + y + 7 = 0

  1  x2 
20. The range of the f  x   cos1  
log4 x 
2
 sin1 
 4x 
 is equal to

     
(A)  0,   (B)  ,  
 2 2 
  2 2 2 
  
(C)  ,  (D)  
6 2 6 

Space for rough work

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n  3n 1
21. If lim n
 where n  N, then the number of integer(s) in the range
n 
n  x  2  n  3 n 1
3 n 3
‘x’ is
(A) 3 (B) 4
(C) 5 (D) infinite

1/m
 x  
7m
22. For any natural number m,  x 2m  x m 2x 6m  7xm  14 dx (where x > 0),
equals
m 1 m 1

(A)
 7x 7m
 2x 2m  14xm  m
c (B)
 2x 7m
 14x 2m  7xm  m
c
14  m  1 14  m  1
m 1 m 1

x
23. Let f: (0, )  R be a continuous function such that F  x    t f  t  dt . If F(x2) = x4 + x 5,
0
12
then  f r 2  is equal to
r 1
(A) 216 (B) 219
(C) 222 (D) 225

x 1
24. The area of the region bounded by the curve C : y  and the line L : y = 1, is
x2  1
1  
(A) 1  ln2  (B) ln 2   1
2 4 4
1  
(C) ln 2   1 (D) ln 2   1
2 4 2

dy 1
25. If
dx
 ey  x   where y(0) = 0, then y is expressed explicitly as

1
(A) ln 1  x 2
2
  (B) ln(1 + x 2)


(C) ln x  1  x 2  
(D) ln x  1  x 2 
Space for rough work

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     
26. If a, b, c are non-zero vectors then value of the scalar  a  b  a  b  a   b  equals
   2   2
 
(A)  a  b a  b (B) a 2 a  b
2  2    2
(C) b a  b  
(D) a  b a  b

1 t
27. The distance between the line x = 2 + t, y = 1 + t, z    and the plane
2 2
 ˆ
 
r i  2ˆj  6kˆ  10 , is
1 1
(A) (B)
6 41
1 9
(C) (D)
7 41

28. Let a relation R in the set N of natural numbers be defined as (x, y)  R if and only if
2 2
x – 4xy + 3y = 0  x, y  N. The relation R is
(A) reflexive (B) symmetric
(C) transitive (D) an equivalence relation

29. The top of a hill observed from the top and bottom of a building of height h is at angles of
 and  respectively. The height of the hill is
hcot  hcot 
(A) (B)
cot   cot  cot   cot 
h tan 
(C) (D) none of these
tan   tan 

30. Mean deviation about mean from the following data

xi : 3 9 17 23 27
f i : 8 10 12 9 5 is
(A) 7.15 (B) 7.09
(C) 8.05 (D) none of these

Space for rough work

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AITS-FT-I-PCM (Sol)-JEE (Main)/15

ANSWERS, HINTS & SOLUTIONS

FULL TEST – I
(Main)
ALL INDIA TEST SERIES

S. No. PHYSICS CHEMISTRY MATHEMATICS

1. A A B
2. B B A
3. A C C
4. A A D
5. A D C
6. A B A
7. A D B
8. A C B
9. B B C
10. D A A
11. D D C
12. D B B
13. A D C
14. C A C
15. D C A
16. C B A
17. A A C
18. C B B
19. A B A
20. A B D
21. B D C
22. B D C
23. B C B
24. D D C
25. A D C
26. D D A
27. D B D
28. C A A
29. A D B
30. A A B

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AITS-FT-I-PCM (Sol)-JEE (Main)/15
2

Physics PART – I

SECTION – A
1. Rate of change of speed is minimum at highest point. Since at y
highest position.
2 v cos
R
 v cos  
v 2 cos2 
g g g


x

x2
3 2
2. x  t  4t  5t  9 x1
O
dx
V
dt

 3t 2  8t  5 
5 x3
V = 0 at t = 1, sec.
3
x1 (t = 0) = 9m; x 2 (t = 1) = 11 m, x3 (t = 5/3)  10.85
d = (x2 x2) + (x2  x3) = 2.15 m

  
d| v | a.v 24
3.  at   =  4.8 ms2
dt |v| 5

4. Situation is same as the motion in vertical circle with initial Y

situation |a| = 2 equivalent to lowest point.
V  5 a  5  2  10  10 ms1
X
10 m
o
45

2
5. | PA  PB | d  g effective   10 4 N/m

d
45o
A B

6. Friction would change the diagram with time as tension in string would increase. So, (a) is correct
option.

10 10
7. VB cos 1 = 10; vB cos 2 = vc; vc = cos 2  m/sec.
cos 1 3

8. Net force would be equal to figure mg when contact force with hemispherical ball is zero.  =
cos1 (2/3)

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AITS-FT-I-PCM (Sol)-JEE (Main)/15

 
9. Angle between V & a is obtuse and centripetal acceleration is directed towards left.

10. I = 10 tan (90  )

dI
 10 cos ec 2  
dt
= 500 m/sec

sini 5
11. 
sinr 2 2
 4
sinr  5 
 

KP cos 45o
12. E1  
Y3 E1
2KP sin 45o E2
E2 
Y3
P cos 45o
o
P sin 45

 q
13. V0   0
0 40R

14. Initially capacitor will offer zero resistance at steady state it will behave like open circuit.

15. By KVL & KCL

dNy
16.   xNx   y Ny
dt

X  2T cos 
17. Y  2 tan 
X (R  d)

Y R

O

18. Torque about O

M2

(/2) (/2)  + k (/2) (/2)  = 12
K M2
 
2 612
6K
2 
M
M
T  2
6K


19. T= 2
g cos 

20. At steady state current through capacitor is zero and EA1RA EB1RB
potential difference across inductor is zero. Circuit will be R
E
like i  .
rA
2rA  rA  rB  R
R  rA  rB since R > 0.

 0i
21. B
2x
d  Bdv
r
 0iv 1 dx   0iv r2 
  n 
2 r2 x  2 r1 

 hc 
   f  ev 
22.  
 hc  f 
eV 
 2 3 
hc
f  ,  0  4
4

23. r particle will go undefected since it does not have ,  particle will deflect in opposite directions

24. X = N1
Y = N2
(x  y)
(N1  N2) =  (x  y)


25. T + V.g = 40 T V.g

So T = 37.12 N

Vg
26. V2  2V1;K 2  4K1

27. | E | (aT  2at)

H   E2 / R dt

28. dh
  h
dt

29. E  BV  80 V

30. Condition of balanced bridge.

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AITS-FT-I-PCM (Sol)-JEE (Main)/15

Chemistry PART – II

SECTION – A

1. Pc  P Vc  V Tc  T

a 8a
. 2
P .3b  V . T
27b 27Rb
 3
      3  1  8R
 

4. [H2CO3] = 2M
Vblood = 10 ml
[NaHCO3] = 5 M
Let V ml 8 NaHCO3 is mixed
 total V = (10 + V) ml
2  10
 [H2CO3]mix =
 v  10 
5 V
[NaHCO3] =
 v  10 
 pH = pKa + log
salt 
acid
5V
 7.4 = -log(7.8  10–7) + log
20
 V = 78.36 ml

8. For a chemical reaction with rise in temperature by 10, the rate of reaction is nearly doubled. The
temperature dependence of the rate of a chemical equation can be explained by Arrhenius
equation:
K  AeE / RT

18. Number of G.I. = 2 4  21  2

Number of OI = 2 4  21  2
n = number of stereocentres

28. P  2  9.11 10 31  KJ   1.602  017

 5.4  1024 kgms 1

h 6.6  10 31 JS
   0.1226
p 5.4  10 24 kgms1
1
29. Bond length 
% characater

30. Ph

CH2 CH2 CH CH CH2

H+
A

Ph
+
CH2 CH2 C CH CH3

CH2 CH2 C CH CH3

H3C CH2 Ph

Mathematics PART – III

SECTION – A

1.
log
For x > 1, let f  x   5 10
x 3
 x2 
The there equation are f(x) = y2, f(y) = z2 and f(z) = x2
 x3 – x2 = x2(x – 1) is increasing, f is an increasing function
If, say x > y then y < z and z < x, yielding a contradiction
Thus we can only have that x = y = z and so log10 x3  x 2  log5 x 2  
Let 2t  log5 x2  t > 0, x2 = 52t  x = 5t
 53t – 52t = 102t  5t – 1 = 4t  5t – 4t = 1
  5 t 
 5t – 4t = 4t     1 is an increasing function of t
 4  
 
 This equation has unique solution i.e. t = 1, x = y = z = 5

2.    
A at12 , 2at1 , B at 22 , 2at 2 , C at 23 , 2at 3   A
P(at1t2, a(t1 + t2)) P
N(at1t3, a(t1 + t3)) N
M(at2t3, a(t2 + t3)) B
Equation of AC y(t1 + t3) = 2x + 2at1t3
M
Line through B parallel to x-axis y = 2at2
C
L  (a(t1t2 + t2t3 – t1t3), 2at2)
  t t  t t  a  t1  2t 2  t 3  
Mid-point of L and N   a 1 2 2 3 , 
 2 2 
  t t  t t  a  t1  2t 2  t 3  
Mid-point of PM   a  1 2 2 3  ,  . Hence, parallelogram
  2  2 

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AITS-FT-I-PCM (Sol)-JEE (Main)/15

etan x  e x ln  sec x  tan x   x

3. For continuity k  f  0   lim f  x   lim  lim
x 0 x 0 tan x  x x 0  tan x  x  3
 3 x
 x 

= lim

e x etan x  x  1   3 lim ln  sec x  tan x   x
x 0 tan x  x x 0 x3
sec x  1 1 3
= 1  3 lim  1 
x0 3x 2 2 2

2008
4.  +  +  = 0,  = 
8
Now, let  +  = A,  +  = B,  +  = C
 A + B + C = 2( +  + ) = 0
3 3 3
Hence, A + B + C = 3ABC
2 3 3
 ( + ) + ( + ) + ( + ) = 3( + )( + )( + ) = –3

5. N a b c d
a can be chosen in 2 ways i.e. 4 or 5
b and c can be chosen in 6 ways i.e. 34, 35, 36, 45, 46, 56
d can be 0 or 5
Hence, total number = 2  6  2 = 24

8 1 8 5 3
6. C3  p3 1  p  
 C5 p 5 1  p 
25
p 5
 1 p   p 
5 6

10 10 10 10
Cr Cr 1 Cr Cr 1
7. r
 r 1
and r

2 2 2 2r 1
8 11
 r r=3
3 3

8. Let G.P. is a, ar, ar2 …..

Now, a + ar = 12 and ar2 + ar3 = 48
 r2 = 4  r = –2 and hence a = –12

dx
dx dt 12t 2 1
9.   
dy dy 12t 3 t
dt
d2 x d  dx  d  1  dt 1 1 1
Now,       2 
dy 2 dy  dy  dt  t  dy t 12t 3 12t 5
d2 x 1
 n5
dy2 12t 5  t
 n
 = constant  n = 5
 dx  1 12
  tn
 dy 

3 2
 1   1   1  
10. Write f  x    x    3 x    4  x    2
 x  x  x 
1
Let x   t (x > 0)
x
Let g(t) = t3 – 3t – 4t2 + 8
g(t) = (t – 3)(3t + 1) = 0  t = 3
g(3) > 0  g(3) = –10
3 2 3
11. Tangent at (t, t – at) is given by y = (3t – a)x – 2t
2
This line cuts the curve again at x = –2t, where slope of tangent is 12t – a
2 2
As required (12t – a)(3t – a) = –1
4 2 2
 36t – 15t a + a + 1 = 0
For positive root, a > 0 and discriminant  0
2 2
i.e. a > 0 and 225a – 144(a + 1) = 9(3a – 4)(3a + 4)  0
4 
 a   , 
3 

sin A a
12. We have 
C sinB bc
sinB sinC c b b sinB  c sinC c 2  b2
     
c b ab ac bc abc
2
b c 2
b  2R sinB   c  2R sinC 
 a 
b sinB  c sinC b sinB  c sinC

 a = 2R  A 
2

13. The roots of A are unit orthogonal vectors

 
 R1  R2  0  2b2 = c2
 
R2  R3  0  a2 – b2 – c2 = 0
 
R3  R3  1  a2 + b2 + c2 = 1
1 2 1 2 1 1
 a2 = ,b = ,c =  |abc| =
2 6 3 6

n
14.  aii  a11  a22  .....  ann
i 1
n
 aii 
2 1100  2100  3100  .....  n100   2 lim 1 100 100
r 
1
2
lim i1  lim   n  = 2 x100 dx 
n  n101 n n 101 n  n 101
r 1 a

15. We have a2 + b2 + c2 + ab + bc + ca  0
 (a + b)2 + (b + c)2 + (c + a)2  0
 a + b = 0, b + c = 0, c + a = 0
a=b=c
4 0 1
Now, 1 4 0  65
0 1 4

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2
16. |z – (1 + i)| = 2
 (x – 1)2 + (y – 1)2 = 2
2 2
 x + y = 2(x + y) ….. (1)
2 2 2  x  iy 
Let w = h + ik =   2 , so
z x  iy x  y2
2x 2y
h 2 2
,k  2
x y x  y2
2x  y
 hk   1 (from 1)
x2  y2
 Locus of w is x – y = 1


17. Point M lies on director circle of hyperbola, hence angle between the tangents is
2
1 1   1 1 
as tan1  tan1  tan1 1   2  tan1  tan1  
3 2 4  3 2 2

18. Let p(a, 4 – a)

The equation of chord of contact AB is A
p(a, 4 – a)
xa + y(4 – a) = 8 ….. (1)
Equation of chord AB with M(h, k) as midpoint
hx + ky = h2 + k2 ….. (2) M(h, k)
(1) and (2) are same
a 4a 8 x+y=4
   2
h k h  k2 B
 x2 + y2 – 2x – 2y = 0

19. Circum-centre is (0, 0)

 5  13 cos   13 sin  12  13 sin   13 cos  
Centroid (G) is  , 
 3 3 
Centroid divides line joining orthocentre and circum-centre in ration 2 : 1
5  13cos   13 sin  2  0  1 h 12  13 sin   13 cos  2  0  1 k
  and 
3 3 3 3
h5 k  12
 sin   cos   and sin   cos  
13 13
Hence, locus is x – y + 7 = 0

2 
20. Domain of f(x) is x = 1 so, f(1) = 0  sin1 
4 6

n  3n 1 1 1
21. lim n
 , so lim n

n 
n  x  2  n  3 n 1
3 n 3 n
 x 2 1 3
  3
 3  n
x2
1   1  –1 < x < 5
3
Hence, possible number of integers in the range = 5
1/m
 x  
7m 1
22. I=  x2m1  xm 1 2x7m  7x 2m  14xm dx

Put 2x7m  7x2m  14xm  t m

1 m 1 m 1/m t m 1
I
14 
t t dt 

14  m  1
c

x2
23.    t f  t  dt  x
F x2  4
 x5 ….. (1)
0
On differentiating 2x(x2)f(x2) = 4x3 + 5x4
5
 
 f x2  2  x
2
….. (2)
12 12
 5 
r 1
 
  f r 2    2  r   216
r 1  2 
12
Hence,  f  r 2   219
r 1

1 x
24. On solving C : y  and the line L : y = 1 gives x = 0, 1
1 x2
1 1
 x 1  1 1 
Required area =   2
2
2 4

 1 dx  ln x 2  1  tan1 x  x  ln2   1
0 x 1  0

dx dx
25.  ey  x   x  ey
dy dy
Hence, integrating factor = ey
1
Solution is xey  e2y  c
2

 y  ln x  x 2  1 
        
26. a  b   a  b  a   b   a  a  b  b  a  a  b  b  a  a  b  b
              3       3
     
= a2 b2 a  b  a2  b2 a  b  a  b a2  b2  a  b      a  b a2  b2  a  b 
   
 
  2 2   2    2
  a
=  a b b  a b  =  a b a b
   
1
z
x  2 y 1 2  t (Let)
27. Line is  
1 1 1

2
Plane : x + 2y + 6z – 10 = 0
 1
Vector along line : V  ˆi  ˆj  kˆ
2

Vector normal to the plane n  ˆi  2jˆ  6kˆ
 
V  n  0  line is parallel to plane
 1 9
Distance of  2, 1,   from the plane x + 2y + 6z – 10 = 0 is
 2  41
28. R is reflexive

29. Let AD be the building of height h and BP be the hill. P

hx x
Then tan   and tan  
y y x
hx hcot 
 tan    x 
x cot  cot   cot  D C
hcot  
 hx  A B
cot   cot  y

30. Mean deviation calculation about mean

xi fi fixi |xi – 15| fi|xi – 15|
3 8 24 12 96
9 10 90 6 60
17 12 204 2 24
23 9 207 8 72
27 5 135 12 60
N   fi  44  fi xi  660  fi xi  15  312
1 660
Mean = X 
N
  fi xi  
44
 15

1 312
Mean deviation = M.D. =  fi xi  15   7.09
N 44

Paper 1

Time Allotted: 3 Hours Maximum Marks: 240

 Pl ea s e r ea d t h e i n s t r u c t i o n s c a r ef u ll y . Yo u a r e a l l o t t ed 5 m i n u t es
s p ec i f i c a ll y f o r t h i s p u r p o s e.
ALL INDIA TEST SERIES

 Yo u a r e n o t a l l o wed t o l ea v e t h e E xa m i n at i o n Ha l l b ef o r e t h e en d o f
t h e t es t .

INSTRUCTIONS
A. General Instructions

1. Attempt ALL the questions. Answers have to be marked on the OMR sheets.
2. This question paper contains Three Parts.
3. Part-I is Physics, Part-II is Chemistry and Part-III is Mathematics.
4. Each part is further divided into three sections: Section-A, Section-B & Section-C
5. Rough spaces are provided for rough work inside the question paper. No additional sheets will be
provided for rough work.
6. Blank Papers, clip boards, log tables, slide rule, calculator, cellular phones, pagers and electronic
devices, in any form, are not allowed.

B. Filling of OMR Sheet

C. Marking Scheme For All Three Parts.

1. Section – A (01 – 04) contains 4 multiple choice questions which have only one correct answer.
Each question carries +3 marks for correct answer and – 1 mark for wrong answer.
Section – A (05 – 08) contains 4 multiple choice questions which have one or more than one
correct answer. Each question carries +4 marks for correct answer and – 1 mark for wrong
answer.
Section-A (09 to 11) contains 1 paragraph having 3 questions. Each question carries +4 marks
for correct answer and – 1 mark for wrong answer.

2. Section – B (01 – 02) contains 2 Matrix Match Type questions containing statements given in 2
columns. Statements in the first column have to be matched with statements in the second
column. Each question carries +8 marks for all correct answer. For each correct row +2 marks
will be awarded. There may be one or more than one correct matching. No marks will be given for
any wrong matching in any question. There is no negative marking.
3. Section-C (01 – 06) contains 6 Numerical based questions with answers as numerical value from
0 to 9 and each question carries +4 marks for correct answer and – 1 mark for wrong answer.

Name of the Candidate

Enrolment No.

Useful Data

PHYSICS
Acceleration due to gravity g = 10 m/s2

Planck constant h = 6.6 1034 J-s

Charge of electron e = 1.6  1019 C

Mass of electron me = 9.1  1031 kg

Permittivity of free space 0 = 8.85  1012 C2/N-m2

Density of water water = 103 kg/m3

Atmospheric pressure Pa = 105 N/m2

Gas constant R = 8.314 J K1 mol1

CHEMISTRY

Gas Constant R = 8.314 J K1 mol1

Atomic No: H=1, He = 2, Li=3, Be=4, B=5, C=6, N=7, O=8,

Physics PART – I

SECTION – A

(Only One Option Correct Type)

This section contains 4 multiple choice questions. Each question has four choices (A), (B), (C)
and (D) out of which ONLY ONE option is correct.

1. Three particles, two with mass m and one with mass M, might be arranged in any of the
four configurations shown below. Rank the configurations according to the magnitude of
the gravitational force on M, from least to greatest (i.e., in increasing order).
m m
d
d
d d d d d
M m m m M m M m M d m

(1) (2) (3) (4)

(A) 1, 2, 3, 4 (B) 2, 1, 3, 4
(C) 2, 1, 4, 3 (D) 2, 3, 4, 2

2. The diagram shows four pairs of large parallel conducting plates. The plates are
separated by equal distance in all cases. The value of the electric potential is given for
each plate. Rank the pairs according to the magnitude of the electric field between the
plates, least to greatest.
20V +70V +20V +70V 10V +90V +30V +90V

1 2 3 4
(A) 1, 2, 3, 4 (B) 4, 3, 2, 1
(C) 2, 3, 1, 4 (D) 2, 4, 1, 3

Space for Rough work

3. An observer ‘O’ at an accelerated trolley observes a projectile Q of mass 5 kg moving

under uniform gravitational field of earth acting towards negative z-axis. He observes the
 
momentum P of the projectile Q as P  15t ˆi  10t 2 ˆj  50tkˆ (kg m/s), where t denotes time
2
in second. Find the acceleration of the trolley at t = 2 sec. (Take g = 10 m/s )
(A) 6iˆ  8 ˆj  20kˆ m/s2 (B) 3iˆ  8ˆj m/s2
(C) 3iˆ  8ˆj m/s (D) 3iˆ  8ˆj m/s
2 2

4. A plane thick wall having uniform surface temperature along planes are 0°K and T0K (T0
= 300K) at x = 0 and x = x 0 respectively. Thermal conductivity varies linearly with
temperature
K = K0 (1 + T) The temperature of wall at the plane x = 2x0 is approximately: (where T is
in kelvin)
(A) 300K (B) 400K
(C) 425K (D) 450K

(One or More than One Options Correct Type)

This section contains 4 multiple choice questions. Each question has four choices (A), (B), (C)
and (D) out of which ONE or MORE THAN ONE are correct.

5. Two identical containers contains liquid of

same mass m but different densities as
shown in the figure given 1 > 2. P1 and P2 1 2
are pressures at the bottoms of the container- H1 H2
1 and container -2 respectively and F1 and F2
are the forces exerted by the liquid on the Container-1 Container-2
walls of container-1 and container -2
respectively. (Neglect the atmospheric
pressure)
(A) P1 > P2 (B) P1 < P2
(C) F1 > F2 (D) F1 < F2

Space for Rough work

6. Two point masses, each of mass M are kept at rest at points A and B respectively. A third
point m is released from infinity with a negligible speed, so that it can move along y-axis
under the influence of mutual gravitational attraction on it due to point masses kept at A
and B respectively as shown in the figure -1. Figure 2 represents the potential energy of
system (includes m, M at A and M at B) with position of m at y-axis. (Neglect any other
forces other than gravity) (given
2
Gm /d = 12 Joule(m = 6 kg). Choose the correct option(s)
y U

C m

A B
M = 2m M = 2m x y
d d U1

(A) Point mass m will perform periodic motion (B) U1 = 24 Joule
(C) Maximum speed of particle is 24 m/s (D) Maximum speed of particle is 4 m/s

7. Choose the correct statements:

(A) The de broglie wavelength of electron, which are accelerated by an electric potential
h
V is  
2me eV
(B) Wave velocity of the matter field is one half of the particle velocity
(C) Wave velocity of the matter field is equal to the particle velocity
(D) None of these

2
–bt 2 –2 abxt)
8. Equation of a travelling wave is y  5e(–ax when x & y is in meter and t is in
sec.
a = 25 m–2 and b = 9 sec–2
(A) Travelling wave is propagated along (+) x direction.
(B) Travelling wave is propagated along (–) x direction.
(C) Speed of wave is 3/5 m/sec.
(D) Maximum displacement of particle is 5 m.

Space for Rough work

Comprehension type (Only One Option Correct)

This section contains 1 paragraph, describing theory, experiments, data etc. Three questions
relate to the one paragraph. Each question has only one correct answer among the four given
options (A), (B), (C) and (D).

Paragraph for Questions 09 & 11

A body of mass 150 gm is dropped from point A, takes time t1 A

M
seconds to reach the point B at the ground. Once more, the same
body is dropped from point A and when body traverses half the H = 160 m

distance, a bullet of mass 50 gm, moving horizontally with speed m C

40 m/s hits the falling body at point C and sticks to it. The total time 40 m/s
of flight of the body in this case is t2 seconds and body strikes the
ground at point D. The air drag should be neglected [Take g = 10 B D
m/s2]. Read above passage carefully and answer the following
questions.

9. The ratio of t1/t2 is

(A) 2 : 3 (B) 2 2 : 3
(C) 3 2 : 4 (D) 2 : 4

10. Distance between B and D is

(A) 160 m (B) 80 m
(C) 40 m (D) 20 m

11. Speed of the body just before striking at point D is

(A) 5 26 m/s (B) 10 26 m/s
(C) 15 26 m/s (D) 20 26 m/s

Space for Rough work

SECTION - B
Matrix – Match Type
This section contains 2 questions. Each question contains p q r s t
statements given in two columns, which have to be matched. The
A p q r s t
statements in Column I are labelled A, B, C and D, while the
statements in Column II are labelled p, q, r, s and t. Any given B p q r s t
statement in Column I can have correct matching with ONE OR
MORE statement(s) in Column II. The appropriate bubbles C p q r s t
corresponding to the answers to these questions have to be
darkened as illustrated in the following example: D p q r s t

If the correct matches are A – p, s and t; B – q and r; C – p and q;

and D – s and t; then the correct darkening of bubbles will look like
the following:

1. An electron in a hydrogen atom makes a transition n1  n2, where n1 and n2 are the
principal quantum numbers of the two states. Assume Bohr model to be valid.
Column I Column II

(A) The electron emits an energy of 2.55 eV (p) n1 = 2, n2 = 1

(B) Time period of the electron in the initial state (q) n1 = 4, n2 = 2

is eight times that in the final state.

(C) Speed of electron become two times (r) n1 = 5, n2 = 3

(D) Radius of orbit of electron becomes 4.77A (s) n1 = 6, n2 = 3

(t) n1 = 8, n2 = 4

Space for Rough work

2. A triangular wave pulse on a string, is moving in the positive x-direction with speed ‘v’.
The tension in string is F and linear mass density is . At t = 0 shape of pulse is give by:
 0 if x  L

h(L  x) / L if L  x  0
y(x,0)  
h(L  x) / L if 0 x L
 0 if xL
Match the column I to that in column II
Column I Column II
(A) for  < x  vt  L (p) Slope of string is negative
Instantaneous power transmitted is non zero
(B) for vt  L  x  vt (q)
constants.
(C) for vt  x  vt + L (r) Particle velocity is positive
(D) for vt + L  x <  (s) Particle velocity is negative
(t) Slope of string is zero

SECTION – C
(One Integer Value Correct Type)

This section contains 6 questions. Each question, when worked out will result in one integer
from 0 to 9 (both inclusive).

1. Two spherical mirror, one of convex and other concave, each of same
radius of curvature R are arranged coaxially at a distance 2R from
each other. A small circle of radius a is drawn on the convex mirror
near the pole as shown in figure. The radii of 3rd image (taking the first 2R
reflection at the concave mirror, then the successive reflection at the
convex mirror and finally the third successive reflection at concave
M1 M2
 a 
mirror again) of the circle is   , then value of x is
 40  x 

Space for Rough work

2. A uniform rod AB of mass 5 kg and length 10m is hinged

at A. A block of mass 10 kg is hanging with a string and
other end of the string is attached to a middle point of the
rod passes over a smooth pulley as shown in the figure. 10 kg 5m
Find the value of cos(/2). A


10 m B

3. Pressure variation due a sonic wave propagating along positive x-direction is given by the
equation P(x) = P0sin  2( t  x) . One end of an open organ pipe is closed by a cap
and held at rest with its axis parallel to x-axis and the open end at the origin. The
minimum value of frictional force (in Newton) between the cap and the pipe so that pipe
will remain closed, is
(Given P0 = 2  103 N/m2 and cross sectional area of the tube is equal to 5  104 m2)

4. Two sound waves of frequencies 100 Hz and 102 Hz and having same amplitude ‘A’ are
interfering. A stationary detector, which can detect waves of amplitude greater than or
equal to A, In a given time interval of 12 seconds, find the total duration in which detector
is active.

5. ABC is an isosceles triangular sheet of negligible thickness, A

made of isotropic material. Mass of the sheet is m. Its moment of 90
a a
inertia about axes perpendicular to its plane and passing through
the points A and B are I1 and I2 respectively. [AB = AC = a and
BC = 2a]. Find I2/I1. B 2a C

6. If the heat dissipated in the circuit after a zero resistance A V

2CV 2
wire is connected across the pairs A & B, is , the C 2C
n
value of n is
B R

Space for Rough work

Chemistry PART – II

SECTION – A
(Only One Option Correct Type)
This section contains 4 multiple choice questions. Each question has four choices (A), (B), (C)
and (D) out of which ONLY ONE option is correct.

1. Calcium cyanamide is treated with ethyl chloride to give a product which on acidic
hydrolysis followed by heating gives another product (A). Product (A) is treated with
excess of CH3I followed by heating with moist Ag2O gives the product (B). Product (B) on
heating strongly gives (C) which is
(A) Ethanol (B) Ethene
(C) Isobutene (D) Tetra allyl ammonium iodide

2. Toluene on heating with succinic anhydride in the presence of anhydrous AlCl3 followed
by reduction with Zn(Hg)/conc. HCl gives a product which on heating with thionyl chloride
followed by heating with anhydrous AlCl3 in CS2 gives dominantly the ultimate product
(P). Hence ultimate product (P) is
(A) CH3 O (B) CH3

(C) CH3 (D) CH3

Space for Rough work

3. At cinema hall has equidistant rows 1m apart the length of cinema hall is 287 m and it
has 287 rows. From one side of cinema hall, laughing gas N2O is released and from the
other side, weeping gas (C6H5COCH2Cl) is released. In which rows, spectators will be
laughing and weeping simultaneously?
(A) 100th row from weeping gas side
(B) 100th row from laughing gas side
(C) 187th row from the weeping gas side
(D) 185th row from the laughing gas side

4. An electron is in excited state in hydrogen atom. The orbital, in which electron is present,
is shown the curve. The value of quantum numbers of electron in this exited state are

2 2
4r R

r
(A) n = 2,   0 , m = 0, s = +1/2 or –1/2 (B) n = 2,   1 , m = 0, s = +1/2 or –1/2
(C) n = 3,   0 , m = 0, s = +1/2 or –1/2 (D) n = 3,   0 , m = 0, s = +1/2 or –1/2

Space for Rough work

(One or More than One Options Correct Type)

This section contains 4 multiple choice questions. Each question has four choices (A), (B), (C)
and (D) out of which ONE or MORE THAN ONE are correct.

5. Identify the correct statement?

Cl
H2C CH C Cl
(A)
Cl
has greater ionization potential due to reverse hyper conjugation and –I effect.
NH2
N
(B)   HN C
N N NH2
is the order of basic strength

  
(C)
is the order of heat of combustion of the compounds.
NH3 NH3 NH3
CH3

(D)  

CH3
is the correct order of acidic strength

Space for Rough work

6. The diagram given below depicts the boiling point as the function of composition of the
mixture of CCl4 and SiCl4 which of the following statements about the diagram is/are
true?
o
TSiCl4
b
a

c
o
B.Pt. TCCl 4

0 1
CCl4
(A) The point a represent the composition of solution and the point b that of the vapour in
equilibrium.
(B) The proportion of CCl4 in the solution is smaller than that in the vapour in equilibrium.
(C) bc represents the condensation of the vapour.
(D) The point c represents the composition of solution and the point b that of the vapour
in equilibrium.

7. Assuming the bond direction to be Z-axis, which of the overlapping of atomic orbitals of
two atoms A and B will result in bonding; (I)s orbital of A and Px orbital of B, (II)s orbital of
A and Pz orbital of B, ((III) Py orbital of A and Pz orbital of B, (IV)s orbitals of A and B:
(A) I (B) II
(C) III (D) IV

8. Which of the following orders is are correct?

(A) CH3NH  CH3O   CH3 CH2  increa sing basicity 
(B) CH3 C  C  CH3 CH  CH  CH3 CH2CH2 Increa sing basicity 
(C) CH3 OCH3  CH3CH2 OH  CH3 CH2OH2 Increa sing acidity 

(D)  H 5C2 O C2 H5
O
(Solubility in water)

Space for Rough work

Comprehension type (Only One Option Correct)

Paragraph for Questions 09 & 11

M  P  B  C  H2O  D   E
Mixture Brown Colourless

Ca  OH CO2 H2O

C 
2
F  G
White ppt

KMnO4 / H
No decolourisation
FeSO 4 H2SO 4
 No brown ring observed
P 

Na2 SO 4
Filter FeSO 4 H2SO 4
Q  R     Brown ring observed

White ppt Solution

B  Yellow in both hot and cold condition

9. What is the composition of original mixture?

(A) ZnCO3 + Zn(NO3)2 (B) Na2CO3 + NaNO3
(C) PbCO3 + Pb(NO3)2 (D) CuCO3 + Cu(NO3)2

10. What is F?
(A) ZnSO4 (B) PbSO4
(C) CaSO4 (D) CaCO3

11. NO + E  Gas D

D + H2O  Product
(A) HNO3 (B) HNO2 + HNO3
(C) N2O5 + N2O3 (D) HNO2

Space for Rough work

If the correct matches are A – p, s and t; B – q and r; C – p and q;

and D – s and t; then the correct darkening of bubbles will look like
the following:

1. Match column I with column II such that reagent present in column II cannot be used for
distinguishing pair given in column I.
Column – I Column – II
(A) Phenol and benzene (p) Br2 water
(B) CH3CHO and (Me3C)2CO (q) Na
(C) HCOOH and CH3CHO (r) Na2CO3
(D) Phenol and ethanol (s) [Ag(NH3)2] OH
(t) NaHCO3

2. Match column (I) with column (II).

Column – I Column – II
(A) 3 (p) Orbital involvd is dx2  y2

CuCl 5 
(B) 2 (q) Orbital involved is dz2
Cu  NH3  
 4
(C) 3 (r) Orbital involved is dxy
FeF6 
(D) Ni  CO   (s) Tetrahedral
 4
(t) Square planner

Space for Rough work

SECTION – C
(One Integer Value Correct Type)

This section contains 6 questions. Each question, when worked out will result in one integer
from 0 to 9 (both inclusive).

1. How many tautomeric structures are possible for the compound 14 which have got a

NO
C group.
O
2. If a solid A B  having ZnS structure is heated so that the ions along two of the axis
passing through the face centre particles are lost and bivalent ion (Z) enters here to
maintain the electrical neutrality, so that the new formula unit becomes A xBy Zc, report
the value of x + y + c.
o o
3. The area and circumference of nth orbit of a H atom are 549 A 2 and 83.05 A
respectively. The maximum number of lines that can be produced when the electron falls
from this orbit to 2th orbit is
 pH  –6
4. Calculate the   at equilibrium point when a solution of 10 M acetic acid treated
 1.428 
with a solution of 10–6 NaOH, Ka for acid = 1.9  10–5 (Answer given in whole integer).

5. Dissolved O2 in water is determined by using a redox reaction:

2Mn2  aq   4OH  aq   O2  g 

2MnO2  s   2H2O   
How many equivalent of O2 will be required to react with 1 mole of Mn+2?

6. At high temperature, the following equilibrium exist in a mixture of carbon, oxygen and
their compound:


C  s   O2  g  
 CO2  g  Keq1  10
2



2C  s   O2  g 
 2CO  g Keq2  10
3



C  s   CO2  g  
 2CO  g  Keq3  X


2CO  g   O2  g  
 2CO2  g  Keq4  Y
What will be the ratio of X and Y?

Space for Rough work

Mathematics PART – III

SECTION – A

(Only One Option Correct Type)

This section contains 4 multiple choice questions. Each question has four choices (A), (B), (C)
and (D) out of which ONLY ONE option is correct.

1. If a, b, c are three distinct non-zero complex numbers such that |a| = |b| = |c| and the
equation
az2 + bz + c = 0 has a root whose modulus is 1 than
(A) b2 = ac (B) c2 = ab
2
(C) a = bc (D) none of these

2. If f: [0, 1]  [0, ) is differentiable function with decreasing first derivative, such that f(0) =
0 and f(x) > 0 then
1 1
dx f 1 dx f 1
(A)  2  (B)  2 
0 f  x  1
f ' 
1 0 f  x  1
f ' 1
1
dx tan1  f 1  1
dx f 1
(C)  f 2  x  1  (D)  f 2  x   1  f ' 1
0
f ' 1 0

3. Let P(x) is polynomial of degree n, having n real and distinct non zero roots. Q(x) is
another polynomial of degree m such that Q(x) = 0 has at least one real root not equal to
any of the root of P(x) = 0. If n, m > 2 then which of the following statements is correct
(I) : x2P(x) + 3xP(x) + P(x) = 0 has n real roots
(II) : P(x)Q(x) + P(x)Q(x) = 0 has at least one real root
(A) only (I) (B) only (II)
(C) both (I) and (II) (D) neither (I) nor (II)

4. Let a  [0, 4]. The maximum area bounded by the curves y = 1 – |x – 1| and y = |2x – a|
is
1 1
(A) (B)
3 4
1
(C) (D) none of these
2

Space for rough work

(One or More than One Options Correct Type)

This section contains 4 multiple choice questions. Each question has four choices (A), (B), (C)
and (D) out of which ONE or MORE THAN ONE are correct.

5. Let f, g and h are function differentiable on some open interval around 0 and satisfy the
1 4 1
equations f '  2f 2 gh  , f  0   1 , g'  fg2h  , g  0   1 and h'  3fgh2  , h  0   1 .
gh fh fg
The function f is given by
1/6 1/12
       
 sin  6x     sin  6x   
 4   4 
(A) f  x   21/12  (B) f  x   21/6 
 2    2  
 cos  6x  4    cos  6x  4  
     
1/12 1/4 1/4
(C) f(x) = (sec 12x) (sec 12x + tan 12x) (D) f(x) = (sec 12x) (sec 12x + tan
1/12
12x)

x2 y 2
6. Let PA and PB are two tangents drawn from point P, outside the ellipse   1 , (a >
a2 b2
b). If H1 and H2 are reflections of F1 and F2 respectively in the tangents PA and PB, then
which of the following option(s) are correct
(A) PH2F1 = PF2H1
(B) mid-points of line joining F1H1 lies on a circle of radius a
(C) PF2H1 and PF1H2 are similar
(D) PF2H1 and PF1H2 are congruent

7. Let f, g and h are quadratic polynomials such that f(x) f(x + 1) = g(h(x)). If 1, 1 are roots
of
f(x) = 0, 2, 2 are roots of g(x) = 0 and 3, 3 are roots of h(x) = 0 then which of the
following is/are true?
(A) 1 + 1 = 2 + 2 (B) 1 + 1 = 3 + 3
(C) 11 = 22 (D) 11 = 33

8. For any real number   1. let f() denote the real solution of the equation x(1 + ln x) = 
then
(A) f() is strictly increasing function in its domain
(B) lim f     

(C) if h(x) = x(1 + ln x), h: [1, )  [1, ) then f() is inverse function of h(x)
f    ln 
(D) lim 1
 

Space for rough work

Comprehension type (Only One Option Correct)

Paragraph for Questions 09 & 11

Read the following write up carefully and answer the following questions:
 
Dot product of two vectors A   a1, a2 , a3  and B   b1, b2 , b3  is defined as
     
A  B  a1b1  a2b2  a3b3 . Also, A  B  a1b1  a2b2 for A   a1, a2  and B   b1, b2  . Using these
definition of dot product and algebra it can easily be proved that following equations/in-equations
holds true
                
(I) A  B  B  A  
(II) A B  C  A  B  A  C    
(III) c A  B  cA  B  A  cB  
 
9. If the dot product of two vectors A   a1, a2  and B   b1, b2  is defined as
 
A  B  2a1b1  a2b2  a1b2  a2b1 . And other algebraic operations vectors remains as
usual then
(A) (II) holds true (B) (II) and (III) do not hold true
(C) all of (I), (II), (III) do not hold true (D) none of these
 
10. If the dot product of two vectors A   a1, a2  and B   b1, b2  is defined as
   
A  B  a1b1  a2b2  2 a2b1  2 a1b2 and modulus of A is defined as A  a1  a2 ,
then which of the following is always true for ai, bi > 0,  i = 1, 2, 3
      3  
(A) A  B  A  B (B) A  B  A  B
2
  3     3 
(C) A  B 
2
A B  (D) A  B  A
2
 
11. If however dot product of two vectors A   a1, a2 , a3  and B   b1, b2 , b3  in defined as
  3
A B   aibi , while all other algebraic operations remains as usual, then
i1
(A) (I) is not true (B) (II) is not true
(C) all (I), (II), (III) are true (D) (II) and (III) are true

Space for rough work

If the correct matches are A – p, s and t; B – q and r; C – p and q;

and D – s and t; then the correct darkening of bubbles will look like
the following:

1. Parabola y2 = 4ax is given. Three normals are drawn from any point (N) inside the
parabola. Tangents drawn at the intersection points (P, Q, R) of parabola and normals
form an equilateral triangle (ABC). Then match the following Column-I with Column-II
Column – I Column – II
(A) Circumcentre of ABC (p)  3a, 2a 3 
(B) Co-ordinates of N (q) (5a, 0)
(C) Co-ordinate of a point at a distance ‘a’ along
negative x-axis from one of the intersection point (P, (r) (0, 0)
Q, R) is
(D) Co-ordinates of intersection points (s) (–a, 0)
(t)  3a,  2a 3 
2. Match the following Column-I with Column-II
Column – I Column – II
(A) If f(x) + f(x) = –x g(x)f(x) and g(x) > 0 x  R then
(p) 0
(f(x))2 + [f(x)]2 has k as its maxima. Then k is
(B) If f2(x) + f(x) = g(x) and g(x) is always increasing then
the minimum value that f(x) can attain so that f(x) also (q) 1
increasing is k. Then (2(k + 1)) is
x
(C) If g(x) = f  x   f  t  dt is non-increasing function and
0
(r) 2
f(x) is continuous in R then f(1) is
(D) If f(x) is differentiable function and f(x) = x2 +
x
t k (s) –1
 e f  x  t  dt then f(–1) = 3 then k is
0
(t) 3

Space for rough work

SECTION – C
(One Integer Value Correct Type)

This section contains 6 questions. Each question, when worked out will result in one integer
from 0 to 9 (both inclusive).

 /2
rC x
r
sin xdx
0
1. If lim  /2
 L , then value of L – C is equal to _____ (C  R, L > 0)
r 
r
 x cos xdx
0

2 2
2. Let f: R  R, satisfy the equation (x – y)f(x + y) – (x + y)f(x – y) = 4xy(x – y )  x, y  R.
If f(1) = –2, then the absolute value of difference between maximum and minimum value
of f(x) on the interval x    3, 3  is _____

3. The number of all the pairs of natural numbers (x, y) which satisfy the equation
2x(xy – 2y – 3) = (x + y)(3x + y) is _____

4. The number of distinct solutions x  [0, ] which satisfy the equation 8 cos x cos 4x cos
k
5x = 1 is k then is equal to _____
2

1
5. n biased coins, with mth coin having probability of throwing head equal to
 2m  1
(m = 1, 2, ….., n), are tossed once. The probability of getting an odd numbers of heads, if
n
results for each coin are independent, is then  +  +  is equal to _____
n  

6. Consider the parabola y = ax – bx2. If the least positive value of a for which there exist
,   R – {0} such that both the point (, ) and (, ) lies on the given parabola is k
then [k] is equal to _____

Space for rough work

ANSWERS, HINTS & SOLUTIONS

FULL TEST– I
(PAPER-1)
ALL INDIA TEST SERIES

Q. No. PHYSICS CHEMISTRY MATHEMATICS

1. B B A

2. D B B

3. D A C
4. C C A

5. A, D A, B, D A, C

6. B, D A, B, C A, B, C, D

7. A, C B, D A, C, D

8. B, C, D B, C, D A, B, C, D

9. B C A
10. D D C

11. B A B
(A)  (q) (A  s, t) (A)  (s)
(B)  (p, q, s, t) (B  p, q, r, t) (B)  (q)
1.
(C)  (p, q, s, t) (C  p, s) (C)  (s)
(D)  (r, s) (D  q, s, t) (D)  (p, r, t)
(A)  (t) (A  q) (A)  (p)
(B)  (q, s) (B  p, t) (B)  (q)
2.
(C)  (p, q, r) (C  p, q) (C)  (p)
(D)  (t) (D  s) (D)  (r)
1. 1 4 3

2. 1 7 4

3. 2 3 3
4. 8 5 5

5. 2 2 4

6. 3 1 3

Physics PART – I

SECTION – A
    
dp ˆ  50kˆ N
3. Fnet   15iˆ  20tjˆ  50kˆ = Freal  Fpseudo and Freal  mg(k)
dt

dT
4. Q = KA
dx
x T
Q
A 
dx = K 0 (1  T)dT
0

0
T
Q  T2 
x = K0  T  
A  2 0

by solving
Q  T2 
x  K 0  T  
A  2 

Q  (300)2 
So, x0  K0  300  
A  2 
So, at x = 2x0 temperature T  425 K

A2
5. m = V  =  A1  A 2  A1A 2  H
F
m
H  mg
 A1  A 2  A1A 2  FB
 
A1
A2  H 
P1 = 1gH1
P2 = 2gH2  P1 > P2
F = mg  FB = mg  PA1
F1 = mg  A1P1,
F2 = mg  A1P2

2GMm GM2
6. U 
d2  y 2 2d

2GMm 2Gm2
 
d2  y2 2d

 2 1
= 2Gm2   
 d2  y2 d 
 
1
= U  mv 2
2

h E E P v 
7. v = f =      Particle 
p h P 2m  2 

a x  b t )2
8. y  5e –(

09-11. h = 160 m
2h 2  160
t10    4 2 sec
g 10
Velocity of system after collision in vertical direction, v 10 = 30 m/s;
Velocity of system after collision in horizontal direction, v 20  ¼  40 = 10 m/s
v120  900 + 2  10  80  v10  50 m/s
50  30
t 20  = 2 sec
10
t10  4 sec
SECTION - B
2. Shape of sting at t = 0 is given by. y
 0 if   x  vt  L
h(L  x  vt) / L if vt  L  x  vt

y(x,t)  
h(L  x  vt) / L if vt  x  vt  L
 0 if vt  L  x   x
O
y

x
O ((vt  L), 0) (vt, 0) ((vt + L), 0)

SECTION – C


2. 10gsin = 5g sin 
2

10 sin = 5 (2 sin /2 cos /2)
2
cos /2 = 1

4. y1 = A sin 1t Ar
y2 = A sin 2t cycle
2A0
 (   1 )   ( 2  1 ) 
yr = 2A cos  2 t  sin t A0
2 2 1s
1/6 s

1/3 s
1/4 s

3/4 s
2/3 s

  
5/6s

Resultant amplitude Ar = 2A|cos()t/2|

t  1
()   t = s
2 2 4
t  1
( )   t = s
2 3 6
In one cycle of intensity of 1/2s, the detector remain idle for
 1 1 1
2    s  sec
4 6 6

 1 1
 In ½ sec cycle, active time is    = 1/3 sec
 2 6
1/ 3 
 In 12 sec interval, active time is 12  = 8 sec
(1/ 2)

(2C)V 2
6. V1   V
3C 3
1 4 2 2CV 2
U1  C V 
2 9 9
V
V2 =
3
1 V2 V 2 CV 2
U2  (2C) C 
2 9 9 9
CV 2
Initial total energy =
3
1
Final total energy = (2C)V 2  CV 2
2
2 4
Charge flow through the battery  2CV  CV  CV
3 3
4
 Wbattery  CV 2
3
4CV 2 2CV 2 2
 Heat dissipated    CV 2
3 3 3

Chemistry PART – II

SECTION – A

1. H3C CH2 H3C CH 2 O

CH3 CH2 Cl N C N  
N C OH  CO
CaCN 2   H3 O 
2


H3C CH2 H3C CH 2

H3C CH2 CH3

N i 2CH3I
H  
 
 H3C CH2 N CH 2 CH3 OH

ii Moist Ag O
2
H3C CH 2 CH3
(A)
(B)
H
H2C CH2  H3C CH2 N CH3  H2O
(C)

2. O
CH3 CH3 O O
H2 C C CH3
O
H2 C C C CH2 CH2 C OH
O

Anhydride AlCl
 
3

(minor) O
O C CH2 CH2 C OH
 i  Zn  Hg /HCl
(major)
 ii  SOCl2 / 
 i  Zn  Hg / HCl
 i SOCl2 / 
Cl
CH3

O
Cl
Anhyd. AlCl3 /   CS2 
O

CH3

Anhyd. AlCl3 /   CS2 

CH3

O
(minor) O

(major)

3. N2O C8H7OCl

287 m
Molar weight of C8H7OCl = 8  12 + 7  1 + 16 + 35.5
= 154.5 g
Molar weight of N2O = 2  14 + 16 = 44 g
According to Graham’s Law of diffusion
rN2O MC8H7OCl 154.5
   3.5  1.87 : 1
rC8H7 OCl MN2O 44
1.87 th
dN2O   287 = 187 row from N2O side.
2.87
1.0 th
dC8H7OCl   287 = 100 row from weeping gas side.
2.87
Therefore, the spectator from the side of N2O in the 187th. Row will be laughing and weeping
simultaneously. Alternatively, the spectator from the side of weeping gas in 100th row will laugh
and weep.
4. The given orbital diagram has two radial nodes, i.e. n    1  2
 The orbital, in which electron is present is 3s.

Cl Cl

5. (A) H2C CH Cl  H3C CH C Cl

Cl Cl
These occurs decrease in electron density and hence the ionization potential increases.
N
(B) Basic strength  electron density over one N-atom has lesser electron density over
N

one N-atom than in

N
1
(C) Heat of combustion 
Stability
stability  No. of  –H
and if no. of –H of same stability  ease of hyperconjugation.
(D) Conjugate acid of weaker base is stronger acid. Basic strength
NH2 NH2 NH2
CH3
 

CH3
6. B. pt. of CCl4 < that of SiCl4. The vapour must always be richer in more volatile component.
Hence, the upper curve represents the composition of the vapour and the lower curve represents
the composition of the solution at corresponding B. pt.
8. The correct basic order for (a) option
CH3O > CH3NH > CH3CH2
(-ve) charge on most (-ve) charge on least
electronegative atom electronegative atom
and it can also explain by its conjugate rest all option are correct.
Solution for the Q. No. 9 to 11.
M – PbCO3
P – Pb(NO3)2
B – PbO
C – CO2
D – NO2
E – O2
F – CaCO3
G – Ca(HCO3)2
Q – PbSO4
R – NaNO3
SECTION –C
1. O O O O

14 14 14 14

N O N O N OH N O

2. ZB  4  2  2
ZA  4
ZC2  1
 A 4B 2Z 1
x+y+z=7

o
3. r2 = 5.49 A 2
o
2r = 83.05 A
o
then r = 13.22 A
0.529  n2
 r
Z
0.529  n2
 13.22 =
Z
n=5
Number of maximum lines possible = 3

CH3 COOH  NaOH  CH3 COONa  H2O

4. Int. No. of mole 106 V 10 6 V 0 0
6
at t  t sec 0 0 10 V
6
10 V
CH3COONa  2V
= 5  10–7< 10-6.
Kw  C
OH  
  Ka

104  107  5
=
1.9  105
= 2.63  10–8 < 10–6
so pOH = –log (2.63  10–8 + 10–7)
pH = 7
pH
 5
1.42

5. Mn2  MnO2  2e 

4e   O2  2H2O
1 Eq of Mn+2 = 1 Eq of O2
(n = 2) (n = 4)
½ mole of Mn+2 = 1 Eq. of O2
2 mole of Mn+2 = 4 Eq. of O2

6. 

2C  s   O2  g   Keq  103
 2CO  g 

CO2  g  

 C  s   O2  g  Keq  10

2



C  s   CO2  g  
 2CO  g  Keq  101  X



C  s   O2  g  
 2C  s   CO2  g Keq  1/ x  10
1



2CO  g   O2  g   1
 2CO2  g  Keq  10  Y
Hence X : Y = 1 : 1

Mathematics PART – III

SECTION – A
1. Let z1, z2 be the two roots with |z1| = 1
c c 1
 z 1z 2 =  z2  1
a a z1
 z1z1  z2 z2  1
b
 z1  z 2   and |b| = |a|
a
 |z1 + z2|2 = 1
 1 1
 (z1 + z2)  z1  z2   1   z1  z2     1
 z1 z2 
 (z1 + z2)2 = z1z2  b2 = ac

2. For x  [0, 1], f(x)  f(1) ( f(x) is decreasing)

f 'x f ' 1
 2  2
f x  1 f x  1
1 1
f ' x  dx
  f 2  x   1 dx  f ' 1  f 2  x   1
0 0
1
1 dx
 tan1 f  x   f ' 1  2
0 f x 1
0

1
dx tan1 f 1
 
 f 2  x   1 f ' 1
0

tan1 f 1 f 1

 tan–1  <    > 0  
f ' 1 f ' 1
1
dx f 1
  f 2  x   1  f ' 1
0

3.  P(x) = 0 has n real roots

 x P(x) = 0 has n + 1 real roots

d

dx
 xP  x   0
 x P(x) + P(x) = 0 has n real roots
 x(x P(x) + P(x)) = 0 has n + 1 real roots
d 2

dx
 
x P '  x   xP  x   0 has n real roots

i.e. x2 P(x) + 3x P(x) + P(x) = 0 has n real roots  (I) is true

Let one of the roots of P(x) = 0 is , and one root of Q(x) = 0 is  as given,   
 P(x) Q(x) + P(x) Q(x) = 0
d

dx
P  x  Q'  x    0
Consider f(x) = P(x) Q(x)
f() = 0, f() = 0
d
Hence, by Rolles theorem
dx
P  x  Q'  x    0 has at least one real root lying between  and 

4. Case-I:
a a
If a  [0, 1], the curves intersect at  ,  and (a, a). The bounded region is contained in the
3 3
 1  1
triangle with vertices (0, 0),  , 0  and (1, 1) with area =
2  4
1
Hence, area can not exceed
4
Case-II:
a a a 
If a  [1, 3]. In this case the bounded region is a quadrilateral with four vertices  ,  ,  , 0  ,
3 3 2 
a2 4a 1 a a 1  4  a  a
 3 , 3  and (1, 1). In this case area bounded = 1  2  3  2  2  3  2  2 
    
2
1  a  2 1
=  
3 6 3
Case-III: If a  [3, 4]. This case is symmetric with case-I

2 (2, 0) (2, 0)
a  1 1 a  a 
 , 0  , 0  , 0
2  2  2 

5. Multiplying the first DE by gh, the second by fh and the third by fg, and adding the equations
gives
(fgh) = 6(fgh)2 + 6
Let f(x) g(x) h(x) = k(x)
We have k(x) = 6(k(x))2 + 6
Integrating and using k(0) = 1 gives
 
k(x) = tan  6x  
 4

f ' x  1
Now 1st DE   2k  x  
f x k  x
f ' x    
  2 tan  6x    cot  6x  
f x  4   4 
Integrating and using f(0) = 1 gives option A and C

6. By reflection property of ellipse PAF2 = QAF1 P H1

 H1 is reflection of F1  QAH1 = QAF1 A
Now PH2 = PF2, PH1 = PF1 H2 B Q
and F2H1 = F2A + AF1 = 2a
and F1H2 = F1B + BH2 = F1B + F2B = 2a F2 F1
Hence, PF2H1 and PF1H2 are congruent as all three
sides are equal
This leads to all the options

7. Let f(x) = a(x – 1)(x – 1)

 f(x) f(x + 1) = a2(x – 1)(x – 1 + 1)(x – 1)(x – 1 + 1)
= a2(x2 + x – 1x – 1x + 11 – 1)(x2 + x – 1x – 1x + 11 – 1)
 f(x) f(x + 1) = a2[(x2 – (1 + 1 – 1) x + 11) – 1]  [(x2 – (1 + 1 – 1) x + 11) – 1]
Hence, g(x) = a2(x – 1)(x – 1) and h(x) = x2 – (1 + 1 – 1) + 11

8. Consider h(x) = x(1 + ln x), h: [1, )  [1, )

Clearly h(x) to increasing function in its domain
 f() denotes the solution of x(1 + ln x) = 
 f: [1, )  [1, ), f is inverse function of h(x)
As lim h       lim f     
 

1 1 1
f '     f '    
h'  x  x  f 2  ln x xf 2  ln  f    
 
f    ln  f    ln  ln 
Now consider L  lim  lim = lim
  

f    1  ln  f       1  ln  f    

1/  f  2  ln f   
By L’Hospitals Rule L  lim  lim  lim
 f '    f     f '     1  ln f   
 2 
 ln f   1 
 L  lim 
   1
  1 
 ln f     1 
 

9. (I) holds true obviously

  
 
For (II): A B  C   a1, a2 b1  c 1, b2  c 2 
= 2a1(b1 + c1) + a2(b2 + c2) + a1(b2 + c2) + a2(b1 + c1)
= (2a1b1 + a2b2 + a1b2 + a2b1) + (2a1c1 + a2c2 + a1c2 + a2c1)
   
= A B  A C
Similarly it can be checked that (III) also holds true
 
10. A  B  a1b1  a2b2  2 a2b1  2 a1b2

   a  b1   a2  b2  3
 A B   1     a2  b1    a1  b2     a1  a2    b1  b2  
 2   2  2
  3  
 A B 
2
A B 
   
11. A  B  a1b1  a2b2  a3b3  B  A
Hence, (I) holds true
  
 
A B  C  a1  b1  c1   a2  b2  c 2   a3  b3  c 3   a1b1  a2b2  a3b3    a1c1  a2c 2  a3c 3 
 a1b1  a2b2  a3b3  a1c1  a2c 2  a3 c 3
 (II) Don’t hold true
Similarly (III) can be checked

SECTION – B
3
1. Normal y + tx = at + 2at P(t1)
 at3 + (2a – h)t – k = 0
N(h, k)
 t1 + t2 + t3 = 0  t3 = –(t1 + t2) ….. (1)
A
2a  h
 t1t2 + t2t3 + t3t1 = ….. (2)
a
k Q(t2) R(t3)
Also, t1t 2 t 3  ..... (3) (–a, 0)
a B
O
 ABC is equilateral
So, G  H  O C
a a 
G    t1t 2  t 2 t 3  t 3 t1 ,  t1  t 2  t 2  t 3  t 3  t1  
3 3 
a 
G    t1t 2  t 2 t 3  t 3 t1 , 0  ..... (4)
 3 
 a 
H   a,  t1  t 2  t 3  t1t 2 t 3   ..... (5)
 3 
Comparing (4) and (5), t1t2t3 = 0 and from (3), k = 0
2a  h a
From equation (2), we get   a
a 3
2a – h = –3a
h = 5a
 O  circumcentre of ABC = (–a, 0)
Image of O will lie on circle
 N  (5a, 0)
 One of the intersection point lie on vertex ( k = 0)
So, t2 = 0
 at3 + (2a – 5a)t = 0
 t3 – 3t = 0  t(t2 – 3) = 0
t = 0,  3
 
 Intersection points  P  3a, 2a 3 , Q  (0, 0), R  3a,  2a 3  

2. (A) (f(x))2 + [f(x)]2 = h(x) h(x)

h(x) = 2f(x) f(x) + 2f(x) f(x)
= 2f(x) (f(x) + f(x))
= 2f(x) {–x g(x) f(x)}
2 O
= –2x(f(x)) g(x)
So, h(x) < 0 of x > 0 and h(x) > 0 for x < 0
Then, x = 0 is a maxima
(B) f 2(x) + f(x) = g(x)
2f(x) f(x) + f(x) = g(x) and g(x) = f(x) {2f(x) + 1}
So, g(x)  0 as g(x) is increasing and for f(x)  0
2f(x) + 1  0
1
Hence, f  x   
2
x 
2 G(x)
(C) Let G  x     f  t  dt  ….. (1) O
 0 
x
G'  x 
G'  x   2f  x   f  t  dt and  g x
0
2
and G(0) = 0 = g(0)
Also, g(x) is non-increasing
Hence, g(x)  0 x  (–, 0), g(x)  0 x  [0, )
i.e., G(x)  0 for x  0, G(x)  0 for x  0 and G(0) = 0
But G(x) (from equation (1))  0
Hence, G(x) = 0 x  R
x

 f  t  dt  0 xR
0
i.e., f(x) = 0 x  R
Hence, f(1) = 0
x
(D) f  x   x 2   e t f  x  t  dt ….. (1)
0
x
f  x   x 2  e x  et f  x  x  t  dt
0
x
f  x   x 2  e x  et f  t  dt
0
x
f '  x   2x  e x e x f  x    e  x  et f f  t  dt
0
x
f '  x   2x  f  x    e   f  t  dt = 2x + f(x) + x2 – f(x)
 x t

0
2
f(x) = 2x + x
2 x3
f(x) = x +  c and from equation (1) f(0) = 0
3
Hence, c = 0
x3
i.e., f(x) = x 2 
3
1 2 k 2
f  1  1   ,  k=2
3 3 3 3

SECTION – C
 /2
1. Let f  r    xr sin xdx
0
 /2  /2
x r 1  /2 xr 1
Now,  xr cos xdx  cos x 0   sin xdx
0
r 1 0
r 1
 /2
f  r  1
  xr cos xdx 
0  r  1
r C f  r  r  1
So, we have lim L
r  f  r  1
r 1

 /2  
2
Now, consider f  r    xr dx   
0
r 1
2x  
Also as sin x 

x  0, 2 
 
r 1

 /2
2  
2
f  r    xr 1dx   
0
 r 2
r 1
r 2 r
  r  f r  
r2  r 1
r 1
2
r   f r 
f r   2  r  1 2
Hence, lim  lim  =
r  f  r  1 r  r  2
 r 
2
r  1   f  r  1

f r  2
Now, lim r C  r  1  L  lim r C  r  1  L
r  f  r  1 r  

2
For positive L we should have C = –1 and L =


 
2. Let x + y =  and x – y =   x  and y 
2 2
2 2 f  f  
Given equation  f() – f() = ( –  )   2   2
 
f x
Hence,  x 2  k (k is some real constant)
x
f(x) = x3 + kx
f(1) = –2  k = –3, f(x) = x3 – 3x
f(x) = 3(x – 1)(x + 1) on the interval x    3, 3  extreme value are –2 and 2

3. The given equation can be re-written as quadratic in y

y2 + (8x – 2x2)y + (3x2 + 6x) = 0
Its discriminant to equal to

= (64x2 – 32x3 + 4x4) – 4(3x2 + 6x) = 4x(x3 – 8x2 + 13x – 6)

2
= 4x(x – 6)(x – 1)
For solutions in integers this discriminant should by perfect square this happens if and only if
z2 = x(x – 6) = (x – 3)2 – 9
9 = (x – 3)2 – z2 = (x + z – 3)(x – z – 3) for name integer z
Checking all factorisations 9 = (–9)  (–1) = (–3)  (–3) = (–1)  (–9) = 9  1 = 3  3 = 1  9
This gives all possible pairs (x, y) as (x, y) = (–2, 0), (–2, 24), (0, 0), (8, 4), (8, 60), (6, 12)
Therefore exactly three pairs of natural number (x, y) are possible


4. As no multiple of  nor any odd multiple of satisfies the equation we can multiply both sides by
4
sin x cos 2x, which gives
sin 8x cos 5x = 2 sin 4x cos 4x cos 5x = 4 sin 2x cos 2x cos 4x sin 5x
= (sin x sin 2x)(8 cos x cos 4x cos 5x) = sin x cos 2x
 sin 13x + sin 3x = sin 3x – sin x
 sin 13x = sin (–x)
 13x = n + (–1)n (–x)
 5 7 11  2 3  4 5 6
Which gives 10 solutions , , , , , , , , ,
12 12 12 12 7 7 7 7 7 7

5. Let Hi  the event of getting a head on ith coin

H  getting odd number of heads on throwing n coins once
Coin Number: 1 2 3 4 …….. m ….. n
1 1 1 1 1 1
P(Hi) ….. ….. (1)
3 5 7 9  2m  1  2n  1
2 4 6 8 2m 2n
 
P Hi
3 5 7 9
……..
 2m  1
…..
 2n  1
….. (2)

Now, P(H) = sum of series which has each term consisting of product of an odd number of terms
from (1) and even number of terms from (2) such that total number of factors in each term is n
 2 1  4 1  6 1   2n 1  1
Consider the product        .....     2n  1
 3 3  5 5  7 7   2n  1 2n  1 
1
 
LHS = P H  P H 
2n  1
Also, P H  P H  1
n
 P H  
2n  1
  = 1,  = 2,  = 1,  +  +  = 4

6. (, ) and (, ) will lie on some line y = –x + 

Solving line and parabola – (–x + ) = ax – bx2 (, )
 bx2 – (1 + a)x +  = 0 ….. (1)
Again, y = a( – y) – b( – y)2
2 2
 by + y(1 + a – 2b) + (b – a) = 0 ….. (2) (, )
Both (1) and (2) has same roots which are  and 
1 a
Hence, 1 + a – 2b = –(1 + a)   
b
Now, for ,  to exist disc of (1) > 0
2
 (1 + a) – 4b > 0
 (a – 3)(a + 1) > 0
 a  (–, –1)  (3, )

Paper 2

Time Allotted: 3 Hours Maximum Marks: 240

 Pl ea s e r ea d t h e i n s t r u c t i o n s c a r ef u ll y . Yo u a r e a l l o t t ed 5 m i n u t es
s p ec i f i c a ll y f o r t h i s p u r p o s e.
ALL INDIA TEST SERIES

 Yo u a r e n o t a l l o wed t o l ea v e t h e E xa m i n at i o n Ha l l b ef o r e t h e en d o f
t h e t es t .

INSTRUCTIONS
A. General Instructions

B. Filling of OMR Sheet

C. Marking Scheme For All Three Parts.

Name of the Candidate

Enrolment No.

Useful Data

PHYSICS
Acceleration due to gravity g = 10 m/s2

Planck constant h = 6.6 1034 J-s

Charge of electron e = 1.6  1019 C

Mass of electron me = 9.1  1031 kg

Permittivity of free space 0 = 8.85  1012 C2/N-m2

Density of water water = 103 kg/m3

Atmospheric pressure Pa = 105 N/m2

Gas constant R = 8.314 J K1 mol1

CHEMISTRY

Gas Constant R = 8.314 J K1 mol1

Atomic No: H=1, He = 2, Li=3, Be=4, B=5, C=6, N=7, O=8,

Physics PART – I

SECTION – A

(Only One Option Correct Type)

This section contains 4 multiple choice questions. Each question has four choices (A), (B), (C)
and (D) out of which ONLY ONE option is correct.

1. A conducting wire is bent in the form of a parabola y2 = 2x y

carrying a current i = 2A as shown in the figure. The wire
 (2m, 0)
is placed in a uniform magnetic field B  –4Kˆ . The
magnetic force on the wire is x
(A) –16iˆ (B) 32iˆ
(C) –32iˆ (D) 16iˆ

2. If M = mass, L = length, T = time and I = electric current, then the dimensional formula of
resistance R will be given by
1 2 1 2 2
(A) [R] = [M L T3I2] (B) [R] = [M L T3I ]
1 2 3 2 1 2 3 2
(C) [R] = [M L T I ] (D) [R] = [M L T I ]

3. A block of mass m 1 = 3 kg is connected with ideal u = 4 m/s

spring of spring constant k = 30000 N/m and kept at k
horizontal frictionless surface as shown in the figure. m2 m1
The block m 2 = 1 kg is moving with velocity u = 4
m/s towards block m 1. v1 and v2 are the final speed
of the block m 1 and m 2 respectively.
Choose the correct statement(s)
(A) v 1 = v2 = 1m/s
(B) The maximum compression in the spring is 2 cm
(C) The maximum compression in the spring is 4 cm
(D) none

Space for Rough work

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4 The diagram shows four systems of charged particles. Each system contains two
charged particles at point A and point B as shown in the figure. A third positive point
charge q0 is kept at point P in each system. The direction of electrostatic force
experienced by the charge q0 will be
A A
Q +Q
30 30
30 P 30 P
q0 q0
+Q +Q
B B
System -1; AP = BP = r and AB is a System -2; AP = BP = r and AB is a
straight line straight line
P P
q0 q0

30 30
30 30
Q +Q Q Q
A B A B

System -3; AP = BP = r and AB is a System -4; AP = BP = r and AB is a

circular arc of radius r and centre at O circular arc of radius r and centre at O
System -1 System-2 System-3 System-4

(A)    

(B)    

(D)    

Space for Rough work

(One or More than One Options Correct Type)

This section contains 4 multiple choice questions. Each question has four choices (A), (B), (C)
and (D) out of which ONE or MORE THAN ONE are correct.

5. In the given figure a wire loop has been bent so that it has y
three segments. Segment AB is a quater circle in xy plane,
segment BC is B
L–shaped wire in yz-plane, and segment CA is straight line A
in xz plane. Here are three choices for a magnetic field (0, 0, a) (a, 0, 0) x
through the loop. C

(1) B1  3iˆ  7jˆ  5tkˆ z

(2) B2  5t ˆi  4jˆ  15kˆ

(3) B3  2iˆ  5tjˆ  12kˆ
  
If the induced current in the loop due to B1,B2 and B3 are i1,i2 and i3 respectively then:
(A) i1  i2 (B) i1  i3
(C) i3  i2  i1 (D) i1  i2  i3

6. The surface tension phenomenon is the result of the tendency of a system to keep total
(A) density of the system is minimum (B) potential energy minimum
(C) volume of the system is minimum (D) surface area of the system is minimum

7. Radiation coming from a radioactive substance is passed

through a cloud-chamber in which strong constant
magnetic field is switched on in the direction shown in
figure. Figure shows path followed by few particles
coming from radioactive substance (shown by dotted
lines) Neglect the interaction between different particles.
Tick the correct statements :
(A) Path 1 can be followed by -radiation only
(B) Path 3 can be followed by -radiation only
(C) Path 4 can be followed by -radiation only
(D) Path 4 can be followed by -radiation

Space for Rough work

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8. A particle of mass ‘m’ starts from origin and move in x-y plane velocity of particle is given
by

v  k1i  k 2 1  y ˆj
where k1, k2 &  are constant and ‘y’ is y-coordinate of position of particle. Choose the
correct statements :
m  k 22
(A) Average force acting on particle
2
(B) Force acting on particle decreases first then increases
(C) Path of particle will be hyperbolic.
(D) Radius of curvature of the path followed by particle will decrease first then increase.

Comprehension type (Only One Option Correct)

Paragraph for Questions 09 & 11

Figure shows a tube whose two limbs are

vertical and one is horizontal of sufficient
length. An ideal gas is enclosed between
two massless non conducting movable
piston of cross-sectional area 10 cm 2. A h
1
liquid of density 1 = 1000 kg/m 3 and 2 = gas h2
3
2000 kg/m are poured in the vertical limbs
of height of h1 = 2m and h 2. Now gas is
slowly heated. Consider the wall of tube is 1 2
piston
insulated.
9. Ratio of h1 and h2 at any instant will be
1 2
(A) (B)
2 1
3
(C) (D) not constant
2

10. If due to heating separation between piston is increased by 3m then displacement of

piston on left will be
(A) 2m (B) 1m
(C) 2.5 m (D) 1.5 m

11. Find the work done by gas

(A) 70 Joule (B) 390 Joule
(C) 90 Joule (D) 350 Joule

Space for Rough work

SECTION - B
Matrix – Match Type
This section contains 2 questions. Each question contains p q r s t
statements given in two columns, which have to be matched. The
A p q r s t
statements in Column I are labelled A, B, C and D, while the
statements in Column II are labelled p, q, r, s and t. Any given p q r s t
B
statement in Column I can have correct matching with ONE OR
MORE statement(s) in Column II. The appropriate bubbles C p q r s t
corresponding to the answers to these questions have to be
darkened as illustrated in the following example: D p q r s t

If the correct matches are A – p, s and t; B – q and r; C – p and q;

and D – s and t; then the correct darkening of bubbles will look like
the following:

1. Terminal A is connected to positive terminal of battery and B is connected to negative

terminal of plate and it send same charge Q in all four combination. Each plate have same
area and separation between two consecutive plate is same.
Column I Column II
1
2 Top surface of plate 1 have
(A) (p)
A 3 zero charge
4 B
1
2 Upper surface of plate 4
(B) A (q)
3 have Q charge
B
4
1
2 Same charge Appear on
(C) A (r)
3 both surface of plate 2
4 B
1
2 Bottom surface of plate 4
(D) (s)
A 3
3
have zero charge
B
4
(t) Total charge on plate 2 is Q

Space for Rough work

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2. Column I shows four physical situation and relates these with column II.
Column I Column II
(A) (p) Mechanical energy is
V1 V2
k conserved
M1 M2

=0
Blocks are given velocities as shown on
smooth horizontal surface
(B) Q1 Q2 (q) Momentum is conserved
V1 V2
Two positive charge particles are given
velocities towards each other in a gravity
free space as shown.
(C) Q1 (r) K.E. of centre of mass will
V1 remain constant
L
Q2
V2
Two positive charge particles are given
velocities along different line towards each
other in a gravity free space as shown.
(D) 2m (s) At maximum or minimum
m separation K.E. in centre of
O mass frame will be zero or
v minimum.
R
2R

Beads are free to move on a smooth circular (t) Angular momentum is

rings placed on a smooth horizontal table. conserved about any point in
inertial frame

Space for Rough work

SECTION – C
(One Integer Value Correct Type)

This section contains 6 questions. Each question, when worked out will result in one integer
from 0 to 9 (both inclusive).

1. In the given circuit the resistance of each sides and diagonals is E

 35 
R   . The equivalent resistance across the points A and B is
 2  D C
(given ABCD is a square)

A B

2. If the sound heard by observer, whose equation is given as

y = 8sin10t cos 200t at x = 0
The number of beat frequency heard by observer is 2k, then the value of k is

3. Point B is the image of point A due to reflection from mirror (which is either spherical
concave mirror, or spherical convex mirror or plane mirror), whose principal axis is x-axis.
y

O x

One square in x-axis represents 5 cm and one square on y-axis represents 1mm.
The co-ordinate of centre of curvature of mirror is (x1, 0) and co ordinate of pole of mirror
is (x2, 0) respectively. If |x1| + |x 2| = 5K cm, then find the value of K.

Space for Rough work

4. Figure shows a circuit containing three resistor X1,

X2 & X3 having resistance R each, an inductor,
L
capacitor and an emf source having inductance
‘L’, capacitance C & emf  respectively (Given C
L
R ). The switch is first connected to
C
position-1. When charge on capacitor becomes
half of it maximum possible value then switch S is S
connected to position-2. Current in resistance X3 X3
just after shifting the switch from position-1 to
position-2, is
(given L = 5mH, C = 2F, R = 10 ,  = 5V)

5. In the figure shown a block of mass m is attached in a light spring

of spring constant K and an identical spring hangs from ceiling.
Initially lower spring is compressed in a state with compression
mg
3mg
equal to from natural length of spring when block is K Equilibrium
K
released it strikes upper spring and sticks to it. Amplitude of
oscillation (in cm) is given mg = 10 N and K  100 7 N/m

6. Figure shows wave fronts coming

10
form a source at equal interval of
3
milli second. Frequency of wave
emitted by source is 300 Hz. XX’ is
line along common diameter of wave
fronts passing through ‘A’. Distance A
between two consecutive wave fronts
along line XX’ is 0.9 m and 1.3 m in
the right and left of point A
respectively. At some instant source
is at point A. Detector is placed at
point ‘Q’. AQ makes an angle 60° with
line XX’ If frequency of wave received
f
by detector is f Hz, then (in Hz)
110
is equal to

Space for Rough work

Chemistry PART – II

SECTION – A
(Only One Option Correct Type)
This section contains 4 multiple choice questions. Each question has four choices (A), (B), (C)
and (D) out of which ONLY ONE option is correct.

1. True or False
(a) In any strong acids solution, the concentration of [OH– ] will be zero.
(b ) If Go of a reaction is positive, then the reaction will not proceed at all, in the forward
direction for any concentration of reactants and products.
(c) When titration curves are drawn for
(i) 1 M HCl (50 mL) with 1 M NaOH and
(ii) 0.01 M HCl (50 mL) with 0.01 M NaOH on the same graph paper they look like

VNaOH

(A) TTF (B) FFT

2. O
OH (excess)
CH3
Pr oduct X; X is
O

(A) CH3 (B) H3C

O
O
(C) (D) None of these

O
CH3

Space for Rough work

3. O O
PhNHNH

2
 Pr oduct X; X is
Ph Oet

(A) (B) Ph
N
Ph N
N
Ph N
O Ph
O
(C) Ph (D) N
Ph
N N
Ph
N O
O
Ph

4. Which of the following statement is false?

(A) In XeOF4, the central atom Xe doesn’t lie on the plane of the square
(B) In PF2Cl3, both the F-atoms lie on the axial bond
(C) In PCl5, the axial bond length is longer than the equatorial bond length
(D) The dipole moment of CH3F is greater than CH3Cl

Space for Rough work

(One or More than One Options Correct Type)

This section contains 4 multiple choice questions. Each question has four choices (A), (B), (C)
and (D) out of which ONE or MORE THAN ONE are correct.

5. Which of the following is correct for lyophilic aquasols?

(A) Its surface tension is lower than that of H2O
(B) Its viscocity is higher than that of water
(C) Colloidal silver
(D) Colloidal antimony

6. Which of the following metals can be obtained by reducing its metal oxide using
aluminium?
(A) Ca (B) Fe
(C) Mg (D) Cr

7. Select wrong statements

(A) Phenol reacts with Na2CO3 and librates CO2 gas
(B) Phenol turn blue litmus to red
(C) Reactivity of methanol with sodium metal is more than that of isopropyl alcohol
(D) Methanol gives iodoform test

8. Borax bead test is given by

(A) Co2+ (B) Zn2+
2+ 2+
(C) Cu (D) Ni

Space for Rough work

Comprehension type (Only One Option Correct)

Paragraph for Questions 09 & 11

Electrolysis is the process in which electrical energy is converted to chemical energy.

In electrolytic cell, oxidation takes place at anode and reduction at cathode. Electrode process
depends on the nature of electrode used for electrolysis. Amount of substance liberated at an
electrode is directly proportional to the amount of charge passed through it. The mass of
substance liberated at electrode is calculated by using the following relation:
ItE
m
96500
Here, E represents the equivalent mass and 96500 C is called the Faraday constant. Faraday
(96500C) is the charge of 1 mole electron, i.e., 6.023 × 1023 electrons. This much charge is used
to liberate one gram equivalent of any substance at each electrode.

9. The platinum electrodes were immersed in a solution of cupric sulphate (CuSO4) and
electric current is passed through the solution. After some time, it was observed that the
colour of copper sulphate disappeared with evolution of a gas at the electrode. The
colourless solution contains:
(A) platinum sulphate (B) copper nitrate
(C) copper sulphate (D) sulphuric acid

10. The passage of current liberates H2 at cathode and Cl2 at anode. The solution is:
(A) copper chloride in water (B) NaCl in water
(C) mercuric chloride in water (D) AuCl3 in water

11. Calculate the volume of gas liberated at the anode at S.T.P. during the electrolysis of a
CuSO4 solution by a current of 1 A passed for 16 minutes and 5 seconds:
(A) 224 mL (B) 56 mL
(C) 112 mL (D) 448 mL

Space for Rough work

SECTION - B
Matrix – Match Type
This section contains 2 questions. Each question contains p q r s t
statements given in two columns, which have to be matched. The
A p q r s t
statements in Column I are labelled A, B, C and D, while the
statements in Column II are labelled p, q, r, s and t. Any given p q r s t
B
statement in Column I can have correct matching with ONE OR
MORE statement(s) in Column II. The appropriate bubbles C p q r s t
corresponding to the answers to these questions have to be
darkened as illustrated in the following example: D p q r s t

If the correct matches are A – p, s and t; B – q and r; C – p and q;

and D – s and t; then the correct darkening of bubbles will look like
the following:

1. Match the following:

Column – I Column – II
(A) When reacts with Br2/CCl4, diastereomer will be (p) H3C H
formed
H3C

H3C
(B) When reacts with O3/H2O, dicarboxylic acid will (q) H3C H
be formed H3C

H3C
(C) Optically active compound (r) H3C H
H3C

H3C
(D) When undergo reductive ozonolysis following (s) H3C H
product will be formed H3C
H3C H
H
H3C H3C

O
H3C

H O
(t) CH3
H
H3C

H3C

Space for Rough work

2. Match the following:

Column – I Column – II
(A) O (p) Final major product has a six
membered ring
H O

3

R R
(B) Cl (q) One of the group will undergo

Ag migration
C
OH
(C) 
(r) Aldol condensation is one of the step
O O H
H 
H
(D) Ph H (s) Final compound has three degree of
C C NaNH2 unsaturation


Ph Br
(t) Contain at least one  bond.

Space for Rough work

SECTION – C
(One Integer Value Correct Type)

This section contains 6 questions. Each question, when worked out will result in one integer
from 0 to 9 (both inclusive).

1. Two aqueous solutions are put in an evacuated chamber. When equilibrium is obtained, it
is found that one solution contains 0.01% of CH3COOH and other 0.014% of urea by
weight, if degree of dissociation is ‘’ then what is the value of 10?

2. Metal element ‘M’ of radius 50 nm is crystallized in FCC format and make cubical crystal
such that face of unit cells aligned with face of cubical crystal. If total number of metal
30
atoms of ‘M’ at face of cubical crystal is 6  10 then area of one face of cubical crystal is
16 2
A × 10 m , the value of A is

3. A 10 L box contains 41.4 gram of a mixture of gases C xH8 and CxH12. The total pressure
of 44°C in flask is 1.56 atm. Analysis revealed that the gas mixture has 87% total C and
13% total H. Find out the value of x.

4. 

In the reaction A  B 
 AB , if the concentration of A and B is increased by a factor of
2, it will cause the equilibrium concentration of AB to change to…………..
–10
5. A solution of SrCO3 is saturated with respect to SrF2. Ksp = 7.9 × 10 . If the fluoride ion
concentration is found to be 4.0  10–2. Then the carbonate ion is found to be 1.4  10–x
M. Find the value of x.

6. Which of the following carbocation have stability greater than an isopropyl cation
 CH3 2 CH
O O O H
CH2 O H2C CH CH2
N
(i) (ii) (iii) (iv)
(v)
O
H
N

(vi)
(vii)

Space for Rough work

Mathematics PART – III

SECTION – A

(Only One Option Correct Type)

This section contains 4 multiple choice questions. Each question has four choices (A), (B), (C)
and (D) out of which ONLY ONE option is correct.

1. If f: R  R is a continuous function satisfying f(0) = 1 and f(2x) – f(x) = x  x  R and

  x 
lim  f  x   f  n    P  x  then P(x) is
x 
  2 
(A) a constant function (B) a linear polynomial in x
(C) a quadratic polynomial in x (D) a cubic polynomial in x

x
2014
2. If f(x) is continuous function such that f(x) > 0  x  0 and  f  x    1   f  t  dt then the
0
2013
value of (f(2014) is equal to
(A) 2013 (B) 2014
(C) 2015 (D) 2016

3. An exam consist of 3 problems selected randomly from a collection of 2n problems,

where n is an integer greater then 1. For a student to pass, he needs to solve correctly at
least two of the three problems. If the student knows to solve exactly half of 2n problems,
then the probability that the student pass the exam is
1 1
(A) (B)
2 3
3
(C) (D) cannot be determined
4

4. If the radius of in-circle and radii of ex-circles of a triangle are consecutive terms of a
geometric progression then the largest angle of the triangle will be
(A) 60º (B) 90º
(C) 120º (D) none of these

Space for rough work

(One or More than One Options Correct Type)

This section contains 4 multiple choice questions. Each question has four choices (A), (B), (C)
and (D) out of which ONE or MORE THAN ONE are correct.

5. A point Q moves on the positive y-axis and a point P follows the point Q such that the
motion of P is always towards Q and distance between P and Q is always equal to some
constant K > 0. If initially Q is at origin and P is at (K, 0) then
dy K 2  x2
(A) the differential equation representing the curve traced by P is 
dx x
(B) y-axis is asymptote to the curve traced by P
 K  K2  x2 
(C) the equation of curve traced by P is y  K ln    K2  x2
 x 
 
(D) the curve traced by P will has x-axis as asymptote

6. If a function y = f(x) is such that f(x) is continuous function and satisfies

x
(f(x))2 = K +   f  t    f '  t  
0
2 2
 dt , K  R , then
+

(A) f(x) is increasing function  x  R (B) f(x) is bounded function

(C) f(x) is neither even nor odd (D) if K = 100 then f(0) = 10
2 2
7. Consider parabola P1  y = x and P2  y = –8x and the line L   x + my + n = 0. Which
of the following holds true (a point (, ) is called rational point if  and  are rational)
(A) if  , m, n are odd integers then the line L can not intersect parabola P1 in a rational
point

(B) line L will be tangent to P1 if m, , n are in G.P.
2
(C) if line L is common tangent to P1 and P2 then  + m + n = 0
(D) if line L is common chord of P1 and P2 then  – 2m + n = 0

8. If Hn denotes the number of ways in which n horses participating in a race can reach the
finishing line, when one or more horses can reach the finishing line at the same time,
then
(A) H2 = 3 (B) H3 = 13
(C) H4 = 75 (D) H3 = 10

Space for rough work

Comprehension type (Only One Option Correct)

Paragraph for Questions 09 & 11

Read the following write up carefully and answer the following questions:
y z x z
A plane p contains the line L1:   1 , x = 0 and is parallel to the line L2:   1 , y = 0
b c a c

9. Equation of plane p is
x y z x y z
(A)    1  0 (B)   1  0
a b c a b c
x y z x y z
(C)    1  0 (D)    1  0
a b c a b c

1 1 1 1
10. If the shortest distance between L1 and L2 is then the value of 2  2  2 equals
4 a b c
(A) 16 (B) 64
(C) 128 (D) 192

 5 8 11 
11. Distance of image of A(a, 0, 0) in the plane p from M   , ,  , where a = b = c = 1, is
 3 3 3
equal to
(A) 1 (B) 2
(C) 3 (D) 4

Space for rough work

SECTION - B
Matrix – Match Type
This section contains 2 questions. Each question contains p q r s t
statements given in two columns, which have to be matched. The
A p q r s t
statements in Column I are labelled A, B, C and D, while the
statements in Column II are labelled p, q, r, s and t. Any given p q r s t
B
statement in Column I can have correct matching with ONE OR
MORE statement(s) in Column II. The appropriate bubbles C p q r s t
corresponding to the answers to these questions have to be
darkened as illustrated in the following example: D p q r s t

If the correct matches are A – p, s and t; B – q and r; C – p and q;

and D – s and t; then the correct darkening of bubbles will look like
the following:

1. Match the following Column-I with Column-II

Column – I Column – II
(A) f(x) = x[x]
(p) f(x) is continuous and
(where [.] denotes the greatest
differentiable everywhere on real
integer function and |.| denotes
domain
modulus function)
(q) f(x) is continuous and
(B) f(x) = x|x|
differentiable at x = 0
(r) f(x) is continuous but not
(C) f(x) = x + |x|
differentiable at x = 0
(s) f(x) is either discontinuous or
(D) f(x) = x + [x] non-differentiable for more than
one x  R
(t) f(x) is a periodic function

2 Match the following Column-I with Column-II

Column – I Column – II
(A) Values of k for which the origin and the radical centre of
three circles described on there sides 3x – 4y + 5 = 0,
(p) 2
x + y + 5 = 0 and 4x + 3y – 10 = 0 of  as diameter lies
opposite to the line 4x – ky + 1 = 0
(B) Values of  for which the pair of lines x2 + 2xy + 2y2 = 0 and
(q) 3
(1 + )x2 – 8xy + y2 = 0 are equally inclined to each other
(C) Find the points on the axis of parabola
y2 – 3x + 6y + 15 = 0 such that three distinct normals can be (r) –2
drawn
(D) Let A and P represents the area and perimeter of the
triangle formed by the common tangents to the circles
A P (s) 6
x2 + y2 – 6x = 0 and x2 + y2 + 2x = 0. Then value of ,
3 3
(t) 4

Space for rough work

SECTION – C
(One Integer Value Correct Type)

This section contains 6 questions. Each question, when worked out will result in one integer
from 0 to 9 (both inclusive).

1. Let f(x) = x + 3x with  > 1,  > 1. The value of ‘a’ for which the area of the figure
bounded by the graph of y = f(x) and straight lines x = 0, x = 1 and y = f(a) have the
greatest value is A and the value for which area is minimum is B. Find the value of A +
2B _____

12 22 32 5002
2. Let S     .....  . If where [.] denotes the greatest integer function
1.3 3.5 5.7 999.1001
then the number of total divisors of [S] is _____


2 120
3. If   sin x  ax    x  dx is minimum for a 

then the value of  +  is _____
0

4. Consider the polynomials P(x) = x6 – x 5 – x3 – x 2 – x and Q(x) = x 4 – x3 – x2 – 1. Given

that z1, z2, z3 and z4 are the roots of the equation Q(x) = 0. The value of P(z1) + P(z2) +
P(z3) + P(z4) is _____
    
5. Let A  ˆi  2jˆ  3kˆ , B  2iˆ  ˆj  kˆ , C  ˆj  kˆ . If the vector B  C can be expressed as a
    
linear combination B  C  xA  yB  zC where x, y, z are scalar, then find the value of
x + y + z _____

6. If w is the imaginary cube root of unity, then the number of ordered pairs of integers (a, b)
such that |aw + b| = 1 is _____

Space for rough work

ANSWERS, HINTS & SOLUTIONS

FULL TEST– I
(Paper-2)

Q. No. PHYSICS CHEMISTRY MATHEMATICS

ALL INDIA TEST SERIES

1. B B B

2. A A B

3. B B A

4. A D B

5. A, B A, B A, B, C

6. B, D B, D A, C¸D

7. A, D A, D A, B, C, D

8. A, D A, C, D A, B, C

9. B D D

10. A B B

11. B B C
(A)  (p, r, s) (A  p, q, r, s) (A)  (r, s)
(B)  (p, q, r, s, t) (B  p, q) (B)  (p, q)
1. (C)  (p, s, t) (C  p, q, r, s, t) (C)  (r)
(D)  (p, s) (D  q) (D)  (s)
(A)  (p, q, r, s, t) (A  p, q, t) (A)  (q, s, t)
(B)  (p, q, r, s, t) (B  p, q, s, t) (B)  (p, r)
2.
(C)  (p, q, r, s, t) (C  q, r, t) (C)  (s, t)
(D)  (p, q, r, s, t) (D  p, q, t) (D)  (q, s)
1. 6 4 3

2. 5 2 4

3. 8 5 6

4. 0 2 6

5. 5 3 1

6. 3 4 6

Physics PART – I

SECTION – A

7. -radiation being neutral is not deflected by magnetic field  &  particle may follow helical path or
straight line depending upon velocity of particle & magnetic field.

8. Comparing the eqn. with velocity of projectile


v  ucos  i  u2 sin2   2gy ˆj
 k 22
Acceleration of particle a y  , ax  0
2
 Motion will be projectile motion and hence the result.
m  k 22
 Force acting on particle will be constant & equal to
2
2
m  k2
 <F>=
2

9. (h11g)A = (h22g)A
h1 2

h2 1

10. (h1 + x1)1g = (h2 + x2)2g

x1/x2 = 2 …(i)
x1 + x2 = 3 …(ii)
x1 = 2 m

4
11.
 
W1  dW1  (p0  x1g)Adx = 260 J
2
2

 
W2  dW2  (p0  x2 g)Adx = 130 J
1
W = W 1 + W 2 = 390 J

SECTION – C

L
4. R  RC = L/R i.e. time constants of RC & LR
C
circuit are same.
After switching from 1 to 2

i0 
2R
from loop rule,
(i0 – i1) R + /2 + i1 R = 0
2i R – 
 i1  0 0
4R

1 2 1 2
5. mgx 2  kx 2  kx 2 k
Velocity is zero
2 2
2 x2 A
1  3mg   3mg  Reference level
 k   mg   Natural position of springs 2 and 1
2  k   k  x1 = 3mg/k
9m2 g2 3m2 g2 3m2g2 x0 = mg/k New equilibrium position
kx12
 mgx 2    x2/2
2k k 2k equilibrium position
2 2 2 2
2k x 2  2mgkx 2  3m g  0
k
2mgk  4m2 g2k 2  24m2 g2k 2
x2 
4k 2
2mgk  2mgk 7 mg
=  [ 7  1]
4k 2 2k
x mg 7 10  7 1
A  0  x2    m  5cm
2 2k 2  100 7 20

6. 0 – Vst = x1
0 + V s t = x 2
x2 – x1 x  x2
Vs = = 60 m/s & 0 = 1 = 1.1 m
2T 2
Speed of sound V = n × 0 = 330 m/s
Frequency detected at point Q
V
n=  300Hz  330Hz
V – Vs

Chemistry PART – II

SECTION – A

2. O O H2 C O
O CH3
CH3 CH3 O
H3 C CH3
OH
 

O

O O
O CH3
OH
 CH 2 


 CH3
O
CH3

3. O O Ph O Ph O

PhNHNH

2
 
Ph Oet N Oet N N
HN Ph
Ph

4. XeOF4 is square pyramidal in shape

O
F F

F F

Due to lone pair-bond pair repulsion the four F atoms move away from lone pair resulting in the
movement of Xe towards lone pair.
F
Cl

Cl P

Cl
F
The most electronegative substituent lies at the axial position to minimize bond pair-bond pair
repulsion.
Cl
Cl

Cl P

Cl
Cl
PCl5 shows sp3d hybridisation with the involvement of axial dz2 orbital. Since the axial Cl is at 90°
hence suffers relatively greater repulsion increasing the axial bond length.

The size of F is smaller than Cl leading smaller C–F bond length decreasing the dipole moment.

5. Surface tension of lyophilic aquasols is lower than that due to force of attraction between
dispered phase and dipersion medium.
Its viscocity is higher due to force of attraction between phase and disperesion medium.

8. White salts or ions giving colourless beads do not give borax bead test.

9. Cu2+ + 2e  Cu
–2 +
Hence SO4 ions are left which form H2SO4 with H from H2O.

 1
11. At anode: 2OH 
 H 2 O  O 2  2e 
2
22.4
Equivalent volume Ve =  5.6 litre
2 2

ItVe 1 965  5.6  10 3

 V  = 56 ml.
96500 96500

SECTION –C
1. At equilibrium relative lowering of vapour pressure of both the solution is same. Then
1    0.01 0.014

60 60
(1 + ) = 1.4
 = 0.4
so
10 = 4

2. Consider the unit cell face

So, atoms on face will be
1 1 1
=    1  1 atom
8 4 2
1
th
8

1
th
2
contribution

No. of atoms at one face of crystal

1
=  6  1030  1030 atm
6
No. of unit cells at one face of crystal = 6  1030 / 6
So no. of unit cells at edge of crystal = 1030 = 1015
4
Now edge length of one unit cell =  50  1015 n
2
 4 
So area of face of crystal =   50  1015  nm2
 2 
16
=  25  102  1030
2
= 2  1034 nm2
= 2  10–18+34 m2
= 2  1016 m2

3. Given P = 1.56 atm, V = 10 L

T = 317 K, R = 0.083 L
PV 1.56  10
Total moles (n) =   0.6 mol
RT 0.082  31%
Let CxH8 be a mol, therefore moles of CxH12 = (0.6 – a) mol. Mass of C in a mol of CxH12 = 129 x
gm.
Mass of C in (0.6 – a) mol of CxH12 = 12 (0.6 – a) gram
Total mass of C in mixture = 12ax + 12x (0.6 – a)g
= 41.4 g

7.2x
% of C in mixture =  100
41.4
Given % of C = 87%
Or
720x
 87
41.4
x=5

4. 

For A  B 
 AB

K
 AB
 A B


When 2A  2B 
 2AB
Now eq. conc. will be 2 times of AB

5. 

SrF2  2 
 Sr  aq  2F  aq 
2
Ksp  Sr 2  F 

Ksp 7.9  1010

Sr 2     4.938  107 M
  2 2
F   4.0  102 
   
This concentration of Sr+2 ions should also satisfy the other expression for SrCO3

SrCO3  s    2  
 Sr  aq   CO3  aq 

Ksp  Sr 2  CO3  

 
10
CO3    Ksp  7.0  10
  Sr 2  4.938  10 7
 
= 1.4  10–3

6. O H
CH2 H2C CH CH2
O
N

Mathematics PART – III

SECTION – A

x x
1. Write the functional equation as f  x   f   
2 2
x
    x x
 f f  
 2  4 4
x x x
 f f  
4 8 8
–––––––––––
–––––––––––

 x   x  x
f  n1   f  n  n
2  2  2
 x  1 1 1 1
Adding up we obtain f  x   f  n   x     .....  n 
2  2 4 8 2 
  x 
Hence, lim  f  x   f  n    x
n 
  2 

2. Differentiating the given equation 2014 (f(x))2013 f(x) = f(x)

 2014 (f(x))2012 f(x) = 1
2014 2013
Integrate
2013
 f  x   xc

2014
 f(0) = 1  c 
2013
2013 2013
Hence,  f  x    x 1
2014
 (f(2014))2013 = 2013 + 1 = 2014

3. Let Ai  the event that student correctly solves exactly i of the three proposed problems
(Where i = 0, 1, 2, 3)
Required probability = P(A) = P(A2  A3) = P(A2) + P(A3)
 Student know how to solve exactly half the problem
P(A0) = P(A3) and P(A1) = P(A2)
Also, P(A0) + P(A1) + P(A2) + P(A3) = 1
1
 P(A) = P(A2) + P(A3) =
2

4. Let r to in-radius and r1, r2, r3 are ex-radii of circle touching the sides BC, CA and AB respectively
of ABC
If is area of triangle and s is semi-perimeter,  = rs = r1(s – a) = r2(s – b) = r3(s – c)
1 1 1 s a sb sc s 1
       
r1 r2 r3     r
So, that r is the smallest term in G.P. Suppose that A  B  C
A B C
then r1  s tan  r2  s tan  r3  s tan
2 2 2
Hence, there is a number t > 1 for which r1 = tr, r2 = rt2, r3 = rt3
and (s – b) = t(s – c), (s – a) = t2(s – c), s = t3(s – c)
4 2
By Herons formula t3r(s – c) = r3(s – c) = rs =  = s  s  a  s  b  s  c   t 6  s  c   t 3  s  c 
C r
Hence, r = s – c, and so tan   1  C = 90º
2 s  c 

dy
5. If the curve traced by P is y = f(x) then = slope of line
dx
joining P and Q as P is always moving towards Q Q
Let Q  (0, ) and P(x, y), (0 – x)2 + ( – y)2 = k2
K
–y= K 2  x2 P

dy   y K 2  x2
  (K, 0)
dx 0  x x
Solving the differential equation with initial condition f(K) = 0 gives

 K  K2  x2 
y  K ln    K 2  x2
 x 
 

6. Differentiating the given equation 2f(x) f(x) = (f(x))2 + (f(x))2  (f(x) – f(x))2 = 0
x
 f(x) = f(x)  f(x) = ce
 f(0) = K  c  K  f  x   Kex

2
7. Point of intersection of P1 and L is given by mx +  x + n = 0
2 
Line is tangent if  = 4mn  m, , n are in G.P.
2
p
If point of intersection is rational (let x  ) where p and q are co-prime
q
2 2
Then mp + lpq + nq = 0 ….. (1)
Now, if one of p and q is even and other is odd then (1) can not hold as sum of an even and an
odd integer can’t be zero
If p, q are odd then (1) can not hold true as sum of three odd numbers can’t be zero
Common tangent to P1 and P2 is 2x – y – 1 = 0
Common chord of P1 and P2 is 2x + y = 0

8. The sequence of two horses reaching the finishing line can be 2 or 1 + 1

 Number of ways = 1 + 2! = 3
The sequence of three horses reaching the finishing line and number of ways are shown below
3  1 way
2 + 1  3C2 = 3 ways
1 + 2  3C2 = 3 ways
1 + 1 + 1  3! = 6 ways
Hence H3 = 1 + 3 + 3 + 6 = 13
The sequence of 4 horses reaching the finishing line and number of ways are shown below
4  1 way
 4! 
2+1+1    3! ways
 2! 1! 1! 2! 
3 + 1  4C3 ways
1 + 3  4C1
2 + 2  4C2
1 + 1 + 1 + 1  4!
Hence, H4 = 1 + 4 + 4 + 6 + 36 + 24 = 75

9.-11. Equations of lines L1 and L2 can be obtained in symmetric form

x yb z
L1:  
0 b c
xa y z
L2:  
a 0 c
x y b z
Equation of plane p: 0 b c 0
a 0 c
x y z
   1 0
a b c
Lines L1 and L2 in vector form
       
 
L1  r  p1  q1  r  bjˆ   bjˆ  ckˆ and L2  r  p2  q2  r  aiˆ  4 aiˆ  ckˆ  
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AITS-FT-I (Paper-2)-PCM(Sol)-JEE(Advanced)/15

a b 0
0 b c
   
p1  p2 q1 q2  a 0 c 1
Shortest distance =    
q1  q2 2 2 2 4
bc    ac    ab 

2abc 1 1 1 1
   2  2  2  64
2 2 2 4 a b c
bc    ca    ab 

SECTION – B
1. (A) f(x) = x[x]
f(x) is continuous at x = 0, but not differentiable at x = 0
f(x) is discontinuous at integers
(B) f(x) = x|x|
 x 2 , x0
f  x   2
 x , x  0
f(x) is continuous and differentiable everywhere
2x , x  0
(C) f(x) = x + |x| = 
0 , x  0
f(x) is continuous everywhere but not differentiable at x = 0
(D) f(x) = x = [x]
f(x) is discontinuous at integers

2. (A) Given lines are

3x – 4y + 5 = 0 ….. (1)
4x + 3y – 10 = 0 ….. (2)
x+y+5=0 ….. (3)
Radical centre of three circles described on sides of  as diameter = orthocentre
As two lines are perpendicular to each other
So, orthocentre is point of intersection of (1) and (2)
Solving (1) and (2) x = 1, y = 2
So, radical centre (1, 2)
4x – ky + 1 = 0 is the given line (0, 0) 0 + 0 + 1 > 0
So, for radical centre to be on opposite side 4 – 2k + 1 < 0
5 – 2k < 0
5
 k
2
2 2
(B) x + 2xy + 2y = 0
x2  y2 xy
Equation of angle bisectors  ….. (1)
1 
(1 + )x2 – 8xy + y2 = 0
x2  y 2 xy
Equation of angle bisectors  ….. (2)
 4
For equally inclined lines (1) and (2) must represent same
1 
So,    2 = 4,  = 2
 4
(C) y2 – 5x + 6y + 15 = 0

 6
y 2  6y  9  5  x  
 5
6
 y  3   5  x  
2

 5
2 5
Y = 4aX, a 
4
6
Xx
5
For three normals to be distinct
X – 2a > 0
6 5
x 
5 2
5 6
x 
2 5
37
x
10
(D) x2 + y2 – 6x = 0
x2 + y2 + 2x = 0 A
The triangle formed by common tangents 3
is ABC 1
CD 1 C
Let C  (h, k),  D(–1, 0) E(3, 0)
CE 3
33
h  3
2 B
k=0
So, C  (–3, 0)
Equation of line thru Q, y = m(x + 3) 1 : –3
 mx – y + 3m = 0 D C C
This represents two tangents with two (–1, 0) (h, k) (3, 0)
difference values of m
m  3m
Length of perpendicular from D = Radius 1
m2  1
4m2 = m2 + 1
2
3m = 1
1
m
3
x x
So, equation of tangents y   3 and y   3
3 3
   
 A  0, 3 , B  0,  3 , C  (–3, 0)
 AB = BC = CA = 2 3
3 2
So, P  6 3 , A 
4

 2 3  3 3

P A
6, 3
3 3

SECTION – C

1. f(x) = x + 3x
f(x) =  x – 1 + 3x – 1
 f(x) > 0  x  (0, 1)
f(x) = ( – 1)x – 2 + 3( – 1)x – 2  x  (0, 1)
 f(x) > 0  x  (0, 1)
Hence, f(x) is monotonically increasing and concave upwards curve
0 1
B ,A=1
2

r2 1 r r 
2. Tr   
 2r  1 2r  1 4  2r  1 2r  1 
1 500  r r 
 S   
4 r 1  2r  1 2r  1 
500
 r r  1  1 2 2 3 3 4 500 500
 4S      =              .....  
r 1  2r  1 2r  1  1 3 3 5 5 7 7 999 1001
r r 1
  1
2r  1 2r  1
500 501000
Combining all such terms, we get 4S  1  499  
1001 1001
125250
S  [S] = 125
1001
Number of divisors = 4

  
2
3. I  a   a2  x 2    x  dx  2a  x    x  sin xdx   sin2 xdx
0 0 0
5
 2 
 Ia  a  8a 
30 2
2
5  120   480
Ia  a  5    5
30    2 
120
Hence, I(a) is minimum for a    = 1,  = 5
5

4. Dividing P(x) by Q(x) gives P(x) = Q(x)(x 2 + 1) + x2 – x + 1

2
4
 4  4
then P  z 
 i   i   i j  zi  4
 z  2 z z 
i 1  i1  i 1
4
Use the formula for sum and products of the roots to obtain  P  zi   1  2  1  4  6
i 1

    
5 B  C  xA  yB  zC
 
Taking dot product with B  C
    2
       B2  C 2  B  C  
  
B  C B  C  x  A B C  x    
 A B C
 

   
 
Similarly taking dot with C  A gives y 
BC C A
  
  
 A B C
 
   
 
and dot with A  B gives z 

BC  A B
  
 
 A B C
 
  
  A B C  12
   
 
 B2  C2  B  C  12
x=1
1 1
 y  and z  
2 2

6  |aw + b| = 1

 |aw + b|2 = 1  (aw + b)  aw  b   a2 – ab + b2 = 1
 (a – b)2 + ab = 1
When (a – b)2 = 0 and ab = 1 then (1, 1), (–1, –1)
When (a – b)2 = 1 and ab = 0 then (0, 1), (1, 0), (0, –1), (–1, 0)
Hence, total six ordered pairs

Time Allotted: 3 Hours Maximum Marks: 360

 Pl ea s e r ea d t h e i n s t r u c t i o n s c a r ef u l l y. Yo u a r e a l l o t t ed 5 m i n ut es
s p ec i f i c a ll y f o r t h i s p u r p o s e.

ALL INDIA TEST SERIES

Yo u a r e n o t a ll o wed t o l ea v e t h e E xa m i n at i o n Ha ll b ef o r e t h e en d o f
t h e t es t .

INSTRUCTIONS

A. General Instructions
1. Attempt ALL the questions. Answers have to be marked on the OMR sheets.
2. This question paper contains Three Parts.
3. Part-I is Physics, Part-II is Chemistry and Part-III is Mathematics.
4. Each part has only one section: Section-A.
5. Rough spaces are provided for rough work inside the question paper. No additional sheets will be
provided for rough work.
6. Blank Papers, clip boards, log tables, slide rule, calculator, cellular phones, pagers and electronic
devices, in any form, are not allowed.

B. Filling of OMR Sheet

C. Marking Scheme For All Three Parts.

(i) Section-A (01 to 30) contains 30 multiple choice questions which have only one correct answer.
Each question carries +4 marks for correct answer and – 1 mark for wrong answer.

Name of the Candidate

Enrolment No.

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Useful Data

PHYSICS
Acceleration due to gravity g = 10 m/s2

Planck constant h = 6.6 1034 J-s

Charge of electron e = 1.6  1019 C

Mass of electron me = 9.1  1031 kg

Permittivity of free space 0 = 8.85  1012 C2/N-m2

Density of water water = 103 kg/m3

Atmospheric pressure Pa = 105 N/m2

Gas constant R = 8.314 J K1 mol1

CHEMISTRY

Gas Constant R = 8.314 J K1 mol1

Atomic No: H=1, He = 2, Li=3, Be=4, B=5, C=6, N=7, O=8,

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Physics PART – I

SECTION – A
Single Correct Choice Type

This section contains 30 multiple choice questions. Each question has four choices (A), (B), (C) and (D)
out of which ONLY ONE is correct.

1. The physical quantity, electric potential (V) has dimensions [Ma Lb Tc Ad] and capacitance (c) has
e f g h
dimensions [M L T A ]. Then a + b + c + d + e + f + g + h is :
(A) zero (B) 1
(C) 2 (D) None of these

2. Mark the CORRECT statement:

(A) Two vectors are equal to each other only if their corresponding components are equal
(B) The cross product of two parallel vectors can be NON-zero
(C) The direction of a zero vector can be determined
 
(D) If we multiply A by the scalar-1, we obtain the vectors  A , which is a vector with the

opposite magnitude but the same direction as A

3. The acceleration of a particle moving only on a horizontal x-y plane is given by a = 2t î + 3t ĵ ,

where a is in meters per second squared and t is in seconds. At t = 0 the particle has the velocity
 
vector V   5.00m / s  î   2.00m / s  ˆj and position vector r  10.00m  î  10.0m  ˆj . At t = 6s,
the position vector of the particle is :
(A) 112m  î  130m  ˆj (B) 102m  î  120m  ˆj
(C)  256m  î   238m  ˆj (D) None of these

Space for Rough work

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4. A child’s toy consists of three cars that are pulled in tandem on small friction less rollers (Fig. A).
The cars have masses m1, m2 and m3 (m1 < m2 < m3). If they are pulled to the right with a
horizontal force P, mark the INCORRECT statement (T = Tension in the string):

(A) TA > TB > TC, (B) TA < 2TB

5. Four conducting plates are arranged with

equal spacing and surfaces of all the
plates are marked as 1, 2, 3, 4, 5, 6, 7 and
8 as shown. Two middle plates are given
charged +Q and –Q, after which S1 and S2
are closed. The charge distribution on the
surfaces of conducting plates from 1 to 8
in order is:

(A) O, –Q, +Q, O, O, –Q, Q, O

(B) O, –Q/3, Q/3, 2Q/3, –2Q/3, –Q/3, O
(C) O, O, O, Q, –Q, O, O, O
(D) O, –2Q/3, 2Q/3, Q/3, –Q/3, –2Q/3, 2Q/3, O

6. Consider a particle P moving on a circle of radius ‘A’ with a

constant angular speed ‘w’. RP and PQ are  to X and Y
axes respectively:
(A) The x-coordinate of the particle at time t is A sin t
(B) The particle has a non-zero displacement in a time
interval 2/
(C) R and Q execute simple harmonic motion and their
phases differ by /2
(D) If the radius of the circle is doubled, the phase
difference between R and Q would also double

Space for Rough work

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7. An infinite non conducting uniformly charged sheet has a hole of radius ‘R’ in it (change density =
). An electron is placed on an axis passing through the centre of hole and perpendicular to the
plane of sheet at a distance ‘R’ from the centre. The speed with which the electron reaches the
centre is ....
( > 0) Given : electric field on the axis of a uniformly charged disc of radius ‘r’
  x 
E(x)  1  
2   x2  r 2 
eR 2eR
(A) (B)
m 0 m 0

8. Two thin rigid rods, AB (mass m, length ) and CD (mass

2m, length 2) are connected to make a ‘T’ as shown in
figure. The assembly is kept vertical on a frictionless
horizontal surface and is restrained to fall sidewise. What
is the least value of F to be applied horizontally at B, such
that B is able to leave contact with the surface :

(A) 3mg (B) (9/4) mg

3 3 3
(C) mg (D) mg
2 2

9. In a two block system shown in the

figure, the two blocks have velocities in
same horizontal direction. Mark the
CORRECT statement about ‘the velocity
of the centre of mass’ of the system with
respect to a stationary frame:
 
(A) Vcm is a function of time and will become O at some instant(s)

(B) Vcm is non-zero at all times
 
(C) Vcm is O at all times
 
(D) Vcm may be O at some instant(s)

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10. A block of mass m is attached to a pulley disc of same mass and radius ‘r’ by means of a block
strings a shown. The pulley is hinged about its centre on a horizontal table and the block is
projected with an initial velocity of 3 m/s. Its velocity when the string becomes taut will be:

(A) 3 m/s (B) 2 m/s

11. Consider a plane surface of length ‘a’ and width ‘b’ which is inside a liquid of density . Find the
force on the entire surface due to the liquid in S.

gab gab
(A)
2
a  2  sin  (B)
2
a    sin2 
gab gab
(C)
2
a    sin2  (D)
4
a  2  sin2 

12. A pressure meter attached to a closed water tap reads 1 bar. When the tap speed is
10 ms–1 and the reading of the pressure meter is
(A) 1.5 × 105 Pa (B) 3 × 105Pa
5
(C) 0.5 × 10 Pa (D) 105 Pa

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13. A film of the liquid is supported in a vertical rectangular

area, the top border of which is a sliding wire. An
external force of 5 mN acts on the sliding wire of length
50 mm and linear mass density of 1.75×10–3 kg/m. Find
the surface tension of the liquid:

(A) 0.083 N/m (B) 0.041 N/m

14. The magnitude of gravitational potential energy of the sun-earth system is u with zero potential
energy at infinity separation. The kinetic energy of the earth with respect to the sum is K.
(A) U < K (B) U > K
(C) U = K (D) None of these

15. Mark the CORRECT option, for stress-strain

curve of a metal wire:
(A) Curve ab represents elastic behaviour and
e denotes fracture
(B) ONLY curve ac represents elastic
behaviour
(C) Curve bd represents plastic behaviour and
e represents permanent set
(D) Point d represents fracture point and the
gap between b and d is respectively large
for brittle materials

16. Mark the CORRECT option for a gas:

(A) Heat is expelled by gas
(B) Heat is absorbed by gas
(C) Process AB is adiabatic
(D) Temperature of the gas decreases
from A to B

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17. A musical instrument has a string which is timed to frequency 400 Hz and has a length of 0.5m.
What is the third longest wavelength of resonance that reaches the ear of the listener? (Vair = 343
m/s):
(A) 0.572 m (B) 0.286 m
(C) 1.7515 m (D) 1.29 m

ART
18. The most probable speed of a gas with temperature ‘T’ and molar Mass ‘M’ is , A is a
M
dimensionless constant and R is universal gas constant in JK–1 mol–1. Four equal charges A×10–
6
C each are fixed at the four corners of a square of side 5 cm. Find the coulomb force
experienced by one of the charges due to the other three :
(A) 20 N (B) 27.5 N
(C) 44.6 N (D) 61.9 N

19. One cannot see through fog, because:

(A) Fog absorbs the light
(B) Light suffers total reflection at droplets
(C) Refractive index of the fog is infinity
(D) Light is scattered by droplets

20. The speed time graph of a car is given below the V(m/s)
car weights 1 Mg.
(A) Distance travelled in first two seconds is 30 m A B
15
(B) Acceleration of the car at t = 4s is 7.5 m/s2
(C) Braking force applied at the end of 5s to bring O 2 5 6
the Car to a stop within 1s is 15 KN t(s)
(D) (A), (B) and (C) are WRONG

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21. Electric

field due to a ‘short’ electric dipole at a point P as shown is at an angle  with respect to
p . Then

 tan    tan  
(A)     tan1   (B)     tan1  
 2   2 
 tan    tan  
(C)     tan1   (D)     tan1  
2 2  5 

22. Mark the INCORRECT statement:

(A) By convention, the charge on a plastic rod rubbed with fur is negative and the charge on a
glass rod rubbed with silk is positive.
(B) The ratio of electric force and gravitational force between a proton and electron is
ke2
 2.4  1039
Gmemp
(C) Conservation of total charge of an isolated system is a property dependent on the scalar
nature of charge.
(D) Quantisation of electric charge is a basic unexplained law of nature, interestingly; there is no
analogues law of quantisation of mass.

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23. A substance when placed in a non-uniform magnetic field is attracted towards stronger magnetic
field. That substance CAN be:
(A) Paramagnetic
(B) Diamagnetic
(C) Either Paramagnetic OR diamagnetic
(D) Neither Paramagnetic NOR diamagnetic

24. Mark the CORRECT option:

(A) A vertical plane containing the vertical axis (orthogonal to ground) and the longitude is called
the geographic meridian
(B) Geographic meridian and magnetic meridian for earth are always same at a point
(C) The dip circle is simply a compass needle pivoted about a fixed vertical axis and free to move
in a horizontal plane
(D) There is a bar magnet busied deep inside earth’s interior which gives rise to earth’s magnetic
field

25. Two resistors R and 2R are connected in parallel in an electric circuit. The thermal energy
developed in R and 2R are in the ratio:
(A) 1 : 2 (B) 2 : 1
(C) 1 : 4 (D) 4 : 1

26. Consider the circuits drawn below:

All the capacitors are identical, then

(A) CCD > CEF = CAB (B) CCD > CEF > CAB
(C) CCD > CAB > CEF (D) CEF > CCD > CAB

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27. A wave front AB passing through a system C

averages as DE. The system ‘C’ CAN be
(A) a slit
(B) a biprism
(C) a prism
(D) a glass slab

28. A convex lens is made of two different materials having

radius of curvature of surface A and surface B as 25 cm and
24 cm respectively. The effective focal length of lens is
(A) 75/4 cm
(B) 75 cm
(C) 50 cm
(D) 30 cm

29. Hydrogen atom is its ground state is excited by means of monochromatic radiation of wavelength
970.6Å. How many different wavelengths are possible in the resulting EMISSION spectrum? [hc
= 1.986×10–25 J-m] :
(A) 3 (B) 4
(C) 5 (D) 6

30. Attenuation, when referring to communication systems, is :

(A) The process of mixing low frequency signal and a high frequency signal
(B) The weakening of a signal as it propagates through a medium
(C) The process of increasing the amplitude of a signal
(D) The process of recovering a low frequency signal from a high frequency signal

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12

Chemistry PART – II

SECTION – A
Single Correct Choice Type

This section contains 30 multiple choice questions. Each question has four choices (A), (B), (C) and (D)
out of which ONLY ONE is correct.
3
1. A particular 100-octane aviation gasoline used 1.00 cm of tetraethyl lead (C2H5)4Pb of density
1.66 g/cc, per litre of product. This compound is made as follows.
4C2H5Cl + 4NaPb  (C2H5)4Pb + 4NaCl + 3Pb. How many gram of ethyl chloride is needed to
make enough tetraethyl lead for 1 L of gasoline.
(A) 1.33g (B) 2.66 g
(C) 9.2g (D) 0.33g

2. The treatment of CH3CH = CHCH3 with NaIO4 produces

(A) CH3CHO only (B) CH3COOH only
(C) CH3CHO & CH3COOH (D) CH3COCH3 + HCOOH

3. Among the following, which exhibits aromaticity?

(A) Cyclopentadienyl cation (B) Cyclopentadienyl radical
(C) Cycloheptatrienyl radical (D) Cycloheptatrienyl cation

4. SN1 characteristics would be exhibited in the hydrolysis of

(A) chlorobenzene (B) 1-chloronapthalene
(C) propenyl chloride (D) 2-chloro-1-propene

5. The ease of dehydration of the following alcohols with H2SO4 is

(I) CH3CH2CH(OH)CH2CH2CH3 (II) (CH3)2C(OH)CH2CH2CH3
(III) CH3CH2CH(OH)CH(CH3)2 (IV) (CH3)2C(OH)CH(CH3)2

(A) III  IV  II  I (B) III  II  IV  I

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6. The solubility product of A2X3 is 1.08  10–23. Its solubility will be

(A) 1  10–3 (B) 1  10–6
–4 –5
(C) 1  10 (D) 1  10

7. When dilute H2SO4 is electrolysed between Pt electrodes, the gas liberated at the anode will be
(A) SO2 (B) SO3
(C) O2 (D) H2

8. 0.7 g of Na2CO3 .xH2O were dissolved in water and the volume was made to 100 mL. 20 mL of
this solution required 19.8 mL of N/10 HCl for complete neutralization. The value of x is
(A) 7 (B) 2
(C) 3 (D) 5

9. The product of reaction of an aq. solution of Bi3+ salt with sodium thiosulphate gives
(A) Bi2S3 (B) Na3[Bi(S2O3)3]
(C) Na[Bi(S2O3)2] (D) [Bi2(S2O3)2]Cl2
–7
10. The basic ionisation constant for hydrazine, N2H4 is 9.6 10 . What would be the percent
hydrolysis of 0.1 N2H5Cl?
(A) 0.016% (B) 3.2%
(C) 1.6% (D) 0.032%

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11. Which of the following combinations will act as a buffer

(A) NH3 + HCl (molar ratio 2:1) (B) HC2H3O, NaOH (Molar ratio 1:2)
(C) HCl + NaCl (D) NaOH + HC2H3O2 (1:1 molar ratio)

12. Gaseous cyclobutane isomerizes to butadiene in a first order process which has a k value at
153C of 3.3  10–4s–1. How many minutes would it take for the isomerisation to proceed 40% to
completion at this temperature?
(A) 26 min (B) 52 min
(C) 13 min (D) None of these

13. Current efficiency is defined as the extent of a desired electrochemical reaction divided by the
theoretical extent of the reaction times 100%. What is the current efficiency of an
electrodeposition of Cu metal in which 9.8 g Cu is deposited by passage of 3A current for 10000
seconds.
(A) 0.9% (B) 49.9%
(C) 99.1% (D) 51.1%

14. A 10 L cylinder of oxygen at 4 atm pressure and 17C developed a leak. When the leak was
repaired, 2.5 atm of O2 remained in the cylinder, still at 17C. The no. of moles of gas escaped.
(A) 1.2 mol (B) 0.30 mol
(C) 0.63 mol (D) 2.4 mol

15. Arrange the following in decreasing order of basicity

(I) Aniline (II) p-nitro aniline (III) N, N dimethyl aniline (IV) N, N, 2,6 tetramethylaniline

(A) I II  IV  III (B) III  IV  I  II

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16. Which of the following is not linear

+
(A) Ag(NH3)2 (B) HgCl2
(C) PbCl2 (D) CS2

17. The order of enol content in the following compounds is

CH3COCH2COOEt (EtOCO)2CH2 PhCOCH2COCH3 PhCOCH2COOEt
(I) (II) (III) (IV)
(A) II  I  III  IV (B) IV  III  II I
(C) III  IV  I  II (D) III  I  IV  II

18. All the halobenzenes are separately treated with HNO3/H2SO4. The quantity of o-product should
be maximum when the substance is
(A) Flurobenzene (B) Chlorobenzene
(C) Bromobenzene (D) Iodobenzene

19. The minimum number of carbon atoms for an alkene hydrocarbon to exhibit optical isomerism is
(A) 5 (B) 6
(C) 9 (D) None of these

20. Bicyclohexane was found to undergo two parallel 1st order rearrangements. At 730 K the first
order rate constant for the formation of cyclohexene was 1.26  10–4 s–1 and for the formation of
methyl cyclopentene the rate constant was 3.8 10–5 s–1. What was the percentage distribution of
methyl cyclopentene.
(A) 77% (B) 23%
(C) 66% (D) 34%

21. Tf / K f has the same value of 1 mol kg-1 for 8% AB2 and 10% A2B by mass of solvent, both AB2
and A2B being non electrolytes.
Atomic masses of A and B will be respectively.
(A) 20, 40 (B) 20, 50
(C) 40, 20 (D) 50, 40

22. What will happen when D-(+)-glucose is treated with methanolic HCl followed by Tollen’s
reagent?
(A) A black ppt. will be formed. (B) A green colour will appear.
(C) A red ppt. will be formed. (D) No characteristic colour or ppt. will be
formed.

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23. The r  / r  ratio of ZnS is 0.402. Which of the following is the incorrect statement?
(A) ZnS is 4 : 4 coordination compound.
(B) ZnS does not crystallize in rock salt type lattice because r  / r  is too small to avoid
2-
overlapping of S .
2+ 2-
(C) Zn ion is too small to fit precisely into the tetrahedral voids of S ions.
2+ 2-
(D) Zn ion is too large to fit into tetrahedral voids of S ions.

24. The wavenumber of last line of Lyman series of hydrogen spectrum is 109674 cm -1. The wave
number of H line in Balmer series He+ is?
-1 -1
(A) 438696 cm (B) 106974 cm
-1 -1
(C) 30465 cm (D) 60930 cm

25. Consider the following statements

(I) NO+ is more stable towards dissociation into atoms than NO.
(II) CO+ is less stable towards dissociation into atoms than CO.
Then which of the following is correct?
(A) I only (B) II only
(C) I & II both (D) None of these

26. Below critical micelle concentration (CMC) sodium oleate in aqueous solution:
(A) exists largely as micelles of anions.
(B) dissolve substances like grease, fat etc. colloidally.
(C) increases the viscocity of water abruptly.
(D) behaves as strong electrolyte.

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27. Given that:

SHo 2  131 JK 1 mol1
S oCl2  223 JK 1 mol1
o
and SHCl  187 JK 1 mol1
The standard entropy change in formation of 1 mole of HCl (g) from H2(g) and Cl2(g) will be:
-1 -1
(A) 20 J K (B) 10 J K
-1
(C) 187 J K (D) 374 J K-1

28. The hybridization of P in PO34 is the same as that of:

(A) N in NO3 (B) S in SO3

(C) I in ICl 2 (D) I in ICl4

29. Among the following compounds which exist as solid at room temperature?
(A) XeF6 (B) OF2
(C) SF4 (D) SF6

30. Magnesium is mainly extracted by

(A) Carbon reduction method
(B) self reduction method
(C) the method of electrolysis
(D) leaching with aq. solution of NaCN followed by reduction

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18

Mathematics PART – III

SECTION – A
Single Correct Choice Type

This section contains 30 multiple choice questions. Each question has four choices (A), (B), (C) and (D)
out of which ONLY ONE is correct.

1. If the lines represented by x 2 – 2pxy – y2 = 0 are rotated about origin through an angle , one in
clockwise direction and other in anticlockwise direction, then equation of the bisectors of the
angle between the lines in the new position is
(A) px2 + 2xy + py2 = 0 (B) px2 – 2xy + py2 = 0
2 2
(C) px – 2pxy – py = 0 (D) px2 + 2xy – py2 = 0

a11 a12 a13

2. Let D0  a21 a22 a23 and let D1 denote the determinant formed by the co-factor of elements of
a31 a32 a33
D0 and D2 denote the determinant formed by co-factor of D1 and so on. If Dn denotes the
determinant formed by the co-factor of Dn – 1, then the value of determinant Dn is
2n n
(A)  D0  (B) D02
2
(C) Dn0 (D) D20

1
3. The maximum value of f  x    t sin  x  t  dt is
0
1 2 1
(A)  4 (B) 2  4
 2
1
(C) 2  4 (D) 2  4
2 2

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4. Number of integers in the range of ‘a’ so that the equation x 3 – 3x + a = 0 has all its roots real and
distinct, is
(A) 2 (B) 3
(C) 4 (D) 5

 2 n  1  f n 
5. f(n) = cot 2  cot 2  .....  cot 2 , (n > 1, n  N) then lim 2 is equal to
n n n n  n
1 1
(A) (B)
2 3
2
(C) (D) 1
3

2 2 2 2
6. Given two circles x + y + 3 2  x  y   0 and x + y + 5 2  x  y   0 , let the radius of a circle
2  1
which touches the two circles and their common diameter be then  equals

(A) 10 (B) 8
(C) 7 (D) 5

7. Let ABCD be a tetrahedron with AB = 41, BC = 36, CA = 7, DA = 18, DB = 27 and DC = 13, then
square of the distance between mid-points of AB and CD equals
(A) 1096 (B) 548
(C) 274 (D) 137

8. In triangle ABC, if 2b = a + c and A – C = 90º, then sin B equals

7 5
(A) (B)
5 8
7 5
(C) (D)
4 3

9. The graph of f(x) = x 2 and g(x) = cx3 intersect at two points. If area of region bounded between
 1 2 1 1 
f(x) and g(x) over the interval 0,  is equal to then the value of   2  is
 c 3 c c 
(A) 20 (B) 2
(C) 6 (D) 12

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10. The value of sec 40º + sec 80º + sec 160º will be
(A) 4 (B) –4
(C) 6 (D) 8

11. Consider the chords of the parabola y2 = 4x which touches the hyperbola x 2 – y2 = 1, the locus of
the point of intersection of the tangents drawn to the parabola at the extremities of such chords is
a conic section having latus rectum , the value of  is
(A) 1 (B) 2
(C) 3 (D) 4

12. The area of the region bounded by lines y = x, y = 0 and

x = sin–1 (a4 + 1) + cos–1 (a4 + 1) – tan–1 (a4 + 1) is
2 a2 2 a2
(A)  (B) 
8 4 8 2
2 2
(C) (D)
16 32

13. If ax3 + by3 + cx2y + dxy2 = 0, represents three distinct straight lines, such that each line bisects
the angle between the other two, then which of the following is true
(A) 3b + c = 0 (B) 3a + d = 0
(C) d + 3a = 0 (D) b + 3c = 0

14. The number of integral value of n (where n  2) such that the equation 2n{x} = 3x + 2[x] has
exactly 5 solution (where [.] denotes the greatest integer function and {x} is fractional part of x) is
(A) 2 (B) 3
(C) 4 (D) 0

15. The rth, sth, tth terms of an A.P. are 6, 8 and 12 respectively. If f(x) = tx2 + 2rx – 2s then f(x) = 0
has
(A) both roots are negative (B) both roots greater then 2
(C) one root negative other greater than 1 (D) exactly one root  (0, 1)

16. Let a, b, c, d  R be such that a2 + b2 + c2 + d2 = 25, then

25
(A) ab + bc + cd + ad  (B) ab + bc + cd + ad  25
2
5
(C) ab + bc + cd + ad  5 (D) ab + bc + cd + ad 
2

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3 sin x  a2  10a  30 x  Q

17. Let f  x    which one of the following statements is correct?
 4cos x x Q
(A) f(x) is continuous for all x when a = 5
(B) f(x) must be discontinuous for all x, when a = 5
3
(C) f(x) is continuous for all x = 2n – tan–1   , n  I, when a = 5
4
4
(D) f(x) is continuous for all x = 2n – tan–1   , n  I, when a = 5
3

18. Number of ways in which 5 alike red balls and 6 alike green balls can be arranged so that exactly
two pairs of green balls are together is
(A) 72 (B) 90
(C) 144 (D) 100

 
19. If z  2  i  z sin   arg z  , then locus of z is
4 
(A) pair of straight line (B) circle
(C) parabola (D) ellipse

20. If x, y, z are real variables such that 3 tan x + 4 tan y + 5 tan z = 20, then the minimum value of
tan2 x + tan2 y + tan2 z is
(A) 10 (B) 15
(C) 8 (D) 12

2
21. If   then cos cos 2 cos 3 ….. cos 1004 is equal to
2009
1
(A) 0 (B) 2008
2
1 1
(C) 1004 (D)  1004
2 2

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xdx
22.  is equal to
3
1 x 2
1  x 2

1 2
(A)
2

n 1  1  x2  c  (B) 3/2
c

3 1 1 x 2


(C) 2 1  1  x 2  c  (D) 2 1  1  x 2  c

x2 y 2
23. There exist two points P and Q on the hyperbola   1 such that OP  OQ (where O is
a2 b2
origin), then number of points in the xy plane from where pair of perpendicular tangents can be
drawn to the hyperbola is
(A) 0 (B) 1
(C) 2 (D) infinite

24. The shortest distance between the lines 2x + y + z – 1 = 0 = 3x + y + 2z – 2 and x = y = z is

1
(A) (B) 2
2
3 3
(C) (D)
2 2

 1 1  i 2
25. If A     (  R) is a unitary matrix then  is
1  i 1 
1 1
(A) (B)
2 3
1 2
(C) (D)
4 9

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26. Solution of equation (2 + x) dy – y dx = 0 represents a curve passing through a fixed point p, then
are of equilateral triangle with p as one vertex and x + y = 0 as its one side is
(A) 2 3 (B) 3
2 4
(C) (D)
3 3

5
 1 1 
27. The term independent of x in the expansion of  2   1  x  x 2  is
 x x 
(A) 381 (B) 441
(C) 439 (D) 359

28. If the standard deviation of the data 1, 3, 5, 7, ….., 2013 is d, then [d] is (where [.] denotes the
greatest integer function)
(A) 580 (B) 581
(C) 582 (D) 583

29. If number of different reflexive relation on set A is equal to number of different symmetric relation
on set A, then n(A) is equal to
(A) 1 (B) 3
(C) 1 and 3 (D) 3 and 7

xn1 yn2
30. If (sin–1 x + sin–1 y)(sin–1 z + sin –1 w) = 2, then the value of can not be equal to (where
zn3 w n2
n1, n2, n3, n4  N)
(A) –2 (B) 0
(C) 1 (D) 2

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ANSWERS, HINTS & SOLUTIONS

FULL TEST –II
(Main)
ALL INDIA TEST SERIES

S. No. PHYSICS CHEMISTRY MATHEMATICS

1. C A D
2. A B B
3. A D B
4. D C B
5. B B B
6. C D B
7. C C D
8. B B C
9. B A C
10. B D C
11. A A A
12. C A D
13. B C A
14. B C B
15. C D D
16. B C B
17. B C C
18. B D B
19. D B C
20. C A C
21. B C C
22. C D D
23. A D A
24. A C A
25. B C B
26. B D C
27. C B A
28. A C B
29. D A B
30. B C C

Physics PART – I

SECTION - A
1. Q = C.V. , [Q] = T’A’
So, a + b + c + d + e + f + g + h = 2

  t t   t t  
3. r  t   r(0)   v.dt , v  t   t0 a.dt  v  0 
t0
t t t
   3t 2  
 v.dt    t 
 5 ˆi    2  ˆj  dt
2

t 0 0   2  
 t3   t3 
=   5t  ˆi    2t  ˆj
3  2 
  t3  ˆ  t3 
r  t     5t  10  i    2t  10  ˆj
3  2 

   TA TB T
4. a1  a2  a3    c
m1  m2  m3  m3  m2 m3
 TA > TB > TC
 m1  m1
TA  TB 1   and 1
 m 2  m 3 m 2  m3

So, TA  2TB
 m  m
TB = TC 1  2  and 2  1
 m3  m3
So, TB < 2TC
 m  m2 
Now, TA  TC 1  1 
 m3 
m  m2
And, m1 + m2 < m3 + m3  1 <2
m3
Or TA < 3TC

5. ‘T’ total charge in the system (initially) = zero We know that when two or more charged
conducting plates are peaced parallel to each other, the outermost surfaces get equal charges ‘p’
T
can be proved using gauss’s law) Where, P  ; So, P = 0
2

Since the middle plates are not connected total change on than would be conserved the charges
on the plates after S1 and S2 are connected are shown in the figure. Again, we utilize the fact that
the facing surfaces must carry opposite charges. Finally the potential difference between surface
1 and 8 must be zero. The electric field is non-zero between the conducting plates and zero
inside them.
 q  Qq  q 
d   d  d 0
 A0   A0   A 0 
d = spacing between the plates A = Area of the plate
 q = Q/3

6. x = A cos wt
 
R =  sin wt : Q = A cos wt = A sin   wt 
2 

7. Electric field at a distance x-on the axis

= Electric field due to an infinite sheet
– Electric field due to a uniformly charged disc of radius ‘R’
   x 
  1  
20 20  2 2
x  r 
 x
E
20 x  R 2
2

ex
F  qE  
20 x 2  R2
du ex
 mu 
dx 2 0 x 2  R 2

u
 
2  1 eR
m 0

8. When the point B is about to leave ground the normal reaction to the rigid rod passes through A.
Gravity would not cause a torque about centre of mass. Also, the torques must balance just
before B leaves ground.

 2  
F    3mg  
 3 2
9mg
F=
4
 
9. Vcm of the system is non-zero initially and would remain so because Fext  0 at all times.

10. Initial angular momentum = mvr

 mr 2 
Final angular momentum =   w  m  wr  r
 2 
 Text = 0 : L is conserved
3
Mur = mwr 2
2
2u
Or, wr 
3
Velocity of mass m first after the string is taut = wr = 2m/s

11. Force on inclined surface in liquid

= Pressure at the centroid of the surface × area of surface
  a 
 g  l   sin   ab
  2 

1 2
12. P1  u  P2
2
1 2
105 Pa   103  10   P 2
2
1
P2   105 Pa
2

P  dg
13. 
2d
P = 5mN,  = 1.75×10–3 kg/m, d = 50 mn

14. Potential energy = – 2 × kinetic energy

Magnitude of potential energy = 2 × K.E.
Clearly, U > K

16. Volume of the gas is increasing (assuming its mass remains constant) and volume of the gas is
directly proportional to T.
k
nCv (dT) = (dV) or V T
m
The process is isobaric and exchange of heat is given by nCp (dT)

2 1
17. 3   m
3 3
Wave speed on the string = 1 (lowest resonant node) ustring = 400 m/s
When a wave passes from one medium to another (the air) of differing wave speed, the
frequency remains the same.
v
Now,  air   string air
v string
343 1
=  m
400 3

 
 2 1 
2 
18. Fq = kq  
 5 
2
 2 5  10 2
 5  10 

9  109  4  1012  1
 4  2  N
25  10  2 

= 27.5 N

20. Distance travelled in first two seconds = 15 m

Acceleration of the Car at t = 4s is zero
Braking force = Ma
= 1 Mg×15 m/s2
= 15 KN

2k  pcos   k  p sin  
21. E1   3 ;E2  
|r| | r |3

E2 tan 
 tan  
E1 2
 tan  
or   tan1  
 2 
 tan  
or     tan1  
 2 

25.
P1 H1
 
V 2 / R1 R
 2 

2R 2

P2 H2 2
V / R2 R 1 R 1

26. CAB = C/2 ; CCD = 2C ; CEF = C

27. A slit would have produced a spherical wave front. A biprism and a glass slab would produce a
parallel wave front.

1  1  1 1  1  1
28.  1.5  1    ;  1.8  1   
f1  25  50 f2  24  30
1 1 1 8 4 75
    ;F  cm
F f1 f2 150 75 4

hc
29. E  12.79eV


En  E1  E  0.81eV
13.6
En    0.81eV;n  4
n2

Chemistry PART – II
SECTION – A

1. The mass of 1 cc of (C2H5)4Pb is = 1  1.66 = 1.66 g and this is the amount needed per litre.
1.66
No. of moles of (C2H5)4Pb needed = = 0.00514 ml
323
1 mole of (C2H5)4Pb requires 4  (0.00514) = 0.0206 ml of C2H5Cl
 Mass of C2H5 Cl = 0.0206  64.5 = 1.33 g

2. The sp3 bound leaving groups can be directly displaced by the nucleophile or the leaving group
can ionise off to form a carbocation (SN1) >C = C – L. However, sp2 bound leaving groups are
very difficult to ionise Halogen attached to allylic system are activated towards SN1 due to stability
of allyl carbocation.

4. Aromatic stabilisation occurs in rings that have an unbroken loop of p-orbitals. Any cyclic
compound will be stable especially when the ring contains (4n + 2) -electron.

5. The alcohols II & III are 3 but III gives more substituted C = C, I & IV are both 2 but IV can give
a more substituted C = C.



A 2 X3  3
 2A  3X
2
6.
2S 3S
Ksp = [2S]2 [3S]3
= 22. 33. S5
1.08  10 23 5
S 5  1 10 25  1 10 5
108

7. 

H2SO 4   2
 2H  SO 4


H2O  
 H  OH


At anode: 2OH 
 H2O  O  2e
O  O 
 O2
OH ions are discharged at anode as OH ions have lower discharge potential then SO24 .

1
8. Meq. of Na2CO3.xH2O in 20 mL = 19.8 
10
1 5
 Meq. of Na2CO3.xH2O in 100 ml = 19.8 
10
0.7 
 1000  19.8 
M 10
2
 M = 141.41
 23  2 + 12 + 3  16 + 18x = 141.41
x=2

9. 3Na2S2O3 + BiCl3 = Na3[Bi(S2O3)3] + 3NaCl

Bi2(S2O3)3 + 3H2O = Bi2S3 + 3H2SO4

10. N2H5+ + H2O N2H4 + H3O+


[N H ][H O ] k w 10 14 x2
Kh = 2 4 3    = 1.04  10–8
[N2H5 ] k b 9.6  10 7 0.1
 x = 3.2  10–5
3.2  105
% hydrolysis =  100 = 0.032 %
0 .1
11. NH3 + H+ Cl  NH4+ Cl
Solution contains NH4Cl in contact with NH3 (excess) & hence buffer

12. 2.303 log C/C0 = – kt

40% to completion means 60% unreacted

2.303 log 0.6 = – (3.3  10–4)t
 t = 26 min

9 .8
13. 9.8 g Cu =  2 = 0.308 mol e–
63.5
1
10000  3  = 0.311 mol e–
96500
If 0.311 mol electrons provided by the current 0.308 mol was used to deposit copper.
0.308
The current efficiency is  100 = 99.1%
0.311
14. Moles escaped = original – final moles
4  10 2.5  10
=  = 0.630 mol
0.082  290 0.082  290

1.26  10 4
20. = 77% cyclohexene
(1.26  10  4 )  (3.8  10  5 )
 methyl cyclopentene = (100 – 77) = 23%

21. Let x and y be the molar mass of A & B respectively then

Tf 8  1000 80
 m 1  … (1)
Kf  x  2y   100 x  2y
Tf 10  1000 100
&  m  1  … (2)
Kf  2x  y   100 2x  y
Solving (1) & (2) x = 40; y = 20.

22. Reaction of D-(+)-glucose with methanolic HCl leads to the formation of methyl glucoside
(C, - OH group is methylated) which, being acetal is not hydrolysable by base, so it will not
respond Tollen’s reagent.

23. For precise fitting of Zn2+ ions into tetrahedral voids of S2- ions packing r  / r  ratio must be 0.225.

 1 1
24.   RZ 2  2  2 
 n1 n2 
For last line of Lyman series of H-specturm; Z = 1, n1 = 1, n2 = .
+
For H line in Balmer series of He ; Z = 2, n1 = 2, n2 = 3.

25. NO+ has lost an antibonding electron where as CO+ has lost bonding electron.

26. Below CMC no micellization takes place. Sodium oleate ionizes almost completely in aqueous
solution.

1 1
27. H2  g  Cl2  g  HCl  g  … (1)
2 2
So   So  product   So  reactants 
o 1 1 
 SHCl   SHo 2  SoCl2 
2 2 
1
 187  131  223   10 JK 1
2

28. In PO34 , P shows sp3 hybridization.

In NO3 , N shows sp2 hybridization’
In ICl2 , I shows sp3 hybridization.

29. OF2  Pale yellow.

SF6 & SF4  Gas.

Mathematics PART – III

SECTION – A

1. Bisectors of the lines in both the cases will remain unchanged

n
2. D1  D02 , D 2  D12  D04 , ..... Dn  D02

1 1
 t  1
3. f  x    cos  x  t     cos  x  t  dt
 0  0
1 2
= cos x  2 sin x
 
1 4 2  4
fmax  2
 4

  2
4. y = x3 – 3x (–1, 2) y
y = –a 2 1
dy x
 3x 2  3
dx y = –a
 x = 1 or –1
For three different roots –a  (–2, 2) (1, –2)
k
5. x  icot , k  {1, 2, 3, ….., n – 1}
n

cosk
x n
Then 
1 k
isin
n
 x  1 2k 2k
   = cos  isin
 x  1 n n
n
 x  1
    cos 2k  isin 2k  1
 x  1
 (x + 1)n – (x – 1)n = 0
 n C1xn1  nC3 xn 3  .....  0
k
Roots of above equation are icot , k  {1, ..... n – 1}
n
2 2 n
n 1
 k   n1  k    p   q  C3
   icot 
n 
   icot 
n 
  2     icot  
n    icot
n  = 0  2 n
C1
k 1  k 1  1p  qn 1 

n  1n  2 
 f n 
3

 5 5 
6. c1 :  ,  c3
 2 2 p
r1 = 5
c1
 3 3 
c2 :  , 
 2 2 c2
r2 = 3

r be radius of third circle and its centre be at c3, then

2
 c 2c 3 2  pc 3 2   c1c 2   2
 c1c 3   pc 3  2
 

 
2
  r2  r   r 2   r1  r2   r 2 
2 2
r1  r 
 
4r r  r  r  15
 r  1 2 1 22 
r1  r2  8
=8

7. D be the origin and DA  a , DB  b and DC  c D

a  b  41 , b  c  36 , c  a  7  
a c
distance between mid-points of AB and CD 
2
b
abc 1 2
d2  
 2
 
 4
2 2

a  b  c  2a  b  2a  c  2b  c  A C

1 2
=
4  2 2 2 2
a  b  c  ab  ac  bc
2

= 137 B

8. 2b = a + c  2 sin B = sin A + sin C

 B B A C AC
 2  2sin  cos   2 sin  cos
 2 2  2 2
B 1 B 7 7
 sin  , cos   sinB 
2 2 2 2 2 2 4

9. Obviously for c  (0, 1), f(x) lies above the g(x), also x2 = cx3
1
 x = 0, x 
c
1/c
1 2
 x 
2
Hence,  cx3 dx  3

0 12c 3
1
 c
2

8 6
10. sec 40º, sec 80º, sec 160º are the roots of  1 0
t3 t
or t3 – 6t2 + 8 = 0
Sum of roots = 6

11. The chord of contact from (h, k) to y2 = 4x is ky = 2(x + h). If it touches the hyperbola
2 2 x2 y 2
 4h + k = 4 locus is  1
1 4

12. x = sin–1 (a4 + 1) + cos–1 (a4 + 1) – tan–1 (a4 + 1) is defined at a = 0 only


 x
4
1   2
 Area A =   
2 4 4 32

13. This can happen, if three lines are real and distinct as well as angle between any two adjacent
2
sides is
3
3 2
 f(m) = bm + dm + cm + a = 0 has three distinct real roots (where m = y/x)
m  m2 m  m3 m  m1
And 1  2  3  3
1  m1m 2 1  m2m3 1  m1m3
 3 + m1m2 + m2m3 + m3m1 = 0 and 3bm2 + 2dm + c = 0 has roots ,  with f() f() < 0
 3b + c = 0

14. 2n {x} = 3x + 2[x]

5x
  x  as 0  {x} < 1
2n  3
2n  3
 0  x  {n  2}
5
2n  3
It has 5 solution, [x] = 0, 1, 2, 3, 4, only if 4  5
5
23
  n  14  12, 13, 14
2

15. Let a is first term and d is common difference of the A.P.

a + (r – 1)d = 6, a + (s – 1)d = 8, a + (t – 1)d = 12
Solving, we get 2r + t = 3s also f(x) = tx 2 + 2rx – 2s
 f(0) = –2s, f(1) = s
 f(0) f(1) = –2s2 < 0 hence, exactly one root in (0, 1)

16. (a – b)2 + (b – c)2 + (c – d)2 + (d – a)2  0

 2(a2 + b2 + c2 + d2)  2(ab + bc + cd + ad)
 ab + bc + cd + ad  25

17. f(x) is continuous where 3 sin x + a2 – 10a + 30 = 4 cos x

or a2 – 10a + 30 = 4 cos x – 3 sin x
LHS  5, RHS  5
 (a – 5)2 + 5 = 4 cos x – 3 sin x
3
 a = 5, x = 2n – tan1   , n  I
4

18. Required ways = 6 C2  4C2  90

19. z = x + iy
2 2 1
  x  2   y  1  xy
2

20. Let a  3iˆ  4 ˆj  5kˆ

b  tan xiˆ  tan yjˆ  tan zkˆ
  2 2 2
 
ab  a b
 tan2 x + tan2 y + tan2 z  8

21. P = cos cos 2 cos 3 ….. cos 1004

Q = sin  sin 2 ….. sin 1004
1004
Then 2 PQ = sin 2 sin 4 ….. sin 2008
= (sin 2 ….. sin 1004)[sin (2 – 1003) sin(2 – 1001) ….. sin (2 – )]
= (sin 2 sin 4 ….. sin 1004)(–sin 1003) ….. (–sin ) = Q
1
 p  1004
2
2 2
22. 1 + x = t  x dx = t dt
tdt dt
I   2 1 t  c  2 1 1 x2  c
2 3 1  t
t t

23. If OP  OQ for P and Q on the hyperbola then b > a

Director circle of the hyperbola is x2 + y2 = a2 – b2, that exist only when a > b

24. Any plane through first line can be written as 3x + y + 2z – 2 + (2x + y + z – 1) = 0

3
It is parallel to x = y = z if   
2
Thus, plane parallel to 2nd line is –y + z – 1 = 0

25.  
A *  A  . If AA* = I, then A is unitary

26. Solution is y = m(x + 2)

Fixed point p is (–2, 0)
2
Area =
3

5
 1 1 2 1 5
27.  2   1  x  x   10 1  x
5
  1  x 5
 x x  x
Now, coefficient of x10 in (1 – x5)5 (1– x)–5 is = 14
C4  5C1  9C4  5 C2  4 C4  381

28. Mean = 1007

1
Variance 2 
1007
 
12  32  .....  20132  1007 2 = 338016

 d    338016  581.39

29. Let n(A) = n

n2 n
n2  n
Then 2  2 2
n=3

30. (sin–1 x + sin–1 y)(sin–1 z + sin –1 w) = 2 is possible only when x = y = z = w = 1 or –1 then

xn1 yn2
= 0, 2 or –2
zn3 w n2

Paper 1

Time Allotted: 3 Hours Maximum Marks: 246

 Pl ea s e r ea d t h e i n s t r u c t i o n s c a r ef u ll y . Yo u a r e a l l o t t ed 5 m i n u t es
s p ec i f i c a ll y f o r t h i s p u r p o s e.
ALL INDIA TEST SERIES

 Yo u a r e n o t a l l o wed t o l ea v e t h e E xa m i n at i o n Ha l l b ef o r e t h e en d o f
t h e t es t .

INSTRUCTIONS
A. General Instructions

B. Filling of OMR Sheet

C. Marking Scheme For All Three Parts.

1. Section – A (01 – 07) contains 7 multiple choice questions which have only one correct answer.
Each question carries +3 marks for correct answer and – 1 mark for wrong answer.
Section – A (08 – 11) contains 4 multiple choice questions which have one or more than one
correct answer. Each question carries +3 marks for correct answer. There is no negative
marking.
Section-A (12 – 16) contains 2 paragraphs. Based upon paragraph, 2 and 3 multiple choice
questions have to be answered. Each question has only one correct answer and carries
+3 marks for correct answer and – 1 mark for wrong answer.

2. Section – B (01 – 02) contains 2 Matrix Match Type questions containing statements given in 2
columns. Statements in the first column have to be matched with statements in the second
column. Each question carries +8 marks for all correct answer. For each correct row +2 marks
will be awarded. There may be one or more than one correct matching. No marks will be given for
any wrong matching in any question. There is no negative marking.
3. Section-C (01 – 06) contains 6 Numerical based questions with answers as numerical value from
0 to 9 and each question carries +3 marks for correct answer. There is no negative marking.

Name of the Candidate

Enrolment No.

Useful Data

PHYSICS
2
Acceleration due to gravity g = 10 m/s

Planck constant h = 6.6 1034 J-s

Charge of electron e = 1.6  1019 C

Mass of electron me = 9.1  1031 kg

2 2
Permittivity of free space 0 = 8.85  1012 C /N-m

Density of water water = 103 kg/m3

Atmospheric pressure Pa = 105 N/m2

Gas constant R = 8.314 J K1 mol1

CHEMISTRY

Gas Constant R = 8.314 J K1 mol1

Atomic No: H=1, He = 2, Li=3, Be=4, B=5, C=6, N=7, O=8,

Physics PART – I

SECTION – A
Straight Objective Type
This section contains 7 multiple choice questions. Each question has four choices (A), (B), (C) and (D)
out of which ONLY ONE option is correct.

 ( yiˆ  xj)
ˆ
1. Electric field at point P in a region whose coordinates are (x, y, z) is give by : E  E0
x2  y2
where E0 is constant. Which of the following can best represent the Electric field lines
corresponding to this field?
(A) y (B) y

O x
O x

O x O x

Space for Rough work

2. Net charge distributed on a ring shown in figure is zero. AA ' is axis of ring.
Then choose the correct statement
(A) Electrostatic field at all points an axis AA ' will be zero.
(B) Component of Electrostatic field along axis must be zero.
(C) Component of Electrostatic field perpendicular to the axis must be zero
(D) Electrostatic potential must be zero at all point inside ring in the plane of
ring

3. Figure shows a irregular shape conductor with

irregular cavity inside it. A charge Q is placed
inside cavity and a charge Q ' is placed outside
conductor. Let Q’ind be charge induced at outside
surface of conductor and Qind be the charge
induced at inside surface of cavity. A 'B ' is an
arbitrary line passing through charge Q and a, b, c
   
be the points on line as shown in figure. Let E, E', Eind and E'ind represent electric field at different
points due to charge Q, Q’ Qind, & Q’ind respectively. If Va,Vb and Vc represent potential at point a,
b, and c. Choose the incorrect statement

(A) Va  Vb  Vc

rb
     
(B) Vb 
 (E  E '  E ind  E 'ind ).dr , where rb = position vector of point b
–

rb
  
(C) Vb 
 (E '  E '

ind ).dr


(D) At point ‘b’, E'  0

Space for Rough work

4. Two wires W 1 & W 2 carry current ‘2i’ and i respectively directed into the plane A

of paper as shown in figure. AA ' is line at some distance from wire. Let B be W1

  2i d
the net magnetic field due to wires. The magnitude of

 B .d will be equal to
W2
(A) 0i /2 (B) 0i /4 i
(C) 0i /8 (D) zero A

5. A conducting rod of square cross section is bent in a semi circular

shape of radius ‘r’ as shown in figure. Length of side of cross
section is equal to ‘a’. Resistivity of material is ‘’. The resistance
of rod across its end will be
 2
(A) (B)
aln 1 a / r  aln 1 a / r 
(r  a) 2  r
(C) 2
(D)
a a2

6. A charge ‘q’ is undergoing S.H.M. about

point ‘A’ along the line AB at distance ‘d’
from the center B of a conducting sphere of
radius R. Amplitude oscillation of charge is
‘a’ angular frequency is ‘’. The sphere is
grounded through conducting wire CC ' .
Choose the correct statement

 Ra 
(A) If a < < d then the maximum value of current in wire CC ' will be q  2 
d 
 qRa2 
(B) If a < < d then the maximum value of current in wire CC ' will be  3 
 d 
 2Ra 
(C) If a < < d then the maximum value of current in wire CC ' will be q  2 
 d 
 2qRa2 
(D) If a < < d then the maximum value of current in wire CC ' will be  3 
 d 

Space for Rough work

7. A hollow cylindrical conductor (inner radius = a, outer radius = b) carries a current i uniformly
spread over its cross section. Which graph below correctly gives B as a function of the distance r
from the center of the cylinder?
(A) B (B) B

a r a r
b b
(C) B (D) B

a r a r
b b

(Multiple Correct Choice Type)

This section contains 4 multiple choice questions. Each question has four choices (A), (B), (C) and (D)
out of which ONE or MORE THAN ONE are correct.

8. A rod is free to rotate in horizontal plane about one of its end hinged
at point ‘O’ as shown in figure. Point P & Q are fixed points located in
horizontal plane in such away that PO is perpendicular to rod & QO
is along length of rod. A bullet hits the rod perpendicularly at some
distance from point ‘O’ and gets imbedded into it. Then :
(A) Linear momentum of rod + bullet system will decrease after
collision.
(B) Linear momentum of rod + bullet system may increase
(C) Angular momentum of rod + bullet system about point ‘P’ will not
change before & after collision.
(D) Angular momentum of rod + bullet system about ‘Q’ may
increase after collision.

Space for Rough work

9. Shape of string transmitting wave along x-

axis at some instant is shown
Velocity of point P is v = 4 cm/s and
  tan1(0.004)
(A) Amplitude of wave is 2 mm
(B) Velocity of wave is 10 m/s
(C) Max acceleration of particle is 80 2
cm/sec2
(D) Wave is traveling in –ve x-direction

10. A small ball of mass m is placed in a circular tube of mass M and radius R in v0
gravity free space. Friction is absent between tube and ball. Ball is given a
velocity v0 as shown. Then,
m
(A) Path of ball from centre of mass of system will be circular.
R
(B) Path of ball from centre of mass of system will be elliptical

MR
(C) Radius of curvature of ball at the time of projection of ball is
mM
2mV02
(D) Normal force between tube and ball if M = 2m, at the time of projection of ball is
3R

D
11. A system of rods is assembled such that each rod has a length 
2k k
and cross-sectional areas S. The mode of heat transfer is
A k C
conduction and the system is in steady state. The temperature of
Junction A is T and that of C is 2T. Now answer the following k 2k
question. B
(A) Temperature of junction B is 1.6 T.
(B) Temperature of junction D is 1.4 T.
1 kTS
(C) The rate of heat flow along BD is .
5 
21kTS
(D) The rate of heat flow along BD is .
5 

Space for Rough work

Comprehension Type

This section contains 2 paragraphs. Based upon one of the paragraphs 2 multiple choice questions
and based on the other paragraph 3 multiple choice question have to be answered. Each of these
questions has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

Paragraph for Questions 12 & 13

A square wire loop having side a, mass m and resistance R is moving y 

B
along positive x-axis with speed v 0 enters into uniform steady magnetic a

ˆ at t = 0 sec as shown in the figure.
field B  B0 (k)
a x
The magnitude of v0 is sufficient so that the loop comes out from region
of magnetic field with v1. Neglect any type of energy loss other than
3a
3B2 a3
heat loss in resistances of wires of square (given v0  )
mR
Read above passage carefully and answer the following questions.

12. The plot of speed of square versus position x will be represented by

(A) v (B) v

O a 2a 3a x O a 2a 3a x

13. The total amount of heat loss in the resistance will be

5 4
(A) mv 02 (B) mv 20
9 9
2 mv 20
(C) mv 02 (D)
9 9

Space for Rough work

Paragraph for Questions 14 & 16

Figure shows intensity versus wavelength graph of X-rays coming from Coolidge-tube with molybdenum
as anticathode:

The two peaks shown in graph correspond to K & K X-rays

14. Wavelength of L X-rays from Coolidge tube will be (approximately)

(A) 5.60Å (B) 4.26 Å
(C) 0.33 Å (D) 1.34 Å

15. Voltage applied across Coolidge tube in (approximately)

(A) 20 kV (B) 16 kV
(C) 31 kV (D) 18 kV

16. In the potential difference across the Coolidge tube is increased such that the cut off wavelength
of x-ray becomes 0.2 Å. Find the value of    in this condition.
(A) 0.02 Å (B) 0.04 Å
(C) 0.06 Å (D) 0.08 Å

Space for Rough work

SECTION - B
Matrix – Match Type
This section contains 2 questions. Each question contains statements given in p q r s t
two columns, which have to be matched. The statements in Column I are
p q r s t
labelled A, B, C and D, while the statements in Column II are labelled p, q, r, A
s and t. Any given statement in Column I can have correct matching with B p q r s t
ONE OR MORE statement(s) in Column II. The appropriate bubbles
corresponding to the answers to these questions have to be darkened as C p q r s t
illustrated in the following example:
D p q r s t
If the correct matches are A – p, s and t; B – q and r; C – p and q; and D – s
and t; then the correct darkening of bubbles will look like the following:

1. In the adjacent circuit shown, all ammeters 0.5 P 5V

+  0.5
A2
and batteries are ideal. Internal resistances 20 V 10 V
of all the batteries are negligible. Point R is 1
grounded. Then, match the following. 5V 20 V
Q +  + 

12.5  10 

A1
R

Column – I Column – II
(A)
Potential of point P (p) 6 SI unit
(B)
Potential of point Q (q) 25 SI unit
(C)
Reading of ammeter A1 (r) 18 SI unit
(D)
Reading of ammeter A2 (s) 47 SI unit

Space for Rough work

2. A lens is made up of a material of refractive index  which is completely immersed in a medium of

refractive index 0. The power of lens is P. Match the properties in column I with the
corresponding type(s) of lens in column II
Column I Column II
(A) P  0 and  > 0 (p) Bi-convex
(B) P  0 and  < 0 (q) Bi-concave
(C) P  0 and  > 0 (r) Convexo concave
(D) P  0 and  < 0 (s) Concavo convex
(t) Plano convex

SECTION – C
(Integer Answer Type)

This section contains 6 questions. Each question, when worked out will result in one integer from 0 to 9
(both inclusive).

1. Conducting shell of radius R carrying charge Q, 3R

another charge Q is placed at 3R distance from
centre of shell. If the potential at point P due to R
11 KQ
charge on shell is , Then the value of n is O
3n R P Q
R/2

2. A liquid is kept in a cylindrical vessel. When the vessel is rotated about its axis, the liquid rises at
its sides. If the radius of the vessel is 0.05 m and the speed of rotation is 2 revolutions per
second, the difference in the heights (in cm) of the liquid at the centre and at the sides of the
vessel is (take g = 10 ms2 and 2 = 10)

Space for Rough work

3. When a resistance R is connected in series with an element A, the electric current is found to be
lagging behind the voltage by V angle 30 When the same resistance is connected in series with
element B, current leads by 60. When R, A, B are connected in series, the current new leads
voltage by  . Which is equal to tan1(K/3), then the value of K is (assume same AC source is
used in all cases)

4. A ball of mass m = 2kg thrown from ground form point ‘O’ between
two walls, W1 & W2 at a distance 5 cm from ‘O’ an shown in figure.
Had the wall w2 not been there range of ball would have been 20 m.
Horizontal component of velocity of ball at the time of projection was
5 m/s. All collisions are elastic. Average force acting on one of the
walls (in newton) is

5. A small object at P oscillates infront of a concave mirror as R

shown in the figure perpendicular to principal axis according to
equation. P C
 
y = 4 sin  t   cm,
 3
3R/2
y is the displacement of particle perpendicular to principal axis.
The amplitude (in cm) of image is

6. A particle moves in a circle with a uniform speed, when it goes from a point A to a diametrically
 
opposite point B, the momentum of the particle changes by PA  PB  2kgm / s(ˆj) and the
 
ˆ where ˆi, ˆj are unit vectors. Then the
centripetal force acting on it changes by FA  FB  8N(i)
angular velocity (in rad/s) of the particle is

Space for Rough work

Chemistry PART – II

SECTION – A
Straight Objective Type
This section contains 7 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for
its answer, out which ONLY ONE option is correct.
 Br  
1. At what does the following cell have its reaction at equilibrium?
CO32  
Ag(s) |Ag2CO3(s)| Na2CO3(aq)| |KBr(aq)| AgBr(s)| Ag(s)
–12 –13
Ksp = 8  10 for Ag2CO3 and Ksp = 4  10 for AgBr

(A) 1 107 (B) 2 107

7 7
(C) 3  10 (D) 4 10

2. A human body excretes [removes by waste discharges, sweating etc] certain material by a law
similar to radio activity. If technitium is injected in some form in the human body, the body
excrets half of the amount in 24 hours. A patient is given an injection confainingTc99. This
isotope is radioactive with a half life 6 hrs. The activity from the body just after the injection is 6
C. How many hours will elapse before the activity falls to 3 Ci?
(A) 24 hours (B) 6 hours
(C) 4.8 hours (D) 12 hours

3. Find the product of the following reaction

R R

CH2
R O
(A) R O (B) R
and H3C CH2 R
R R
O R
(C) R (D) R
and H2C CHO R CH2
R R
R
O

4. A big RED spherical balloon (radius = 6a) is filled up with gas. On this balloon six small GREEN
spherical balloons (radius = a) are stuck on the surface in a specific manner. As RED balloon is
slowly deflated, a point comes when all these six GREEN balloons touch and green balloons
arrange themselves in a 3-D closed packing arrangement. At that stage the radius of the RED
balloon would have reduced by approximately.
(A) 14.5 times (B) 1.414 times
(C) 6.0 times (D) 2.42 times
Rough Work

5. Which of the following graphs represents the radial charge density for the electron of outer most
+2
sub shell of Cu ?

4r2 R2 4r2 R2

(A) (B)
r r

4r2 R2 4r2 R2

6. The order of reactivity towards Perkin reaction of the following aromatic aldehydes is
CHO CHO CHO CHO
CH3

N(CH3)2 NO2

(a) (b) (c) (d)

(A) (d) > (a) > (b) > (c) (B) (c) > (a) > (b) > (d)
(C) (b) > (c) > (a) > (d) (D) (d) > (a) > (c) > (b)

7. Which of the following represents the correct order for the wavelength of absorption in the visible
region for the following
[Ni(NO2 ) 6 ]4 [NiF6 ]4  [Ni(H 2O) 6 ]2  [Ni(NH 3 ) 6 ]2 
(I) (II) (III) (IV)
(A) I  IV  III  II (B) II  IV  III  I
(C) II  III  IV  I (D) I  III  II  IV
Rough Work

Multiple Correct Choice Type

This section contains 4 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for its
answer, out which ONE OR MORE is/are correct.

8. A + B   C + D is a stoichiometrically balanced reaction. The initial rate of the reaction is

doubled if the initial concentration of A is doubled, but is quadrupled if the initial concentration of
B is doubled.
Select the correct statement(s)
(A) The reaction is first order in B and second order in A
(B) The reaction is first order in A and second order in B
(C) The reaction can not be a single-step reaction
(D) The overall order of the reaction is 3

9. In which of the following cases will a violet colouration be observed

(A) An alkaline solution of sodium nitroprusside is treated with a solution of Na2S
(B) A solution of sodium cobaltinitrite is treated with KCl
(C) A solution of Mn(NO3)2 is treated with sodium bismuthate or red lead in the presence of
concentrated HNO3
(D) A solution of sodium nitroprusside in aqueous, NaOH is heated ith Na2SO3

?
10. CH 3  CH  NH   CH 3CH 2  NH 2 for this conversion suitable reagent
(A) Na  Hg H 2 O (B) H 2 Pd
(C) LiAlH 4 Et 2 O (D) NaBH 4 EtOH

11. X  H 2SO 4  Y (a colourless gas with pungent smell)

Y  K 2Cr2O 7  H 2SO 4  Green solution
(A) X is SO32– (B) X is CO3–2
(C) Y is H2S (D) Y is SO2

Rough Work

Comprehension Type
This section contains 2 paragraphs. Based upon the first paragraph 2 multiple choice questions and
based upon the second paragraph 3 multiple choice questions have to be answered. Each of these
questions has four choices A), B), C) and D) out of WHICH ONLY ONE IS CORRECT.

Paragraph for Question Nos. 12 to 13

The process of eudiometry is used to calculate the volume of the various reacting gases present together
in a closed vessel, it is done by sparking the content and noting the volume change or pressure change
depending upon the conditions mentioned. Various reagents like alkaline pyrogallol, turpentine oil, caustic
potash, conc. H2SO4 are used to absorb the selective gases like O2, O3, CO2 and H2O (v) respectively.
A 2.0 mole mixture of H2(g), O2(g) and He(g) are placed together in a closed container at pressure equal
to 50 atm. An electric spark is passed and pressure is noted as 12.5 atm after cooling the content to
original temp. (room temp). Oxygen gas is introduced for pressure to change to 25 atm, keeping volume
and temp constant. Again electrical spark is passed and pressure drops to 10 atm under original temp.
conditions.

12. The moles of He(g) present in the mixture sample is

(A) 0.2 (B) 0.1
(C) 0.5 (D) 1.4

13. The mole fraction of H2(g) present in the initial mixture is:
(A) 0.05 (B) 0.7
(C) 0.5 (D) 1.25
Rough Work

Paragraph for Question Nos. 14 to 16

An unknown organic compound (A) with the molecular formula C9H12O does not decolourize Br2 in CCl4,
reacts with sodium to give a colourless, odourless gas and can be oxidised by hot alkaline KMnO4 to give
benzoic acid. Then, following test were performed on the compound (A) is given sequence:
Test (a): The colour of the solution changes from orange to green when K2Cr2O7 is added to it.
Test (b): The compound can be resolved.
Test (c): No yellow precipitate of CHI3 is formed when reacted with I2 and OH–.
Test (d): Oxidation of the compound with CrO3 in pyridine gives another compound that gives
Tollen’s test.

14. Which of the following would be the structure of compound if test (a) is negative?
(A) PhCH(OH)CH2CH3 (B) PhC(OH)(CH3)2
(C) PhCH2CH(OH)CH3 (D) PhCH(CH3)CH2OH

15. Which of the following would not be the structure of compound (A) after test (b)?
(A) PhCH(OH)CH2CH3 (B) PhCH2CH(OH)CH3
(C) PhCH2CH2CH2OH (D) PhCH(CH3)CH2OH

16. The structure assigned to compound (A) after the test (d) would be
(A) PhCH(OH)CH2CH3 (B) PhCH2CH(OH)CH3
(C) PhCH(CH3)CH2OH (D) PhC(OH)(CH3)2
Rough Work

SECTION-B
(Matrix-Match Type)
This section contains 2 questions. Each question contains statements given p q r s t
in two columns, which have to be matched. The statements in Column I are
A p q r s t
labelled A, B, C and D, while the statements in Column II are labelled p, q, r,
s and t. Any given statement in Column I can have correct matching with B p q r s t
ONE OR MORE statement(s) in Column II. The appropriate bubbles
corresponding to the answers to these questions have to be darkened as C p q r s t
illustrated in the following example:
D p q r s t
If the correct matches are A – p, s and t; B – q and r; C – p and q; and D – s
and t; then the correct darkening of bubbles will look like the following:

1. Match the following metals given in Column I with the appropriate metal extraction process(es)
listed below in Column II.
Column – I Column – II

(A) Silver (p) Fused salt electrolysis

(B) Lead (q) Cyanide process

(C) Iron (r) Carbon monoxide reduction

(D) Magnesium (s) Self reduction

(t) Frothflotation

Rough Work

2. Column – I contain reaction and Column – II consist of reagent used in these reaction. Match the
following.
Column – I Column – II

(A) CHO CHO (p) X = Na/liq. NH3

 
X  
Y
Y = cold dil. KMnO4/OH
   

(B) OH CHO (q) X = Br2, CHCl3

Y = CH3ONa/CH3OH
  X  Y
   

(C) (r) X = O3/H2O, Zn

Me Y = NaOH/

 
X  Y HC Me COCH3
   

CH(CH3)2

(D) O (s) X = HIO4

Y = NaOH/
H3 C C C CH 2 C CH3 X
  Y
  Racemic Mix.

(t) X= H2/Pd
Y=Zn-Hg/ conc. HCl

Rough Work

SECTION –C
Integer Answer Type
This section contains 6 questions. Each question, when worked out will result in one integer from 0 to 9
(both inclusive).

1. How many of the following reactions are addition-elimination reactions of carbonyl compounds?
(I) Cyanohydrin formation
(II) Acetal formation
(III) Semicarbazone formation
(IV) Aldol formation
(V) Hemiacetal formation
(VI) Oximeformation
(VII) Ketal formation
(VIII) Imine formation
(IX) 2,4-DNPderivative formation

2. Total number of stereoisomers of the H

compound given below is x
C
C COCH3

And number of chiral carbons are there CH3CO

in the product ‘P’is y
H3C
 Br2 /CCl4
  ?  P

then the value of (x+y)/2 is?

3. A 50 ml 1.92% (W/V) solution of a metal ion Mn+ (atomic weight = 60) was treated with 5.332 g
hydrazinehydrate (90% pure) and mixture was saturated with CO2 gas when entire metal gets
precipitated as a complex [M(H2N – NHCOO)n]. The complex was filtered off and the filtrate was
M
titrated with KIO3 in the presence of conc. HCl according to the following equation:
10
N2H4  IO3  2H  2Cl   ICl  3H2O  N2 
M
The volume of KIO3 solution needed for the end point to arrive was 480 ml. Find the value of
10
n.
Rough Work

4. How many statements are true in the given question of about?

(i) orthoboric acid It is a strong tribasic acid.
(ii) orthoboric acid does not act as a proton donor but behaves like a Lewis acid by accepting OH
ion.
(iii) orthoboric acid is prepared by the action of mineral acid on borax solution.
(iv) orthoboric acid has a layer structure in which hydrogen bonds unite B(OH)3 units.

(v) If the number of revolutions made by electron in1.0 s in H atom in its nth orbit is twice of the
number of revolution made by electron in 1.0 s in the 2nd orbit of Hatom, then n is 1.

(vi) Forms a colourless solution with coloured precipitate(r)CrCl3(s) + dil HNO3(aq)

(vii) Forms a coloured solution with no precipitate(s)K2CrO4(aq) + Ba(NO3)2(aq)

(viii) F > N > C > Si >Ca - metallic character

(ix) Saccharin (pKa = 2) is an artificial zero calorie sweetner added to toothpaste
O

C
+
N Na

S
O O
Sodium Saccharin Structure .Two flasks A&B both contain 18.3 gm of Saccharin dissolved in 100
ml solution. 2 grams of solid NaOH is added to A and 4 grams of solid NaOH is added to B. This
resulted in the formation of the above structure in both flasks. The difference in pH of the two
flasks after the experiment is: 6
(x) C > Si > P > N - electron affinity
(xi) O > N > F > C - second ionisation potential
(xii) I2 <F2  Br2  Cl2 Bond energy ORDER

(xiii) mixture of HCl and CH3COOH titrated against NaOH the correct graphical representation
is
Conductance

Vol. of titre

(xiv) The CORRECT ORDER OF basic character of the following alcohols is

CH2OH CH2OH CH2OH
CH 2OH
> > >

OCH3 Cl NO2

Rough Work

5. In the given graph, the area of circle A

and B are 40 unit and 36 unit respectively,
total work done ‘X’ unit.
A B
P

V
When a system is taken from state A to B C B
along path ACB, as shown in the figure, 5
unit of heat flows into the system and the
system does 3 unit of work. If ‘y’unit heat
flows into the system along the path ADB,
and the work done by the system is 2 unit A D
?

Then the value of x + y is

6. Alkaline hydrolysis of chloral hydrate produces a compound (A) which is used as a solvent and
preservative for anatomical specimens and also has anesthetic properties. (A) is prone to aerial
oxidation in presence of sunlight to give a poisonous compound (B) which may be made non
poisonous by adding dilute solution of ethanol which converts it into a non poisonous compound
(C). How many moles of CH3MgBr will be added to (C) followed by acidic hydrolysis to give
alcohol?
Rough Work

Mathematics PART – III

SECTION – A

Straight Objective Type

This section contains 7 multiple choice questions numbered 1 to 7. Each question has 4 choices (A), (B),
(C) and (D), out of which ONLY ONE is correct

1. The number of 6-digit numbers whose digits are selected from set {5, 6, 7, 8, 9} such that any
digit that appears in the number appears at least twice is
(A) 1100 (B) 1200
(C) 1400 (D) 1405

2. The equation of common tangent to the curve |z – 2i| = 2 and |z – 1 – 3i| + |z + 1 – 3i| = 2 2 is
(A) z  z  4 (B) z  z  8i
(C) z  z  8 (D) z  z  4i

n n
3. Let f  n    k Cr then total number of divisors of f(9) is equal to
r 0 k r
(A) 5 (B) 6
(C) 7 (D) 8

100 50
100   
4. 
Let 2 1  x 3    ai xi  cos  2  x  i    . If  a2i  2k
i0   i0
then the value of k is

(A) k = 100 (B) k = 200

5. Let M = {(x, y) : y  x2} and N = {(x, y) : x 2 + (y – a)2  1}. The necessary and sufficient condition
for M  N = N is
5 5
(A) a  (B) a 
4 4
(C) a  R+ (D) a  (0, 1)

Space for rough work


3
x loge  sin x  dx
0
6. The value of the ratio 
is
x
2
loge  
2 sin x dx
0

 2
(A) (B)
2 3
3 3
(C) (D)
4 2
2
7. A parabola y = ax + bx + c crosses the x-axis at (, 0) and (, 0) both to the right of the origin. A
circle also passes through these two points the length of tangent from the origin to the circle is
a c
(A) (B)
b a
2
b
(C) bc (D) 2
c

Multiple Correct Choice Type

This section contains 4 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for its
answer, out of which ONE OR MORE is/are correct

8. Positive integers are written on all the faces of a cube, one on each. At each corner (vertex) of
the cube, the product of the numbers on the faces that meet at the corner is written. The sum of
the numbers written at all the corner is 2004. If T denotes the sum of the number on all the faces,
the possible value of T are
(A) 174 (B) 175
(C) 339 (D) 505

9. If the equation x2 + (a – 2)x + 1 = 3|x| has exactly three distinct real solutions in x, the value of a
is/are
(A) 1 (B) 2
(C) 3 (D) 4

Space for rough work

10. In an acute triangle ABC, O is circumcentre, H is orthocentre and G is the centroid. Let OD be
perpendicular to BC and HE be perpendicular to CA, with D on BC, E on CA. Let F be the mid-
point of AB. Suppose the area of triangle ODC, HEA and GFB are equal the possible value of
angle C is
 
(A) (B)
3 4
 
(C) (D)
6 2

11. A player throws an ordinary dice, whenever he throws 1, he gets an additional throw. Then
probability to get sum of score n is
1  64  1  1  65  1 
(A)  4  is n = 5 (B)  5  if n = 6
5 6  5 6 
1  65  1 
(C)   if n = 7 (D) 0 if n = 1
30  65 

Comprehension Type

This section contains 2 paragraphs. Based upon one of paragraphs 2 multiple choice questions and
based on the other paragraph 3 multiple choice questions have to be answered. Each of these questions
has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct

Paragraph for Question Nos. 12 to 13

Read the following write up carefully and answer the following questions:
A trapezium ABCD, in which AB is parallel to CD, is inscribed in a circle with centre O. Suppose the
diagonal AC and BD of trapezium intersect at M and OM = 2

12. If AMB = 60º then difference between length of parallel sides is

(A) 3 (B) 2
(C) 2 3 (D) 4

13. If AMD is 60º then difference between length of parallel sides is

(A) 2 (B) 3
(C) 2 3 (D) 4 3

Space for rough work

Paragraph for Question Nos. 14 to 16

Read the following write up carefully and answer the following questions:
An ellipse whose major axis is parallel to x-axis such that segment of a focal chord are 1 and 2 units. The
x y 1
line    0 are the chords of the ellipse such that a, c, b are in H.P. and are bisected by the point
a b c
at which they are concurrent, the equation of auxiliary circle is 2x 2 + 2y2 + 4px + 4qy – 16p + 1= 0 then

14. The centre of ellipse is

(A) (1, –1) (B) (1, 1)
 1 1  1 1
(C)  ,  (D)   ,  
 2 2  2 2

15. Equation of director circle is

2 2 37 2 2 37
(A) x + y – x – y – 0 (B) x + y + x + y + 0
6 6
37 37
(C) x2 + y2 + x + y – 0 (D) x2 + b2 – x – y + 0
6 6

16. Eccentricity of ellipse is

1 1
(A) (B)
3 3
2 2
(C) (D)
3 3

Space for rough work

FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942
website: www.fiitjee.com
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AITS-FT-II-(Paper-1)-PCM-JEE(Advanced)/15

1. Match the following Column-I with Column-II

Column – I Column – II
(A) The number of all triples (a, b, c) such that all three term a,
b, c are in harmonic progression in which a = 20 and b
(p) 66
divides c (where a, b, c are strictly increasing positive
integers)
(B) The number of solutions of m + n + p = 10, (m, n, p are
(q) 20
non-negative integers)
(C) The number of ordered pair (a, b) satisfying the equation
(r) 4
ab + b – a + 1 = 0 are where a, b  I
(D) The number of ordered pairs (x, y) of real numbers such
(s) 5
that (x + iy)20 = x – iy is (where i = 1 )
(t) 22

2. Match the following Column-I with Column-II

Column – I Column – II
(A) Let f(x) = x 4 + ax3 + bx2 + cx + d (where a, b, c, d are real
coefficient) and f(x) = 0 has real roots, If |f(i)| = 1 and (p) 0
a = b = c = d = k then k can be (where i  1 )
x x2 x3 x4
(B) If x5 = 1 (x  1) then    equals (q) 1
1  x 2 1  x 4 1  x 1  x3
 1
(C) Let f  x   ln cos x   (where [.] denotes the greatest
 2
x2    f  x  n  

integer function), then  lim  2   dx is (r) 2
 n  x  tan2 x  
x1 
  
  
(where x1, x 2    ,  )
 6 6
(D) Let ABCD be a cyclic quadrilateral inscribed in a circle of
radius 2, such that BD = 2 3 , AB = 1. If BAD is acute (s) 3
and AD = a  b 5 , then the value of a + b is equal to
(t) 4

Space for rough work

SECTION – C

(Integer Answer Type)

This section contains 6 questions. Each question, when worked out will result in one integer from 0 to 9
(both inclusive).

1. Let X be the set of all positive integers greater then or equal to 8 and let f : X  X be a function
such that f(x + y) = f(xy) for all x  4, y  4. If f(8) = 9 then f(9) is equal to _____

2. If f(x) and g(x) be two function such that f(x) = 2f(x) – 2f(x), g(x) = f(x) – f(x). Let F(x) be defined
as F(x) = [f2(x) + g2(x)][f 2(–x) + g2(–x)]. If F(0) = 1 then the value of F(1) is _____

5
3. Let f(x) = tan–1 (cot x – 2 cot 2x) and  f  r   a  b where a, b  N, then the value of b is _____
r 1

4. The least possible area of convex quadrilateral having two vertices on two branches of hyperbola
xy = 2 and other two vertices on the two branches of xy = –2 is _____

200 200
5. Consider two polynomial f(x) and g(x) given by f  x    r xr and g  x    r xr such that r = 1
r 0 r 0
200
 100  r  200 and f(x + 1) = g(x). Let A   r , then the remainder when A is divided by 15 is
r 100
equal to _____

6. Consider a square ABCD of diagonal length 2 3 . The square is folded along the diagonal AC so
that the plane of ABC is perpendicular to the plane of ADC. The shortest distance between AB
and CD is _____

Space for rough work

ANSWERS, HINTS & SOLUTIONS

FULL TEST–II
(PAPER-1)

Q. No. PHYSICS CHEMISTRY MATHEMATICS

1. C B D
ALL INDIA TEST SERIES

2. B C B

3. D D D

4. A A A

5. A A A

6. A A D

7. C C B
B, C, D B, C, D A, B, C, D
8.
A, B, C, D A, C A, C
9.
A, C, D A, B, C, D A, B
10.
A, B, C
11. A, D A, B, C, D
12. A B C
13. B B C

14. A B D

15. C C C
16. D C B
(A) (r), (B) (q), (A → q) (B → s) (A)  (s) (B)  (p)
1. (C → r) (D → p)
(C) (p), (D) (p) (C)  (r) (D)  (t)
(A)  (p, s, t), (A)  (q), (B)  (s), (A)  (p) (B)  (r)
1. (B)  (q, r), (C)  (C)  (r), (D)  (p) (C)  (p) (D)  (r)
(q, r), (D)  (p, s, t)
1. 5 4 9
2 7 1
2.
2 n=3 5
3.
5 9 8
4.
2 8 1
5.
4 3 2
6.

Physics PART – I
SECTION – A

1. Electric field is tangential direction at all points (E.rˆ  0) and decreasing with increase in r. (r =
magnitude of radius vector of a point)

2. Since, Electrostatic potential is zero at all points on axis,

v
So, along axis  0  Er = 0
r

3. Va  Vb  Vc because electric field inside conductor is zero.

 
rb rb
         
Vb   (E  E '  E ind  E ind ).dr   ' ).dr , because E ' E ind
(E '  E ind '  0 for all points inside
 
conductor or outside the conductor.

4. An infinite wire can be taken as half of closed loop.


   0i
Hence  B  d 
2


5. Let us consider a strip of thickness dx at a distance ‘x’

 . x
dR 
adx
ra
1 1 a dx 
Reff
  
dR  
r
x
 Reff 
 a
aln 1 
 r

q K qind
6. VA  Vind  0  K .  0
(d  a sin t) R
R.q dqind Rqa cos t
 qind  –  
(d – a sin t) dt (d  a sin t)2
8. Force applied by hinge on rod may be towards P, away from P or zero depending upon distance
from point ‘O’. Accordingly linear momentum of rod + bullet system may decrease increase or
remains constant respectively. Similarly angular momentum of system about ‘Q’ may decrease,
increase or remain constant respectively.

9. =1m
vmax = A = 4
y / t 4 10 2
 v   10m / s
y / x 4  103
–3 v
A = 2 × 10 m also    10Hz and amax  A2  80  cm / s2


10. From centre of mass of system ball and centre of tube moves in circular path.
MR
Radius of curvature of ball at the time of projection of ball is
mM
Normal force between tube and ball if M = 2m, at the time of projection of ball is
mV0
Vcm =
mM
  V  V 2 
N  m 0 cm 
 r1 
 

11. Equivalent circuit is D

R 2R

A 2R C

2R R
B

1 1 1 M
14.   L
 k   k   L L
K K
k  .  k 
 L    5.60Å K
k  –  k 

hc hc 6.626 10 34  3 10 8

15. eV   V   31103 V
 cut off e cut off 1.6  10 19  0.4 10 10

SECTION - B
1. Using Kirchof’s first law at point P

SECTION – C
1. VO = VP
KQ KQ 2Kq
  Vshell 
R 3R 5R
KQ  1 2 
Vshell  1 
R  3 5 
KQ  4 2 
 
R  3 5 
KQ  20  9  11 KQ
= 
R  15  15 R
2. Using Bernoulli’s theorem
1 2 v2
v  p  gh or h  ….. (1)
2 2gh
Put : V = r = r (2) in (1)

1 1 1
5.  
v u f

1 1 2
 
v (3R / 2) R
1 2 2 4
  
v R 3R 3R
3R
v= 
4
 3R 
v   2 1
4 
m      
u  3R  4 2
 
 2 
 1
amplitude of image = 4     2
 2

Chemistry PART – II
SECTION – A

1. Ag | Ag2CO3  s  | Na2CO3  aq. || KBr  aq. | AgBr  s  | Ag  s 

Anode : 2Ag  s   CO32 
 Ag2CO3  s   2e
Cathode : AgBr  s   e 
 Ag  s   Br   aq.
CO32  aq.  2AgBr  s  
 2Br   aq.  Ag2CO3  s 
2
o 0.059 Br  
Ecell  E cell  log
2 CO32 
At eqm.
2
0.059 Br  
0 E o
cell  log  2
2 CO3 

o
Br  
0 E cell  0.059 log
2
CO3 
Br  
0.059log   EoC  EoA
2
CO3 
EoAg/ AgCl/ Cl  E oAg / Ag  0.059 log K spAgBr
K sp Ag2CO3
EoAg/Ag CO 2  EoAg  0.059 log
3 /CO3
2
K sp AgBr
Br   K AgBr
log   log sp
CO23 K sp Ag2CO3
Br  4  10 13
  2  10 7
CO32  8  10 12

2. 1N  2N  N
ln 2 ln 2
1  2 
24 6
No  6
No e t  3
et  2
t  ln 2
 ln2 ln2 
 24  6  t  ln 2
 
t = 4.8 hours

3. R R

R CH2
O

R R

R + CH2
O

R R
R R
CH2 R O

H2C CH CH2
R O

r
4.  0.414 for octahedral void fitted with our red spherical balloon.
r
r
 0.414

r   0.414
Initial r  radius = 0.6
Final r  radius = 0.414
Initial r 6
  14.5
Finalr 0.414

5. The number of maxima in the radial charge density curve is (n  l)

For 3d electron, n = 3 and d = 2
Hence, n  l = 3  2 = 1
There will be only one maximum in the curve for 3 d electron.

6. In Perkin reaction, the carbanion attacks the carbonyl carbon of aromatic aldehyde to produce
finally the unsaturated acid
The electron withdrawing groups increase reactivity of aldehydes in this reaction by increasing
the electron deficiency of carbonyl atom. Thus NO2 increases the reactivity and the groups like 
N(CH3)2 and CH3 decreases the reactivity.
– –
7. The order of ligand strength in the spectrochemical series F  H2O  NH3  NO2 . A strong ligand
causes a larger degree of splitting resulting in high value of E (energy). Therefore corresponding
 hc 
low value of   E 
  

Protonation occurs on the OH which can produce more stable carbocation and while
rearrangement phenyl has higher migratory aptitude over o-substituted phenyl.

Solutions for Q. No. 12 to 13

Let the no. of moles of H2, O2 and He are x, y and z respectively
 x + y + z = 2.0
Pressure exerted by 2 moles of gaseous mixture is 50 atm.
After the first electric spark decrease in pressure = 50 – 12.5 = 37.5 atm
 decrease in no. of moles of gaseous mixture is 1.5
But from the given information limiting reagent in first step is O2
1
 from H2  g  O 2  g  H2O   
2
 H2 and O2 reacted are in the ratio of 2 : 1
 0.5 mole of O2 should be reacted with 1.0 mole of H2
 No. of mole of O2 in the mixture (y) = 0.5
When again O2 is passed, pressure increased from 12.5 to 25.0 atm
 Change in pressure = 12.5 atm
12.5  2
 No. of moles of O2 added =  0.5 mole
50
Now H2 will be completely reacted
No. of moles H2 actually left after first electric spark = x – 1
1
H2  O2  H2O
2
x  1
 x  1  
 2 
Now change in pressure in 25 – 10 = 15 atm
15  2
 change in no. of moles is  0.6
50
 x  1
  x  1     0.6
 2 
 x  1.4
 no. of moles of H2 = 1.4
 no. of moles of He = 2.0 – (1.4 + 0.5) = 0.1
1.4
 mole fraction of H2 = = 0.7
2.0

14. Possible structure after test (a) would be that of a 1 or a 2 alcohol and not that of 3 alcohol.

15. Only non-resolvable compound is a 1 alcohol out of the given choices.

16. Oxidation of compound with CrO3 in pyridine gives a compound, that gives Tollen’s test. This
implies that it is a primary alcohol.
SECTION-B

1. Silver Ore is dissolved by complex formation using CN in presence of air followed by
precipitation with zinc.
Lead  Involves roasting of PbS followed by self reduction.
Iron  Involves calcination followed by carbonmonoxide reduction.
Magnesium  Involves calcination (of MgCl2.6H2O in presence of dry HCl gas) followed by fused
salt electrolysis.
SECTION –C

2. COCH3
H3C

P=
Br
Br

1.92
3. nMn  mol = 0.016 mol.
2  60
nhydrazinehydrate =  0.9  5.332 mol = 0.096 mol.
50
1
neqKIO3 = 4  N  0.48L = 0.192  neqhydrazine reacted with KIO3
10
0.192
 nhydrazine left after reaction with Mn+ = mol = 0.048 mol
4
n+
 nhydrazine reacted with M = (0.096 – 0.048) mol = 0.048 mol
nM 1 0.016 1
  
nhydrazine n 0.048 3

4. FTTTTFFFTTFTTT
(IX) A is a buffer solution; pH of A = pKa = 2
1
B is an aqueous solution of SacNa; pH of B =  pK w  pK a  logc  = 8
2

5. Total work done = Area of circle A (anticlockwise) – Area of circle B (clockwise)

= 40 – 36 = 4 unit
heat flows into the system along the path ADB, = 5 – 3+2 = 4 unit

6. CCl3–CH(OH)2 + NaOH  CHCl3 + HCOO–Na+ + H2O

Chloral hydrate (A)
1 h
CHCl3 + O2  COCl2 + HCl
2
(A) (phosgene)
(B)
Cl OC2H5
O=C + 2HOC2H5 O=C + 2HCl
Cl OC2H5
(diethyl carbonate)
(C)
O O O
–OC2H5
H5C2OC–OC2H5 + CH3MgBr H5C2OC–OC2H5 H5C2OC–CH3
CH3 CH3MgCl

OH O
CH3MgCl
CH3–C–CH3 CH3–C–CH3
H3O+
CH3

Mathematics PART – III

SECTION – A

1. Possibilities:
6!
(i) three digit occur twice = 5 C3   900 ways
2!2!2!
(ii) two digit occur three times = 200 ways
(iii) one digit occur four times and other twice = 300 ways
(iv) all digits are same = 5
 Total possibilities = 1405

2. |z – 2i| = 2 represent a circle with centre (0, 2), radius 2 unit and |z – 1 – 3i| + |z + 1 – 3i| = 2 2
represent an ellipse they touch each other at (0, 4)
n
3. Let f  n    n1Cr 1  2n1  1
r 0
f(9) = 1023 = 3.11.31
Total divisors are = 8
100       
4. 
2 1  x3 
 a0  a1x  a 2 x 2  .....  a100 x100   cos x  cos  x    .....  cos  x  100  
 2  2 2   2 2 
Replacing x by –x and adding
100 100  50 x 
1  x3 
 1  x3  
 a0  a2 x2  .....  a100 x100    cos
 2
cos 
2 
50
Put x = 1  2100   a2i  k = 100
i 0

5. The desired condition is to have the circle N inside the convex region of the parabola y = x 2
clearly a  1, otherwise circle will cross x-axis also for circle to be inside the parabola the number
of points of intersection must be less than or equal to 2
 Equation y + (y – a)2 = 1, has one or less than one solution
D0
5
a
4

 
3 3
6. IN =  x loge  sin x  dx      x  loge  sin x  dx
0 0
 
3 2 3 2 3
= x loge 2 dx  x loge  sin x  dx = ID
2 0 2 0 2
IN 3
 
ID 2

7. OT2 = OA · OB
T

O A B

8. Let ABCDPQRS be a cube, the numbers a, b, c, d, e, f be written on the faces

 ade + abe + abf + adf + cde + bce + bcf + cdf = 2004
2
 (e + f)(a + c)(b + d) = 2 · 3 · 167
Possible combination are 4 · 3 · 167, 2 · 6 · 167, 2 · 3 · 334, 2 · 2 · 501
T = a + b + c + d + e + f = 174, 175, 339, or 505
2
9. If x  0, x + (a – 5)x + 1 = 0
2
x < 0, x + (a + 1)x + 1 = 0
Equation has exactly three distinct solution if (1)
Either (a – 5)2 > 4 and (a + 1)2 = 4
Or (a – 5)2 = 4 and (a + 1)2 > 4

10. ODC = HEA = GFB

1 1
 R2 sin A cos A  2R 2 sinC cosC cos2 A  R 2 sin A sinB sinC
2 3
First two equality gives, tan A = 4 sin C cos C
From first and last, 3 cos A = 2 sin2 C (1 + 4 cos2 C) cos A
 3 = 2 sin2 C (1 + 4 cos2 C)
3 1
 sinC  or
2 2

11. Score of 5 can be obtained as 5 or (1, 4) or (1, 1, 3) or (1, 1, 1, 2)

1 1 1 1
P (scoring 5) =  2  3  4 similarly
6 6 6 6
1 1 1 1 1
P (scoring 6) =  2  3  4  5
6 6 6 6 6
1 1 1 1 1
P (scoring 7) = 2  3  4  5  6
6 6 6 6 6
P (scoring 1) is impossible event

12.-13. Let OM = d D L C
MDC ~ AMB x x
AK2 + OK2 = AO2 = DO2 = DL2 + OL2
M
 k2x2 sin2  + (kx cos  – d)2 = x2 sin2  + (x cos  + d)2 O
 (k2 – 1)x2 = 2xd (k + 1) cos  as k + 1 > 0 kx  kx
 (k – 1)x = 2d cos 
Now, AB – CD = 2(AK – LD) = 2(k – 1)x sin  = 2d sin 2 A K B
If AMB = 60º,  = 30º
3
If AMD = 60º,  = 60º in either case sin2 
2
AB – CD = 2 3

14.-16. a, c, b are in H.P.

2 1 1
  
c a b
x y 1  1 1
Line    0 passes through a fixed point
a b c
 2,  2 
 
 1 1
So centre of ellipse is   ,  
 2 2
Which is also centre of auxiliary circle
1 1
 p  , q  radius of auxiliary circle = 2
2 2
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 Length of major axis = 4 also it segment of focal chord are 1, 2

1 1 4 8
   2  b2 
1 2 b 3
2 2
 1  1 8
Director circle  x     y     4
 2  2 3

SECTION – B

1. (A) 20, b, c are in H.P.

1 1 2
  
20 c b
This can be written as (40 – b)( c + 20) = 800 we must have 20 < b < 40 or 0 < 40 – b < 20,
consider factorisation of 800 in which one term is less that 20
(40 – b)(c + 20) = 800 = 1  800 = 2  400 = 4  200 = 5  160 = 8  100 = 10  80 = 16  50
Thus pair (b, c) = (39, 780), (38, 380), (36, 180), (35, 140), (32, 80), (30, 60), (24, 30) but b
divides c
 Only 5 triples satisfy required requirement
(B) Solutions of m + n + p = 10, are 3 10 1C3 1
(C) ab + b – a + 1 = 0
 (a + 1)(b – 1) = –2
 (a + 1, b – 1)  {(1, –2), (–1, 2), (2, –1), (–2, 1)}
 (a, b)  {(0, –1), (–2, 3), (1, 0), (–3, 2)
(D) Let z = x + iy then (x + iy)20 = x – iy
 |z|20 = |z|  |z|(|z|19 – 1) = 0
 |z| = 1 or 0
 |z| = 0
 Ordered pair (0, 0)
Let z  0 then (x + iy)20 (x + iy) = x2 + y2
 z21 = |z|2 = 1  z21 – 1 = 0, which are 21 roots

2. (A) f(x) = x4 + ax3 + bx2 + cx + d

, , ,  are real roots of equation f(x) = 0
 f(x) = (x – )(x – )(x – )(x – )
 |f(i)| = 1  (1 + 2)(1 + 2)(1 + 2)(1 + 2) = 1 this is possible if  =  =  =  = 0
a=b=c=d=0
(B) x5 – 1 = 0
4 3 2
 (x – 1)(x + x + x + x + 1) = 0
2
x x x3 x4 x x2 x3 x4
       
1  x 2 1  x 4 1  x 1  x3 1  x2 1  x 4 x5  x x5  x3

=
x

x2

x2

x
= 2
 x

x2 
=

2 x  x 5  x2  x 4
2

 2 
1  x 2 1  x 4 1  x 4 1  x2  1 x 1 x4  1 x2  x4  x6
 1 1 1 1 3
(C) cos x    0  cos x   1 and   cos x  
 2 2 2 2 2
x2
 1
 cos x    1  f(x) = 0 
 2  f x  0
x1

(D) Let AD = x, BAD =  D

BD C
In ABD,  2R
sin  x 2 3
2 3 
 sin     = 60º
4 A 1
x 2  1  BD 2 B
Using cosine rule cos 60º 
2x
1 3 5
 BD 
2
1 3
 a , b
2 2
SECTION – C
1. f(9) = f(4 + 5) = f(4.5) = f(20)
= f(16 + 4) = f(16.4) = f(64)
= f(8.8) = f(8 + 8) = f(16)
= f(4.4) = f(4 + 4) = f(8)
2. g(x) = f(x) – f(x) ….. (1)
g(x) = f(x) – 2f(x) ….. (2)
F(x) = (2f(x)f(x) + 2g(x)g(x))(f2 (–x) + g2(–x)) + (f 2(x) + g2(x))(–2f(–x)f(–x) – 2g(–x)g(–x)) ….. (3)
Using (1) and (2) in equation (3), we get F(x) = 0
 F(x) is constant function  F(0) = 1 = F(1)
3. f(x) = tan–1 (tan x)
5
 f r   f 1  f  2   f  3   f  4   f  5  = 1 + 2 –  + 3 –  + 4 –  + 5 – 2 = 15 – 5
r 1
a = 15, b = 5
4. Minimum area will be in case of square with vertices
 2, 2 ,  2, 2 ,   
2,  2 and  2,  2  
 Minimum area = 8s
200 200
r
5.  r x r   r 1  x 
r 0 r 0
200
 0  1x  .....   200 x 200  0  1 1  x   .....  200 1  x 
201 100

= 0  1 1  x   .....  99 1  x 
99 1  x   1  x 

x
100200 101
Equating coefficient of x , x , ….., x respectively we get
201 201 201
100 = C101, 101 = C102, ….., 200 = C201
200
 A  r  100  101  .....   200  2200
r 100
200
When 2 is divided by 15 we get remainder 1
6. Take B as the origin, BC and BA as x and y axis y
and z axis perpendicular to the plane of ABC  3 3 
   3
0, 6, 0 A D ,
 2 2
, 3

 3ˆ  
 BA  6ˆj , CD    6  ˆi  j  3kˆ
 2  2
  
 n  BA  CD = 3k  3 2i  3 2iˆ  kˆ
ˆ ˆ  
 n 2 x
The minimum distance d = BC     3 2 B(0) C
n 3  6, 0, 0 
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FULL TEST – II
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Paper 2

Time Allotted: 3 Hours Maximum Marks: 246

 Pl ea s e r ea d t h e i n s t r u c t i o n s c a r ef u ll y . Yo u a r e a l l o t t ed 5 m i n u t es
s p ec i f i c a ll y f o r t h i s p u r p o s e.
ALL INDIA TEST SERIES

 Yo u a r e n o t a l l o wed t o l ea v e t h e E xa m i n at i o n Ha l l b ef o r e t h e en d o f
t h e t es t .

INSTRUCTIONS
A. General Instructions

B. Filling of OMR Sheet

C. Marking Scheme For All Three Parts.

2. Section – B (01 – 02) contains 2 Matrix Match Type questions containing statements given in 2
columns. Statements in the first column have to be matched with statements in the second
column. Each question carries +8 marks for all correct answer. For each correct row +2 marks
will be awarded. There may be one or more than one correct matching. No marks will be given for
any wrong matching in any question. There is no negative marking.
3. Section-C (01 – 06) contains 6 Numerical based questions with answers as numerical value from
0 to 9 and each question carries +3 marks for correct answer. There is no negative marking.

Name of the Candidate

Enrolment No.

Useful Data

PHYSICS
2
Acceleration due to gravity g = 10 m/s

Planck constant h = 6.6 1034 J-s

Charge of electron e = 1.6  1019 C

Mass of electron me = 9.1  1031 kg

2 2
Permittivity of free space 0 = 8.85  1012 C /N-m

Density of water water = 103 kg/m3

Atmospheric pressure Pa = 105 N/m2

Gas constant R = 8.314 J K1 mol1

CHEMISTRY

Gas Constant R = 8.314 J K1 mol1

Atomic No: H=1, He = 2, Li=3, Be=4, B=5, C=6, N=7, O=8,

Physics PART – I

SECTION – A
(Straight Objective Type)
This section contains 7 multiple choice questions. Each question has four choices (A), (B), (C) and (D)
out of which ONLY ONE option is correct.

1. A neutral particle at rest in a uniform magnetic field, decays into two charged particles of different
masses at point P as shown in the figure. The energy released goes to their kinetic energy and
particles move in the plane of the paper. Magnetic field is into the plane of paper. Select the
diagram which describes path followed by the particles most appropriately.
(A) (B)

2. Choose the correct statement about centre of mass.

(A) For calculation of any quantity associated with extended body we can assume total mass of
body to be concentrated at centre of mass of the body.
(B) Centre of mass of a body moves in the same way as if all external forces are acting on it and
total mass of body is concentrated at it.
(C) Torque of centrifugal force about centre of mass on rigid body undergoing rotational motion
is always zero.
(D) All above

Space for Rough work

3. A block of mass m carrying charge q is suspended through non

conducting spring in equilibrium spring elongation is x0. Now another
charge is moves very slowly from infinity until it is in the original position of
suspended block. In this process spring acquire its natural length. Work
done by external agent.
X0
3 1
(A) (mgx0 ) (B) (mgx0 )
2 2 q Q

5
(C) (mgx0 ) (D) (mgx 0 )
2

4. Rod of mass m and length  carrying one charge in length x +++++++++++     

A B
and negative charge remaining portion. Linear charge density E
x
 is same for both charges. Rod is hinged at end A. A
uniform E perpendicular to rod is switched on and rod remain
in equilibrium. Find value of x. (consider gravity g is absent)
 
(A) x  (B) x 
2 2 2
3
(C) x =  (D) x 
2

5. Initially spring is relaxed and m1is released from rest. When m1 come
to rest for a moment string AB is cut at the same amount. Find the
acceleration of m2 just after cutting thread. M2
m   2m  B
(A) a   1  1 g (B) a   1  1 g
 m2   m2  A
M1
 m   3m 
(C) a   1  1 g (D) a   1  1 g
 2m2   m2 

Space for Rough work

6. In the adjacent figure, a block ‘A’ of mass m is kept on a

wedge ‘W’ of mass ‘M’, which is lying at rest on a smooth
horizontal surface and is connected to a rigid support
through a light spring of spring constant ‘K’. Initially block
and wedge are at rest and spring is at relaxed state. The
system (block + wedge) is set free. Calculate maximum
possible compression in the spring. All surfaces are
frictionless and inclination of wedge is .
m2 g 2Mm g
(A) 2
(B)
K (M  m sin ) K (M  m sin2 )
mgsin2 2mgcos 
(C) (D)
K K

7. If an electric dipole is rotating about its centre with uniform angular velocity in anticlockwise
direction in a uniform magnetic field which is in the direction of angular velocity then :
(A) Net magnetic force as well as torque due to the magnetic field on the dipole is zero
(B) Net magnetic force as well as torque due to the magnetic field on the dipole is non-zero
(C) Net magnetic force on dipole is zero but net torque due to the magnetic field on dipole is non-
zero
(D) Net magnetic force on dipole is not zero but net torque due to the magnetic field on dipole is
zero.

Space for Rough work

(Multiple Correct Choice Type)

This section contains 4 multiple choice questions. Each question has four choices (A), (B), (C) and (D)
out of which ONE or MORE THAN ONE are correct.

8. The shown figure, there is a conical shaft rotating on a bearing of very Motor
small clearance t. The space between the conical shaft and the bearing, is 
filled with a viscous fluid having coefficient of viscosity . The shaft is
having radius R and height h. If the external torque applied by the motor is t m
 and the power delivered by the motor is P working in 100% efficiency to
rotate the shaft with constant . Then

2 R 3 R 2  h2 R 3 R 2  h2
(A) P = (B)  =
2t 2t
2 R3 h R3 h2
(C) P = (D)  =
2t 2t R2  h2

9. A metal wire of length L, area of cross section A and Young modulus Y is stretched by a variable
force F such that F is always slightly greater than the elastic forces of resistance in the wire.
When the longation of the wire is , then choose the correct option(s).
1 YA 2
(A) the work done by F is 
2 L
1 YA 2
(B) the elastic potential energy stored in the wire is 
2 L
(C) no heat is produced during the elongation
YA 2
(D) the work done by F is 
L

Space for Rough work

10. Two parallel resistance less rails are connected by an inductor of inductance B
L at one end as shown in the figure. A magnetic field B exists in the space
which is perpendicular to the plane of the rails. Now a conductor of length  J

and mass m is placed transverse on the rails and given an impulse J towards
the rightward direction. Then choose the correct option(s).
(A) Velocity of the conductor is half of the initial velocity after a displacement of the conductor
3J2 L
d .
4B 2 2 m
(B) Current flowing through the inductor at the instant when velocity of the conductor is half of the
3J2
initial velocity is i 
4Lm
(C) Velocity of the conductor is half of the initial velocity after a displacement of the conductor
3J2L
d
B2  2 m
(D) Current flowing through the inductor at the instant when velocity of the conductor is half of the
3J2
initial velocity is i 
mL

11. A particle of mass ‘m’ is attached to the rim of a uniform disc of y

mass ‘m’ and radius R. The disc is rolling without slipping on a Am
stationary horizontal surface, as shown in the figure. At a
particular instant, the particle is at the top most position and O
v0
D  C
centre of the disc has speed v 0 and its angular speed is .
Choose the correct regarding the motion of the system (disc + x
O B
particle) at that instant.
(A) v0 = R
11 2
(B) kinetic energy of the system is mv 0
4
(C) speed of point mass m is less than 2v 0
  
(D) | v C  vB | = | vB  vD |

Space for Rough work

Comprehension Type

Paragraph for Questions 12 & 13

Two moles of a diatomic gas are carried through the cycle ABCDA shown P A
in the figure PV diagram as shown in figure. The segment AB represents
an isothermal expansion, the segment BC an adiabatic expansion. The B
pressure and temperature at A are 5 atm and 600 K respectively. The
D
volume at B is twice that at A. The pressure at D is 1 atm. C

V
12. What is the pressure at B?
(A) 2.5 atm (B) 3 atm
(C) 3.5 atm (D) None of these

13. Find the T versus V graph of the above cyclic process.

(A) (B)
A B T A B
T
C
D C
D
V V
(C) (D)
A B A B
T T
C C

D D
V V

Space for Rough work

Paragraph for Questions 14 & 16

Image formation from a thin prism: For a thin prism, at near normal incidence, A
the angle of deviation does not vary considerably with angle of incidence and
consequently we may write angle of deviation as  = ( 1)A, where A is the
angle of the prism and  is the refractive index of the material of the prism.
Consider a thin prism ABC and a source of light S. Suppose we want to trace
S B C
the image of source S, for sake of simplicity we draw a simple ray diagram as
shown in the figure.

Suppose we take S as the origin, S as the image of S and the horizontal and A
S
vertical lines passing through S as x and y axis, and the distance SB as a (SB
is horizontal). Now using geometry and using the fact that angle CAB is small, 
we can find the co-ordinates of S. Now answer the following questions.
S
B 
a
14. The x-coordinate of S will be
(A) zero (B) a/2
(C) 2a (D) a

15. The y-coordinate of S will be

1
(A) (  1)Aa (B) (  1)Aa
2
3
(C) 2( 1)Aa (D) (  1)Aa
2

16. If an identical prism is joined at the base of the first prism as

shown in figure and a new image S is formed due to it. Then the S

separation between S and S will be S

(A) ( 1)Aa (B) 2(  1)Aa
S
  1 Aa 3    1 Aa
(C) (D)
4 4

Space for Rough work

SECTION - B
Matrix – Match Type

This section contains 2 questions. Each question contains statements given p q r s t

in two columns, which have to be matched. The statements in Column I are
A p q r s t
labelled A, B, C and D, while the statements in Column II are labelled p, q, r,
s and t. Any given statement in Column I can have correct matching with B p q r s t
ONE OR MORE statement(s) in Column II. The appropriate bubbles
corresponding to the answers to these questions have to be darkened as C p q r s t
illustrated in the following example:
D p q r s t
If the correct matches are A – p, s and t; B – q and r; C – p and q; and D – s
and t; then the correct darkening of bubbles will look like the following:

1. An electron in a hydrogen atom makes a transition n1  n2, where n1 and n2 are the principal
quantum numbers of the two states. Assume Bohr model to be valid.
Column I Column II

(A) The electron emits an energy of 2.55 eV (p) n1 = 2, n2 = 1

(B) Time period of the electron in the initial state (q) n1 = 4, n2 = 2

is eight times that in the final state.

(C) Speed of electron become two times (r) n1 = 5, n2 = 3

(D) Radius of orbit of electron becomes 4.77A (s) n1 = 6, n2 = 3

(t) n1 = 8, n2 = 4

Space for Rough work

2. Choose the correct equation of current in the Column- II as a function of t through the circuit
element ‘ab’ of the circuits in the column – I
List – I List – II
(A) 2 a
 
t

1  e  2104 sec 

100 F
(p) i = (5A)
 
10V  
b
(B) 2 2 a t
  
  4104 sec  
(q) i = (5A) 1  e
(200/3) F  
2
30V  
b
(C) 2 a 
t
 210 4
sec 
0.4 mH
(r) i = (5A) e
10V
b
(D) 2 2 a 
t
 410 4
sec 
(s) i = (5A) e
0.6 mH
2
30V
b

SECTION – C
(Integer Answer Type)

This section contains 6 questions. Each question, when worked out will result in one integer from 0 to 9
(both inclusive).

1. Two rods of same length  but there masses and coefficient of friction , m
 F
with table are m, 2m and , 2 respectively are joint at point O as x
shown in the figure. Find the position x from joint of two rod where O
force should apply perpendicular to the length of the rod to drag the

K 2, 2m
rod with constant velocity is  . Find the value of K.
10

Space for Rough work

2. A string of length 0.4 m and mass 102 kg is tightly clamped at its one end
and other end is free to move as shown in the figure. The tension in the
string is 1.6 N identical pulses are produced at one end at equal intervals at
time t. The minimum value of t which allows constructive interference
between successive pulses is

3. A uniform rod OAB is bent in L shape to form right angle at A. Length B

L
of OA is L and that of AB is respectively as shown in figure. The
2 L/2
rod is hinged at the end O and is free to rotate in a vertical plane O
about O. It is set free from rest, when larger section of it is in A
L
65 g
horizontal position. Maximum angular acceleration of rod is .
K L
The value of K is

4. For an equilateral prism, it is observed that when a ray strikes grazingly at one face it emerges
grazingly at the other. Its refractive index will be

5. A block of mass (M = 1 kg) is kept on smooth floor of a truck. M a0

One end of a spring of force constant 100 N/m 2 is attached to k O
the block and other end is attached to the body of truck as
shown in the figure. At t = 0, truck begin to move with
constant acceleration 2 m/s2. The amplitude of oscillation of
block relative to the floor of truck is /100. Find the value of 

6. A uniform rod of mass M is released from horizontal position P M

as shown in the figure. The rod is free to rotate about a
horizontal frictionless axis passing through hinge P. The L
reaction from the hinge just after the release is Mg/K. Find
the value of K.

Space for Rough work

Chemistry PART – II

SECTION – A
Straight Objective Type
This section contains 7 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for
its answer, out which ONLY ONE option is correct.

1. OH
CH 2 N 2

(1eqv)
A. A is
HO
NO2 OH
O OH
CH3

(A) (B) CH3

HO HO
NO2 OH NO2 OH
OH OH

(C) H3C (D) HO

O
NO2 O
NO2 OH CH3

2. Choose the correct statement(s) from among the following

 an 2 
(A) The corrected pressure  P   in vander waals’ equation (symbols have their usual
 v2 
meanings) is the pressure which the gas would exert if it was ideal.
(B) Above their respective Boyle’s temperature, N2 shows throughout positive deviation
(z  1) while O2 shows negative deviation (z  1) followed by positive deviation (z  1).
(C) A gas shows negative deviation when long range attractive intermolecular forces are
dominating
(D) The intercept of PV vs. P isotherm of any gas at 27°C is equal to 24.6 L atm.

3. The ZnS (zinc blende) structure is cubic. The unit cell may be described as a face-centred sulfide
ion lattice with zinc ions in the tetrahedral voids. Then among the following which is incorrect:
(A) The nearest neighbours of Zn2+ are four
o
(B) The angle made by the lines connecting any Zn+2 to any two of its neighbours is 109 28’.
(C) The nearest neighbours of S–2 are four
(D) The nearest neighbours of S–2 are eight

Rough Work

4. Some organic compounds are given below

CH2 OH CH2 OH CH2 OH
O O O
H H H H H H
H H H
CH2 OH
OH H OH H OH H
OH OH OH OH C O
(CH2 )3
H OH H OH H OH
O CH3
I CH2 OH HOH2 C
O IV
O O
H H
H H
OH H H OH
OH H OH H

H OH OH H
II III

Select the correct statement about the above compounds from the following
(A) Compounds I and III would reduce Fehling’s solution and will also mutarotate
(B) Compound IV will not reduce Tollen’s reagent and will not mutarotate
(C) Compound II will not be reducing but mutarotating
(D) The product(s) of the acid catalysed hydrolysis of both II and III will be reducing but not
mutarotating

H2 O CuO
5. Mg 3 N 2   A (g) 

 B  C(g)  H 2 O
C(g) can be obtained by heating
(A) NH4NO3 (B) (NH4)2SO4
(C) NH4ClO4 (D) NaNO2 + Zn + NaOH (Conc.)

6. Which of the following statements are incorrect?

(A) Sulfate and thiosulphate ions are isostructural.
(B) Thiosulphate on reaction with HCl precipitates sulphur along with SO2 evolution.
2
(C) In S2 O3 the oxidation states of the two ‘S’ atoms are -2 and +6.
2 2
(D) Both SO 4 and S2 O3 can decolourise I2.

7. Difference in PKa values of compounds A and B can be best explained by

H Cl
H Cl Cl H
Cl H
COOH COOH

(A) (B)
Ka Ka
P = 6.07 P = 5.67
(A) Inductive effect (B) Electromeric effect
(C) Field effect (D) Resonance effect

Rough Work

Multiple Correct Choice Type

This section contains 4 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for its
answer, out which ONE OR MORE is/are correct.

8. Which of the following statements are correct?

(A) Nylon 6 and Nylon 66 are examples of condensation polymerization
(B) D-glucose and D-galactose are epimers
(C) Polyethene is produced via addition polymerization
(D) Teflon has a low coefficient of friction

9. Consider the reaction



2SO2(g)  O2(g) 
 2SO3(g)  Heat
What will happen if 0.5 mole of Helium gas is introduced into the vessel so that the temperature
and pressure remain constant?
(A) the equilibrium concentration of SO2 increases
(B) the equilibrium will shift to left
(C) the equilibrium concentration of O2 increases
(D) equilibrium constant will increase

10. The standard enthalpy of formation of CO2 will be given by :

(A) standard enthalpy of combustion of diamond
(B) standard enthalpy of combustion of graphite
(C) standard enthalpy of combustion of CO
(D) sum of standard enthalpy of formation and standard enthalpy of combustion of CO.

11. The lithium enolate base formed from cyclopentanone reacts with alkyl halides, often in different
ways. As shown here, methyl iodide and tert-butyl bromide react to give different organic
products, I and II, together with lithium halides. What are the products from these reactions?
+
CH3 O Li
H3C C Br
CH3
H3CI
II
THF
 
THF
I
(A) I is 1-methoxycyclopentane
(B) II is 1-t-butoxycyclopentane
(C) I is 2-methylcyclopentanone
(D) II is a mixture of cyclopentanone and 2-methylpropene
Rough Work

Paragraph For Question No.12 and 13

Heat of neutralization is defined as the enthalpy change when 1 mole of acid/base is completely
neutralization by base/acid in dilute solution.
H   aq   OH   aq   H 2 O  l  ;  r H 0  55.84 kJ / mol
0
H ionization of aqueous solution of strong acid and strong base is zero.
When a dilute solution of a weak acid or base is neutralized, the enthalpy of neutralization is some what
less because of the absorption of heat in the ionization of the weak acid or base, for weak acid/base
H 0neutralisation  H 0ionization   r H 0  H   OH   H 2O 

0
12. What is H for complete neutralization of strong diacidic base A(OH)2 by HNO3 ?
(A) –55.84 kJ (B) –111.68 kJ
(C) 55.84 kJ (D) 111.68 kJ

13. Under the same conditions how many mL of 0.1 M NaOH and 0.05 M H2A (strong diprotic acid)
solution should be mixed for a total volume of 100 mL produce the highest rise in temperature.
(A) 25 : 75 (B) 50 : 50
(C) 75 : 25 (D) 66.66 : 33.33
Rough Work

Paragraph For Question No.14 and 16

A green ore (A) of a metal present as a double/mixed compound is treated with HCl and then H2S is
passed in the solution. A black precipitate (B) is obtained, that is insoluble in yellow ammonium sulphide.
The precipitate is dissolved in HNO3 and then excess of NH4OH is added. The solution becomes coloured
but this colour is discharged upon addition of KCN in excess due to the formation of compound (C). The
solution of (A) in H2O liberates a colourless and odourless gas on reaction with dilute H2SO4 and solution
of (A) gives white precipitate on addition of BaCl2 solution.

14. The green ore (A) is

(A) CuSO4.Cu(OH)2 (B) CuCO3.Cu(OH)2
(C) PbSO4.Pb(OH)2 (D) PbCO3.Pb(OH)2

15. The compound (C) and its structure is

(A) [Cu(CN)4]2–, tetrahedral (B) [Cu(CN)4]2–, square planar
(C) [Cu(CN)4]3–, tetrahedral (D) [Cu(CN)4]3–, square planar

16. The procedure used to estimate the metal ion of ore (A) gravimetrically, is
(A) treatment of ore with excess KCN.
(B) treatment of solution of ore with Ag metal.
(C) treatment of ore with excess ammonia.
(D) treatment of ore with excess KSCN.

Rough Work

1. Column – I (Compounds) Column – II (Magnetic moment)

(A) [Fe(H2O)5NO]SO4 (p) 24 B.M.
(B) K2[Fe(CN)5NO] (q) 8 B.M.
(C) [Mn(CN)6]3– (r) Zero B.M.
(D) [Co(SCN)6] 3– (s) 15 B.M.
(t) 3 B.M.

2. Column – I Column – II
(A) N2O3 (p) Contains N – O – N linkage
(B) N2O4 (q) Neutral oxide
(C) N2O5 (r) Colored
(D) N2O (s) mixed Anhydride
(t) Basic oxide
Rough Work

SECTION –C
Integer Answer Type
This section contains 6 questions. Each question, when worked out will result in one integer from 0 to 9
(both inclusive).

1. How many optically active isomers are possible for CH3CH(Br)CH(Br)CH(Br)CH3?

2. A certain saturated hydrocarbon effuses about half as fast as H2O vapour. Number of carbons in
the formula of this hydrocarbon are?

3. 2.505 g of hydrated dibasic acid requires 35 ml of 1N NaOH solution for complete neutralization.
When 1.01 g of the hydrated acid is heated to constant weight 0.72 g of the anhydrous acid is
obtained. The degree of hydration (number of water molecules) of the hydrated dibasic acid is
approximately ______.

4. The number of mole of bromine required for complete bromination of 276 g of salicylic acid in
aqueous medium are___________.

5. 1 g of an acid C6H10O4 requires 0.768 g of KOH for complete neutralization. Determine the
basicity of acid.

6. The structural formula of diphenyl methane (C13H12 ) is CH2

How many structural isomers are possible when one of the hydrogens is replaced by a chlorine
atom?
Rough Work

Mathematics PART – III

SECTION – A

Straight Objective Type

This section contains 7 multiple choice questions numbered 1 to 7. Each question has 4 choices (A), (B),
(C) and (D), out of which ONLY ONE is correct

27 1
1. A gun can hit the target A with probability and target B with probability , it has 47 bullets,
28 28
each of which can be fired either at A or at B and each bullet may hit the target, independently of
1
the other bullet, with probability . How many bullets must be fired at A to hit the target with
3
maximum probability
(A) 20 (B) 25
(C) 30 (D) 35

2. A polynomial P(x) of nth degree satisfies P(k) = 2k for k = 0, 1, 2,.., n, then the value of P(n + 1) is
(A) 2n (B) 2n + 1
n+1
(C) 2 –1 (D) 2n + 2

3. The number of all polynomial whose coefficients are all 1, that have only real roots are
(A) 6 (B) 8
(C) 10 (D) 12

4. For positive integer n, define f(n) = 1n + 2n – 1 + 3n – 2

+ ….. + (n – 2)3 + (n – 1)2 + n then the
f  n  1
minimum of is
f n 
8
(A) 3 (B)
3
(C) 2 (D) 1

Space for rough work

1  sin x  2  sin x 
5. The value of  1  sin x  2  sin x  dx is
t 3  1
t  3  1
(A) ln    2 tan t  c (B) ln    tan t  c
t 3 t 3
t 3  1 2t
t 3  1
(C) 3 ln    tan c (D) 3 ln    2 tan t  c
t 3  1 t2 t 3
2  sin x
(where t = )
2  sin x

1/3 1/3
6. Let x be  108  10    108  10  then x is equal to
(A) 10 (B) 8
(C) 4 (D) 2

7. A circle of radius 320 units is touches internally a circle of radius 1000. The smaller circle touches
a diameter of larger circle at the point p, least distance of the point p from the circumference of
the larger circle is
(A) 300 (B) 360
(C) 400 (D) 420

Multiple Correct Choice Type

This section contains 4 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for its
answer, out of which ONE OR MORE is/are correct

8. The roots of a quadratic equation ax 2 + bx + c = 0 are   z2  z 4  .....  z2n and

 2 
i 
  z  z3  .....  z2n 1 where z  e  2n 1 
, n  z+ then
(A)  +  = 1
(B)  ·  = 1
1   
(C) equation can be written as x 2  x  sec 2  0
4  2n  1 
(D)  +  = –1

Space for rough work

1 1 x x 1
9. Let f 1(x) = x, f 2(x) = 1 – x, f3  x   , f4  x   , f5  x   , f6  x   suppose that
x 1 x x 1 x
f6  fm  x    f4  x  and fn  f4  x    f3  x  then
(A) m = 5 (B) n = 5
(C) m = 6 (D) n = 6

x x
10. If a continuous function f(x) satisfy the relation  t f  x  t  dt   t f  t  dt  sin x  cos x  x  1 , for all
0 0
real number x, then which of the following does not hold good
(A) f(0) = 1 (B) f(0) = 0


(C) f(0) = 2 (D)  f  x  dx  e
0

1
11. Let, f(0, )  R be a strictly increasing function such that f  x     x > 0 and
x
 1
f  x  f  f  x     1  x > 0, then
 x
1 5 1 5
(A) f 1  (B) f 1 
2 2

Space for rough work

Comprehension Type

Paragraph for Question Nos. 12 to 13

Read the following write up carefully and answer the following questions:
A line through the origin has the vector equation r   s ,  is a scalar parameter. Consider a family of line
    
L with s  cos   3 ˆi  2 sin  ˆj  cos   3 kˆ then 

12. The line through origin and making acute angle with family of lines of the L–family are in the
6
form of
(A) (i – j) (B) (i – k)
(C)  (i + j) (D) (i + k)

13. If the lines in L meet the plane x – z = 4 3 at P. If  varies and P describes a circle on the plane
with centre on M then radius of this circle is
(A) 2 (B) 2
(C) 4 (D) 2 2

Space for rough work

Paragraph for Question Nos. 14 to 16

Read the following write up carefully and answer the following questions:
While finding sine of a certain angle x, an absent minded professor failed to notice that his calculator was
not in correct angular mode. However he was lucky to get the right answer. The two least positive value
m p
of x for which the sine of x degree is the same as sine of x radians were found by him as and
n q
mn
where m, n, p, q are positive integers suppose be denoted by L, then
pq

14. The value of m + n + p + q is equal to

(A) 720 (B) 900
(C) 1080 (D) 1260

BC
15. If x measured in radian and lim
x 
 
Ax 2  Bx  Cx  L , the value of
A
equals (A, B, C  R)
(A) 4 (B) 2
1
(C) (D) none of these
2

16. Assume that f is differentiable for all x, the sign of f is a follows

f(x) > 0 on (–, –4)
f(x) < 0 on (–4, 6)
f(x) > 0 on (6, )
Let g(x) = f(10 – 2x) the value of g(L) is
(A) positive (B) negative
(C) zero (D) the function g is not differentiable

Space for rough work

1. Match the following Column-I with Column-II

Column – I Column – II
(A) Real value of x satisfying
x/2 x/2
 x 2  5x  6  x 2  5x  4   x 2  5x  6  x 2  5x  4  (p) 0
x4
= 2 4 is
(B) Integral value of ‘a’ for which the equation
(x2 + x + 2)2 – (a – 3)(x2 + x + 2)(x 2 + x + 1) + (a – 4)(x2 + x + 1)2 = 0 (q) 1
has at least one real root is
(C) If Pr is the coefficient of xr in the expansion of
2 2 2
x x x (r) 2
1  x  1    1  2   1  3  ..... then P1 is equal to
2

 2  2   2 
(D) If the number of positive integral solution of x + y + z = n be denoted by
Pn Pn1 Pn 2
Pn and   Pn1 Pn 2 Pn 3 then || is equal to (s) 4
Pn 2 Pn3 Pn 4
(t) 6

2. Match the following Column-I with Column-II

Column – I Column – II
(A) Let z = (cos 12º + i sin 12º + cos 48º + i sin 48º)6, then Im(z) is (p) 0
(B) Let f(x) be a continuous and differentiable function in (a, b). If
2
lim f  x    , lim f  x    and f(x) + (f(x)) = –1 then
x a x b
(q) 1
ba
minimum value of is

(C) The number of solution of
 1   1  (r) 3
tan   cos1 x   tan   cos1 x   1 is
4 2  4 2 
(D) The value of
1  2 3  2  2 3  (s) 4
tan2  tan2  tan2 cot  cot 2  cot 2 is
21  7 7 7 
 7 7 7 
(t) 5

Space for rough work

SECTION – C

(Integer Answer Type)

This section contains 6 questions. Each question, when worked out will result in one integer from 0 to 9
(both inclusive).

f  x  dx
1. Let f(x) is a quadratic function such that f(0) = 1 and  x 2  x  13 is a rational function, then the

value of f(0) is _____

2. If 0  [x] < 2, –1  [y] < 1 and 1  [z] < 3, (where [.] denotes the greatest integer function) then the
 x  1  y z
maximum value of the determinant  x   y   1  z  is _____
x  y   z  1

 36  4  8  cot   cot   cot 

3. Let   sin1   ,   cos 1 ,   tan1   and A  ,
 85  5  15  cot   cot   cot 
B = tan  · tan  + tan  · tan  + tan  · tan  then A + B is _____

1
4. Let f(x) = x 3 + 3x2 – 6x + 11, g(x) = ln|x|. If A = f(g(–1)), B = g(f(–1), C =  f '  x  g'  f  x   dx ,
2
t2
d
D g  x  dx evaluated at t = 3 and A(B + C) – D = kln(3) then the value of k is _____
dt 1

5. Let r1, r2 ,....., rn be the sides of a regular polygon inscribed in a circle of limit radius if
r1  r2  r2  r3  .....  rn  r1 = r1  r2  r2  r3  .....  rn  r1 then smallest possible value of n is _____

x 2 y2
6. A ray of light emerging from the point p(0, 4) along the tangent of the ellipse   1 and
25 16
strike a circle of radius 65 units and reflected from it. The reflected ray entering inside the
ellipse through the focus (–3, 0) and strike lower surface of ellipse and gets reflected from it. If the
reflected ray returns to the point p(0, 4), then ordinate of the centre of circle is _____

Space for rough work

ANSWERS, HINTS & SOLUTIONS

FULL TEST–II
(PAPER-2)
Q. No. PHYSICS CHEMISTRY MATHEMATICS
1. B C B
ALL INDIA TEST SERIES

2. B A C
3. A D D

4. B A B

5. A C D

6. C D D
7. D C C
A, B A, B, C, D C, D
8.
A, B, C A, B, C B, C
9.
A, B B, D A, B, D
10.
A, B, D C, D A, C
11.

12. A B B

13. A B D
14. A B B
15. A C A
16. B D C
(A)  (q), (A) – (s), (B) – (r), (A)  (p, q, s)
(B)  (p, q, s, t), (C) – (q), (D) – (p) (B)  (t) (C)  (s)
1.
(C)  (p, q, s, t), (D)  (q)
(D)  (r, s)
(A) (r), (B)  (r), (A – r), (B – s ), (A)  (p) (B)  (q)
2. (C)  (p), (D)  (p) (C – p, ), (D – q) (C)  (p)
(D)  (t)
1. 3 2 3

2. 2 5 4

3. 7 2 2
4. 2 6 7

5. 5 2 8

6. 4 4 5

Physics PART – I
SECTION – A

1. Momenta and charged are equal and opposite, so sense of the motions are opposite.
Mv
R=  their radii are equal.
qB

2. We can not assume total mass to be concentrated at centre of mass for calculation of all the thing e.g.
for calculation of moment of inertia, calculation of torque of centrifugal force on a rod about hinge as
shown.
(i) (ii)

6. Kx = (m + M) a + ma’ cos 
Block: a’ + a cos  = g sin 
...(ii)
From (i) & (ii)
Kx  mgsin  cos 
a  K1x  K 2
M  msin2 
 Max. compression
2mgsin  cos 
= Block + Wedge
K

7. Net magnetic moment of loop formed is zero, hence torque due to magnetic field is zero.

11. K = K D + KP A
1  1 1 v
= mv 20  1    m(2V0 )2 C
2  2  2
11 O
= mV02
4

SECTION – C
3. Let A, B, and C be C.M. of larger
section , smaller section and total rod
L L
CC '  , AC ' 
12 6
1
2 2
 L L   L   2
OC =        =
 2 6   12  
65
L
12
Moment of inertia about ‘O’
7
I  ML2  Maximum torque of about ‘O’
8
3Mg
max =  (OC)
2
 65  g 
max = max   
I 7 L 

sin 900
4. = 2
sin 300 900
300

5. kx0 = ma0

 L ML2  3 g
6.  = /I =  Mg 
 2 3  2L
Mg  N
 acm = L/2 = 3g/4 =
M
Mg
 N=
4

Chemistry PART – II

SECTION – A

1. Diazomethane being a base will pick up the hydrogen which is most acidic. The H of the OH group
ortho to –NO2 group is most acidic
OH OH



H O O

NO2 OH NO2 OH
CH 2  N  N


CH 3  N  N

2. Choice (B) is incorrect as all the gases above their respective Boyle’s temperature show throughout
+ve deviation.

4. All reducing sugars are mutarotating. Although IV is an -hydroxy ketone and hence reducing but it
can’t mutarorate as it is not a carbohydrate/can’t form ring. In II the glycosidic bond is in between two
anomeric carbons and hence ring opening can’t occur. Thus non-reducing as well as non-
mutarotating.
5. Mg 3 N 2  6H 2O 
 3Mg(OH) 2  2NH3
(A)

2NH 3  3CuO 
 3Cu  N 2  3H 2O
(C)

NaNO2  Zn  NaOH  Na2 ZnO2  NH3  H2O

Ammonium salts containing anions which are oxidizing in nature, on heating liberates N2(g) instead of
NH3.

9. At constant pressure, the addition of inert gas shifts the equilibrium towards larger no. of moles.

14. Since, it liberates a gas on reaction with dilute H2SO4, it cannot be a sulphate salt, it would be a
carbonate salt. Malachite ore of copper, CuCO3. Cu(OH)2 is green in colour.
2
15. Cu NH3  4   2KCN  Cu  CN 2   2K   4NH3

Blue colour Yellow ppt

decomposes
2Cu  CN 2    2CuCN    CN 2 

White ppt
3
CuCN  3KCN  Cu  CN 4   3K 
Colourless
Cu1 has a d10 configuration, so it is tetrahedral in shape with sp3 hybridization.

16. Gravimetric estimation of Cu2+ would be possible only when the given procedure gives an insoluble
compound on treatment with a given reagent.

 Cu  SCN 2  2K 
Cu2   2KSCN 

Black ppt
decomposes
2Cu  SCN2    2CuSCN    SCN2 
White ppt

With other reagents, either there is no reaction or even if it reacts, then the product is a soluble
compound.
SECTION –C

1. Only two optically active isomers are possible.

CH3 CH3
H Br Br H
H Br Br H
Br H H Br
CH3 CH3
rCnH2n 2 1 MH2 O
2.  
rH2 O 2 MCnH2n2

 MCnH2n2 = 4  18 = 72

 12  n + 1  (2n + 2) = 72
 14n + 2 = 72
n=5
3. Let formula of hydrated dibasic acid be H2 A .x H2 O
Equivalent of dibasic acid = Equivalent of NaOH
2.505  2 35
 1
M 1000
2.505  2  1000
M  143.14
35  1

H2 A.xH2 O   H2 A  xH2 O

Mole of H2 O formed  x  mole of hydrated dibasic acid

(1.01  0.72) 1.01
 x
18 143.14
x  2.28
degree of hydration is  2.
OH OH
COOH Br Br
4. + 3Br2(aq) + CO2+3HBr
(Molar mass = 138) Br
276
Moles of salicylic acid  2
138
Moles of Br2 required = 3 × 2 = 6.
5. Let basicity of acid be n.
Equivalent of acid = Equivalent of KOH

1 n 0.768

146 56
0.768  146
n 2
56
6. Three in the ring and fourth on the CH2 group as shown
1 2
4
CH2 3

Mathematics PART – III

SECTION – A

1. Let x bullet are fired at A

E1 – the target is at A, E2 – the target at B
E – the target is hit
27 1 E
P E1   , P  E2   , P   = probability that target is hit when x bullet are fired at A
28 28  E1 
x
 1
= 1  
3
47  x
 E   1
P   1  
 E2  3
x 47  x
A  A  27   1   1   1  
P E   P E1   P    P E2   P   =  1      1    
 E1   E2    
28   3   28   3  

27  x 1 x  47
= 1 3  3
28 28
d 27  x 1 x  47

dx
 P E   
28
3 ln3 
28
3 ln3

d2 27 1
dx2
P E     28 ln3 2 3 x  28 ln32 3x  47  0
d
For maximum P(E) 
dx
P E    0  27 · 3–x = 3x – 47  x = 25

x  x  1 .....  x  r  1
2. For 0  r  n, the polynomial x Cr  is of degree r
r!
x x x
 Q(x) = C0 + C1 + ….. + Cn is polynomial of degree n
Q(k) = 2k for k = 0, 1, 2, ….., n
 P(x) = Q(x)  x then
 P(n + 1) = Q(n + 1) = 2n + 1 – 1

3. If a polynomial a0xn + a1xn – 1 + ….. + an is such a polynomial,

We can assume a0 = 1, let r1, r2, ….., rn are its roots
 r12  r22  .....  rn2  a12  2a2 , r12r22 ..... rn2  an2
a12  2a2 2/n
 AM  GM    an 
n
 a1, a2 =  1, we must have a2 = –1, and n  3
 Polynomials are
(x – 1), (x + 1), (x2 + x – 1), (x2 – x – 1), (x3 + x2 – x – 1), (x3 – x2 – x + 1)

f  n  1 8 22 65 209 732
4. For n = 1, 2, 3, 4, 5, 6, = 3, , , , ,
f n  3 8 22 65 209
8 f  n  1 8
The minimum of these is , for n > 6, 3
3 f n  3

5. Let 1 + sin x = y

1 3y 3y
I dy again let  t2
y 1 y 1 y
t2 t 3
I = 8 dt  3 ln  2 tan1 t  c
t 2

 3 t 1 2
 t 3

1/3 1/3
6. Let a   108  10  , b  108  10 
a – b = x, a3 – b3 = 20, ab = 2
3 3 3 3
x = (a – b) = a – b – 3ab(a – b) = 20 – 6x
3
 x + 6x – 20 = 0
 x = 2 is only one real solution

7. c1c2 = 1000 – 320 = 680

2 2
c 1p = c1c 2    320   600 , PQ = r1 – c1p = 1000 – 600 = 400

8. S =  +  = –1
2
 z2n  1 
3 z 1 1   
P =  = z  2   2
  sec 2  
 z 1  1
 z  1 z   2 4  2n  1 
z

1
9. f6  fm  x    f4  x  
1 x
fm  x   1 1
f6  fm  x    
fm  x  1 x
x 1
Let f m(x) = k  fm  x    f6  x   m = 6
x
1
Assign, f n(f4(x)) = f 3(x) =
x
 1  1 1 t 1
 fn     let 1  x = t  x = t hence, n = 5
 1 x  x

x x x
10. Given equation can be written as  xf  t  dt   t f  t  dt   f  t  dt  sin x  cos x  x  1 differentiating
0 0 0
x
x f  x    f  t  dt  x f  x   f  x   cos x  sin x  1 again differentiating
0
f(x) = f(x) – sin x – cos x which is linear differential equation of 1st order whose solution gives
–x –x
ye = –e cos x + c
x = 0, y = 0  c = 1
x
f(x) = e – cos x

11. Let f(1) = t

 1
Put x = 1 in f  x  f  f  x     1 ….. (1)
 x
We get
1
tf(t + 1) = 1  t  0 and f(t + 1) =
t
1 1 
Put x = t + 1 in equation (1), we get f     t  f 1
 t t  1
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1 1 1 5
  1  t 
t t 1 2
1 5 1
But for t  , t > 1 – f(1) < f(1 + t) =  1 a contradiction
2 t

12. Let M (line be r    ai  bj  ck  then acute angle between L and M is

 a  c  cos   2b sin    a  c  3
cos  
2 2 a2  b2  c 2

 , a + c = 0, b = 0, c = –a
6
M   i  k 

13.  
M meets the given plane in P1, 2 3, 0,  2 3 line L meets the plane in


P 2cos   2 3, 2 2 sin , 2 cos  – 2 3 
2
Thus (PP1) = 8
 P describes a circle of radius 2 2

14. In radians we can say

x  x   x 
sin x  sin  sin    or sin  2 
180  180   180 
180 360
x or x 
180   180  
p m
On comparing with and
q n
We have, m = 360, n = 180, p = 180, q = 180
 m + n + p + q = 900

15. lim
A  C  x 2 2
 Bx
 2 for existence of limit A = C
2
x   B 
x  A   C
 x 
BC B
Hence,  (using A = C2)
A C
B
 L 2
A C
If C = – A then limit does not exist hence C  A
B
 4
C

16. g(x) = –2f(10 – 2x)  g(L = 2) = –2f(6) = 0

SECTION – B

x/2
1. (A)  x 2  5x  6  x 2  5x  4  A
x/2
 x2  5x  6  x 2  5x  4  B

x4 x x
Now, A + B = 2 4  2  2 4 , AB  2 2
A–B=0A=B
Since power of A and B are same
 x = 0, x2  5x  4  0  x = 0, 4, 1
2
 x2  x  2   x2  x  2 
(B) Equation can be written as  2   a  3  2   a  4   0
 x  x 1  x  x 1 
x2  x  2  7
t 2 as t   1, 
x  x 1  3
 7
Equation reduces to t2 – (a – 3)t + (a – 4) = 0 at least one root of equation lies in  1, 
 3
t = a – 4, 1
 7 7  19 
 One root lies in  1,   1  a  4   a   5, 
 3  3  3
2 2
2 x  x 
(C) 1  x   1   .....  1  3  .....  P0  P1x  .....  Pr x r ..... ….. (1)
 2   2 
x
Replace x by
2
2 2 2
 x  x   x  x n
 1    1  2  .....  1  n   P0  P1    .....  Pr x  .....
 2   2   2  2
 
Equating the coefficient of xr
22  Pr 1  Pr  2 
 Pr  putting x0 in (1), we get P0 = 1
2r  1
1 1
 P1  2  1   2  .....  4
2 2
(D) x + y + z = n, x  1, y  1, z  1 number of solution is Pn  n 1Cn3
n 1 n n 1
Cn  3 Cn  2 Cn1  n  1n  2  n n  1 n n  1
n n 1 n 2 1
 Cn  2 Cn 1 Cn  n n  1 n n  1 n  1n  2 
8
n 1
Cn 1 n 2
Cn n 3
Cn1 n  n  1  n  2    n  3  n  2 
n  1

1 n  n  1 n
1
On applying C3  C3 – C2, C1  C1 – C2 and C1  C1 + C3 = 0 2n 1 = –1
2
0 2 n  1 1
 || = 1
6
2. (A) z = (2 cos 30º · cos 18º + i·2 sin 30º · cos 18º)
= 26 cos6 18º (cos 180º + i sin 180º) = –26 cos6 18º
2 dy
(B) f(x) + f (x) = –1   y 2  1  y = tan (c – x)
dx
 
 We must have c – a =  2n  1 and c – b =  2n  1 for some n  I
2 2
ba
 min   1
  
   
(C) Let cos–1 x =  then tan     tan     1
 4 2  4 2
 2 sec  = 1

1
 sec  no solution
2

(D)    7 =   4 =  – 3  tan 4 = –tan 3
7
 t3 – 21t2 + 35t – 7 = 0, t = tan2 
 2 3
Above equation has the roots tan2 , tan2 , tan2 and sum of roots = 21
7 7 7
1
Replacing t =
r
7r3 – 35r + 21r – 1 = 0 has three roots
 2 3
cot 2 , cot 2 , cot 2 and sum of roots = 5
7 7 7

SECTION – C

f  x  dx
1. Let g  x    x 2  x  13 ….. (1)

A B C D E 
=   x  x 2  x  1   x  12   x  13  dx
 
B D E
= A ln x   Cln 1  x   
x 1  x 2  x  12
Since g(x) is a rational function hence A = C = 0
B D E 
g x    2  2
 3
 dx ….. (2)
x  x  1  x  1 
 
Comparing (1) and (2), we get f(x) = (B + D)x 3 + (3B + D + E)x2 + 3Bx + B
 f(x) is quadratic
 B + D = 0, f(0) = 1 gives B = 1  D = –1
 f(x) = (2 + E)x2 + 3x + 1
f(0) = 3

1 0 1
2. 0 1 1 R1  R1  R 3 , R 2  R 2  R 3 on solving [x] + [y] + [z] + 1, taking maximum value, we
 x y z  1
get 4

 36  3  8  
3.       tan1    tan1    tan1   
 77   4  15  2

  cot    cot  and hence A 

 cot   1
 cot 
and  tan  tan   1  A + B = 2
4. A = f(–1) = 19
B = g(–9) = 2 ln 3

C=  3
1

3 x 2  2x  2 
dx  ln9  ln27   ln3
2
2 x  3x  6x  11

t2
d
D= g  x  dx  6g  9   12 ln 3
dt 1
A(B + C) – D = 7 ln 3  k = 7

2 O
5. r1  r2  r1 r2 sin
n
A3
2 2
r1  r2  r1 r2 cos
n n r2
2 2
The given equality reduces to tan 1
n n
2  8 A1 r1 A2
  m   n  m  z  , least n = 8
n 4 1  4m

6. Using reflection property of ellipse equation of reflected

ray SP is B P
4x + 3y = 12 ….. (1)
Solving equation (1) with ellipse, we get
 75 32  S S
A  , 
 17 17 
A
Equation of reflected ray BS is
16x + 63 + 48 = 0 ….. (2)
Equation of incident ray PB is
y=4 ….. (3)
Equation of bisector of (2) and (3) with negative slope gives normal to circle at 3 which is
4x + 32y – 53 = 0
 75 
Point B is   , 4  , using parametric equation of line ordinate of centre is 5
 4 

Time Allotted: 3 Hours Maximum Marks: 432

 Pl ea s e r ea d t h e i n s t r u c t i o n s c a r ef u l l y. Yo u a r e a l l o t t ed 5 m i n ut es
s p ec i f i c a ll y f o r t h i s p u r p o s e.

ALL INDIA TEST SERIES

Yo u a r e n o t a ll o wed t o l ea v e t h e E xa m i n at i o n Ha ll b ef o r e t h e en d o f
t h e t es t .

B. Filling of OMR Sheet

C. Marking Scheme For All Three Parts.

(i) Section-A (01 to 03 and 10 to 12) contains 6 multiple choice questions which have only one
correct answer. Each question carries +8 marks for correct answer and – 2 mark for wrong
answer.

Section-A (04 to 09 and 13 to 30) contains 24 multiple choice questions which have only one
correct answer. Each question carries +4 marks for correct answer and – 1 mark for wrong
answer.

Name of the Candidate

Enrolment No.

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Useful Data

PHYSICS
Acceleration due to gravity g = 10 m/s2

Planck constant h = 6.6 1034 J-s

Charge of electron e = 1.6  1019 C

Mass of electron me = 9.1  1031 kg

Permittivity of free space 0 = 8.85  1012 C2/N-m2

3 3
Density of water water = 10 kg/m

Atmospheric pressure Pa = 105 N/m2

Gas constant R = 8.314 J K1 mol1

CHEMISTRY

Gas Constant R = 8.314 J K1 mol1

Atomic No: H=1, He = 2, Li=3, Be=4, B=5, C=6, N=7, O=8,

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Physics PART – I

SECTION – A
Single Correct Choice Type

This section contains 30 multiple choice questions. Each question has four choices (A), (B),
(C) and (D) out of which ONLY ONE is correct.

1. The unit vector perpendicular to ˆi  2ˆj  kˆ and 3iˆ  ˆj  2kˆ is

5iˆ  3ˆj  7kˆ 3iˆ  5ˆj  7kˆ
(A) (B)
83 83
5iˆ  3 ˆj  7kˆ 3iˆ  5 ˆj  7kˆ
(C) (D)
83 83

2. A ball is projected at an angle 60o with the horizontal with speed 30 m/s. After certain
time it makes an angle 45o with the horizontal. The speed of ball at that moment is
(A) 30 m/s (B) 15 2 m/s
15
(C) m/s (D) 30 2 m/s
2

3. A block of mass 1 kg rests inside a cubical vessel, as v

shown in the figure which is moving with a velocity of
v  29 t m/s, where t is in sec. All the surfaces are
smooth. The block is at rest w.r.t. the cube, the total force F
exerted by the cube on the block
(A) 129 N (B) 130 N
(C) 10 N (D) 29 N

4. A ball falls vertically on to a floor, with momentum p, and then bounces repeatedly, the
coefficient of restitution is e. The total momentum imparted by the ball to the floor is
p
(A) p 1  e  (B)
1  e 
(C) p
1  e   1
(D) p  1  
1  e   e 

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5. A particle of mass 4m which is at rest explodes into three fragments. Two of the
fragments each of mass m are found to move with a speed v each in mutually
perpendicular directions along x and y axis. Calculate momentum of the fragment 2m
after explosion, the energy released in the process of explosion.
v v  3 v v  3
(A) 2m  î  ˆj  , mv 2 (B) 2m  î  ˆj  , mv 2
 2 2  2  2 2  2
v v  3 v v  3
(C) 2m  î  ˆj  , mv 2 (D) 2m  î  ˆj  , mv 2
 2 2  4  2 2  4

6. A solid sphere of mass M and radius r slips on a rough v0

horizontal surface. At some instant, it has translational 
r 2r
 v 
velocity v0 and angular velocity    0  about its center v0
 2r 
as shown in figure. The translational velocity after the
sphere starts pure rolling motion is
v 2v 0
(A) 0 (B)
7 7
5v 0 6v 0
(C) (D)
7 7

7. A uniform rod of length l and mass M is suspended

from two inextensible string as shown in figure.
Tension in left string at the instant, when right string
snaps, is

mg
(A) (B) mg
2
mg mg
(C) (D)
4 8

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k
8. In a certain region at space gravitational field is given by I   . Taking the reference
r
point to be at r = r0 with V = V0, the potential V(r) is given by
r r
(A) V r   K ln  V0 (B) V r   K ln  V0
r0 r0
r r
(C) V r   K ln 0  V0 (D) V r   K ln 0  V0
r r

9. A planet of mass m moves along an ellipse around the sun so that its maximum and
minimum distances from the sun are equal to R and r respectively. The angular
momentum of this planet relative to the centre of the sun (mass M) is
GMmrR GMmR
(A) 2 (B) 2R
r R r R
GMrR 2GMrR
(C) 2m (D) m
r R r R

10. A piston of a syringe pushes a liquid with a speed of 1 cm/sec. The radii of the syringe
tube and the needle are R = 1 cm and r = 0.5 mm respectively. The velocity of the liquid
coming out of the needle is

r = 0.5 mm

R = 1 cm

(A) 2 cm/sec. (B) 20 cm/sec.

11. Two cylinders A and B fitted with pistons contain equal amounts of an ideal diatomic gas
at 300 K. The piston A is free to move while that of B is held fixed. The same amount of
heat is given to the gas in each cylinder. If the rise in temperature of the gas in A is 30 K,
then the rise in temperature of the gas in B is
(A) 30 K (B) 18 K
(C) 50 K (D) 42 K

12. In a room where temperature is 30oC, a body cools from 61oC to 59oC in 4 minutes. The
time taken by the body to cool from 51oC to 49oC will be
(A) 4 min (B) 6 min
(C) 5 min (D) 8 min

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13. The motion of a particle varies with time according to the relation
y  a  sin t  cos t 
(A) the motion is oscillatory but not SHM
(B) the motion is SHM with amplitude a
(C) the motion is SHM with amplitude a 2
(D) the motion is SHM with amplitude 2a

14. If x, v and a denote the displacement, the velocity and the acceleration of a particle
executing SHM of time period T, then which of the following does not change with time?
aT
(A) 3a2 T2  42 v 2 (B)
x
aT
(C) aT  2v (D)
v

15. A sound wave of wavelength  travels towards horizontally with a velocity v. It strikes
and reflects from a vertical plane surface, traveling at a speed v 0 towards the left. The
number of positive crests striking in a time interval of 3 sec. on the wall is
3  v  v0  3  v  v0 
(A) (B)
 

16. A section of an infinite rod of charge having linear charge density  which is constant for
all points on the line. Electric field E at a distance r from the line is
 
(A) (B)
0r 20r
 2
(C) (D)
40r 0r

17.  
Potential in the x-y plane is given as V  5 x 2  xy volts. The electric field at the point
(1,-2) will be
(A) 3 ˆj V/m (B) -5 ˆj V/m
(C) 5 ˆj V/m (D) -3 ˆj V/m

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18. What is the mechanical work done in pulling the slab out of the capacitor after
disconnecting it from the battery?
1 2
(A) E2C  r  1 (B) E C
2
(C) E2C  r  1 (D) none of these

19. Calculate the steady-state current in the 2 2

resistor shown. The internal resistance of the
battery is negligible and the capacitance of the A B
capacitor is 0.2 F. 3
(A) 0.6 A
(B) 0.7 A 0.2F
4
(C) 0.8 A
(D) 0.9 A

6V 2.8 

20. In the secondary circuit of a potentiometer a cell of internal resistance 1.5  gives a
balancing length of 52 cm. To get a balancing length of 40 cm, how much resistance is to
be connected across the cell?
(A) 5 . (B) 6 .
(C) 7 . (D) 4 .

21. The mass of the three wire of copper are in the ratio 1 : 3 : 5, and their lengths are in
ratio 5 : 3 : 1. The ratio of their electrical resistance is
(A) 1 : 3 : 5 (B) 5 : 3 : 1
(C) 1 : 15 : 125 (D) 125 : 15 : 1

22. A point P lies on the axis of a flat coil carrying a current. The magnetic moment of the coil
is  . The distance of P from the coil is d, which is large compared to the radius of the
coil. The magnetic field at P has magnitude
       
(A) 0  3  (B) 0  3 
2  d  4  d 
      
(C) 0  2  (D) 0  2 
6  d  8  d 

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23. A winding wire which is used to frame a solenoid can bear a maximum 10 A current. If
–2
length of solenoid is 80 cm, then length of its winding wire is (B = 0.2 T, r = 3  10 m)
(A) 1.2  10 2 m (B) 4.8  102 m
3
(C) 2.4  10 m (D) 6  103 m

24. An inductor of inductance L = 400 mH and resistors of

E
resistance R1 = 2  and R2 = 2  are connected to a battery
of emf 12 V as shown in the figure. The internal resistance of L
the battery is negligible. The switch S is closed at t = 0. The R1
potential drop across L as a function of time is
12 3t R2
(A) 6e5t V (B) e V S
t
 
(C) 6 1  e t /0.2 V (D) 12e5t V

25. A person looking through a telescope T just sees

the point A on the rim at the bottom of a N T
Liquid
cylindrical vessel when the vessel is empty. r
  1.5 C
When the vessel is completely filled with a liquid
   1.5 , he observes a mark at the centre, B, i
of the bottom without moving telescope or the h
vessel. What is the height of the vessel if the
diameter of its cross-section is 10 cm? B
A
(A) 5.5 cm D
(B) 6.6 cm 5 cm
(C) 7.5 cm
(D) 8.45 cm

26. The focal length of a plano convex lens when curved surface is silvered and the object is
in front of plane surface is
R 2R
(A) (B)
2 
R 2R
(C) - (D) 
2 

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27. At what distance from the objective should an object be placed to focus it properly so that
the final image is formed at the least distance of clear vision?
(A) 1.1 cm (B) 2.2 cm
(C) 3.3 cm (D) 4.4 cm
o
28. The wavelength of 10 keV electron beam is 0.1227 A . when these waves are diffracted
from a metal foil having d = 0.55 Ao the first maxima would occur at an angle  where
sin  is
(A) 0.1116 (B) 0.2218
(C) 0.4464 (D) 0.8928

29. The wavelength of the photoelectric threshold for silver is  0 . The energy of the electron
ejected from the surface of silver by an incident light of wavelength     0  will be
hc
(A) hc   0    (B)
0  
h 1 1   
(C)    (D) hc  0 
c   0   0  

30. If resistivity of pure silicon is 3000m and the electron and hole mobilities are
0.12 m2 V 1s1 and 0.045 m2 V 1s1 respectively. The resistivity of a specimen of the
19 3
material when 10 atoms of phosphorous are added per m is
 Given e = 1.6  10 19
c, =3000m,  e  0.12 m2 V 1s1, h  0.045 m2 V 1s1 
(A) 2.21 m (B) 3.21 m
(C) 4.21 m (D) 5.21 m

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Chemistry PART – II

SECTION – A
Single Correct Choice Type
This section contains 30 multiple choice questions. Each question has four choices (A), (B),
(C) and (D) out of which ONLY ONE is correct.

1. In the given reaction,


H /H O
 A   RMgX   3 2
Alcohol
(excess)
(A) cannot be
(A) R— C — Cl (B) R— C — OC2H5
|| ||
O O
(C) O (D) O
|| ||
R— C — R R— C — H

2. Tautomerism is not exhibited by

(A) CH CH OH (B) O O

O
(C) O (D)
O
O

3. Match the following:

Column-I Column-II
2
(A) Boric acid (P) sp hybridised boron
(B) Diborane (Q) Ring having 6 e– delocalised
(C) Borazine (R) Reacts with water
(D) Boron Nitride (S) Bond having partial double bond character
3
(T) sp hybridised boron

(A) APRS BRT CPQRS DPQRS

(B) APR BPR CPQR DPQR
(C) ARS BRT CPQRS DPQRS
(D) ARST BPRS CPQR DPRS

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– +
4. I2(s) | I (0.1M) half cell is connected to a H (aq) | H 2 (1 bar) | Pt half cell and e.m.f. is found
o
to be 0.7714V. If EI   0.535V , Then the pH of H+/H2 half cell is :
2 /I

(A) 1 (B) 3
(C) 5 (D) 7

5. In the given reaction

NBS Mg CO2
(X)
ether H / H2O
(X) will be :

(A) CHO (B) OH

(C) COOH (D) COOH

COOH

6. Equivalent mass of As2S3 and KMnO4 during the formation of H3AsO4, H2SO4 and Mn2+
ions are respectively :
(A) M/2, M/5 (B) M/6, M/5
(C) M/28, M/5 (D) M /14, M/3

7. What will be the maximum value of  (n +  + m) for all unpaired electrons of chlorine in its
second excited state ?
(A) 25 (B) 20
(C) 28 (D) 27

8. A gaseous compound A reacts by three independent first order processes with rate
constant
–3 –3 –3 –1
2×10 , 3×10 and 1.93×10 sec for products B, C and D respectively. If initial pressure
of pure A was 8 atm taken in the closed container. Then the partial pressure of B (in atm)
after 100 sec from starting:
(A) 0.288 (B) 0.577
(C) 1.154 (D) 2.15

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9. Which of the following statements for crystals having frenkel defect is not correct?
(A) Frenkel defects are observed where the difference in sizes of cations and anions is
large
(B) The density of crystals having frenkel defect is lesser than that of a pure perfect crystal
(C) In an ionic crystal having frenkel defect may also contain schottky defect
(D) Usually alkali halides do not have frenkel defect

10. Which of the following is true in respect of chemical adsorption?

(A) H < 0, S > 0, G > 0 (B) H > 0, S > 0, G < 0
(C) H < 0, S < 0, G < 0 (D) H > 0, S < 0, G > 0

11. 0.1 M NaCl and 0.05 M BaCl2 solutions are separated by a semi-permeable membrane in a
container. For this system, choose the correct answer :
(A) There is no movement of any solution across the membrane
(B) Water flows from BaCl2 solution toward NaCl solution
(C) Water flows from NaCl solution towards BaCl2 solution
(D) Osmotic pressure of 0.1M NaCl is lower than the osmotic pressure of BaCl2

12. Glucose reacts with acetic anhydride to form :

(A) hexaacetate (B) tetraacetate
(C) monoacetate (D) pentaacetate

13. Among the following the coloured compound is :

(A) CuCl (B) K3[Cu(CN)4]
(C) CuF2 (D) [Cu(CH 3CN)4] BF4

14. Identify the incorrect statement :

(A) Liquid oxygen is an important constituent of rocket fuel
(B) When SO2 gas is passed through cupric chloride solution, then the solution becomes
colourless and a white precipitate is formed
(C) Radon is formed from the decay of Radium
(D) Hydrofluoric acid is preserved in glass bottles

15. Which of the following ores is best concentrated by froth floatation process:
(A) Malachite (B) Cassiterite
(C) Galena (D) Magnetite

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16. The IUPAC name of [Ni (NH3)4] [NiCl4] is:

(A) Tetrachloronickel (II) tetra ammine nickel (II)
(B) Tetraammine nickel (II) tetra chloronickelate (II)
(C) Tetraamminenickel (II) tetrachloronickel (II)
(D) Tetrachloronickel (II) tetra amminenickelate (O)

O
LiAlH4
17. O C NH2

OH
(A) (B) HO NH2

HO NH2

O
(C) O NH2 (D)
HO NH2

18. Which of the following is not correctly matched ?

heat
(A) Phenol + CHCl3 + NaOH 
 Salicylaldehyde
heat
(B) Phenol + Phthalic anhydride 
 Phenetole
H2SO4

– H3PO2
(C) C6H5 — N  NCl  C6H6
H2O
anhyd
(D) C6H6 + CH3COCl  C6H5 COCH3  HCl
AlCl3

OH OH
COOH COOCH3
19. and can be differentiated by :

(A) NaOH (B) Na metal

20. The monomers of Buna-S polymers are :

(A) Vinyl chloride and Vinylidene (B) Styrene and butadiene
(C) Acrylonitrile and butadiene (D) Isobutylene and isoprene

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21. In the reaction, 4A + 2B + 3C  A4B2C3, what will be the number of moles of product
formed, starting from 1 mol of A, 0.6 mol of B and 0.72 mol of C?
(A) 0.25 (B) 0.3
(C) 0.24 (D) 2.32

22. The incorrect statement among the following is

(A) In TI3 the oxidation state of T is +3.
(B) B2O3 reacts with P2O5 to form BPO4.
(C) B(OH)3 solution is acidic in nature.
(D) Borax solution can be used as a buffer.

23. Sodium oxide Na 2O is a crystalline solid. It does not have.

(A) Antiflourite structure
(B) Na+ ions are present at body diagonals
(C) Na + ions are present at tetrahedral voids
2–
(D) O ions are present at octahedral voids

24. The oxidation number of carbon in H — C  N and H – N  C are respectively :
(A) +2, +3 (B) +2, +4
(C) +2, +2 (D) +3, +3

25. 0.56 litre each of three samples of H2O2 labelled as 10 vol, 15 vol and 20 vol are mixed.
The resultant normality of H2O2 solution is :
(A) 1.68 (B) 2.68
(C) 4.5 (D) 8.0

26. Calculate the vapour pressure of water at 300K if its heat of vaporisation is 540 cal/gm
(A) 3.162 mm (B) 0.3192 mm
(C) 31.92 mm (D) 319.2 mm

27. Select the incorrect statement :

(A) O3 and O2– both are diamagnetic
+ +
(B) Out of O2, O2 , O3 , least O – O bond length is in O2
+ –
(C) Out of O2, O2 , O2 , only O2 is paramagnetic
(D) Out of O2, O2+, O2–, maximum spin magnetic moment have O2

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28. Which systematic diagram represents the correct sequence of reaction?

 
(A) CaCO3   CO2  CaO (B) CaCO3   CO 2  CaO
H2O ex H2O
ce
ss
CO2 of
CO
Ca(OH)2 2 Ca(OH)2

(C) CaCO3  CaO  CO2 (D) CaCO3  CO2  CaO


Ca H2O (cold) SO H2O

+H Cl 2
Cl( 2
aq
) Ca(OH)2
H2CO3

29. How many of the following addition reactions are syn addition reaction:
(a) H3C CH3 (b) H3C CH3
Br H2
C C 
2
 C C 
CCl4 Ni

H H D D

(e) BH3 .THF

(f)
  cold KMnO 4
H2O2 ,–
OH  

(g) CH3 (h)

Br2 H2O
1.PhCO3H  



2.H /H2O

CH3
(A) 3 (B) 6
(C) 5 (D) 4

30. Column-I Column-II



(A) (P) Hyperconjugation


(B) (Q) All carbon atom are sp2 hybridized

(C) (R) Aromatic

(D)  CH3 3 — C (S) Diamagnetic

(A) APQR BPQR CPQR DPQR

(B) AQRS BQRS CRS DPQR
(C) AQRS BQRS CQRS DP,S
(D) AQRS BQS CQR DP,S

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Mathematics PART – III

SECTION – A
Single Correct Choice Type

This section contains 30 multiple choice questions. Each question has four choices (A), (B),
(C) and (D) out of which ONLY ONE is correct.

1. A and B throw a pair of dice alternately till one of them wins. If A wins by throwing 7 while
B wins by throwing 5 as a sum respectively, then the probability that one of the dice
shows 3 on the last throw if A starts the game is
11 11
(A) (B)
36 28
7
(C) (D) none of these
22

8
2. The sum of rational term(s) in  33245  is equal to
(A) 3150 (B) 336
(C) 3486 (D) 3592

sin x sin x
3. e e  4a  0 will have exactly four different solutions in [0, 2] if
 e 1
(A) a    ,   (B) a  R
 4 4 
  1  e2 
(C) a   ,  (D) none of these
 4e 

2 
1 1
4. If 2 sin x  2  a  2  2 sin x  8a  0 for at least one real x, then
 1
(A) a   ,    2,   (B) a  R – {2}
 8
1
(C) a < 2 (D) a2
8

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5.  
Let f(x) = sin x cos x  cos2 x  cos2  , where ‘’ is a given constant, then maximum
value of f(x) is
(A) 1  cos2  (B) |cos |
(C) |sin | (D) 1  sin2 

6. The number of integral values of K for which the equation 7 cos x + 5 sin x = 2k + 1 has a
solution
(A) 5 (B) 6
(C) 7 (D) 8

7. If sin–1 x + sin–1 y + sin–1 z = , then x4 + y4 + z4 + 4x 2y2z2 = k(x 2y2 + y2z2 + z2x2), where k
is equal to
(A) 1 (B) 2
(C) 4 (D) none of these

8. If the variable takes values 0, 1, 2, 3, ….., n with frequencies proportional to

n n n n
C0, C1, C2, ….., Cn respectively, the variance is
n 2n
(A) (B)
3 5
n
(C) (D) none of these
4

9. Let a relation R in the set N of natural numbers be defined as (x, y)  x2 – 8xy + 7y2 = 0
 x,
y  N. The relation R is
(A) reflexive (B) symmetric
(C) transitive (D) none of these

10. Suppose f(x) = x 3 + ax 2 + 2bx + c, where a, b, c are chosen respectively by throwing a

dice three times. Then the probability that f(x) is a decreasing function for some x is
1 13
(A) (B)
3 36
2
(C) (D) none of these
9

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x2 y2
11. A line intersects the ellipse 2
 2  1 at A and B and the parabola y2 = 4a(x + 2a) at C
4a a
and D. The line segment AB subtends a right angle at the centre of the ellipse. Then, the
locus of the point of intersection of tangents to the parabola of C and D is
5 2
(A) y2 – a2 =  x  4a  (B) y2 – 2a2 = 10(x – 4a)2
4
5 2
(C) y2 + a2 =  x  4a  (D) y2 + 4a2 = 5(x + 4a)2
2

9x  1   2   3   4029 
12. If f  x   , then f  f  f   .....  f   is equal to
9x  9  2015   2015   2015   2015 
4029
(A) 1007 (B)
2
(C) 2014 (D) 2015

x  ai
13. If pi  , (where i = 1, 2, 3, ….., p) and a1 > a2 > a3 ….. ap, then lim  p1p2 ..... pp  , 1
x  ai x  aq

 q  p is equal to
(A) (–1)q + 1 (B) (–1)q – 1
q
(C) (–1) (D) does not exist

14.  x  3  sin1(ln x)  cos1(ln x)  dx is equal to


(A)  x  3 3/2  C (B) 0
3
(C) DNE (D) none of these

    tan x   3 sin 2x 
15. If x    ,  , then the value of tan1    tan1   is
 2 2  4   5  3cos 2x 
(A) 2x (B) 3x
x
(C) (D) x
2

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16. P is any point on the circum-circle of ABC other than the vertices. H is the orthocentre of
ABC, M is the mid-point of PH and D is the mid-point of BC. Then,
(A) DM is parallel to AP (B) DM is perpendicular to AP
(C) AP is opposite side of DM (D) none of these

17. Let f(x) = (x – 3)3xn, n  N. Then, f(x) has a

(A) minimum at x = 0, if n is even (B) maximum at x = 0, if n is even
(C) maximum at x = 3  n  N (D) none of these

1 1
18. If a function y = f(x) is defined as y  2
and t  , t  R. Then, f(x) is
t t6 x2
discontinuous at
2 7 3 7
(A) 2, , (B) 2, ,
3 3 2 3
3 5
(C) 2, , (D) none of these
2 3

1

 1  x  x  e
1 x x
19. dx is equal to
1 1
(A) x  e x  x (B)  x  e x  x  c
1 1
(C)  x  1 e x  x  c (D)  x  1 e x  x  c

20. The co-ordinates of four vertices of a quadrilateral are (–2, 4), (–1, 2), (1, 2) and (2, 4)
taken in order. The equation of the line passing through the vertex (–1, 2) and dividing
the quadrilateral in two equal areas is
(A) x + 1 = 0 (B) x + y = 1
(C) x – y + 3 = 0 (D) none of these

n
r n 1  r loge 10
21.   1 Cr r
equals
r 0

1  loge 10n 
(A) 1 (B) –1
(C) n (D) none of these

22. If a, b, c are distinct positive numbers such that b + c – a, c + a – b and a + b – c are

positive, then the expression (b + c – a)(c + a – b)(a + b – c) – abc is
(A) positive (B) negative
(C) non-positive (D) non-negative

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23. PSQ is a focal chord of a parabola whose focus is S and vertex is A. PA, QA are
produced to meet the directrix in R and T. Then angel RST is equal to
(A) 90º (B) 60º
(C) 45º (D) 30º

24. If z1 and z2 are lying on |z – 3|  4 and |z – 1| + |z + 1| = 3 respectively. Then, A = |z1 – z2|

satisfies
15 15
(A) 0  A  (B) 0  A 
2 2
17 17
(C) 0  A  (D) 0  A 
2 2

25. If four points be taken on a rectangular hyperbola such that the chord joining any two is
perpendicular to the chord joining the other two, and if , , ,  be the inclinations to
either asymptote of the straight line joining these points to the centre, then tan  tan  tan
 tan  is equal to
(A) 1 (B) 2
(C) –2 (D) –1

26. The sum of squares of the perpendiculars drawn from the points (0, 1) and (0, –1) to any
tangent to a curve is 2. Then, the equation of the curve is
(A) 2y = c(x + 2) (B) y = c(x  1)
(C) y = c(x + 2) (D) y = c(x  2)
2 2
27. The area of the region bounded by the curves y = x , y = |2 – x | and y = 2 which lies to
the right of the line x = 1, is
 12  20 3   20  2 
(A)   sq. units (B)   sq. units
 2  3
   
 20  12 2   12  20 2 
(C)   sq. units (D)   sq. units
 3  3
   

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e 2iA e iC e iB

28. If A, B, C are angles of a triangle, then e iC e 2iB e iA is equal to
e iB e iA e2iC
(A) 4 (B) 0
(C) –4 (D) –1

29. Two persons who are 500 m apart, observe the direction and the angle of elevation of a

balloon at the same instant. One finds the elevation to be and direction being South-
3

West, while the other finds the elevation to be and direction being west. Height of the
4
balloon is
3 3
(A) 500 m (B) 250 m
4 6 4 6
3 3
(C) 250 m (D) 500 m
4 6 4 6

30. Let p and q be two statements, then ~(~p  q)  (p  q) is logically equivalent to

(A) q (B) p  q
(C) p (D) p  ~ q

Space for rough work

ANSWERS, HINTS & SOLUTIONS

FULL TEST – IV
(Main)
ALL INDIA TEST SERIES

Q. No. PHYSICS CHEMISTRY MATHEMATICS

1. B D B
2. B B D
3. A A D
4. C B A
5. B D A
6. D C D
7. C A B
8. A C C
9. D B A
10. D C A
11. D B D
12. B D B
13. C C D
14. B D C
15. A C D
16. B B B
17. B D B
18. A B B
19. D C A
20. A B C
21. D C D
22. A A B
23. C D A
24. D C C
25. D B A
26. C C B
27. A C C
28. B A C
29. D D A
30. D C C

Physics PART – I

SECTION – A
   
1. A  B is a vector perpendicular to both A and B
 
  
Now, A  B  ˆi  2jˆ  kˆ  3iˆ  ˆj  2kˆ 
3iˆ  5 ˆj  7kˆ
 
A B
Now, n̂   
A B
3iˆ  5ˆj  7kˆ

32  52  72
3iˆ  5ˆj  7kˆ

83

2. Horizontal velocity remains same

30o cos60o  v cos 45o
 v = 15 2 m/s

3. v  29 t
dv N
a  29
dt F R
R  ma  29 N
Fnet  N2  R2 N
mg
 10 2   29   129 N

st
4. p after 1 impact  ep   p 
 p 1  e 
nd
Similarly p after 2 impact = ep 1  e 
So, pnet  p 1  e  1  e  e2  ........
p 1  e 

1  e 
5. Force of explosion is internal and system is initially at rest
Let the velocities of the first two fragments are v ˆi & vjˆ and that of the fragment 2m be
v1ˆi  v 2 ˆj  v 3kˆ ,
  
So, p1  p2  p3  0

 
 mv ˆi  mvjˆ  2m v 1ˆi  v 2 ˆj  v 3kˆ  0
v v
 v1   , v 2   and v 3  0
2 2

 v v 
p3  2m  ˆi  ˆj 
2 2 
1 1 1
2 2 2

So, K f  mv 2  mv 2   2m  v 12  v 22  v 23 
3
 mv 2
2
Energy released in explosion
E  K f  K i
3 3
 mv 2  0  mv 2
2 2

6. Angular momentum about point of contact will be conserved

Li  L f
 2 2  v0 2 v
 Mr   Mrv 0   Mr 2   Mrv
5  2r 5 r
6
 v  v0
7

7. mg  T  ma and
mgl ml2 a
 I  
2 3 l T
 
2
3g a
a
4

mg mg
So, T 
4

K dV
8. I 
r dr
V r
dr
  dV  K  r
V0 r0

r
 V  K ln  V0
r0

9. By conservation of angular momentum, V1

mv1R  mv 2r
 v1R  v 2r
By conservation of energy, R r
GMm 1 GMm 1
  mV12    mV22 V2
R 2 r 2
2GMr
 V1 
R R  r 
2GMrR
Now, mv1R  m
r R

10. By equation of continuity

Av  av '
A  R 2 
 v'   v   2  v
a  r 
 v '  400 cm/sec.

11. For cylinder A ;

Q  nCp T1
For cylinder B, Q  nCv T2
Hence, nCp T1  nCv T2
 Cv  R  30  Cv T2
For diatomic gas,
5
Cv  R
2
 T2  42 K

dT d
      0 
12. dt dt
 dT  Kdt
In first case,
dT  61  59  2o C ;   30o C , dt = 4 min.
For second case, dT = 2oC;   20o C
dT 2 1
K  
dt 30  4 60
dT 2
dt    6 min.
K 1
 20
60

13. y  a  sin t  cos t 

 1 1 
 a 2  sin t  cos t 
 2 2 
 
 a 2 sin  t  
 4
The motion is SHM with amplitude a 2

aT 2 xT 42
14.   2 T
x x T
2
4
  Constt.
T

15. Relative velocity is  v  v 0 

 v  v0 
Opponent frequency is

The no. of positive crests striking per sec. is same as frequency.
 v  v0 
In 3 sec. 3  
  


16. From symmetry, E due to a uniform linear charged can only be radially directed. As a Gaussian
surface, we can choose a circular cylinder of radius r and length. L, closed at each end by plane
caps normal to the axis.
r E
 +++++ +++++ +++++ +++++ +++++ +++ 

 
o  E. ds  qin
   
o   E.ds   E.ds  qin
 
Cylindrical Plane Surface
oE  2rl  E.ds cos 90o  l
l 
E 
o 2rl 20 r

The direction of E is radially outwards for a line of positive charge.

V
17. Ex     10x  5y   10  10  0
x
V
Ey    5x  5
y

 E  5ˆj V/m.

18. Work done = change in potential energy

 U2  U1
U1  1/ 2  E 2C
2
1  EC  1 E 2 C2
U2   r  1/ 2  E2 Cr
2 C' 2 C
 Work done = 1/ 2  E 2C  r  1 .

19. The resistance of the parallel combination of 2 and 3 resistors is given by

1 1 1 5
    R  1.2
R 2 3 6
This resistance is in series with 2.8 giving a total effective resistance  1.2  2.8  4.
In the steady state, charge on the capacitor C has stablised and hence no current passes
through 4 resistor which is in series with the capacitor.
Thus the current through the circuit = 6/4 = 1.5 A,
VAB  1.5  1.2  1.8V, I through 2 resistor = 1.8/2 = 0.9 A.

20. r  1.5, l1  52cm, l2  40cm

r l1  l2

R l2
rl2
resistance R =
l1  l2 

1.5  40

52  40 
1.5  40
  5
12

l l 2 l 2 l 2 d l 2
21. R    
A Al V m/d m
l2
or R 
m
l12 l22 l32 25 9 1
R1 : R2 : R3  : :  : :  125 : 15 : 1
m1 m2 m3 1 3 5

22. The magnetic field at P due to the flat

n i
coil of n turns, radius r, carrying P
current i is
r i
 nir 2 0 nir 2 d
B 0.  .  d  r 
2 d2  r 2 3/ 2 2 d3
 
0 n  r  i 
2
0 
 .  . .
2 d3 2 d3

0Ni
23. B where N = Total number of turns, l = length of the solenoid
l
4  10 7  N  10
 0.2 
0.8
4
4  10
N

Since N turns are made from the winding wire so length of the wire
L   2r  N2r  length of each turns
4  104
 L  2  3  102   2.4  103 m


E 12
24. l1    6A
R1 2
dl2 E
E L  R2  l2
dt
L
E 12
 
l2  l0 1  e t / t0  l0 
R2

2
 6A R1

L 400  10 3
t0   0.2 R2
R 2 S

l2  6 1  e  t /0.2 

Potential drop across L = E - R2l2  12  2  6 1  ebt  12 e 5t

25. It is mentioned in the problem that on filling the

vessel with the liquid, point B is observed for the N T
Liquid
same setting; this means that the image of point r
B, is observed at A, because of refraction of the   1.5 C
ray at C. For refraction at C, i
sin r h
a l  1.5
sin i
Now, B
A
D
AD 10
sin r =  5 cm
AC 102  h2
Where h is the height of vessel.
BD 5
sin i= 
BC 5  h2
2

10 25  h2
 .  1.5
100  h2 5
25  h2 9
 2

100  h 16
 400  16h2  900  9 h2
 7h2  500
 h  8.45 cm

26. In this case

1 1 1     1
    1   
fL   R R
 R 
and FM 
2
1    1
so PL  
fL R
1 2
and PM   
fM R
and hence power of system
P=PL  PM  PL  2PL  PM
2    1 2 2
P  
R R R
1 R
F    
P 2
i.e. the lens will be equivalent to a converging mirror of focal length  R / 2 

27. For eyepiece,

vE = –25 cm, fE = +5 cm
 uE  4.17 cm  4.2 cm
L  v 0  uE  12.2 cm here v 0  8 cm
f0  1 cm  u0  1.1 cm

28. Wavelength associated with a particle is given by

h

2mE
h 150 o

2meV

V
A  for electron M = 9.11 10 31
kg 
150
  0.122 A o
10  103
For minima
dsin   n 
For first minima n = 1
0.55 sin   0.122
0.122
sin    0.2218
0.55

29. hv  hv 0  E
E  hv  hv 0
c
But v =

1 1   0   
E  hc     hc  
  0   0  

30. The resistivity of pure Si is given by

1 1 1
  
 e  ne e  nhh  en1  e  h 
1 1
or ni   19
e   e  h  1.6  10  3000  0.12  0.045 
16 3
 1.26  10 m
When 1019 atoms of phosphorous (donor atoms of valence five) are added per m3, the
semiconductor becomes n – type semiconductor.
 ne  n h  ne  Nd  1019  nh  1.26  1016
1 1
Re sitivity =   5.21 m
ne e e 1.6  10  1019  0.12
19

Chemistry PART – II

SECTION – A

1. Aldehyde form 2° alcohol with RMgX.

3
2. It does not have -hydrogen atom connected to sp carbon.

3. H3BO3  B2H6 
H H H H
O O
H B B
H H H
B
B

O
H
Borazine  BN
H
sp2 B B
B
N N N
H N N H

B B B
H B B H
N N
N

4. The cell reaction is

H2(g) + I2(g)  2H+ (aq) + 2I–(aq)
2 2
0.0591 H  I 
0.7714 = 0.535  log
2 pH2
pH = 3

Br MgBr COOH

5. NBS Mg CO2
ether +
H /H 2O

6. As2S3  H3AsO4 + H2SO4

3+ 5+ –
(As )2  2As + 4e
2– 6+ –
(S )3  3S + 24e
____________________________
As2S3  2As5+ + 3S6+ + 28e–
E = M/28
Mn7+ + 5e–  Mn2+  E = M/5

7. Cl17  3s2 3p5

st 2 4 1
Cl17 (1 excited state) 3s 3p 3d
nd
Cl17 (2 excited state)
3s2 3p3 3d2
     
n – 3 3 3 3 3
 1 1 1 2 2
m  – 1  1 0 2  1
 (n +  + m ) = 25

–3
8. The overall rate constant, K = K1 + K2 + K3 = 6.93 × 10
0.693
t1/2   100 sec
6.93  10 –3
After half life, total pressure, P = 4 atm

PB K 1
Now, 
P K
2  10–3  4
PB   1.154 atm
6.93  103
11. In osmois water flows from hypotonic solution towards hypertonic solution through SPM
12. In glucose there are 5 – OH groups
13. In CuF2, Copper is in +2 oxidation state
14. Hydro fluoric acid is not preserved in glass bottles because it reacts with silicon dioxide of glass
15. Galena is sulphide ore
O
17. LiAlH4 does not reduce ||
—C — NH2
OC2H5

18. Phenetole is

21. 4A + 2B + 3C  A 4B 2C 3
Given 1 mole 0.6 mol 0.72 mol ?
In the above, C is the limiting reagent hence it will give least product on complete consumption.
 3 mol C give 1 mol product
 0.72 mol C give 0.24 mol product
22. (A) Oxidation state of T in TI3 is +1
(B) B2O3  P2 O5 
 2BPO4



(C) B  OH3  H2 O 
 B  OH 4   H


 2Na B  OH  4   2HBO3

(D) Na2B 4 O7  7H2 O 
Salt Weak acid

23. Na2O has antiflourite structure in which O2– ions are present at ccp and Na+ ions are occupy all the
tetrahedral voids which are located at body diagonals.

24. In both oxidation state of carbon is +2.

25. Vol. Strength of H2O2 = 5.6 × N
10
 Normality of 10 vol = N
5.6
20
Normality of 20 vol = N
5.6
15
Normality of 15 vol  N
5.6
0.56  10 0.56  15 0.56  20
 Total equivalent of H2O2 =   = 1 + 1.5 + 2.0 = 4.5
5.6 5.6 5.6
4.5
 NH O   2.68
2 2
3  0.56
26. By using the formula
P2 H  T2  T1 
2.303 log   
P1 R  T1T2 
For water boiling point = 373 K and thus P = 760 mm
Also H = 540 × 18 cal/mol
760 540  18  373  300 
2.303 log   
P1 2  373  300 
P1 = 31.92 mm

27. O2, O2+ and O2– all are paramagnetic in nature.


28. CaCO 3   CaO  CO 2
 Ca  OH2
CaO  H2O 
Ca(OH)2  CO 2 
 CaCO 3
 2Ca HCO3 2
2CaCO3  2H2O  2CO2 
excess
29. Option (b), (c), (e) & (f) shows syn addition.


30.  sp2 carbon, aromatic and diamagnetic


 sp2 carbon, aromatic and diamagnetic

 sp2 carbon, aromatic and diamagnetic

Mathematics PART – III

SECTION – A

9 5
1. P  A wins   ; P  B wins  
14 14
9 2 5 2 11
P  required     
14 6 14 4 28

8 8! x y z
2. General term in the expansion of  33245  is
x! y! z!
 3  2  5
3 4

(Where x + y + z = 8)
For rational term(s), we have the following cases:
(i) x = 2, y = 6, z = 0
(ii) x = 4, y = 0, z = 4
(iii) x = 0, y = 0, z = 8
(iv) x = 8, y = 0, z = 0

3. Let, t  e sin x  1, e 

 t2 + 4at + 1 = 0
This should have two distinct roots in [1, e]

 
 1
 1

4.  2 sin x  4   2 sin x  2a   0
  
  

1  1
2 sin x   0,    4,  
 4

5. Here, f2(x) cot2 x – 2f(x) cot x + f2(x) – cos2  = 0 for which D  0

7 5 
6. y = 7 cos x + 5 sin x = 8  cos x  sin x  = 8 sin ( + x)
8 8 
2k  1
 sin(   x) 
8
2k  1 2k  1 9 7
1   1 . 1   1,   k 
8 8 2 2

–9/2 –0 –3 –2 –1 0 1 2 3 7/2
Hence number of integral values are 8

7. Here, x 1  y 2  y 1  x 2  z

n 2
 n 
 r 2  n Cr   r  Cr
n

8. Variance = r 0
n
 n
r  0 
 
 n Cr   Cr
n

r 0  r 0 

10. For f(x) to be strictly decreasing f(x) = 3x2 + 2ax + 2b < 0 for some x
i.e., f(x) must have distinct real roots
2
Thus, D > 0  a > 6b
This includes 12 cases for
(a, b), i.e., (3, 1), (4, 1), (4, 2), (5, 1), (5, 2), (5, 3), (5, 4), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5)
1
Thus, P  required 
3

11. Let P(h, k) be the point of intersection

Then, equation of chord of contact is ky – 2ax = 2a(h + 4a)
2
x2 y 2  ky  2ax 
Homogenising it with ellipse, 2
 2   0
4a a  2a  h  4a  
represents pair of straight lines from origin to AB which are perpendicular to each other

12. Here, f(x) + f(2 – x) = 1

13. Here, L.H.L. = (–1)q–1 and R.H.L. = (–1)q

14. Since the given function is not defined for any real number so its domain is  so that the
integration will not exist
   
16. Let a, b, c and p be position vectors of A, B, C and P respectively
  
Then, P.V. of orthocentre H = a  b  c
   
abc p
P.V. of M 
2
 a  p
DM 
2
 

 DM  PA  0

17. Here, f(x) = xn – 1(x – 3)2 [x(n + 3) – 3n]

3n
 f(x) = 0 at x = 0, 3,
n3

18. Here, t is not defined at x = 2 and y is not defined at t = –2, 3

3 7
Now, t = –2 at x  and t = 3 at x 
2 3

1
x  x 1  1  x  x 1
 1  x  x  e
1 xx
19. dx = e dx   x  1  2
 x
e

dx
1
x  x 1 1 1
= e dx  x  ex  x   e x  x dx = x  ex  x  c

20. Let points be A(–2, 4), B(–1, 2), C(1, 2) and D(2, 4)
Let BE be required line such that DE = 
1
Then area of quadrilateral BCDE =  Area of trapezium ABCD
2

n
r n 1 r x
21. Let loge 10 = x. Then,   1 Cr r
r 0 1  nx 
n r r 1
 1  nx n 1 
=   1
r
 nCr      1r 1  n 1Cr 1  
r 0  1  nx  1  nx r 1  1  nx 
n n 1
 1  nx  1 
= 1    1   0
 1  nx  1  nx  1  nx 

b  c  a    c  a  b  1/2
22. Here,   b  c  a  c  a  b 
2
 c  a  b  a  b  c  1/2
  c  a  b  a  b  c 
2
a  b  c   b  c  a 1/2
and   a  b  c b  c  a 
2
Multiplying, abc > (a + b – c)(b + c – a)(c + a – b)

23.    
Let P and Q be at12 , 2at1 and at 22 , 2at 2 respectively for the parabola y2 = 4ax
Then, t1t2 = –1
 2a   2a 
Also, R and T are  a,   and  a,   respectively
 t1   t2 
1 1
 Slope of RS  and slope of TS 
t1 t2

24. Here, |z – 3|  4 is circle with centre at (3, 0) and radius 4

|z – 1| + |z + 1| is ellipse with foci at (–1, 0) and (1, 0) and length of major axis 3

 c  c  c  c
25. Let the four points be  ct1,  ,  ct 2 ,  ,  ct 3 ,  and  ct 4 , 
 t1   t2   t3   t4 
1 1 1 1
 tan   2 ; tan   2 ; tan   2 ; tan   2
t1 t2 t3 t4
1
Also,  1
t1t 2 t 3 t 4

dy
26. Equation of tangent to curve at (x, y) is Y  y  X  x
dx
Let, lengths of perpendiculars be D1 and D2
Then D12  D22  2
2
 dy   dy 
dx
 

   x2  1  2xy    y2  0
 dx 

dy xy  y
 
dx x 2  1

2 2
27. Required area y = x y = |x – 2|
= Area of region APQA + Area of region AQRA y=2
Q
2 2
C R(2, 2)
 x   2  x  dx    4  x  dx
2 2 2
=
1
(0, 2)
2
P(1, 1)


B  2, 0  A  2, 0 
x=1

28. Here, A + B + C = 
 e–i(B + C) = –eiA; e–i(C + A) = –eiB; e–i(A + B) = –eiC
eiA eiB  eiC
Now taking eiA, eiB and eiC common from R1, R2 and R3 =  eiA eiB  eiC
eiA eiB eiC

29. Here, AQ = h P
h
BQ =
3 h
S–W
Using cosine rule in AQB 
A  W
3 4  Q
h  500 m 4
4 6 3
B

Paper 1

Time Allotted: 3 Hours Maximum Marks: 243

 Pl ea s e r ea d t h e i n s t r u c t i o n s c a r ef u ll y . Yo u a r e a l l o t t ed 5 m i n u t es
s p ec i f i c a ll y f o r t h i s p u r p o s e.
ALL INDIA TEST SERIES

 Yo u a r e n o t a l l o wed t o l ea v e t h e E xa m i n at i o n Ha l l b ef o r e t h e en d o f
t h e t es t .

INSTRUCTIONS

A. General Instructions
1. Attempt ALL the questions. Answers have to be marked on the OMR sheets.
2. This question paper contains Three Parts.
3. Part-I is Physics, Part-II is Chemistry and Part-III is Mathematics.
4. Each part is further divided into three sections: Section-A, Section-B & Section-C
5. Rough spaces are provided for rough work inside the question paper. No additional sheets will be
provided for rough work.
6. Blank Papers, clip boards, log tables, slide rule, calculator, cellular phones, pagers and electronic
devices, in any form, are not allowed.
B. Filling of OMR Sheet
1. Ensure matching of OMR sheet with the Question paper before you start marking your answers
on OMR sheet.
2. On the OMR sheet, darken the appropriate bubble with black pen for each character of your
Enrolment No. and write your Name, Test Centre and other details at the designated places.
3. OMR sheet contains alphabets, numerals & special characters for marking answers.
C. Marking Scheme For All Three Parts.
1. Section – A (01 – 06) contains 6 multiple choice questions which have only one correct answer.
Each question carries +3 marks for correct answer and – 1 mark for wrong answer.
Section – A (07 – 09) contains 3 multiple choice questions which have one or more than one
correct answer. Each question carries +4 marks for correct answer. There is no negative
marking.
Section-A (10 – 14) contains 2 paragraphs. Based upon paragraph, 2 and 3 multiple choice
questions have to be answered. Each question has only one correct answer and carries
+3 marks for correct answer and – 1 mark for wrong answer.
2. Section – B (01 – 02) contains 2 Matrix Match Type questions containing statements given in 2
columns. Statements in the first column have to be matched with statements in the second
column. Each question carries +8 marks for all correct answer. For each correct row +2 marks
will be awarded. There may be one or more than one correct matching. No marks will be given for
any wrong matching in any question. There is no negative marking.
3. Section-C (01 – 05) contains 5 Numerical based questions with answers as numerical value from
0 to 9 and each question carries +4 marks for correct answer. There is no negative marking.

Name of the Candidate

Enrolment No.

Useful Data

PHYSICS
Acceleration due to gravity g = 10 m/s2

Planck constant h = 6.6 1034 J-s

Charge of electron e = 1.6  1019 C

Mass of electron me = 9.1  1031 kg

Permittivity of free space 0 = 8.85  1012 C2/N-m2

Density of water water = 103 kg/m3

Atmospheric pressure Pa = 105 N/m2

Gas constant R = 8.314 J K1 mol1

CHEMISTRY

Gas Constant R = 8.314 J K1 mol1

Atomic No: H=1, He = 2, Li=3, Be=4, B=5, C=6, N=7, O=8,

Physics PART – I

SECTION – A
Straight Objective Type
This section contains 6 multiple choice questions. Each question has four choices (A), (B), (C)
and (D) out of which ONLY ONE is correct.

1. Find the angular frequency of oscillation of the show

system (the pulley is massless).
K K 2K
(A) (B)
M 3M
K 2K
(C) (D)
2M 3M

K M

2. An object is placed at a distance of 20 cm from an equi-concave lens of focal length 30

cm whose one side is silvered. If a virtual image is formed at 20/3 cm from the lens then
what is the radius of curvature of the curved surface?
(A) 30 cm (B) 40 cm
(C) 60 cm (D) 15 cm

3. Find the current i in the circuit.

E E R R R
(A) i  (B) i  i
2R R E
E 2E
(C) i  (D) i 
4R R R R

R
E

Space for Rough work

4. Four particles each of mass ‘m’ are located at equal distance from each other. If the
separation between any two particles is a then their gravitational potential energy is:
4Gm2 Gm2
(A) (B)
a 8a
Gm2 6Gm2
(C) (D)
6a a

5. A parallel beam of light is incident on a perfectly absorbing small cube of side a. The
beam is directed along one of the body diagonals of the cube and has sufficiently large
aperture. What is the total radiation force experienced by the cube. I is the intensity of
radiation.
I I
(A) a2 (B) 3a2
c c
I I
(C) 3a2 (D) 2a2
c c

6. A cylinder is released from top of an inclined plane

which itself rests on a smooth horizontal surface. The m
initial velocity of the cylinder as well as the incline is
zero. (Assume no slipping in each case where the
incline is rough). M
(A) As the cylinder reaches the bottom most point, the
inclined plane will acquire a greater speed if it
were rough as compared to the case if it were 0
smooth.
(B) As cylinder reaches the bottom-most point of the inclined plane it will acquire a
greater speed if it were smooth as compared to the case when it were rough.
(C) The cylinder will have a lower kinetic energy i the case when the inclined plane is
rough as compared to the case when it is smooth.
(D) Inclined plane will have a higher kinetic energy in the case when it is rough as
compared to the case when it is smooth.

Space for Rough work

Multi Correct Choice Type

This section contains 3 multiple choice questions. Each question has four choices (A), (B), (C)
and (D) out of which ONE or MORE THAN ONE is/are correct.

7. Four stationary observers have positions (a, 0), (0, a), (–a, 0) and (0, –a) respectively. A
moving source, moving along +x-axis is crossing origin at t = 0. Which of the followings
are correct at t = 0?
(A) Wavelength is same for all the observers.
(B) Speed of wave is same for all observers.
(C) Frequency is same for all observers.
(D) Frequency observed is more than emitted by source for three observers.

8. A parallel plate capacitor is dipped

in an insulated liquid, vertically so
that half of capacitor is in liquid. A A'
Plates of capacitor are joined to
an ideal battery of emf , A, B, A’,
B’ are four points on plate at B'
facing surface as shown. Which of
B
the following is/are correct?
(A) VA – VB’ = VA – VA’
(B) VB – VB’ = VB – VA’
(C) Electric field around A in air = Electric field around B’ in liquid.
(D) Electric field around B in liquid = Electric field around B’ in liquid.

9. Two identical solenoids each of self inductance

L are joined and placed coaxially and
overlapping in such a way that they carry equal
and opposite currents. This arrangement is
joined to a resistance R and battery of emf .
Switch is closed at t = 0. Identify the correct
statements: R
(A) Energy stored in the magnetic field increases
with time.  S
2L
(B) Time constant of the circuit is .
R
(C) Time constant of the circuit is zero.
(D) Power delivered by battery remains constant.

Space for Rough work

Comprehension type (Only One Option Correct)

This section contains 2 paragraphs. Based upon one of paragraphs 2 multiple choice
questions and based on the other paragraph 3 multiple choice questions have to be
answered. Each of these questions has four choices (A), (B), (C) and (D) out of which ONLY
ONE is correct.

Paragraph for Question Nos. 10 and 11

A particle of mass m is fixed on a massless ring which rests

on a sufficiently rough surface so as to prevent slipping. The
system is released from the shown configuration. Then: R
m 

10. Immediately after releasing from the shown configuration what is the angular acceleration
of the ring?
gsin  gsin 
(A) (B)
 2R
4R sin2  
2
 
gsin  gsin 
(C) (D)
R R(1  cos )

11. If  be the  anticlockwise angular acceleration immediately after release from the shown
position, then what is the normal reaction immediately after release:
(A) N = mg – mRcosθ (B) N = mg
(C) N = m[g – Rsinθ] (D) None of these

Space for Rough work

Paragraph for Question Nos. 12 and 14

A large cylindrical chamber has a massless piston fitted at the top. An acoustic source is situated
at the centre of the bottom of the cylinder, while an observer is situated at the bottom
circumference (inside the chamber). The observer is moving directly towards the source with
constant velocity and hears a frequency f.

12. If the chamber were to be isobarically heated then the observed frequency will be
(A) same as f (B) higher than f
(C) lower than f (D) data insufficient

13. If the chamber were to be isochorically cooled then the observed frequency will be:
(A) same as f (B) higher than f
(C) lower than f (D) Date insufficient

14. If the chamber were to be isothermally heated then, the observed frequency will be:
(A) same as f (B) higher than f
(C) lower than f (D) Data insufficient

Space for Rough work

SECTION - B
Matrix – Match Type
This section contains 2 questions. Each question contains p q r s t
statements given in two columns, which have to be matched. The
A p q r s t
statements in Column I are labelled A, B, C and D, while the
statements in Column II are labelled p, q, r, s and t. Any given p q r s t
B
statement in Column I can have correct matching with ONE OR
MORE statement(s) in Column II. The appropriate bubbles C p q r s t
corresponding to the answers to these questions have to be
darkened as illustrated in the following example: D p q r s t

If the correct matches are A – p, s and t; B – q and r; C – p and q;

and D – s and t; then the correct darkening of bubbles will look like
the following:

1. Initially there are N0 number of radioactive nuclei of species A while zero nuclei of
species B and C. A decays to B with a decay constant 1 and B decays to C with a decay
2. C is not radioactive. Then match the column:
Column I Column II
(A) NB + NC (p) Continuously increases.
(B) NB (q) Continuously decreases.
(C) NA + NC (r) First increases then decreases.
(D) NA + NB (s) First decreases then increases.
(t) Remains constant.

2. A black box having 6 points (A, B, D, F, G, A i H

H) contains a network of capacitors, real R1 R
inductors (with resistance) and resistors. E1 Black Box
There is no battery inside the black box. E1 B L G
and E2 are the EMFs of ideal batteries. If we
choose E1 = E2 = 1 volt, then steady state R2
current i is 5A. If we choose E1 = 2 volt and D F
E2
E2 = 1 volt, the steady state current i = 7A.
Now match the column:
Column I Column II
(A) E1 = 3 volt, E2 = 1 volt (p) i = 10A
(B) E1 = 1 volt, E2 = 2 volt (q) i = 11A
(C) E1 = 1 volt, E2 = 3 volt (r) i = 8A
(D) E1 = 4 volt, E2 = 1 volt (s) i = 9A
(t) i = 20 A

Space for Rough work

SECTION – C
(One Integer Value Correct Type)
This section contains 5 questions. Each question, when worked out will result in one integer
from 0 to 9 (both inclusive).

1. A hydrogen atom in ground state absorbs a photon of energy corresponding to energy

difference of first orbit and fourth orbit. After this, electron de-excite from excited state. As
a result what maximum number of photons may be obtained?

2. A capillary tube is a glass cylinder of radius r = 1 mm. It is dipped vertically in a liquid of

–3
surface tension T, T = 80 × 10 N/m, and angle of contact θ = 60°, liquid has normal
capillary rise in the tube [mass of the tube and its material volume is negligible. Then
force required to hold the tube is P × 10–4 N. Calculate P.

3. Two bodies A and B are joined through a conducting rod of high thermal conductivity with
insulated cover. A and B are thermally insulated from surrounding. Initial temperature of
A and B is 100°C and 20°C respectively. After one hour temperature of A is 80°C and B
is 40°C. After further one hour temperature of B is x × 10°C, calculate x?

4. A ring of radius (R = 10 cm) is fixed in a horizontal (Top view)

plane. A sleeve of mass m = 300 gm can move on
circumference of ring without friction. Sleeve is joined
N C
to a spring of spring constant K  200 and natural
m 3
R
 3  4
length R, other end is fixed at P CP  R  . Sleeve
 4  P
is released from rest. Calculate the net normal force
(in N) exerted by ring on sleeve just after releasing.
2
[g = 10 m/s ]

5. A ball is thrown form ground towards a vertical

wall in such a way that motion plane of ball
makes an angle of 30 with normal to wall.
Distance of point of projection form wall is
4 3 m. Collision between ball and wall is elastic. P
At what distance (in metre) from point of
projection ball falls.
Q
A
O

Space for Rough work

Chemistry PART – II

SECTION – A
Straight Objective Type
This section contains 6 multiple choice questions. Each question has four choices (A), (B), (C)
and (D) out of which ONLY ONE is correct.

1. The major product formed in the following reaction:

H3C CH3
C C  : CH2  sin glet  
H H
(A) (B)
H3C CH3 H3C H

H H H CH3
(C) 50 : 50 mixture of above two compounds (D) H3C CH2 CH3
C C
H H

2. O

MeI excess

KH excess
 81% yield, product of the reaction is

(A) O (B) O

Space for Rough work

3. Which of the following exhibits aromaticity?

(A) Cyclopentadienyl cation (B) Cyclopentadienyl radical
(C) Cycloheptatrienyl radical (D) Cycloheptatrienyl cation

4. H < E for the reaction:

(A) N2(g) + 3H2(g)  3H2(g) (B) Ag2O(s)  2Ag(s) + ½ O2(g)
(C) C (s) + ½ O2 (g)  CO (g) (D) C (s) + O2 (g)  CO2 (g)

5. Which of the following is isoelectronic and isostructural?

NO3 , CO32 , ClO3 , SO3
(A) NO3 , CO32 (B) SO3 , NO3
(C) ClO3 , CO32 (D) CO23 , ClO3

6. A particular 100-octane aviation gasoline used 1.00 cm3 of tetraethyl lead (C2H5)4Pb of
density 1.66g/cc, per litre of product. This compound is made as follows.
4C2H5Cl + 4NaPb  (C2H5)4Pb + 4NaCl + 3Pb. How many gram of ethyl chloride is
needed to make enough tetraethyl lead for 1 L of gasoline.
(A) 1.33g (B) 2.66 g
(C) 9.2g (D) 0.33g

Space for Rough work

Multi Correct Choice Type

This section contains 3 multiple choice questions. Each question has four choices (A), (B), (C)
and (D) out of which ONE or MORE THAN ONE is/are correct.

7. The d-orbitals involved in sp3d2 or d2sp3 hybridisation of the central metal ion are:
(A) dx2 – y2 (B) dxy
(C) dyz (D) dz 2

8. Which of the following statement(s) is/are correct when a soid mixture of NaCl and K2Cr2-
O7 is gently warmed with conc. H2SO4?
(A) A deep red vapour is evolved
(B) The vapour when passed into NaOH solution gives a yellow soluton of Na2CrO4
(C) Chlorine gas is evolved
(D) Chromyl chloride is formed

9. Which of the following compound/s shall undergo condensation by reacting with

NaOEt/EtOH.
(A) O (B) O

O O
(C) O (D) O

O O

Space for Rough work

Comprehension type (Only One Option Correct)

This section contains 2 paragraphs. Based upon one of paragraphs 2 multiple choice
questions and based on the other paragraph 3 multiple choice questions have to be
answered. Each of these questions has four choices (A), (B), (C) and (D) out of which ONLY
ONE is correct.

Paragraph for Question Nos. 10 and 11

Variation of equilibrium constant K with temperature is given by Van’t Hoff equation

Sor Hro
lnK  
R RT
From this equation, Hro can be calculated if equilibrium constant K1 and K2 at two temperature
T1 and T2 are known.
K  Hro  1 1 
log  2     
 K1  2.303R  T1 T2 

10. 

For an isomerization X (g) 
 Y (g), the temperature dependency of equilibrium
constant is given by
1000
lnK  2 
T
the value of Sro at 300 K is
(A) 2R (B) 2/R
(C) 1000 R (D) None of these

11. Select the correct statement

(A) value of Kc always increases with increasing temperature
(B) for exothermic reaction value of Kc increase with decreasing in temperature
(C) for endothermic reaction value of Kc increases with decreasing in temperature
 1
(D) for exothermic reaction slope is  log K vs  negative
 T 

Space for Rough work

Paragraph for Question Nos. 12 to 14

Consider the following reaction

light
 X2   mono halogenated product

12. Which halogen will give the best yield of a single mono halogenation products
(A) F2 (B) Cl2
(C) Br2 (D) I2

13. How many mono halo derivatives are possible (excluding stereo-isomers)?
(A) 3 (B) 4
(C) 5 (D) 6

14. Among the following halogens which one is least reactive towards halogenation reaction?
(A) F2 (B) Cl2
(C) Br2 (D) I2

Space for Rough work

SECTION - B
Matrix – Match Type
This section contains 2 questions. Each question contains p q r s t
statements given in two columns, which have to be matched. The
A p q r s t
statements in Column I are labelled A, B, C and D, while the
statements in Column II are labelled p, q, r, s and t. Any given p q r s t
B
statement in Column I can have correct matching with ONE OR
MORE statement(s) in Column II. The appropriate bubbles C p q r s t
corresponding to the answers to these questions have to be
darkened as illustrated in the following example: D p q r s t

If the correct matches are A – p, s and t; B – q and r; C – p and q;

and D – s and t; then the correct darkening of bubbles will look like
the following:

1. Match the following:

Column – I Column – II

(A) Chalcopyrite → copper (pure) (p) Froth floatation

(B) Zinc blende → zinc (pure) (q) Hall process

(C) Bauxite → aluminium (pure) (r) Roasting

(D) Galena → lead (pure) (s) Self reduction

(t) Basemerisation
2. Match the following:
Column – I Column – II
(A) K P > Qp (p) Non-spontaneous
(B) Go  RT loge Qp (q) Equilibrium
(C) K P = Qp (r) Spontaneous and endothermic
H
(D) T (s) Spontaneous
S
(t) Spontaneous and exothermic

Space for Rough work

SECTION – C
(One Integer Value Correct Type)
This section contains 5 questions. Each question, when worked out will result in one integer
from 0 to 9 (both inclusive).

1. Number of stereo centres in a given molecule?

CH3
H3C
C C
H
H H

2. Number of stereoisomers in given molecules [Ma3b2C].

3. Number of possible products formed in given reaction

H3C
CCl2
C CH C2H5  Br2   Pr oduct
H
–4 –1
4. The conductivity of 0.001 M Na2SO4 solution is 2.6  10 S cm and increases to 7.0 
–4 –1 –4 –1
10 S cm and increases to 7  10 S cm , when the solution is saturated with CaSO4.
The molar conductivity of Na and Ca are 50 and 120 S cm 2 mol–1, respectively.
+ 2+
–x 2
Neglect conductivity of used water the solubility product of CaSO4 is y  10 M . What is
the value of x.

234
5. 90 Th disintegrate to give 82 Pb206 as the final product. Total number of  particles
emitted out during this process are:

Space for Rough work

Mathematics PART – III

SECTION – A
Straight Objective Type
This section contains 6 multiple choice questions. Each question has four choices (A), (B), (C)
and (D) out of which ONLY ONE is correct.

1. The locus of a point which moves so that the distance between the feet of perpendiculars
drawn from it to the lines 3x2 – 6xy + y2 = 0 is 3 3 is a circle with radius
(A) 2 15 (B) 15
(C) 6 5 (D) none of these

n  n  1
2. The sum to (n + 1) terms of the series cos   ncos 2  cos3  ..... is
2!
   
(A) 2n sinn cos  2  n  (B) 2n cosn cos  2  n 
2 2 2 2

(C) 2n cos  2  n  (D) none of these
2
2
3. A tangent is drawn to the curve x + 2x – 4ky + 3 = 0 at a point whose abscissa is 3. The
tangent is perpendicular to the line x – 2y + 3 = 0. Then the area bounded by the curve,
this tangent, the x-axis and the ordinate x = –1, is
108 109
(A) sq. units (B) sq. units
47 48
111 104
(C) sq. units (D) sq. units
47 40

Space for rough work

x2 y2
4. Given an ellipse   1 (a > b). A circle is drawn which passes through the foci and
a2 b2
the extremities of the minor axis of this ellipse and let this circle be the auxiliary circle for
a second ellipse whose vertices are the foci of the given ellipse. If any tangent to the
second ellipse meets its auxiliary circle at two points such that it subtends a right angle at
the origin, then the locus of the point of tangency, is (if eccentricity of both the ellipse are
same)
(A) x 2 + 2y2 = a2 (B) 2x2 + y2 = a2
2 2 2
(C) 2x + y = a (D) x 2 + y2 = 2a2

1/ 2 1/ 3 1/ 4 1/n
2x  3x  4x  .....  n  x 
5. The value of lim is
x 
 2x  3 1/2   2x  3 1/3  .....   2x  3 1/n
(A) 0 (B) 2
1
(C) 2 (D)
3

4 2

 e  x  x  2x  e dx
x 3 5 x
6. is equal to
1 x2 x4 1 2 x4
(A) xe e  c (B) x e c
2 2
1 2 4 1 2 4
(C) e x e x  c (D) x 2 e x e x  c
2 2

Multi Correct Choice Type

This section contains 3 multiple choice questions. Each question has four choices (A), (B), (C)
and (D) out of which ONE or MORE THAN ONE is/are correct.

1  x a 1  x b 1
a b
7. If f x  1 1  x  1  x  , then
b a
1  x  1 1  x 
(A) constant term of f(x) = a + b (B) constant term of f(x) = 0
(C) coefficient of x in f(x) is zero (D) none of these

Space for rough work

1
x cos a  1
8. If f  a  
 ln x dx , ‘a’ being a real number other than an odd multiple of , then
0
(A) f(a) = ln(cos a) (B) f(a) = ln(1 + cos a)
a a
(C) f(a) = tan (D) f(a) =  tan
2 2
3 2
9. If f(x) = x – x + 100x + 2002, then
 1   1 
(A) f(1000) > f(1001) (B) f    f  2001 
 2000   
(C) f(x – 1) > f(x – 2) (D) f(2x – 3) > f(x)

Comprehension type (Only One Option Correct)

This section contains 2 paragraphs. Based upon one of paragraphs 2 multiple choice
questions and based on the other paragraph 3 multiple choice questions have to be
answered. Each of these questions has four choices (A), (B), (C) and (D) out of which ONLY
ONE is correct.

Paragraph for Question Nos. 10 and 11

Read the following write up carefully and answer the following questions:
Consider a triangle ABC, where x, y, z are the length of perpendicular drawn from the vertices of
the triangle to the opposite sides a, b, c respectively. Let the letters R, r, s,  denote the
circumradius, in-radius, semi-perimeter and area of the triangle respectively

bx cy az a2  b2  c 2
10. If    , then the value of k is
c a b k
(A) R (B) s
3
(C) 2R (D) R
2

 1 1 1
11. If cot A + cot B + cot C = k  2  2  2  , then the value of k is
x y z 
2
(A) R (B) rR
2 2 2
(C)  (D) a + b + c

Space for rough work

Paragraph for Question Nos. 12 and 14

Read the following write up carefully and answer the following questions:
If the locus of the circumcentre of variable triangle having sides x-axis, x = 3 and px + qy = 4,
2
where (p, q) lies on the parabola x = 4ay is a curve C, then

12. Co-ordinates of the vertex of this curve C is

 9 81   9 81 
(A)  ,  a  (B)   ,  a 
4 8   4 8 
9 9   9 81 
(C)  ,  a  (D)   , a 
4 8   4 8 

13. The curve C is symmetric about the line

9 9
(A) x   (B) x 
4 4
9 9
(C) y   (D) y 
4 4

14. The length of smallest focal chord of this curve C is

1 1
(A) (B)
2a 4a
1 1
(C) (D)
12a 16a

Space for rough work

SECTION - B
Matrix – Match Type
This section contains 2 questions. Each question contains p q r s t
statements given in two columns, which have to be matched. The
A p q r s t
statements in Column I are labelled A, B, C and D, while the
statements in Column II are labelled p, q, r, s and t. Any given p q r s t
B
statement in Column I can have correct matching with ONE OR
MORE statement(s) in Column II. The appropriate bubbles C p q r s t
corresponding to the answers to these questions have to be
darkened as illustrated in the following example: D p q r s t

If the correct matches are A – p, s and t; B – q and r; C – p and q;

and D – s and t; then the correct darkening of bubbles will look like
the following:

1. Match the following Column-I with Column-II

Column – I Column – II
(A) If A is an idempotent matrix and I is an identity matrix of the
(p) 9
same order, then the value of n, such that (A + I)n = I + 127A is
–1 2 7 n
(B) If (I – A) = I + A + A + ….. + A , then A = O where n is (q) 10
(C) If A is a matrix such that a ij = (i + j) (i – j) then A is singular if
(r) 7
order of matrix is
–1
(D) If a non-singular matrix A is symmetric and A is also
(s) 8
symmetric, then order of A can be
(t) 6

2. ‘n’ whole numbers are randomly chosen and multiplied then match the following Column-I
with Column-II
ColumnI ColumnII
8n  4n
(A) The probability that the last digit is 1, 3, 7 or 9 is (p)
10n
5  4n
n
(B) The probability that the last digit is 2, 4, 6 or 8 is (q)
10n
4n
(C) The probability that the last digit is 5 is (r)
10n
10n  8n  5n  4n
(D) The probability that the last digit is zero is (s)
10n
n n
8 4
(t)
10n

Space for rough work

SECTION – C
(One Integer Value Correct Type)
This section contains 5 questions. Each question, when worked out will result in one integer
from 0 to 9 (both inclusive).

1. Let the circle (x – 1)2 + (y – 2)2 = 25 cuts a rectangular hyperbola with transverse axis
along y = x at four point A, B, C and D having coordinates (xi, yi), (where i = 1, 2, 3, 4)
respectively, O being the centre of the hyperbola. If  = x1 + x2 + x3 + x 4, m =
x12  x 22  x 23  x 24 , n = y12  y 22  y 23  y 24 , then find value of n – m – 4  _____

2. If x  [0, 2] for which 2cos x  1  sin 2x  1  sin2x  2 has solution set
   
x   ,  , then  –  is _____
4 4

3. Given (1 – 2x + 5x2 + 10x 3)(1 + x)n = 1 + a1x + a2x 2 + ….. and that a12  2a2 , then the
value of n is _____

2 n f  x 2 n
4. If f(x) = a0 + a1x + a2x + ….. + anx + ….. and = b0 + b1x + b2x + ….. + bnx + ….. If
1 x
11
a0 = 1 and b1 = 3 and b10 = k – 1, then k is _____ (Given that a0, a1, a2, ….. are in G.P.)

 cos 2 sec 2  cos 2 sec 2 

5. For tan1 
 
1 2
 2 1
  tan tan      tan       tan 1

 the

value of  is _____

Space for rough work

ANSWERS, HINTS & SOLUTIONS

FULL TEST– IV
(Paper-1)

Q. No. PHYSICS CHEMISTRY MATHEMATICS

ALL INDIA TEST SERIES

1. B A C
2. C C B
3. A D B
4. D A A
5. C A C
6. B A D
7. B, D A, D B, C

8. A, B, C, D A, B, D B, D
9. C, D A, B, C B, C
10. A A C
11. C B C
12. C C C
13. B B B
14. A A A
A→p A → p, r, s, t (A)  (r)
B→r B → p, r (B)  (s)
1.
C→s C→q (C)  (p, r)
D→q D → p, r, s (D)  (p, q, r, s, t)
A→s A→s (A)  (r)
B→r B→p (B)  (t)
2.
C→q C→q (C)  (q)
D→q D→q (D)  (s)

1. 3 3 4

2. 5 3 6
3. 5 4 6
4. 5 6 2
5. 8 7 2

Physics PART – I

SECTION – A
1. If the pulley comes down by a displacement x then the block comes down by 3x and the string
connecting the block has tension Kx.
3d2 x
 Kx  M  2
dt
K
or   .
3M

1 1 1
2.  
v u Feff
1 1 1
   or Feff = +10 cm
20 / 3 20 Feff
1 1 2
Now  
Feff fm f0
1 1 2
or  
10 fm 30
1 1 2 1
or   
fm 10 30 30
or f m = 30 cm or R = +60 cm.

3. Use KCL/KVL or suppose the current due to each battery. (Forms wheatstones by
removing 1 battery at a time).

4. There is a tetrahedral configuration with a total of 6 interactions.

ˆi  ˆj  kˆ
5. Say the beam is incident along Î  ˆ  ˆi then its
and the normal of one of the faces is N
3
2
ˆ ˆ  a . There are three such faces on which the radiation falls.
projection area is AN.I
3
 Total projection area = 3a2
I
 Force = . 3a2 .
c

6. Mechanical energy and momentum both have to be conserved. In case of rough incline, the
sphere will also have some rotational KE. Thus lesser energy available as translational KE of
incline as well as the cylinder.

7. Speed is characteristic of the medium is same for all observers. Here wavelength is different for
different directions and thus frequency.
For observer (a, 0), (0, a) and (0, –a) observed frequency is more than emitted taking into
account time lag.

8. Potential difference between plates is same as .

VA – VB =  = E.d in both air and liquid.
VA = VB and VA’ = VB’
Complete plate is at same potential.

9. Since coils carry current in opposite sense magnetic field is always zero in solenoid even if
current is there. Thus net inductance = 0, it is simply  and R joined in series.

10. Instantaneous axis has zero.

instantaneous acceleration  = 0
Taking  about bottom most point:
M R
Mg Rsin θ = m(2Rsinθ/2)2 . 
Solving we get, Mg
gsin 


4R sin2  
 2

11. mg – N = mRsinθ
N= m(g - Rsinθ)

12-14. Use Doppler’s effect, the speed of sound increases with increase in temperature.

SECTION - B
1. NA + NB + NC = N0.
B will first have more production rate than decay and finally there will be no B.

2. In steady state all capacitors G have as open circuits and inductors as short circuits.
 purely resistive network:
Assume, i = aE1 + bE2
We get, a = 2, b = 3.
 i = 2E1 + 3E2.
SECTION – C
1. There is only a single atom, thus maximum number of photons obtained are
4 → 3, 3 → 2 and 2 → 1.

2. Surface tension force is acting downward at both inner and outer surface at contact point on the
tube.
F = 2rTcosθ + 2rTcosθ = 4rTcosθ
–t
3. (TA – TB) = T0e
Same fractional change takes same time, where T0 = 80°C is initial temperature difference. Since
in fixed one hour temperature loss of A = temperature gain of B, means heat capacities are same,
means they will always undergo same loss and same gain. After one hour temperature difference
1
become half = 40°C = (80) . In further one hour temperature difference becomes further half.
2
TA + TB = 120°C
2
 1
TA – TB =    T0  20 C
 2
 TB = 50°C  x = 5

4. Vertical component of normal reaction balances NV

mg and horizontal component of normal NH N
reaction balances component of spring force
towards centre.
Nv = mg
kR 4 k 
NH = Fs cosθ =   R mg Fs C
4 5 5
N  N2v  NH2  5 N. P

5. Ball collide at highest point of projectile Motion, Q

after collision, motion is mirror image of motion Q'
8m
that would have been without collision.
Top view of motion is as shown in diagram OQ
is required distance. 8m
30 B
OQ = 2 (OB) sin 30 = 8 m
8m

O 4 3m

Chemistry PART – II

SECTION – A
1. Singlet carbene gives syn addition on alkene. syn addition on cis alkene gives meso compound.

4. H = E + nRT,  H <  E if  n < 0

5. O
O
N C
O O O O
sp 2 sp 2

6. The mass of 1 cc of (C2H5)4Pb is = 1  1.66 = 1.66g and this is the amount needed per litre.
1.66
No. of moles of (C2H5)4Pb needed = = 0.00514 ml
323
1 mole of (C2H5)4Pb requires 4  (0.00514) = 0.0206 ml of C2H5Cl
 Mass of C2H5 Cl = 0.0206  64.5 = 1.33 g

8. When a solid mixture of NaCl, K2Cr2O7 and conc. H2SO4 is heated, the products obtained are
4KCl  K 2Cr2O7  6H2SO 4  2CrO2Cl2  6KHSO4  3H2O
Red vapour

CrO2Cl2  4NaOH  Na2CrO 4  2NaCl  2H2O

Yellow solution

9. Claisen condensation can be shown only when at least 2  H are there.

SECTION-B
2. (A) KP > Q the reaction will proceed in forward direction spontaneously.
(B) Go  RT loge Q then G   ve then non spontaneous.
(C) K P  Q  equilibrium
(D) G  H  T.S
SECTION – C
3. H C2 H5 Br H
C C CCl4
 Br2  H C C C 2H 5
x x
H3C H Br
H3C
Optically active
2 products
H3C C2 H5 Br C2H5
C C CCl4
 Br2  H C C H
H H Br
H3C
Optically active
2 products
total = 2 + 2 = 44. Conductivity of Na2SO4 = 2.6  10–4
1000  2.6  104
m Na2SO4    260 S cm2
0.001
 
m SO24  m Na2SO4   2 m Na  
2 –1
= 260 – 2  50 = 160 S cm mol

Coductivity of CaSO4 solution

= 7  10–4 – 2.6  10–4 = 4.4  10–4 S cm–1

m  CaSO 4    m Ca2   m SO24   
2 –1
= 120 + 160 = 280 S cm mol
1000  K 1000  4.4  104
Solubility, S = 
m 280
–3
= 1.57  10 cm
K sp  Ca2  SO42    0.00157  0.00157  0.001
total
–6 2
= 4  10 M
–x 2
= y  10 M

Mathematics PART – III

SECTION – A

3
1. Here, tan   A
2 P(h, k)
Thus, locus of P is
2 2 
x + y = 180

Hence, radius = 6 5 O
 B
0

i i n
2. Sum = Re(e (1 + e ) )

dy 2
3. For the given curve,   2  k = –1
dx x 3 k
 The curve is, x2 + 2x + 4y + 3 = 0
and the tangent is 4x + 2y – 3 = 0

x2 y 2
4. Equation of first ellipse is  1
a2 b2
Vertices of second ellipse are (ae, 0)
a2
 Equation of second ellipse is x 2  2y 2 
2
a2
 Equation of tangent to it at (h, k) is xh  2yk 
2
2
2 
Homogenising x2 + y2 – b2 = 0 with this straight line x 2  y 2  b2  2 xh  2yk   0
a 
 
4
a
These are perpendicular, hence h2  2x 2  2  a2
2b
 Locus of (h, k) is x 2 + 2y2 = a2

1/2 1/3 1/4 1/n

2x   3x  4 x   .....  n  x 
5. Given, lim 1/2 1/3 1/n
x 
 2x  3    2x  3   .....   2x  3 
1/2 1/3 1/n
2 h   3 h   .....  n h  1
= lim 1/2 1/3 1/n
[Put x  ]
h 0 1/2
h  2  3h   h1/3  2  3h   .....  h1/n  2  3h  h
 1 1  1 1  1 1
 3   4   n 
2  3h 2 
 4h 2 
 .....  nh 2 
= lim  1 1  1 1
h 0
1/2  3  1/3  n  1/n
 2  3h   h 2 
 2  3h   .....  h 2 
 2  3h 
2  0  0  .....  0
=  2
21/2  0  0  .....

6. Substitute x2 = t
1 1
2
2
 
2 2

 2
 I   e t 1  t  2t 2 e t dt   et  t e t  et  2t 2 et  dt
2 
  
1 1 1 2
2 2
 2

2
4
=  et  f  t   f '  t   dt  e t t et  c = e x x 2 ex  c  
2 3
7. Let, f(x) = A + Bx + Cx + Dx + …..
Now, f(0) = 0 and f(0) = 0

1
x cos a ln x   sina  1
a
8. Here, f '  a    dx =   xcos a sina dx =  tan  
0
ln x 0 2
sina
 f a    da  ln 1  cosa 
1  cos a
2
9. Here, f(x) = 3x – 2x + 100 > 0
 f(x) is increasing

bx cy az b2  c 2  a2
10. Here,    b sinB  c sinC  a sin A 
c a b 2R
 k = 2R

R
11. cot A + cot B + cot C =
abc

b2  c 2  a2  c 2  a2  b2  a2  b2  c 2 
R  1 1 1
=
abc

a2  b2  c 2    2  2  2 
x y z 
k=

12.-14. Let (h, k) be the circumcircle of ABC

4  3p 4  3p
Then, h  ; k
2p 2q
4 4h  12
 p ; q
2h  3 k  2h  3 
This (p, q) lies on x2 = 4ay
2
1  9   9
 Locus of (h, k) is  y  a  h  
2a  8   4

SECTION – B
1. (A) A is idempotent, A2 = A3 = A4 = ….. = A
n
(A + I) = I  n C1A  nC2 A 2  .....  nCn A n  I  nC1A  n C2 A  .....  nCn A
= I   n C1  nC 2 A 2  .....  n Cn  A  I   2n  1 A
 2n  1  127  n = 7
2 7 2 7 2 8
(B) (I – A) (I + A + A + ….. + A ) = (I + A + A + ….. + A ) – (A + A + ….. A )
8 8
= I – A = I (if A = 0)
(C) A is skew symmetric  A  A T  ( 1)n A
 A I  (1)n   0 as n is odd, hence |A| = O  A is singular
(D) A is symmetric then A–1 is also symmetric for matrix of any order

2. (A) Required event will occur if last digit in all the numbers is 1, 3, 7 or 9. Thus, required
4n
probability =
10n
8n  4n
(B) Required probability = P (last digit is 1, 2, 3, 4, 6, 7, 8, 9)  P (last digit is 1, 3, 7, 9) =
10n
5n  4n
(C) P (1, 3, 5, 7, 9)  P (1, 3, 7, 9) =
10n

(D) P (required) = P (0, 5)  P (5) =


10n  8n  5n  4n

  
10n  8n  5n  4n
10n 10n

SECTION – C

2  c
1. Let hyperbola be xy = c . Then points of intersection is  ct, 
 t
 c2t4 – 2ct3 – 20t2 – 4ct + c2 = 0
If t1, t2, t3 and t4 are its roots
2 20 4
Then,  t1  ;  t1t 2   2 ;  t1t 2 t 3  and t1t2t3t4 = 1
c c c
  = 2, m = 44, n = 56
2
2. Let, y = 1  sin 2x  1  sin 2x  y = 2 – 2|cos 2x|
   3 5   7 
If x  0,  or  ,  or  , 2  , cos 2x is non-negative
 4 4 4  4 
   7 
 y = 2|sin x|  cos x  |sin x| except for x in 0,  and  , 2 
 4  4 
 3 5  1
So, it holds for  ,  in which sin x 
 4 4  2
  3   5 7 
If x   ,  or  4 , 4  then cos 2x < 0  y = 2|cos x|
4 4   
So, both inequalities hold
  7 
Thus, solution set is x   , 
4 4 
–=6

3. Here, (1 – 2x + 5x2 + 10x3)  n

 2
C0  n C1x  n C2 x2  ..... = 1 + a1x + a2x + …..
n  n  1
 a1 = n – 2; a2   2n  5
2
Since, a12  2a2  n = 6
–1 2 2
4. f(x)(1 – x) = f(x)(1 + x + x + …..) = b0 + b1x + b2x + …..
 (a0 + a1x + a2x2 + ….. + anxn + …..)(1 + x + x 2 + ….. + xn + …..) = b0 + b1x + b2x2 + …..
n
Comparing coefficient of x on both sides, a0 + a1 + a2 + ….. + an – 1 + an = bn
Comparing coefficient of xn – 1 on both sides, a0 + a1 + a2 + ….. + an – 1 = bn – 1
 bn – bn – 1 = an
Also, b0 = a0 and b1 = a0 + a1
 b0 = 1; a1 = 2

 b10 = a0 + a1 + a2 + ….. + a10 = 211 – 1

 tan2      tan2       1
5. Here, R.H.S. = tan1  2 2 
1  tan      tan      
 sin2      sin2       cos2      cos 2      
= tan1  2 2 2 2 
 cos      cos       sin      sin      
  cos 2  cos2 2   cos2  cos2 2 
1 
= tan  
 2cos 2 cos2 
 cos2 sec 2  cos 2 sec 2 
= tan1  
 2 
=2

Paper 2

Time Allotted: 3 Hours Maximum Marks: 243

 Pl ea s e r ea d t h e i n s t r u c t i o n s c a r ef u ll y . Yo u a r e a l l o t t ed 5 m i n u t es
s p ec i f i c a ll y f o r t h i s p u r p o s e.
ALL INDIA TEST SERIES

 Yo u a r e n o t a l l o wed t o l ea v e t h e E xa m i n at i o n Ha l l b ef o r e t h e en d o f
t h e t es t .

INSTRUCTIONS

A. General Instructions
1. Attempt ALL the questions. Answers have to be marked on the OMR sheets.
2. This question paper contains Three Parts.
3. Part-I is Physics, Part-II is Chemistry and Part-III is Mathematics.
4. Each part is further divided into three sections: Section-A, Section-B & Section-C
5. Rough spaces are provided for rough work inside the question paper. No additional sheets will be
provided for rough work.
6. Blank Papers, clip boards, log tables, slide rule, calculator, cellular phones, pagers and electronic
devices, in any form, are not allowed.
B. Filling of OMR Sheet
1. Ensure matching of OMR sheet with the Question paper before you start marking your answers
on OMR sheet.
2. On the OMR sheet, darken the appropriate bubble with black pen for each character of your
Enrolment No. and write your Name, Test Centre and other details at the designated places.
3. OMR sheet contains alphabets, numerals & special characters for marking answers.
C. Marking Scheme For All Three Parts.
1. Section – A (01 – 06) contains 6 multiple choice questions which have only one correct answer.
Each question carries +3 marks for correct answer and – 1 mark for wrong answer.
Section – A (07 – 09) contains 3 multiple choice questions which have one or more than one
correct answer. Each question carries +4 marks for correct answer. There is no negative
marking.
Section-A (10 – 14) contains 2 paragraphs. Based upon paragraph, 2 and 3 multiple choice
questions have to be answered. Each question has only one correct answer and carries
+3 marks for correct answer and – 1 mark for wrong answer.
2. Section – B (01 – 02) contains 2 Matrix Match Type questions containing statements given in 2
columns. Statements in the first column have to be matched with statements in the second
column. Each question carries +8 marks for all correct answer. For each correct row +2 marks
will be awarded. There may be one or more than one correct matching. No marks will be given for
any wrong matching in any question. There is no negative marking.
3. Section-C (01 – 05) contains 5 Numerical based questions with answers as numerical value from
0 to 9 and each question carries +4 marks for correct answer. There is no negative marking.

Name of the Candidate

Enrolment No.

Useful Data

PHYSICS
Acceleration due to gravity g = 10 m/s2

Planck constant h = 6.6 1034 J-s

Charge of electron e = 1.6  1019 C

Mass of electron me = 9.1  1031 kg

Permittivity of free space 0 = 8.85  1012 C2/N-m2

Density of water water = 103 kg/m3

Atmospheric pressure Pa = 105 N/m2

Gas constant R = 8.314 J K1 mol1

CHEMISTRY

Gas Constant R = 8.314 J K1 mol1

Atomic No: H=1, He = 2, Li=3, Be=4, B=5, C=6, N=7, O=8,

Physics PART – I

SECTION – A
Straight Objective Type
This section contains 6 multiple choice questions. Each question has four choices (A), (B), (C)
and (D) out of which ONLY ONE is correct.

1. A capacitor of capacitance C is connected between every two vertices of a cube

(including the body diagonals and the face diagonals). The effective capacitance of the
system across any edge is:
(A) C (B) 3C
(C) 2C (D) 4C

2. In the shown circuit, the switch is closed at t = 0. What 3mH 2

should be the value of L if the galvanometer shows
zero deflection?
(A) 3 mH G
(B) 4 mH
(C) 6 mH L 4
(D) 1.5 mH

100V K

3. Two balls moving in a gravity free space collide with each other inelastically then choose
the correct statement: (0 < e < 1)
(A) An any inertial frame the total pre collision kinetic energy is same.
(B) An any inertial frame the total post collision kinetic energy is same.
(C) The total velocity of separation of the balls will be different inertial frames.
(D) The loss in kinetic energy during collision is same in every inertial frame.


4. If I is the intensity at the central maximum of YDSE then what is the intensity at y  ?
4
( is fringe width)
I I
(A) (B)
4 2
I
(C) 0 (D)
3

Space for Rough work

5. An uncharged conducting sphere of radius R, has a

R
hollow cavity sphere of radius , as shown. Which of R
2 R
the following is/are correct about electrostatic energy? 2
Q

(A) Interaction energy of induced charges with charge placed at centre of cavity is
 1   Q2 
  
 4 0   R 
(B) Interaction energy of induced charges with charge placed at centre of cavity is
1  Q2 
 
40  R 
1  Q2 
(C) Electrostatic energy outside sphere is  
40  9R 
 1   Q2 
(D) Electrostatic energy outside sphere is   .
 40   R 
6. A self luminous point object is placed at focus of a L
thin lens. Which of the following is INCORRECT
about shape of wave fronts?
(A) Before lens it is spherical if lens is converging.
(B) Before lens it is spherical if lens is diverging.
(C) After lens it is spherical if lens is converging.
(D) After lens it is spherical if lens is diverging. f

Space for Rough work

Multi Correct Choice Type

This section contains 3 multiple choice questions. Each question has four choices (A), (B), (C)
and (D) out of which ONE or MORE THAN ONE is/are correct.

7. Two vertical strings of different lengths are joined

to a uniform rod at points A and B, rod is at rest.
Now the string joining B is burnt. Which of the
following are correct just after this?
(A) acceleration of COM of rod is vertical. A
(B) acceleration of COM of rod is along rod.
(C) acceleration of particle at A is zero. B
(D) acceleration of particle at A is horizontal.

8. A charged particle is projected in combined electric and magnetic field. Electric

field is uniform along x-axis and magnetic field which is non-uniform but not time
varying also along x-axis. Initial velocity of particle is 10i  10ˆj m/s. After certain
 
time x component of velocity is 20 m/s. Which of the following is correct at that
time?

(A) Velocity of particle is 20iˆ  10ˆj m/s. 
(B) Speed of particle is 10 5 m/s.
(C) With given information we cannot calculate y component and z component of
velocities.
(D) Vy2  Vz2  100 m2/s2.

9. In a photo electric effect frequency of incident radiation is halved. As a result,

what is possible?
(A) Kinetic energy of most energetic electrons, becomes less than half, if
electrons are emitted.
(B) Kinetic energy of most energetic electrons becomes more than half, if
electrons are emitted.
(C) Kinetic energy of most energetic electrons becomes half, if electrons are
emitted.
(D) May be no photo electrons are obtained.

Space for Rough work

Comprehension type (Only One Option Correct)

This section contains 2 paragraphs. Based upon one of paragraphs 2 multiple choice
questions and based on the other paragraph 3 multiple choice questions have to be
answered. Each of these questions has four choices (A), (B), (C) and (D) out of which ONLY
ONE is correct.

Paragraph for Question Nos. 10 and 11

A boatman starts from point A and always

heads boat towards point B (i.e. Vbr is B
always towards B).
Vbr is the velocity of the boat in still water
and W is the width of the river. It is given W Vr River
that Vbr > Vr.
A

10. What is the time taken by the boat to cross the river?
WVr W
(A) (B)
Vbr2 Vbr  Vr
WV W
(C) 2 r 2 (D)
Vbr  v r Vbr2  Vr2

11. If the boat is observed from the ground frame as it tries to reach point B:
(A) Its speed appears to increase with time
(B) Its speed appears to be constant with time
(C) Its speed appears to decrease with time
(D) None of the above.

Space for Rough work

Paragraph for Question Nos. 12 and 14

A diathermic piston free to move without

friction divides an adiabatic cylinder in two A B F
equal parts. The two parts contain equal
amount of ideal gas. Piston is joined to a string
which comes out passing through right boundary. [Assume that no matter and energy
flows through hole that passes the string]. Now the string is pulled slowly to right.

12. Temperature of gas A

(A) increases (B) Decrease
(C) first decrease then increase (D) first increase then decrease

13. Temperature of gas B

(A) increases (B) Decrease
(C) first decrease then increase (D) first increase then decrease

14. Work done by gas A is

(A) negative (B) positive
(C) zero (D) none of the above

Space for Rough work

SECTION - B
Matrix – Match Type

This section contains 2 questions. Each question contains p q r s t

statements given in two columns, which have to be matched. The
A p q r s t
statements in Column I are labelled A, B, C and D, while the
statements in Column II are labelled p, q, r, s and t. Any given B p q r s t
statement in Column I can have correct matching with ONE OR
MORE statement(s) in Column II. The appropriate bubbles C p q r s t
corresponding to the answers to these questions have to be
darkened as illustrated in the following example: D p q r s t

If the correct matches are A – p, s and t; B – q and r; C – p and q;

and D – s and t; then the correct darkening of bubbles will look like
the following:
3
1. A cube of volume 1 m and specific gravity 2 is T
very slowly being lowered in to a lake with help of A
a massless string tied to one of its vertices as
shown.
Initially the vertex B is just touching the surface of
water and finally, vertex A is just beneath the
surface of water. For the process of immersion,
match the columns: (Ignore change in KE of
–2 B
water). (g = 10 ms ).

Column I Column II
(A) Work done by gravity on water + cube (p) 3 3
system.  104 Joule.
2
(B) Work done by tension on the cube. (q) 3 3
 10 4 Joule.
2
(C) Work done by gravity on water. (r)  3
 104 Joule.
2
(D) Work done by gravity on the cube. (s) 2 3  104 Joule.
(t) 40 3  104 Joule.

2. Match the columns according to the particle released in the reaction:

Column I Column II
(A) 14 14 
C  N  e  ____ (p) Neutrino
6 7
(B) 23 23
Mg 11 Na  e  ____ (q) Anti Neutrino
12
(C) 81
Kr  e 81 (r) Deutron
36 35 Br  ____
(D) 137
Cs 137  (s) -particle.
55 56 Ba  e  ____
(t) -particle

Space for Rough work

SECTION – C
(One Integer Value Correct Type)
This section contains 5 questions. Each question, when worked out will result in one integer
from 0 to 9 (both inclusive).

1. In a different type of Vernier calliper without zero error, five vernier division is equal to
four main scale division. One main scale division is equal to 2 mm
Readings for a rod are.
Main scale reading = 8
Vernier scale reading = 3
If length of the rod is = [14 + x × 0.4] mm, what is the value of x?

2. A uniform rod of mass m and length 3r is free

to rotate about a horizontal axis through O. A m
disc of mass m, radius r, is free to rotate r
about another horizontal axis fixed to other O
end of the rod (say C). [r = 1.2 m]. The m
A
system is released from rest when rod is c
horizontal. What is the velocity of particle at 3r
A (in m/s) when rod becomes vertical?

Space for Rough work

3. There is infinite ladder of network as 1 x 2 x 3 x 4

shown. 1 and 1’ are joined to ideal
battery of emf 16 V. The potential
difference across 4 and 4’ is found to 2 2 2 2
16v
be 2V. What is the equivalent
resistance between 1 and 1’ (Voltage
across the successive 2 resistance 1' 2' 3' 4'
decreases by a factor of 2 as we
move towards right)

4. A convex lens L1 and a concave lens L2 are arranged with separation 10 cm as shown.
Their focal lengths are respectively 20 cm and 10 cm. optical centre of L1 is treated as
origin and optical axis as x-axis as shown. A self luminous point object is placed at [–20
cm, (6 mm)] Y coordinate of the final image in mm is?

6mm

20cm
x
O

L1 L2
10cm

5. Position vector of a particle is given as


r  A cos2 tiˆ  B sin 2 tjˆ , [A = 6 and B = 8]
Find the amplitude of motion.

Space for Rough work

Chemistry PART – II

SECTION – A
Straight Objective Type
This section contains 6 multiple choice questions. Each question has four choices (A), (B), (C)
and (D) out of which ONLY ONE is correct.

1. Which will show greater degree of retention on hydrolysis?

Cl Cl
(A) Et S (B) Et O
CH3 CH3
H3C CH3 CH3
(C) (D) H3C Cl
Cl CH3

2. Which of the following react with HBr at faster rate?

(A) (B) OH
OH

CH3

H3C

3. The value of universal gas constant R depends on

(A) Temperature of gas (B) Volume of gas
(C) Number of moles of gas (D) Units volume and pressure

Space for Rough work

4. When iodine is dissolved in aqueous potassium iodide, the shape of the species formed
is:
(A) Linear (B) Angular
(C) Triangular (D) See-saw

5. Among the following complexes

(i) [Ru(bipyridyl)3)+
(ii) [Cr(EDTA)]+
3–
(iii) trans-[CrCl2(Oxalate)2]
3–
(iv) cis-[CrCl2(ox)2]
The ones that shows chirality are
(A) (i),(ii), (iv) (B) (i), (ii), (iii)
(C) (ii), (iii), (iv) (D) (i), (iii), (iv)

6. The difference nth and (n + 1)th Bohr’s radius of H-atoms is equal to (n-1)th Bohr’s radius.
Hence the value of ‘n’ is:
(A) 1 (B) 2
(C) 3 (D) 4

Space for Rough work

Multi Correct Choice Type

This section contains 3 multiple choice questions. Each question has four choices (A), (B), (C)
and (D) out of which ONE or MORE THAN ONE is/are correct.

7. A binary solution of liquid A and B will show positive deviation from Roult’s law if it fulfils
the following condition:
(A) PA  X APAo and PB  XBPBo
(B) The intermolecular forces of A–B < A–A, B–B
(C) H mixing is positive
(D)  V mixing is negative

8. 

For the reaction: Cl2  g  3F2  g 
 2ClF3  g  ; H  329KJ dissociation of ClF3(g)
will be favoured by:
(A) Increasing the temperature
(B) Increasing the volume of the container
(C) Adding of F2 gas
(D) Adding of inert gas at constant pressure

9. Select the correct statement(s):

(A) When T   or Ea  o then K = A
(B) A positive catalyst can change H of the reaction
(C) A mixture of reactants may be thermodynamically unstable but kinetically stable
(D) A negative catalyst increase the activation energy of the reaction

Space for Rough work

Comprehension type (Only One Option Correct)

This section contains 2 paragraphs. Based upon one of paragraphs 2 multiple choice
questions and based on the other paragraph 3 multiple choice questions have to be
answered. Each of these questions has four choices (A), (B), (C) and (D) out of which ONLY
ONE is correct.

Paragraph for Question Nos. 10 and 11

Read the following paragraph and answer the questions given below:
A proper control by pH is very essential for many industrial as well as biological processes.
Solutions with a definite pH can be prepared from single salts or mixtures of acids/bases and their
salts. We also require solutions which resist change in pH and hence have a reserve value. Such
solutions are called buffer solutions. Henderson gave a theoretical equal for preparing acidic
buffers of definite pH. The equation is:

pH  pK a  log
Salt
 Acid
A similar equation is used for basic buffers. The pH of aqueous solution of single salts is
calculated by using an expression whose exact form depends on the nature of the salt.
Considering all that in mind, answer the following questions:

10. Calculate pH of a mixture containing 100 ml of 0.10 M NaOH mixed with 100 ml of 0.05 M
CH3COOH solution
(A) 10.4 (B) 11.7
(C) 12.4 (D) 13.4

11. Calculate pH if 100 ml of 0.1 M NaOH is mixed with 100 ml of 0.1 M CH3COOH solution
[Ka (CH3COOH) = 1.8 × 10– 5) log 1.8 = 0.255]
(A) 8.71 (B) 7
(C) 6.25 (D) 9.37

Space for Rough work

Paragraph for Question Nos. 12 and 14

Molar conductivity (m) is defined as conducting power of the ions produced by 1 mole of an
electrolyte in a solution. m = k/c where  is conductivity (in S cm2 mol–1) and c is molar
3 2 –1
concentration (in mole/cm ) the molar conductivity of 0.04 M solution of CaCl2 is 200 S cm mol
2
at 298 K. A cell with electrode that are 2.0 cm in surface area and 0.50 cm apart is filled with
CaCl2 solution.

12. Conductance of CaCl2 solution is:

(A) 8  10–3 S (B) 32 S
(C) 0.032 S (D) None of these

13. How much current will flow when the potential difference between the two electrode is 5.0
V?
(A) 156.25 A (B) 0.16 A
(C) 160 A (D) None of these

14. The cell constant of the following cell will be

(A) 0.4 (B) 0.8
(C) 0.25 (D) 0.6

Space for Rough work

SECTION - B
Matrix – Match Type
This section contains 2 questions. Each question contains p q r s t
statements given in two columns, which have to be matched. The
A p q r s t
statements in Column I are labelled A, B, C and D, while the
statements in Column II are labelled p, q, r, s and t. Any given p q r s t
B
statement in Column I can have correct matching with ONE OR
MORE statement(s) in Column II. The appropriate bubbles C p q r s t
corresponding to the answers to these questions have to be
darkened as illustrated in the following example: D p q r s t

If the correct matches are A – p, s and t; B – q and r; C – p and q;

and D – s and t; then the correct darkening of bubbles will look like
the following:

1. Match the contents of Column – I with contents of Column – II for 1 mole of gas.
Column  I Column – II
(A) Raw egg   Boiled egg p. H = + Ve
(B) CaCO3(s)   CaO(s) + CO2(g) q. S = +Ve
(C) P(white)  P(red) r. G = –Ve
solid solid
(D) 

Pd  H2  s.
 adsorbed gas Allotropic change
t. S = – Ve

2. Match the contents of Column – I with contents of Column – II for 1 mole of gas.
Column  I Column – II
(A) Tetragonal (p) a ≠ b ≠ c
(B) Monoclinic (q)  ≠  ≠  ≠ 90º
(C) Rhombohedral (r) a ≠ b ≠ c,  ≠  ≠  ≠ 90º
(D) Triclinic (s) a = b ≠ c,  ≠  ≠  ≠ 90º
(t) a = b = c,  ≠  ≠  ≠ 90º

Space for Rough work

SECTION – C
(One Integer Value Correct Type)
This section contains 5 questions. Each question, when worked out will result in one integer
from 0 to 9 (both inclusive).
1
1. An -particle moving with velocity th times of velocity of light. If uncertainty in position
30
is
3.31
pm, then minimum Uncertainty in kinetic energy is y × 10–16J. Calculate value of ‘y’.


2. The vander Waal’s constants for a gas are a = 3.6 atm L2 mol–2, b = 0.6L mol–1, If R =
–1 –1
0.08 L atmk mol . If the Boyle’s temperature (K) is Tb of this gas, then what is the value
T
of b ?
15

3. Amongst the following, the total number of compounds whose aqueous solution turns red
litmus paper blue is NaCN, KCl, CH3COONH4, NaH2PO4, ZnCl2, Na3PO4, Fe(NO3)3,
Na2CO3, NH4Cl, NaHCO3, Na2C2O4, Na2HPO4 .
Given :
Acid Ka1 Ka2 Ka3
–3 –8 –12
H3PO4 10 10 10
–6 –1
H2CO3 10 10 -

H2C2O4 10–2 10–5 -

4. 

G° for the hypothetical reaction x + y 
 z is – 4.606 kcal. The equilibrium constant
2
of the reaction at 227° is (x  10 ). Find the value of x.

5. Find out number of compounds which are more acidic than benzoic acid from the following

COOH COOH COOH COOH

O
||
H — C — OH , H 2C 2O 4 ,
,

CH3 NO2 N(CH3)2 CF3

Space for Rough work

Mathematics PART – III

SECTION – A
Straight Objective Type
This section contains 6 multiple choice questions. Each question has four choices (A), (B), (C)
and (D) out of which ONLY ONE is correct.

1. A straight line ‘L’ cuts the lines AB, AC and AD of a parallelogram ABCD at points B1, C1
      1
and D1 respectively. If AB1  1 AB , AD1   2 AD and AC1  3 AC , then is equal to
3
1 1 1 1
(A)  (B) 
1  2 1  2
(C) –1 + 2 (D) 1 + 2

cos x  sin x 0 
2. If f(x)   sin x cos x 0  then det (f( + )) is
 0 0 1
(A) 0 (B) 1
2 2
(C) –1 (D) cos  – sin 
     
3. Let a, b and c be three non zero and non coplanar vectors and p, q and r be three
           
vectors given by p  a  b  2c ; q  3a  2b  c and r  a  4b  2c . If the volume of the
  
parallelepiped determined by a, b and c is v 1 and that of the parallelepiped determined
  
by p, qand r is v 2 then v 2 : v 1 is
(A) 1 : 5 (B) 5 : 1
(C) 15 : 1 (D) 1 : 15

4. The point A divides the join of P(–5, 1) and Q(3, 5) in ratio K : 1. The two values of K for
which area of ABC, where B is (1, 5) and C is (7, –2), is equals to 2 units in magnitude
31
are K or then K is
K2
(A) 2 (B) 3
(C) 5 (D) 7

Space for rough work

sin x cos x tan x 1 ak

5. If    k , then bc   is equal to
a b c ck 1  bk
 1 1 1
(A) k  a   (B)  a  
 a k a
1 a
(C) 2 (D)
k k

6. The area () and an angle () of a triangle are given. When the side opposite to the given
angle is minimum, then the lengths of the remaining two sides are
2 3 2 2
(A) , (B) ,
sin  sin sin  sin
4 2 6 6
(C) , (D) ,
sin  sin sin  sin

Multi Correct Choice Type

This section contains 3 multiple choice questions. Each question has four choices (A), (B), (C)
and (D) out of which ONE or MORE THAN ONE is/are correct.

 tan2 x
 2 2
for x  0
 x  x

7. Given a real valued function f such that f  x   1 for x  0 (where [.]

 x cot x for x  0


denotes the greatest integer function and {x} is fractional part of x), then
(A) lim f  x   1 (B) lim f  x   cot1
x 0 x 0
2


(C) cot 1 lim f  x 
x 0
 1 
(D) tan1 lim f  x  
x 0
 4

 2 9 
 log3 x    log3 x  5 
2
8. The equation x  
 3 3 has
(A) at least 1 real solution (B) exactly three real solution
(C) exactly 1 irrational solution (D) complex roots

9. Which of following functions have the maximum value unity?

6 1 1 
(A) sin2 x – cos2 x (B)  sin x  cos x 
5 2 3 
(C) cos6 x + sin6 x (D) cos2 x + sin4 x

Space for rough work

Comprehension type (Only One Option Correct)

This section contains 2 paragraphs. Based upon one of paragraphs 2 multiple choice
questions and based on the other paragraph 3 multiple choice questions have to be
answered. Each of these questions has four choices (A), (B), (C) and (D) out of which ONLY
ONE is correct.

Paragraph for Question Nos. 10 and 11

Read the following write up carefully and answer the following questions:
x
Let f(x) be a non-positive continuous function and F  x    f  t  dt  x  0 and f(x)  cF(x) where c
0

dg  x 
> 0 and let g: [0, )  R be a function such that  g  x   x > 0 and g(0) = 0
dx

10. The total number of root(s) of the equation f(x) = g(x) is/are
(A)  (B) 1
(C) 2 (D) 0
2
11. The number of solution(s) of the equation |x + x – 6| = f(x) + g(x) is/are
(A) 2 (B) 1
(C) 0 (D) 3

Paragraph for Question Nos. 12 and 14

Read the following write up carefully and answer the following questions:
If 4-letter words are formed using the letters of the word ‘MORADABAD’, then

12. The probability that at least two letters are repeated is

120 133
(A) (B)
313 313
123
(C) (D) none of these
310

13. The probability that D comes in the 4-letter word is

209 120
(A) (B)
313 313
189
(C) (D) none of these
310

14. The probability that D comes exactly once in the 4-letters word is
73 144
(A) (B)
313 310
146
(C) (D) none of these
313

Space for rough work

SECTION - B
Matrix – Match Type
This section contains 2 questions. Each question contains p q r s t
statements given in two columns, which have to be matched. The
A p q r s t
statements in Column I are labelled A, B, C and D, while the
statements in Column II are labelled p, q, r, s and t. Any given p q r s t
B
statement in Column I can have correct matching with ONE OR
MORE statement(s) in Column II. The appropriate bubbles C p q r s t
corresponding to the answers to these questions have to be
darkened as illustrated in the following example: D p q r s t

If the correct matches are A – p, s and t; B – q and r; C – p and q;

and D – s and t; then the correct darkening of bubbles will look like
the following:

1. Match the following Column-I with Column-II

Column – I Column – II
(A) The number of solutions of the equation (p) No
1 solutio
cot x  cot x  (0  x  ) is
sin x n
1
(B) If sin   sin   and cos  + cos  = 2, then value of
2 1
(q)
 3
cot   is
 2 
   
(C) The value of sin2   sin     sin     is (r) 1
3  3 
(D) If tan  = 3 tan , then maximum value of tan2      is (s) 2
3
(t)
4

2. Match the following Column-I with Column-II

Column – I Column – II
4
x 1   x 2  1  x 4  1 
(A)  x2 dx (p) ln  C
x 4  x2  1  x 
x2  1 1  x 4  1  2x 
(B) x dx (q) C ln  
1 x4 2   x 2  1 
1  x2  1 
(C)  1  x 2  dx (r) C  tan1  1  1 
4 4
1 x  x 
dx
(D)  x4  x2  1
(s) C
1  x 4  1  x4  x2 x
  x 2  1  x 4  1 
(t) ln  C
 x 

Space for rough work

SECTION – C
(One Integer Value Correct Type)
This section contains 5 questions. Each question, when worked out will result in one integer
from 0 to 9 (both inclusive).

4
1. If x  satisfies the in-equation loga(x2 – x + 2) > loga(–x2 + 2x + 3), then sum of all
9
possible distinct values of [x] is (where [.] denotes the greatest integer function) is _____
  
2. Let v  2iˆ  ˆj  kˆ and w  ˆi  3kˆ . If u is a unit vector and P is the maximum value of the
  
scalar triple product u v w  , then the value of p2 – 50 is _____

3. If

ex 2  x2  
 1 x 
dx  e x 
   c , then 2( + ) is equal to _____
1  x  1 x 2  1 x 

sin 2A sinC sinB

4. If A, B and C are the angles of a triangle and sinC sin2B sin A   sin A sinBsinC ,
sinB sin A sin 2C
then  is _____

5. The number of solutions of the equation log x 1 x 2  1   sin x  cos x  (where [.]
denotes the greatest integer function) is _____

Space for rough work

ANSWERS, HINTS & SOLUTIONS

FULL TEST– IV
(Paper-2)

Q. No. PHYSICS CHEMISTRY MATHEMATICS

ALL INDIA TEST SERIES

1. D A A
2. C B B
3. D D C
4. B A D
5. A A B
6. C D B
7. A, D A, B, C A, C, D
8. B, C, D A, B, D A, B, C
9. A, D A, C, D A, B, C, D
10. C B B
11. C A C
12. A C B
13. A B A
14. B C C
A→p A → (p, q, r) (A)  (r)
B→q B → (p, q, r) (B)  (p)
1.
C→r C → (p, s) (C)  (t)
D→s D → (p, r, t) (D)  (q)
A→q A → (p) (A)  (s)
B→p B → (t) (B)  (p)
2.
C→p C → (s) (C)  (q)
D→q D → (r) (D)  (r)
1. 8 5 0
2. 9 5 9

3. 1 6 3

4. 3 1 0
5. 5 4 1

Physics PART – I

SECTION – A
1. There are 8 points in space and every pair connected through C. Without loss of generality we
can choose 6 of them exactly at the mid-point of the remaining two: (1, 2, 3, 4, 5 and 6 are at
same potential).
C
Acceleration Ceff =  6  C  4C .
2
2. For zero P.D. across galvanometer.
i1R1 = i2R2
di1 di
and L1  L2 2
dt dt
Now, i2  di2
and i1  di1
L R 3 2
 1  1 
L 2 R2 L 4
or L = 6 mH.

3. For any system of particles,

1 2 1 2
KE  MVCM   Mi ViCM
2 2
First term of RHS is frame dependent, second term is not frame dependent.
Loss in KE is any frame is same as loss of KE in C.M. frame. The first term doesn’t change during
collision.
4. I’ = I cos2(D/2)
2 2 dy 2y 2  
D  .Dx  .   . 
  D   4 2
  I
 I'  Icos2   
4 2
5. Charge –Q is uniformly distributed at inner surface of cavity and +Q is uniformly distributed at
outer surface. Outside energy is same as shell.
6. For a point source wave front is always spherical, after passing through lens shape becomes
plane in case of convex lens as object lies on focus, but in case of concave lens it remains
spherical with greater curvature.
7. FBD of rod just after the string is burnt is shown. O1
Net force is vertical so the acceleration is vertical.
Motion of point A is circular. Initial angular
velocity is zero thus the radial acceleration which T
is vertical is zero.
A
C
B

8. Force of electric field is always along x-axis but force of magnetic field is always in y-z plane.
Speed in y-z plane remains constant as magnetic field does not change speed.

9. kmax = hv - 
hv
k’max = 
2
hv
electrons are ejected only if 
2
k max  k max
and k 'max   
2 2 2
10. Considering the velocities in river frame, point B appears to move left with a velocity Vr while boat
has a velocity Vbr always towards A.

Velocity of approach = Vbr – Vr cosθ  Vr

T B
  V
0
br  Vr cos   dt  W .........(1)
Vbr
Also:
T
Vr T   Vbr cos dt .....................(2)
0
( equating x-coordinate).
From (1) and (2):
Vr2
 Vbr T  TW
Vbr
WV
or T  2 br 2
Vbr  Vr

11. As time passes, θ decreases, initial θ is 90°. The net speed is given by:
Vb  Vbr2  Vr2  2Vr Vbr cos(180  )
= Vbr2  Vr2  2Vr Vbr cos  .

12-14. [A and B are always at same temperature as piston is diathermic].

For whole gas system ΔQ = 0, ΔW = –Ve(as work done by gas in right is more –ve than +ve
work done by gas in left)  ΔU = +Ve

SECTION – B

1. C.M. of the cube comes down through a height of 3 m while C.M. of displaced water goes up by
3
a height of m.
2
 Loss in PE of cube = 2000 × 10 × 3 Joule.
= 2 3  104 Joule.
3 3
Gain in PE of water = 1000  10    10 4 Joule.
2 2
Using work energy theorem:
W T + W Gravity = 0

 3  3 3
or W T = –W Gravity = ΔPE =   2 3   10 4 =  10 4 Joule.
  2
 2 

SECTION – C

2. Motion of disc is only a pure translational motion, as torque on

disc about C is always zero, even if C is accelerated. When
rod becomes vertical, A remains horizontally to the right of C.
Velocity of whole disc is same as C. Net velocity of disc be V.
Loss in potential energy = Gain in kinetic energy
2
 3r  1  m(3r)2   V  1 2
mg    mg3r     3r   2 mV
2
  2  3  
A
27
V gr  9 m/s.
4
3. V22’ = V11’ 1 2
V33’ = V22’ = 2V11’
So on, V44’ = 3V11’
2R 2R
3 1
 2 =  (16)   =
2
1' 2'
It means potential get equally divided in consecutive sections.
The equivalent arrangement is
4. After crossing L1, rays become parallel and crossing L2 the rays diverge to create image point at
it’s focal plane which is x = 0. Ray crossing optical centre of L2 passes undeviated and y
coordinate of this ray is focal plane of L2 gives y coordinate of image, which is
1
(6mm)  3mm .
2
L1
L2
O
I

  1  cos2 t  ˆ  1  cos2 t  ˆ
5. r  A  i  B j
 2   2 
A ˆ B ˆ  A ˆ B ˆ   A
= i  j   i  j  cos t = r0   sin( t  )
2 2 2 2  2
 A 2  B2
Amplitude is    5.
2

Chemistry PART – II

SECTION – A

1. CH3
Cl OH
 Et S H2 O
 
Et S CH3 Et S CH3

2. OH

CH3

Give 3° carbocation as intermediate while other gives 2° carbocation.

4. KI  I2  KI3

0.529  n2 o
6. r A
1
 0.529 [(n + 1)2 – n2] = 0.529 (n - 1 )2
on solving, n = 4

m  M 200  0.04
13.    m .C  
1000 1000
= 8  10–3 S cm–1
  0.50 
 = G    8  10–3 = G  
A  2 
G = 0.032 S, V = IR  I/G
I = 5  0.032 = 0.16 A
SECTION –C
h
1. d(kE) = mv.dv = mv
4m x
3  108 6.62  10 34
 
3 3.31
4  10–12

= 5 × 10–16J

a 3.6
2. TB    75k
Rb 0.08  0.6
75
 5K
15

3. NaCN, Na3PO4, Na2CO3, NaHCO3, Na2C2O4, Na2HPO4

4. G° = – 2.303 RT log K

5. Fact

Mathematics PART – III

SECTION – A
      
1. Let AB  a , AD  b , then AC  a  b D C
      
Given, AB1  1a , AD1   2b , AC1  3 a  b  
    
B1D1  AD1  AB1   2b  1a D1 C1
 
Since, vectors D1C1 and B1D1 are collinear A B
  B1
D1C1  kB1D1
    
  
 3 a  b  2b  k  2b  1a 
 3 = –k1 and 3 – 2 = k2
3 3   2
 k   1 2 = 1 3 + 2 3
1 2
1 1 1
  
 3 1  2

cos(   )  sin(   ) 0
2. det det  f(   )   sin(   ) cos(   ) 0 = cos2 (  )  sin2 (  )  1
0 0 1

   
3. Given p  a  b  2c
   
q  3a  2b  c
   
and r  a  4b  2c

Given v1 =  abc  ….. (1)
1 1 2
  
v2 = pqr  = 3 2 1 abc 
 
1 4 2

v2 = 15  abc  ….. (2)
From (1) and (2) v 2 : v1 = 15 : 1

1 5 1
1
4. 7 2 1  2 where (, ) divides PQ in K : 1
2
  1

cos x tan x 1 sin x sin x cos x(1  cos x)  sin2 x a 1 a 1

5.    2  =   
k2 tan x 1  cos x k sin x(1  cos x) k sin x k ak

6. Let  be the angle opposite to side c of ABC

2 2 2 2
 c = a + b – 2ab cos  = (a – b) + 2ab(1 – cos )
1 4
Also,   ab sin   2ab =
2 sin 

2 1  cos  2 
 c 2   a  b   4 =  a  b   4 tan
sin  2
For c to be minimum, a = b
2 4
 2ab = 2a =
sin 
2
 ab
sin 

tan2 x
7. Here, lim f  x   lim 1
x 0 x 0 x


and lim f  x   cot1  lim x  1
x 0 x 0

8. Take log on both sides and log3 x = t
 t  1  t  1   t  3   0 , x  3 , 3, 27
 2

9. sin2 x  cos2 x   cos2x  1

6 1 1  3 2
 sin x  cos x   sin x  cos x = sin x · sin  + cos x cos 
5 2 3  5 5
3 2
Where sin   , cos   = cos(x  )  1
5 5
3 3 3
cos6 x  sin6 x   cos 2 x    sin2 x  = 1 – 3 sin x cos x = 1 
2 2
 sin2x 2  1
4
 sin2x 2
cos2 x  sin4 x  1  1
4

10. f(x)  0 and F(x) = f(x)

d  cx
 f(x)  cF(x) 
dx

e .F  x   0 
 e–cs F(x) is an increasing function
 e–cs F(x)  e–c(0) F(0)
 F(x)  0  f(x)  0
 f(x) = 0
d x
Also,
dx

e gx   0 
 e–x g(x) is a decreasing function
 e–x g(x) < e–(0) g(0)
 g(x) < 0 (as g(0) = 0)
Hence, f(x) = g(x) has one solution x = 0

11. |x2 + x – 6| = f(x) + g(x) = g(x)

 No solution

12.-14. Total number of words

(1) 3 like, 1 different = 20
(2) 2 like, 2 like = 6
(3) 2 like, 2 different = 240
(4) all different = 360

SECTION – B

1
1. (A) cot x  cot x 
sin x

If 0 < x <  cot x > 0
2
1 1
So, cot x = cot x +  = 0 (no solution)
sin x sin x
 1
If  cot x   ,  cot x  cot x 
2 sin x
2cos x 1
 0
sin x sin x
2
1 + 2 cos x = 0 and sin x  0  x 
3
     3
(C) sin2   sin      sin     = sin2   sin2  sin2  
3  3  3 4
(D) tan  = 3 tan 
tan   tan  2tan  2
tan(  )   =
1  tan  tan  1  3 tan2  cot   3 tan 
cot   3 tan 
Max if tan  > 0,  3 (using AM  GM)
2
2 1
  cot   3 tan    12  tan2      
3

x4  1 x4  x2  1
2. (A)  x3 dx , Let t
x 4  x2  1 / x x

x2  x4  x2  1  x4  1 
 dt ,   dx  dt
x3  t  2 4 2
 x  x  x 1
x 4  x2  1
 dt ,  x 4  1 dx  x2  x 4  x 2  1 dt
xt
t x4  x2  1
 t dt ,  dt  t  C = x
C


  x 2  1  x 2  1 1  1
(B)  dx ,  dx , Let x   t ,  1  2  dx  dt
 x 1 x 4 
 x2 x2  1 x  x 
 2
x
 2 
dt
= ln  t t 2
2  C = ln  x  1   x  1
  2   C
 t2  2    
 x  x 
  x 2  1  1  x 4 
ln  C
 x 

 1  1 
1  x 2  1  2  dx  1  2  dx
(C)  dx =  x   x 
1  x2  1  x 4   1 1 2=  2
x 1 2  2  x  1  1
 x  x x   x    2
 x  x
 1  1 dt
Let  x    t ,  1  2  dx  dt , t
 x  x  t2  2
2 2
Let t + 2 = y , 2t dt = 2y dy
dy dy 1 y 2 1  t2  2  2 
 t2 =   y2  2   2  ln   C = ln  C
2  y  2  2 2  t 2  2  2 
2
 2 
1  t  2  2  2
  C = 1 ln  t  2  2   C

= ln  
2 2  t2  2  t 
 2 
  x  1   2  2
1   x  1  x 4  1  2x 
= ln  C = ln    C
2   1
 2   x 2
 1 
x  
  x 
dx dx
(D)   
1  x  4
1 x4  x2  1 1
x5  1  4  1  4  1
 x  x
1 1  4  dx 1
Let 1 4
 1  t2   5  dx  2tdt  5   t 1  4 dt
x 1 x  x x
2 1 4
x
1
t 1  dt  
 x4  
dt
 
dt
  tan1 t  C  C  tan1  1
1
 1 
  1  1  t  1
2
 x 4

1  4  t 1
 x  x4

SECTION – C

4
1. On substituting x  , we get 0 < a < 1
9
 loga(x2 – x + 2) > loga(–x2 + 2x + 3)  0 < x2 – x + 2 < –x2 + 2x + 3
3  17 3  17
 x
4 4
        
2. u v w   u   v  w   u v  w cos 
For maximum value,  = 0º
 
 p  v  w  59

 1 1  x 
3. I   ex   dx
 1  x  1  x 2 1 x 
 
d  1 x  1
Now,   
dx  1  x  1  x  1  x
2

1/2
 1 x 
 I  ex   c
 1 x 
1
  = 1,  
2

2ka cos A kc kb
4. LHS = kc 2kb cosB ka (From sine rule)
kb ka 2kc cosC
2a cos A c b
= k3 c 2bcosB a
b a 2c cosC
cos A a 0 a cos A 0
3
= k cosB b 0 b cosB 0  0
cosC c 0 c cosC 0
=0

5. Here, log x 1 x 2  1  1

 x2  1  x  1
 x = 1, 0, –2 but x  0, 1
 Number of solutions is one

Time Allotted: 3 Hours Maximum Marks: 360

 Pl ea s e r ea d t h e i n s t r u c t i o n s c a r ef u l l y. Yo u a r e a l l o t t ed 5 m i n ut es
s p ec i f i c a ll y f o r t h i s p u r p o s e.

ALL INDIA TEST SERIES

Yo u a r e n o t a ll o wed t o l ea v e t h e E xa m i n at i o n Ha ll b ef o r e t h e en d o f
t h e t es t .

INSTRUCTIONS

A. General Instructions
1. Attempt ALL the questions. Answers have to be marked on the OMR sheets.
2. This question paper contains Three Parts.
3. Part-I is Physics, Part-II is Chemistry and Part-III is Mathematics.
4. Each part has only one section: Section-A.
5. Rough spaces are provided for rough work inside the question paper. No additional sheets will be
provided for rough work.
6. Blank Papers, clip boards, log tables, slide rule, calculator, cellular phones, pagers and electronic
devices, in any form, are not allowed.

B. Filling of OMR Sheet

C. Marking Scheme For All Three Parts.

(i) Section-A (01 to 30) contains 30 multiple choice questions which have only one correct answer.
Each question carries +4 marks for correct answer and – 1 mark for wrong answer.

Name of the Candidate

Enrolment No.

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Useful Data

PHYSICS
Acceleration due to gravity g = 10 m/s2

Planck constant h = 6.6 1034 J-s

Charge of electron e = 1.6  1019 C

Mass of electron me = 9.1  1031 kg

Permittivity of free space 0 = 8.85  1012 C2/N-m2

Density of water water = 103 kg/m3

Atmospheric pressure Pa = 105 N/m2

Gas constant R = 8.314 J K1 mol1

CHEMISTRY

Gas Constant R = 8.314 J K1 mol1

Atomic No: H=1, He = 2, Li=3, Be=4, B=5, C=6, N=7, O=8,

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Physics PART – I

SECTION – A
Single Correct Choice Type

This section contains 30 multiple choice questions. Each question has four choices (A), (B), (C) and (D)
out of which ONLY ONE is correct.

1. A ring of mass m can freely slide on a smooth vertical rod. The ring
is symmetrically attached with two springs (as shown in the figure).  

Each of spring constant k. Each spring makes an angle  with the

horizontal. If the ring is slightly displaced vertically, its time period is: m

m m
(A) 2 (B) 2
2k k
2m m
(C) 2 (D) 2 .
k sin2  2k sin2 

2. A man of height H stands in front of a plane mirror. Water has

been filled upto a height 3H/4. The man can just see his foot
in the mirror placed parallel in front of him. What is the
refractive index of water?
20 5
(A) (B)
3 17 3 17
4 5
(C) (D)
3 17 17
H

3. A 590 kg block at A moves along the horizontal 300N

C
surface with negligible friction under the action
of the constant 300 N force in the cable. The
block is released from rest at A, with the spring
to which it is attached extended an initial
amount x1 = 0.3 m. The spring has a stiffness 0.9m
constant, K = 80 N/m. The kinetic energy of the
block as it reaches position B is:
(A) 86.4 J A 1.2m
(B) 180 J B
(C) 93.6 J x1
(D) None of these

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4. A mass-spring system as shown in the figure oscillates

such that the mass moves on a rough surface having
coefficient of friction . It is compressed by a distance a
from its normal length and, on being released. It moves m
to a distance b from its equilibrium position. The
decrease in amplitude for one half-cycle (–a to b) is:

a b

mg 2mg
(A) (B)
K K
g
(C) (D) None of these
K

5. In a series LR circuit, the frequency of the AC source increases. Which of the following options is
correct?
(A) The phase difference between the current and the voltage increases.
(B) The power factor increases.
(C) The amplitude of the current increases.
(D) The amplitude of potential drop across the inductor decreases.

6. In the diagram shown, when the switch is at 2C R L/2

(2)
position 1 then the frequency of maximum
C R L
current from the battery is . What will be
the frequency of maximum current from the
(1)
battery when the switch is at 2?
(A) 
(B) 2
(C) / 3
(D) None of these

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7. Consider the velocity versus time graph of particle moving on V(m/s2)

st
a straight line. The average velocity on the 1 one second is
nd VA
V1m/s and the average velocity on the 2 one second is V2
m/s. VB
What is the average velocity on the last one second?
(A) V2 – V1
(B) (V2 + V1)/2
(C) 2V2 1 2 3 t(s)
(D) None of these

8. The velocity versus time graph of a particle is Vm/s)

given.
If the total displacement of the particle is 4m V0
then what is the initial velocity of the particle?
(A) 1.0 m/s
(B) 1.1 m/s 6 8 time(s)
(C) 1.5 m/s V
(D) None of these  0
3

9. The potential energy of harmonic oscillator of mass 2 kg in its mean position is 5J. If its total
energy is 9J and its amplitude is 0.01m, its time period will be:
 
(A) s (B) s
100 50
 
(C) s (D) s
20 10

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10. A crate of mass m falls from a height h onto the end of a m

platform, as shown in the figure. The spring is initially
unstretched and the mass of the platform can be
neglected. Assuming that there is no loss of energy, the
maximum elongation of the spring is:
h
mg  m2 g2  2mghK x x
(A)
K
mg  m 2 g2  2mghK
(B)
K
K
(C) K
m2 g2  2mghK
(D) None of these

11. A uniform disc of radius r and mass m is charged uniformly Disc

B
with the charge q. This disc is placed flat on a rough
horizontal surface having coefficient of friction . A uniform a
magnetic field is present in a circular region of radius a (> r) r
but varying as kt3 as shown in figure. Find the time in second
after which the disc begins to rotate. (Given r=1 m, m=18 kg,
q=1C, =0.1, K=4, g=10 m/s2)
(A) 2 (B) 4
(C) 8 (D) None of these

12. A conducting ring of radius r and resistance R

rolls on a horizontal surface with constant
velocity v. The magnetic field B is uniform normal
to the plane of the loop.
(A) The induced emf, between O and Q is 2Bvr.
2Bvr
(B) An induced current   flows in the
R
clockwise direction.
2Bvr
(C) An induced current   flows in the anticlockwise direction.
R
(D) None of these

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13. The P-V diagram of a system undergoing thermodynamic P

transformation as shown in the figure. The work done by the
system in going from A  B  C is 30J and 4J heat is given to C
the system. The change in internal energy between A and C is
(A) 26 J
(B) – 26 J
(C) 34 J
(D) – 34 J
A B
V

14. A sample of gas expands from volume V1 to V2. The amount of work done by the gas is greatest
when the expansion is:
(A) Isothermal (B) Isobaric
(C) Adiabatic (D) Equal in all cases.

15. In a given process for an ideal gas, dW = 0 and dQ < 0. Then for the gas:
(A) The temperature will decrease (B) The volume will increase
(C) The pressure will remain constant (D) The temperature will increase.

16. In the diagram shown, the length of the mirror is half of the
height of the man. What length of his image can the man
see in the mirror? H
(A) H/3
(B) H/2
(C) H
(D) zero
H/3

17. Starting from rest at the same time, a coin and a ring roll down an incline without slipping. Which
reaches the bottom of the incline first?
(A) The ring reaches the bottom first
(B) The coin reaches the bottom first
(C) They arrive at the bottom simultaneously
(D) The winner depends on the relative masses of the two

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18. In the given figure F = 10 N, R = 1 m, mass of the body is 2

kg and moment of inertia of the body about an axis passing
through O and perpendicular to plane of body 4kg-m2. O is
the centre of mass of the body. If the ground is smooth what
is total kinetic energy of the body after 2 seconds
(A) 100 J (B) 75 J
(C) 50 J (D) 25 J

19. A solid ball of mass m and radius r rolls without slipping A m

along the track shown in the fig. The radius of the circular
part of the track is R. The ball starts rolling down the track
from rest from a height of 8R from the ground level. When h O P
the ball reaches the point P then its velocity will be R

(A) gR (B) 5gR

20. Three measurements are made as 18.425 cm, 7.21 cm and 5.00 cm. The addition should be
written as:
(A) 30.635 cm (B) 30.64 cm
(C) 30.63 cm (D) 30.6 cm

21. Two identical charged particles enter in a uniform magnetic field with same speed but at angles
30° and 60° with field. Let a, b and c be the ratio of their time periods, radii and pitches of the
helical paths then:
(A) abc = 1 (B) abc > 1
(C) abc < 1 (D) ab = c

1
22. The acceleration due to gravity on the surface of the moon is that on the surface of earth and
6
the diameter of the moon is one-fourth that of earth. The ratio of escape velocities on earth and
moon will be:
6
(A) (B) 24
2
3
(C) 3 (D)
2

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23. The greatest angle through which a man can oscillate on a swing the ropes of which can support
twice his weight when at rest is:
(A) 90° (B) 60°
(C) 120° (D) 30°

24. A bead of mass m is fitted on a rod and can move on it without friction.
Initially the bead is at the middle of the rod and the rod moves
a0
translationally in a horizontal plane with an acceleration a0 in
horizontal direction forming angle  with the rod. The acceleration of
bead with respect to rod is:
(A) g sin  (B) (g + a0) sin 
(C) g sin  + a0 cos  (D) g sin  – a0 cos 

25. An insect of mass m crawls along the hanging thread with an

g
acceleration a  . The reaction offered by ground on the
2
block of mass 2m is:
3mg g
(A) (B) mg a
2 2
mg m
(C) 2mg (D) 2m
2

26. Two balls are dropped from same height h

on a smooth plane and the other on a h h
rough plane having same inclination  with
horizontal. Both the planes have same
coefficient of restitution. If range and time T1 R2 T2
R1
of flight of first and second balls are R1, T1
and R2, T2 respectively, then:  
(A) T1 = T2, R1 = R2 Smooth plane Rough plane
(B) T1 < T2, R1 < R2
(C) T1 = T2, R1 > R2
(D) T1 > T2, R1 > R2

27. An iron nail is falling from a height h. If it penetrates through a

distance x into the sand, the average resistance of the sand is: m
h h 
(A) mg (B) mg   1
x x 
h
x
(C) mg (D) mg
h

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28. A ray of light traveling in a medium of refractive index n1 is incident on a medium of refractive
index n2 ( n1 ) at an angle of incidence ‘’. The reflected and refracted rays make an angle of
120° with each other. Then the critical angle for the boundary separating the two media is:
1   1 
(A) sin1  tan   (B) sin1  
2   3  tan  
 2 tan  
(C) sin1   (D) sin1  2 cot  
 3  tan  

29. A thin wire is bent into the shape of a parabola in the vertical Y
plane as shown. A small bead ‘P’ can freely slide along the P
wire.
The coefficient of friction between the bead and the wire is ‘µ’.
The condition that bead does not slip on the wire is x
2a O
(A) y  2a (B) y

2a
(C) y  2a (D) y . 2
Y =4ax


30. In a nuclear reactor, energy is produced by the following process :

2 2
1H  1H   1H3  P
3
1H  1H2  2He 4  n
The reactor produces a power of 1010 MW with 50% efficiency. Then the number of deutrons
required to run the reactor for one year will be approximately :
m(1H2 )  2.014u, m(p)  1.007u, m(n)  1.008u, m( 2 He4 )  4.001u
(A) 5 × 1035 (B) 5 × 1032
38
(C) 5 × 10 (D) 5 × 1040

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Chemistry PART – II

SECTION – A
Single Correct Choice Type

This section contains 30 multiple choice questions. Each question has four choices (A), (B), (C) and (D)
out of which ONLY ONE is correct.

1. Lattice energies of BeF2, MgF2, CaF2 and BaF2 are –2906, –2610, –2459 and –2367 KJ/mol
2+ 2+ 2+ 2+ –
respectively. Hydration energies of Be , Mg , Ca , Ba and F are –2494, –1921, –1577, –
–1
1305 and –457 KJ mol respectively. Which of the fluorides is not soluble in water?
(A) BeF2 (B) MgF2
(C) CaF2 (D) BaF2

2. The stability of dihalides of Si, Ge, Sn and Pb increases steadily in the sequence:
(A) GeX2 < SiX2 < SnX2 < PbX2 (B) SiX2 < GeX2 < PbX2 < SnX2
(C) SiX2 < GeX2 < SnX2 < PbX2 (D) PbX2 < SnX2 < GeX2 < SiX2

3. Pressure Vs. density curve for an ideal gas at three different temperatures T1, T2 and T3 is shown
below, which is correct relation here?
T1

P
T3

Which of these is correct?

(A) T1 > T2 > T3 (B) T1 > T3 > T2
(C) T3 > T2 > T1 (D) Can’t be said

4. A metal oxide has the formula Z2O3. It can be reduced by hydrogen to give free metal and water.
0.16 g of the metal oxide requires 6 mg of hydrogen for complete reduction. The atomic weight of
metal is
(A) 27.9 (B) 159.6
(C) 79.8 (D) 55.8

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5. Give the major product from the following reaction sequence:

Br SO H
 2
FeBr

 3
H SO
 
FeBr
 
HO
 A major 
3 2 4 3 2

(A) Br (B) Br

SO 3H
(C) Br (D) Br

HO3S

6. C N

H O
CH3MgBr  Q 
3
P

OCH 3
The product ‘P’ in the above reaction is
(A) OH (B) OC CH3
HC
CH3

OCH 3

(C) CHO (D) COOH

OCH3 OCH3

7. In an open-end manometer, one end filled with mercury is attached to a gas-filled container and
the other end is open to the atmosphere. If the gas pressure in the container is less than
atmospheric pressure.
(A) The Hg level will be higher in the arm open to the atmosphere
(B) Hg will be forced out of the open and of the U-tube.
(C) The difference between the Hg levels in the two arms will be greater than 76 cm.
(D) The Hg level will be higher in the arm connected to the container.

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8. 100 ml of NaHC2O4 required 50 ml of 0.1 M KMnO4 solution in acidic medium. Volume of 0.1 M
NaOH required by 100 ml of NaHC2O4 is:
(A) 50 ml (B) 100 ml
(C) 125 ml (D) 150 ml

9. Ratio of  and  bonds is maximum in

(A) Naphthalene (B) Tetra Cyanomethane
(C) Enolic form of urea (D) Equal in all

10. Which of the following statement is true for CsBr3?

(A) It is a covalent compound (B) It contains Cs3+ and Br–1
+ + – 2
(C) It contains Cs and Br3 (D) It contains Cs , Br and lattice Br molecule
11. (y)
OH
(W)
HO

OH
(z)
(x) HO
Decreasing order of acidic strength of different (- OH) group is:
(A) w > x > y > z (B) w > z > x > y
(C) z > w > x > y (D) z > x > w > y

12. 
LiAlH H
 A   4
 B  



What is (A)?
(A) O (B) O

(C) CH2 (D) O

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13. O
KOH
P  



KOH
 Q  

 Ph CH 2 OH  Ph COO K
O3
R  
P  Q
What is the structure of (R):
(A) Ph CH CH CH3 (B) CH3
Ph CH C
CH3
(C) CH3 (D) CH3
Ph C CH CH3 H3C C CH2

14. CH3 CH3

H3C CH3 
NaNO2
A
HCl
OH NH2
Product (A) is:
(A) O CH3 (B) CH3 CH3

H3C C C CH3 H3C CH3

CH3 OH OH

(C) (D) CH3 CH3

H3C CH3
Cl Cl

15. HNO3 in pure state is colorless but it is often of yellow color. It is due to
(A) Unstable structure of HNO3 which immediately changes to NO2
(B) Photochemical decomposition in presence of sunlight
(C) Interaction of atmospheric gases with HNO3
(D) Conversion of HNO3 into NO2

16. When K2CrO4 is added to CuSO4 solution, there is formation of CuCrO4 as well as CuCr2O7.
Formation of CuCr2O7 is due to
(A) Basic nature of CuSO4 solution which converts CrO24 to Cr2O72
(B) Acidic nature of CuSO4 solution which converts CrO24 to Cr2O72
(C) CuSO4 has the typical property of converting CuCrO4 formed to CuCr2O7
(D) 10 CuCr2O7 is formed

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17. A large excess of MgF2(s) is maintained in contact with 1 lit of pure water to produce a saturated
solution. When an additional 1 lit. of pure water is added to the solid-liquid mixture and
equilibrium re-established, compared to its value in the original saturated solution Mg2   will be:
(A) The same (B) twice as large
(C) three times as large (D) half as large

18. A mixture of 2 colourless substances was dissolved in water. When gaseous Cl2 was passed
through the solution, a deep brown color developed. Addition of BaCl2 or NaOH to the original
solution give a white precipitate. The mixture contained.
(A) NaNO3 and AlCl3 (B) NaBr and KCl
(C) MgI2 and MgSoO4 (D) BaSO4 and PbCl2

melt and Cool layer X

19.  Ag  Pb  alloy 
add zinc
 Ag  Pb  Zn  melt  
layer Y
Select correct statement based on above scheme.
(A) Layer X contain zinc and silver
(B) Layer Y contains lead and silver but amount of silver in this layer is smaller than in the layer X
(C) ‘X’ and ‘Y’ are immisible liquids
(D) All are correct

20. What is the molarity of HCl in a solution prepared by dissolving 5.5 g HCl in 200 g ethanol if the
density of the solution is 0.79 g/ml?
(A) 2.1 M (B) 0.93 M
(C) 1.7 M (D) 0.58 M

21. Select the incorrect statement among the folloing:

(A) Nearest neighbour distance in NaCl = a/2
a 3
(B) Nearest neighbour distance in CaF2=
4
a
(C) Nearest neighbour distance in CsCl =
2
a 3
(D) Nearest neighbour distance in Na2O =
4

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22. In the electrolysis of AgNO3, the mass of silver deposited is plotted against the charge:

Mass of
silver
deposited Slope of line (M) = tan 


(Charge)
Slope of the line gives:
(A) The equivalent mass of silver (B) Electrochemical equivalent of silver
(C) the value of Faraday (D) the current passed through the cell

23. Which of the following is not correctly matched?

(A) CH3 H3C Br (B) CH2 H3C Br

 HBr ,  HBr ,

(C) CH3 H3C Br (D) CH2 H3C Br

D D

 DBr ,  DBr ,

24. ONO OCH2 CH2 CH3 OCH2 CH2 CH3 OCH2 CH2 CH3

A Conc. HNO C D

  3
  B  
 


NH2 NHCOCH3
In the given reaction
(I) A is CH3  CH2  CH2  Br
O CH2 CH2 CH3

(II) B is

NO2
(III) C is Sn/HCl
(IV) D is  CH3CO 2 O
Out of these which are correct?
(A) (I) and (II) (B) (I) and (III)
(C) (I), (III) and (IV) (D) (I), (II), (III) and (IV)

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25. Match List-I and List – II and select the correct answer using the codes given below the lists:
(a) -D and -D glucose (1) Enantiomers
(b) D-glucose and D-galactose (2) Anomers
(c) Erythose and threose (3) Epimers
(d) D-glyceraldehyde and L-glyceraldehyde (4) Diasteromers
(A) (a  3), (b  2), (c  1), (d  4) (B) (a  2), (b  3), (c  4), (d  1)
(C) (a  3), (b  2), (c  4), (d  1) (D) (a  2), (b  3), (c  1), (d  4)

26. Which of the following statements is wrong for inter halogen?

(A) The value of n is AXn (inter halogen) can be 1, 3, 5 or 7
(B) The value of n is AXn (inter halogen) can be 2, 4 or 6
(C) A can never be fluorine as it is most electronegative halogen
(D) X can never be iodine as it is least electronegative halogen

27. What are the products formed in the reaction of Xenon hexafluoride with silica?
(A) XeSiO4 + HF (B) XeF2 + SiF4
(C) XeOF4 + SiF4 (D) XeO3 + SiF4

28. The straight chain polymer is formed

(A) hydrolysis is CH3SiCl3 followed by condensation polymerisation
(B) hydrolysis of  CH3  4 Si followed by addition polymerisation
(C) hydrolysis of  CH3 2 SiCl2 followed by condensation polymerisation
(D) hydrolysis of  CH3 3 SiCl followed by condensation polymerisation

29. O

NH2
NaOCl

H O
  X  80%  ,
2

NO 2 O
Product X will be
(A) COOH (B) NH2

NH2 COOH
NO2
NO 2
(C) COOH (D) NH2

COOH
O2N COOH
NO2

30. Which of the following is a covalent carbide?

(A) CaC2 (B) Al4C3
(C) WC (D) SiC

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Mathematics PART – III

SECTION – A

Straight Objective Type

This section contains 30 multiple choice questions numbered 1 to 30. Each question has 4 choices (A),
(B), (C) and (D), out of which ONLY ONE is correct

2
1. The range of function f(x) where f(x) + 2f(1 – x) = 1 – x ,  x  R is
(A) (–, 1] (B) [1, )
(C) (–1, 1) (D) [–1, 1]

2. Six digit numbers are formed using 1, 2, 3, 4, 5, 6, (repetition is not allowed). The number of ways
such that even digits are in increasing order and odd digits are in decreasing order, is
(A) 20 (B) 40
(C) 60 (D) 120

3. The average of 2 sin 2º, 6 sin 6º, 10 sin 10º, ….. 178 sin 178º is
(A) cot 1º (B) 2 cosec 2º
(C) 2(1 + tan 2º) (D) 2(1 + 2 tan 2º sin 44º)

4. The number of points with integral co-ordinates that lie in the interior of the region common to the
circle x2 + y2 = 16 and the parabola y2 = 4x is
(A) 10 (B) 16
(C) 17 (D) 13

5. The projection of the line segment joining the points (1, –2, 3) and (3, 1, 0) on the line whose
direction ratios are (2, 1, 3) is
1 2
(A) (B)
14 14
3 5
(C) (D)
14 14

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6. Set A has 6 elements and its power set has n elements. Let m be the sum of distinct cardinal
numbers of elements of power set of A, then the value of n – m is
(A) 43 (B) 40
(C) 35 (D) 50

7. Number of solutions of the equation cos 9x cos 3x = cos 18x cos 12x, x  [0, ] is
(A) 12 (B) 30
(C) 28 (D) 32

8. If the roots of the equation x 2 + mx + n = 0 are cubes of the roots of the equation
x2 + px + q = 0, then
(A) m = p3 + 3pq (B) m = p3 – 3pq
3
m p
(C) m + n = p3 (D)  
n q

9. The equation of the hyperbola whose foci are symmetrically situated on the x-axis with respect to
4
the origin, the equation of the asymptotes as y =  x and distance between foci being 20, is
3
x2 y2 x2 y 2
(A)  1 (B)  1
64 36 16 9
x2 y 2 x 2 y2
(C)  1 (D)  1
9 16 36 64

 1
 ln  x
 
 x
10. The value of   dx is
0 1 x 2

 
(A) (B) ln2
2 2

(C) ln 4 (D) none of these
2
  
11. Let a  î  ˆj  kˆ , b  î  4ˆj  kˆ , c  î  ˆj  2kˆ and ŝ be unit vector the magnitude of vector
        
 a  sˆ   b  c   b  sˆ   c  a   c  sˆ   a  b  is equal to
(A) 1 (B) 2
(C) 3 (D) 4

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12. The number of solutions of the equation |x – 1| + |x – 10| + x = 9 + [x] is, (where [.] denote the
greatest integer function)
(A) 10 (B) 2
(C) 4 (D) 20

13. Two dice are thrown at a time, the probability that the difference of the numbers is 2, is
1 2
(A) (B)
9 9
4 5
(C) (D)
9 9

14. The four sides of a quadrilateral are given by equation (xy + 12 – 4x – 4y)2 = (2x – 2y)2. The
equation of a line with slope 3 which divides the area of quadrilateral in two equal parts is
(A) y  3  x  4  (B) y  3x  4
(C) y  3  x  4   4 (D) y  3  x  4 

1
1
15. Let I   dx then which of following is correct
0 1  x5

(A) I < (B) I > 1
4
(C) I > 3 (D) none of these

16. The arithmetic mean of a set of observation is X . If each observation is divided by  and then is
increased by 10, then the mean of the new series is
X X  10
(A) (B)
 
X  10
(C) (D)  X + 10


17. Let w  1 and w13 = 1. If a = w + w3 + w4 + w–4 + w–3 + w–1 and

b = w2 + w5 + w6 + w–6 + w–5 + w–2, then the quadratic equation whose roots are a and b is
(A) x2 + x + 3 = 0 (B) x2 + x + 4 = 0
2
(C) x + x – 3 = 0 (D) x2 + x – 4 = 0

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x2 y 2
18. On the ellipse   1 , the point M nearest to the line 2x – 3y + 25 = 0 is
18 8
 8
(A) (–3, 2) (B)   2, 
 3
(C) (3, 2) 
(D) 3 2, 0 
19. The value of lim  4 x 4  3x3  3 x3  2x2  is
x 
1 1
(A) (B)
6 18
1 1
(C) (D)
12 24

20. In any triangle ABC, a, b, c be the sides opposite to angles A, B and C respectively and
 B C B C
b + c = 3a. Then the value of cot cot  tan tan  is, (where [.] denotes the greatest integer
 2 2 2 2
function)
(A) 1 (B) 2
(C) 3 (D) 4

21. Number of solutions of the equation sin–1(sin x) + cos–1(cos x) = x,  x  (0, ] are

(A) 0 (B) 1
(C) 2 (D) 3

22. Let f (x) = x3 + x2 + 2x  1. The minimum value of [x] that satisfy the f(f(x)) > f (2x + 1) is
_________. (Where [.] denotes the greatest integer function).
(A) 0 (B) 1
(C) 2 (D) 3

23. If the two feet of normals drawn from a point to the parabola x 2 – 6x – 4y + 5 = 0 be (7, 3) and (–
1, 3), then the third foot is
(A) (1, 0) (B) (–3, 8)
(C) (3, –1) (D) (5, 0)

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24. If , ,  are number of elements in sets A, B and C respectively and sum of number of all
subsets of A, B and C is 28, then maximum number of ordered triplets (, , ) is
(A) 1 (B) 2
(C) 4 (D) 6

25. A flag staff of 5 m height stands on a building 25 m high. At an observer at a height of 30 m, the
flag staff and the building subtend equal angles. The distance of the observer from the top of flag
staff is
5 3 3
(A) (B) 5
2 2
2
(C) (D) None of these
3

1 1 1
26. Sum of series    ..... upto 10 terms is
1.3 3.5 5.7
10 1
(A) (B)
21 2
11 9
(C) (D)
21 20

27. 
Solution of the differential equation x 2  6y  dy
dx
 2  xy  1  0 is

(A) yx2 + 2x + 3y2 = c (B) xy2 + 2x + 3y = c

28. The value of (1 – cot 3º)(1 – cot 7º)(1 – cot 11º)(1 – cot 34º)(1 – cot 38º)(1 – cot 42º) is equal to
(A) 3 (B) 2
(C) 27 (D) 8

29. An incident ray (L1) is reflected by the mirror (L2) 3x + 4y = 5 and the equation of reflected ray (L3)
is x + y = 1, the equation of L1 is
(A) x – y = –3 (B) 17x + 31y = 45
(C) 17x – 31y = –77 (D) none of these
m
30. If m is a natural number, then m(m + 1)(m + 2) + (7 – 1) is always divisible by
(A) 5 (B) 9
(C) 6 (D) 12

Space for rough work

ANSWERS, HINTS & SOLUTIONS

FULL TEST –V
(Main)
ALL INDIA TEST SERIES

S. No. PHYSICS CHEMISTRY MATHEMATICS

1. D D A
2. B C A
3. C A B
4. B D D
5. A C B
6. A B A
7. A D C
8. C C B
9. A C D
10. A C C
11. A A C
12. A D A
13. B B B
14. B A C
15. A B D
16. D B C
17. B A C
18. C C A
19. C D C
20. D D B
21. A C B
22. B B B
23. B D C
24. D C D
25. D B B
26. C B A
27. B C A
28. C C D
29. B B B
30. A D C

Physics PART – I

SECTION – A

x
1. For small displacement y 
sin   
2kx0 sin   mg ........... (1)
If ring is displaced y
y 
mg – 2k  x 0  y sin   sin  = ma y
x
2k sin2 
a y
m

2. 1 cos 1   2 cos 2

1
2

3. W F = U + K Length AC
1 1
 
300 1.5  0.9  K x 22  x12  mV 2   1.2 2   0.9 2
2 2

1 2 1
4. Ka  mg  a  b   Kb2
2 2
 a  b   2 mg / K.

L
5. tan  
R
R
cos  
R  XL2
2

0  XL
i0  , VL  XL 
2
R  XL2 R 2  XL2

1
6. 
2 LC

0  VA
7.  V1
2
VA  VB
 V2
2
VB  0
 V3
2

V0
2
V0  6 3
8.  4
2 2
3
V0  m/s
2

1
9. Kinetic energy at mean position  4J  mA 2 2    200 Rad / s.
2

1 2
10. mgh  kx  mgx
2

x dB
11. E
2 dt
3Kxt 2 x
E
2 dx
2
3Kxt 2xdx
d   q.x
2 r 2
r
3Kt 2 q 3
 x dx
r 2 0
3Kq.t 2 2
 .r …(i)
4
torque due to friction force
d  dmgx
r
qm 2 2
  2g 2 
x dx  mgr …(ii)
r 0 3
3Kq.t 2r 2 2
 mgr
4 3
8mg
t = 2 seconds.
9Kqr

12. About point O Ring is in pure rotation.

13.   W  U
U – Q  W

14. Can be understood by graph.

15. dQ  mST  0  T  0

16. Light from any part of body can not reach the eye of man.

g sin  gsin 
17. a , aring 
   2
1  2 
 mr 
2
acoin  gsin 
3

18. Frat  0
2
F.2R  F.R     = 2.5 Rad/s
 = t
1
K  2
2

1 1
19. mgh  mv 2  2
2 2
1 1 2
7mgR  mv 2   mv 2
2 2 5
mv 2  10mgR  v  10gR

20. Number of significant figures in the answer must be three.

T1
21. a 1
T2
r1 sin 1 1 P1 cos 1
 b  c 3
r2 sin 2 3 P2 cos 2

22. Escape velocity, v  2gR

vE 2gERE
   6  4  24
vM 2gMRM

23. Tension in the string at the lowest point in vertical circle is given by
mu2
TL  mg 

mu2 
2mg  mg 
 u P
u  g O
Using the law of conservation of energy, K.E. at O = P.E. at P
1
mu2  mg 1  cos  
2
1
mg  mg 1  cos  
2
1
cos      60
2

24. Along the rod

mg sin  – ma0 cos  = ma
 ma0

25. If the tension in the string is T then for insect

T  mg  ma
3mg
T
2
For block T  N  2mg

26. For the velocity perpendicular to incline after collision will be same for both so time for both will
remains.
For the velocity along the incline after collision, velocity of the ball on smooth surface will be
greater than that of rough surface.

27. mg(h+x) – F.x = 0

28. n1 sin  = n2 sin r = n2 sin (60 – )

n sin  2 sin  r
sin c = 2   n2
n1 sin  60 –   3 cos   sin  120
n1
 

29. For bead to be stationary

dy
slope of tangent on the curve  tan   
dx
2a
 .
y

30. For each reaction involving both the processes, mass defect
= 3 × 2.014 – 4.001 – 1.007 = 0.026 u
931.5  0.026
Energy released per deutron  MeV
3
Let n be the number of deutrons required
931.5  0.026 P  t 1016  365  24  60  60
Total energy   n MeV  
3  0.5
n  5  1035.

Chemistry PART – II

SECTION – A

1. BaF2 Hsolution = LE + Hhydratio

= 2367 – (457 2 + 1305)
= 148
BeF2 Hsolution = 2906 – (457  2 + 2494)
= –502
For solubility : Hsolution should be negative.

3. At constant density, P  T
T1
T2
P T3

D
so, T1 > T2 > T3

5. Br Br

Br SO Friedel Craft alkylation

 2
FeBr

 3
Br    
2

Br SO3H SO3H

H

H O

2

6.
C N H3C C NMgBr H3C C O

3 CH MgBr H O

 3

OCH 3 OCH 3 OCH 3

9. O N
19, 5 N C C C N 8, 8
C N

14. CH3 CH3 CH3 CH3 CH3 CH3

H3C CH3 
NaNO2
 H3C CH3  H3C CH3 
CH3  Shift

HCl
OH NH2 OH N 2 Cl OH

H
O C O C

H
C C C  C C C

C C

h
15. 4HNO3 
 4NO2  O2  2H2O
Yellow

W  mass 
205.5
20. V   260.13 ml
d 0.79
W  1000 5.5  1000
M B   0.58 M
mB  V 36.5  260.13

22. According to Faraday’s 1st law:

QE
W
F
E
Slope 
F

27. 2XeF6  SiO2  2XeOF4  SiF4

29. Hoffmann-degradation and Haloform reaction.

Mathematics PART – III

SECTION – A
2
1. f(x) + 2f(1 – x) = 1 – x ….. (1)
 2f(x) + f(1 – x) = 1 – (1 – x)2 ….. (2)
 x  2 2
 f  x   1 {on solving (1) and (2)}
3
 Range of f(x) is (–, 1]
6!
2. Required number of ways =  20
3! 3!

2 sin2º 6 sin6º 10 sin10º .....  178 sin178º

3. Average =
45
180 sin2º 180 sin 6º .....  90 sin90º
=
45
 sin 44º sin 44º 
180    90
 sin2º 
=
45
 1  cos88º 
= 2  1
 sin2º 
= 2(cosec 2º – 1 + 1)
 Average = 2 cosec 2º
4. Interior region is closed figure OABC A
Integral co-ordinates will be corresponding to
x = 1, x = 2 and x = 3 only
Integral co-ordinates are (1, –1), (1, 0), (1, 1) B
(2, –2), (2, –1), (2, 0), (2, 1), (2, 2), (3, –2), (3, –1) O 1 2 3
(3, 0), (3, 1), (3, 2)
 Total such points are 13
C

 3  1  2  1   2   1  0  3   3 439
5. Projection = 
2 2 2 14
2 1  3
2
 Projection =
14
6. n = 26 = 64
m = 0 + 1 + 2 + 3 + 4 + 5 + 6 = 21
 n – m = 64 – 21 = 43
7. cos 9x cos 3x = cos 18x cos 12x
 cos 12x + cos 6x = cos 30x + cos 6x
 cos 30x = cos 12x
 30x = 2n  12x
 18x = 2n, 42x = 2n
n n
 x ,x 
9 21
Number of solutions = 10 + 22 – 4 = 28

 2
( 0, , ,  will be common in both)
3 3
2
8. Let  and  be the roots of the equation x + px + q = 0
  +  = –p,  = q
 3 + 3 = –m, ()3 = n
3 3 3
 –m =  +  = ( + ) – 3( + )
3
 –m = –p + 3pq
3
 m = p – 3pq

x2 y 2
9. Hyperbola will be of the form  1
a2 b2
2ae = 20
b 4
 ae = 10, 
a 3
 b = 4k, a = 3k
 b2 = a2(e2 – 1)
 b2 + a2 = a2e2
k=2
 b = 8, a = 6
x 2 y2
  1
36 64

10. Put x = tan 

1 1 1
11. 1 4 1  1 8  1  1 3   11  4  = 9 – 3 – 3 = 3
1 1 2

12. |x – 1| + |x – 10| = 9 – {x}

As LHS  9 and RHS  9
 LHS = RHS = 9
 {x} = 0  x  [1, 10]
 10 solutions

13. Let x1 and x2 appears on the dice

(x1, x2)  {(1, 3), (2, 4), (3, 5), (4, 6), (3, 1), (4, 2), (5, 3), (6, 4)}
8 2
 Probability = 2 
6 9

14. The four sides of quadrilateral are given as x = 2, x = 6, y = 2, y = 6, which intersect at the points
(2, 6), (2, 2), (6, 2), (6, 6)
The line which divides the area of this square in two equal parts will pass through its centre i.e.
(4, 4)
 y  4  3  x  4
 y  3  x  4  4

16. Let x1, x2, ……, xn be n observations, then

1
X   xi
2
x
Let yi = i  10 , then

n
1 11  1 1 X  10
Y =  yi   xi   (10n)  Y  X  10  .
n i1  n  n  
17. w13 = 1
2 3 4 5 6 –6 –5 –4 –3 –2 –1
Now, a + b = w + w + w + w + w + w + w + w + w + w + w + w
2 6 7 8 12
= (w + w + ..... + w ) + (w + w ..... w ) = –1
3 4 –4 –3 –1 2 5 6 –6 –5 –2
ab = (w + w + w + w + w + w )(w + w + w + w + w + w )
2 3 12
= 3(w + w + w + ….. + w ) = –3
 Quadratic equation with a and b as roots
= x2 – (a + b)x + ab = 0
 x2 + x – 3 = 0
18. Let M be  
18 cos , 8 sin  . As shortest distance lies along common normal
 Given line will be parallel to the tangent at M
x cos  y sin 
   1 is parallel to 2x – 3y + 25 = 0
18 8
8 cos  2
    tan  = –1
18 sin  3
3
   point M is (–3, 2)
4
20. As b + c = 3a
sinB  sinC
 3
sin A
BC BC
2 sin   cos  
 2   2  3

A A
2 sin cos
2 2
B C B C
cos cos  sin sin
 2 2 2 2 3
B C B C
cos cos  sin sin
2 2 2 2
B C
cot cot  1
 2 2 3
B C
cot cot  1
2 2
B C
 cot cot  2
2 2
B C 1
 tan tan 
2 2 2
 B C B C  1
 cot cot  tan tan   2    2
 2 2 2 2  2

21. cos–1(cos x) = x,  x  (0, ]

 sin–1(sin x) = 0  x =  is the solution

22. f (x) is increasing function

So, x3 + x2 + 2x  1 > 2x + 1
3 2
x +x 2>0
(x  1) (x2 + 2x + 2) > 0.

23. Equation of normal to (x – 3)2 = 4(y + 1) at (2t + 3, t2 – 1) is (x – 3) = –t(y + 1) + 2t + t3

As these normals pass through (h, k)
3
 t + (1 – k)t + (3 – h) = 0
 t1 + t2 + t3 = 0, 2t1 + 3 = 7, 2t2 + 3 = –1
 t1 = 2, t2 = –2
 t3 = 0
 Point is (3, –1)

24. 2 + 2 + 2 = 28
 ,   can (2, 3, 4), (2, 4, 3), (3, 2, 4), (3, 4, 2), (4, 2, 3) and (4, 3, 2)

x
5 30 O  C
25. We have tan = and tan2 =  5m
x x
B
30
 tan2 =
5 cot 
 tan2 = 6tan  3 – 3tan2 = 1 30m
25m
2 3
 tan =  x = 5cot = 5 m
3 2
A

9
1 9
1   2r  3    2r  1 
26.   2r  1 2r  3   2   2r  1 2r  3  

r 0 r 0  
9
1 1 1  1 1  10
      2 1  21  21
r 0 2  2r  1 2r  3   

dy dy
27. x2  2xy  2  6y 0
dx dx
d 2 dy

dx
 
x y  2  6y
dx
0
2 2
 yx + 2x + 3y = c


28. A B 
4
 cot (A + B) = 1
cot A cotB  1
 1
cot A  cot B
 (1 – cot A)(1 – cot B) = 2

3  3
 m 1     m
29. tan   4   4 x+y=1
 3  3  
1     m 1   1   
 4  4 A 3x + 4y = 5
1

4m  3
  4
3m  4 7
4
 7(4m + 3) = (4 – 3m)
17
 31m = –17, m  
31
A = (–1, 2)
 Equation of line (L1) is 17x + 31y = 45

30. m(m + 1)(m + 2) is product of three consecutive numbers, hence divisible by 6 and (7m – 1) is
divisible by 6. Hence, given number is always divisible by 6

Paper 1

Time Allotted: 3 Hours Maximum Marks: 234

 Pl ea s e r ea d t h e i n s t r u c t i o n s c a r ef u ll y . Yo u a r e a l l o t t ed 5 m i n u t es
s p ec i f i c a ll y f o r t h i s p u r p o s e.
ALL INDIA TEST SERIES

 Yo u a r e n o t a l l o wed t o l ea v e t h e E xa m i n at i o n Ha l l b ef o r e t h e en d o f
t h e t es t .

INSTRUCTIONS

B. Filling of OMR Sheet

C. Marking Scheme For All Three Parts.

Name of the Candidate

Enrolment No.

Useful Data

PHYSICS
2
Acceleration due to gravity g = 10 m/s

Planck constant h = 6.6 1034 J-s

Charge of electron e = 1.6  1019 C

Mass of electron me = 9.1  1031 kg

2 2
Permittivity of free space 0 = 8.85  1012 C /N-m

Density of water water = 103 kg/m3

Atmospheric pressure Pa = 105 N/m2

Gas constant R = 8.314 J K1 mol1

CHEMISTRY

Gas Constant R = 8.314 J K1 mol1

Atomic No: H=1, He = 2, Li=3, Be=4, B=5, C=6, N=7, O=8,

Physics PART – I

SECTION – A
Straight Objective Type
This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D)
out of which ONLY ONE option is correct.

1. Two blocks of mass 3 kg and 6 kg respectively are 2 m/s

placed on a smooth horizontal surface. They are 1 m/s
connected by a light spring of force constant k=200
3 kg 6 kg
N/m. Initially the spring is unstretched. The
indicated velocities are imparted to the blocks. The
maximum extension of the spring will be
(A) 30 cm (B) 25 cm
(C) 20 cm (D) 15 cm

2. Two identical rods of mass M and length  are hinged together as M B

shown in the figure and lies at rest on smooth horizontal plane. A force 
F acts on end A of rod OA. The magnitude of instantaneous
acceleration of hinge O is O
F 2F
(A) (B) 
M M
F
(C) (D) None of these A
2M F M

3. Transverse wave are set in a thin rod clamped at one end O. y

The wave velocity is v. The rod vibrates in the 4th overtone
mode with amplitude y0. The equation of motion of the point at
x = 2/3 is O  X
5 5vt
(A) y  y o sin cos
3 2
7 7vt
(B) y  y o sin cos
3 2
10 9 vt
(C) y  y 0 sin cos
3 2
(D) y = 0

Space for Rough work

4. A rigid box contain one mole of monatomic ideal gas. The walls of the box have total surface area
A, thickness b and thermal conductivity K. Initially the gas is at temperature T 0 and pressure P0.
The temperature of surroundings is ½ T0
2KAt
T   
(A) The temperature as a function of time is T  0 1  e bR 
4  
2KAt
 T0
 
(B) The temperature as a function of time is T  1  e 3bR 
 4 
2KAt
P   
(C) The pressure of gas as a function of time is P  0 1  e 3bR 
4  
(D) None of these

5. A smooth uniform straight rod of mass m 1 and length L is m1, L

placed on a smooth horizontal table and hinged at a distance
‘a’ from centre. A particle of mass m 2 moving horizontally with
speed u hits the rod at angle  from normal. Coefficient of a
restitution between rod and particle is e. The angular velocity
of rod just after collision. b

u

m2 e
m2 (a  b)(e  1)ucos  m1 (a  b)(e  1)ucos 
(A) (B)
  L2     L2  
m1   a2   m2 (a  b)2  m 2   a2   m2 (a  b)2 
  12     12  
(C)  = 0 (D) None of these

6. In an LC circuit shown in the figure, C = 1F and L = 4H, at t = 0 charge + –

on capacitor is 4C and it is decreasing at a rate of 5 C/sec. If
C
maximum charge and current in the circuit are qmax and imax then the
value of imax/qmax is L
(A) 2s–1 (B) 0.5 s–1
2 –1
(C) s (D) none of these
3

Space for Rough work

7. A particle of mass ‘m’ is dropped on a triangular wedge of same m

mass. Immediately after collision velocity of particle is in
horizontal direction. If the surfaces are smooth. The maximum
possible value of  so that the above mentioned phenomenon m
takes place is 

 1   1 
(A)   tan1   (B)   tan1  
 2  2
(C)   45  (D)   30 

8. In an R-L-C circuit v = 20 sin (314 t + 5/6) and i = 10 sin (314 t + 2/3). The power factor of the
circuit is
(A) 0.5 (B) 0.966
(C) 0.866 (D) 1

Multiple Correct Choice Type

This section contains 4 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for its
answer, out which ONE OR MORE is/are correct.

 
9. Bichromatic light of wavelength 1  5000 A and  2  7000 A are used in YDSE. Then
(A) 14th order maxima of 1 will coincide with 10th order maxima of 2
(B) 21st order maxima of 2 will coincide with 15th order maxima of 1
(C) 11th order minima of 1 will coincide with 8th order minima of 2
(D) 3rd order minima of 1 will coincide with 4th order minima of 2

10. P, Q and R are three particles of a medium which lie on the x-axis. A sine wave of wave length 
is traveling through the medium in the x-direction. P and Q always have the same speed, while P
and R always have the same velocity. The minimum distance between
(A) P and Q is /2 (B) P and Q is 
(C) P and R is /2 (D) P and R is 

Space for Rough work

11. Which of the following statements is/are true?

(A) In vacuum the speed of red color is more than that of violet color
(B) An object in front of a mirror is moved towards the pole of a spherical mirror from infinity, it is
found that image also moves towards the pole. The mirror must be convex.
(C) There exist two angles of incidence in a prism for which angles of deviation are same except
minima deviation.
(D) A ray travels from a rarer medium to denser medium. There exist three angles of incidence
for which the deviation is same.
-3 3
12. Ten gm of oxygen at STP is expanded to a volume of 14x10 m . Then
(A) The final pressure is 0.5 atm in case of isothermal process
(B) The final pressure is 1/27/5 atm in case of adiabatic process
(C) The final temperature is greater than 273K in case of adiabatic process
(D) The final temperature is 273K in case of isothermal process

Comprehension Type

This section contains 3 paragraphs, each describing theory, experiments, data etc. Six questions relate
to the three paragraphs with two questions on each paragraph. Each question has only one correct
answer among the four given options (A), (B), (C) and (D).

Paragraph for Questions 13 & 14


When a surface is irradiated with light of wavelength 4950 A, a photocurrent appears which vanishes if a
retarding potential greater than 0.6 volts is applied across the phototube. When a second source of light
is used, it is found that the retarding potential is changed to 1.1 volt.

13. The work function of the emitting surface is:

(A) 2.2 eV (B) 1.5 eV
(C) 1.9 eV (D) 1.1 eV

14. The wavelength of the second source is:

 
(A) 6150 A (B) 5150 A
 
(C) 4125 A (D) 4500 A

Space for Rough work

Paragraph for Questions 15 & 16

The atomic masses of the hydrogen isotopes are:
1
Hydrogen 1H = 1.007825 amu
Deuterium 1H2 = 2.014102 amu
Tritium 1H3 = 3.016049 amu

15. The number of fusion reactions required to generate 1 kWh is nearly

(A) 108 (B) 1018
28 38
(C) 10 (D) 10

16. The mass of deuterium, 1H2 that would be needed to generate 1 kWh
(A) 3.7 kg (B) 3.7 g
(C) 3.7 × 10–5 kg (D) 3.7 × 10–8 kg

Paragraph for Questions 17 & 18

A man of mass 50 kg is running on a plank of mass 150 kg
with speed of 8 m/s relative to plank as shown in the figure.
(both were initially at rest and the velocity of man with respect
to ground any how remains constant). Plank is placed on D
smooth horizontal surface. The man, while running whistles =0
with frequency f 0. A detector (D) is placed on plank detects
frequency. The man jumps off with same velocity (with respect
to ground) from point d and slides on the smooth horizontal
surface [Assume coefficient of friction between man and
horizontal is zero]. The speed of sound in still medium is
330 m/s. Answer following questions on the basis of above
situations.

17. The frequency of sound detected by detector D, before man jumps off the plank is:
332 330
(A) f0 (B) f0
324 322
328 330
(C) f0 (D) f0
336 338

18. The frequency of sound detected by detector D, after man jumps off the plank is:
332 330
(A) f0 (B) f0
324 322
328 330
(C) f0 (D) f0
336 338
Space for Rough work

SECTION – C

Integer Answer Type

This section contains 5 questions. Each question, when worked out will result in one integer from 0 to 9
(both inclusive).

1. To a man walking at the rate of 3 km/h the rain appears to fall vertically. When he increases his
speed to 6 km/h
 it appears
 to meet him at an angle of 45 with vertical. The speed of rain is
n 2 km / hr. VRM  VR  VM Find the value of n.

2. A wedge of mass m3 is placed on a smooth horizontal m2

surface. Blocks of mass m 1 and m2 are placed on the m1
smooth inclined faces of m3. Angles of inclination are as m3
shown. It is found that m 3 does not move. Find ratio of 45 15
m2/m1.

3. A flat car of mass 50 kg starts moving to the right

due to a constant force 100 N as shown in figure. S
and spills on the flat car from a stationary hopper.
The velocity of loading is constant and equal to 1 100 N
kg/s. The velocity of the car after 10 seconds. Is
100/n. Find n.

4. For the circuit arrangement shown in the figure, find 60 

the potential difference (in V) across C2 in the steady
30 
state condition.
20 
C1 = 20F 30  C2 = 10F

10  +– K
12 V

5. A charged particle is projected in a magnetic field B = (3i + 4j)  10–2 T. The acceleration of the
 8
particle is found to be a  (  ˆi  yjˆ ) m/s2. Find the value of y.
3

Space for Rough work

Chemistry PART – II

SECTION – A

Straight Objective Type

This section contains 8 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for
its answer, out which ONLY ONE option is correct.

1. Which of the following is true for hydroxyl amine?

(A) More basic than ammonia.
(B) Disproportionates slowly in acidic solution producing N2O and NH4 .
(C) Disproportionates rapidly in acidic solution producing N2 and NH3.
(D) Reacts with 2-methyl cyclohexanone to give cyclic amide of caproic acid.

2. The molecule of P4S9 contains:

(A) 4 p - d phosphorus – sulphur bonds.
(B) P – S – P linkages.
(C) Three phosphorus are tetrahedral unit and one phosphorus as pyramidal unit.
(D) Has 4 P = S bonds.

3. How many moles of HSCN will react with iodine, so that the produced I will react with neutral
10 ml of 0.2 M KMnO4?
(A) 0.16 × 10-4 (B) 1.6 × 10-4
-4
(C) 1.2 × 10 (D) 0.12 × 10-5

Space for Rough work

4. O CH3

H /H O
  A  & B 
2

C6H5CHO / OH

C 
N H /OH
2 4
D 
Major

The structure of compound (C) and has many enantiomeric pairs are possible for (D)?

(A) H5C6 OH (B) O

,4
O OH
,2
C6H5
(C) H5C6 OH (D) C6H5
OCH3
OH ,0
,1

5. CoCl2 gives a buff coloured precipitate with KCN, but with excess of KCN, the precipitate
dissolves due to the formation of:
(A) Co(CN)2 (B) K4[Co(CN)6]
(C) K3[Co(CN)6] (D) K4[Co(CN)4Cl2

6. The emf of the following cell is – 0.46 V.

Pt(H2), HSO3  0.4 M ,SO 32  6.4  10 3 M  || Zn 2 0.3M  | Zn . If EoZn2 /Zn  0.76 V, calculate pKa of
HSO3 , i.e. for equilibrium

HSO3   2
 H  SO3
(A) 7.13 (B) 6.13
(C) 8 (D) 5.13

Space for Rough work

7. O
C
MeOH PCl3 MeNH2
O   A     B    C
C
O
The product C is:

(A) O (B) O

NH2 NH2
Cl ONH2

O O
(C) O (D) O

NMe NH

O O

8. The ratio of masses of photons corresponding to transition with shortest wavelength of Lymann
series of H-atom and lowest frequency of Paschen series of He  ion are:
(A) 7 : 36 (B) 5 : 9
(C) 9 : 5 (D) 36 : 7

Space for Rough work

Multiple Correct Choice Type

This section contains 4 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for its
answer, out which ONE OR MORE is/are correct.

9. Which of the following statements are not correct?

(1) From a mixed precipitate of AgCl and AgI, ammonia solution only dissolves AgCl.

(2) B  OH3  NaOH  
 NaBO2  Na B OH 4   H2O can be made to proceed forward by
addition of trans-diol.
(3) On passing H2S gas in II group sometimes the solution turns milky indicating the presence of
reducing agent.
(4) CuSO4 decolourises on addition of excess KCN, due to the formation of [Cu(CN)4]3-.

(A) 1 is correct (B) 2 is incorrect

10. The reaction given below shows the enthalpy change for the chemical reactants between N2, O2,
NO and NO2 (taking the given data into account).
N2  g  O2  g  2NO  g  H  110 kJ / mol
2NO  g  O2  g  2NO2  g H  200 kJ / mol

Select the correct statements:

(A) The standard heat of formation of NO2 is 90 kJ mol-1.
(B) The oxidation reaction from N2 to NO2 is exothermic.
(C) The standard heat of formation of NO is – 55 kJ mol-1.
(D) The oxidation reaction of NO to NO2 is endothermic.

11. Which of the following is correct order of basic strength:

(A) BuNH2  Bu2NH  Bu3N, order of basic strength having chlorobenzene as solvent.
(B) m  MeOC6H4NH2  o  MeOC6H4NH3  p  MeOC6H5NH2 order of basic strength in gaseous
phase.
N
(C) N > NH2 > N

Pyrimidine

(D) NH
> (CH3CH2)2NH
Pyrrolidine

12. Which of the following are correct statements:

(A) Entropy of system increases when an ideal gas is expanded isothermally.
(B) Gas with higher  has high magnitude of slope in a P(y-axis) v/s V(x – axis) curve.
(C) Final temperature in reversible adiabatic expansion is lesser than in irreversible adiabatic
expansion.
(D) Vapourisation at boiling point, G  0.

Space for Rough work

Comprehension Type

This section contains 3 paragraphs each describing theory, experiment, data etc. Six questions relate to
three paragraphs with two questions on each paragraph. Each question of a paragraph has only one
correct answers among the four choices (A), (B), (C) and (D).

Paragraph for Question Nos. 13 to 14

Chlorine on reaction with compound (X) in basic medium, according to the given equation:
IO3  xOH  Cl2   A   yH2O  zCl
o o
100 C 200 C
Compound (A) on heating in acidic medium  A  
4H2O B   C

13. Identify the compound (C)?

(A) I2O3 (B) I2
(C) I2O7 (D) I2O5

14. The stoichiometric coefficient in the above (x), (y), (z) respectively are:
(A) 6, 3, 2 (B) 5, 4, 1
(C) 3, 4, 2 (D) 6, 2, 3

Space for Rough work

Paragraph for Question Nos. 15 to 16

O

i LiAlH MeMgI H 

2
4
 ii H /H O  A  
H /H O
  B  
2
C
O

15. The structure of compound (A) and (B) respectively are:

(A) OH
HO
, OH
OH

O
(B) O
OH
,
CH3

OH
OH
(C) O

,
O OH
OH
(D) O
OH
,
O
OH

16. Which of the following statement is correct for compound (C)?

(A)

O
 C  is and is optically active.

(B)

O
 C  is and is optically inactive.

Paragraph for Question Nos. 17 to 18

The vapour density of equilibrium mixture for



N2O 4  g  
 2NO 2  g  is 28.75 at 1 atm.
This mixture is allowed to diffuse out pressure for 10 seconds.

17. Calculate the respective mole fraction of the two gases in N2O4 and NO2 in the effusing out
mixture?
(A) XN2O4  0.188, XNO2  0.812 (B) XN2O4  0.812, XN2O4  0.188
(C) XN2O4  0.212, XNO2  0.788 (D) XNO2  0.788, XNO2  0.212

18. Calculate new KP of reaction?

(A) 1.875 (B) 2.2
(C) 1.42 (D) 2.8

Space for Rough work

SECTION – C

Integer Answer Type

This section contains 5 questions. Each question, when worked out will result in one integer from 0 to 9
(both inclusive).

1. How many halogen atoms are bridging in As2Cl82 ?

CrO  C H N 

OH
3 5 5

CH2Cl2
2
B 

ArO O

O
The number of stereoisomer possible for the compound (B) are:

3. The total number of tetrahedral units in the monomers of Buna-N?

4. While studying the decomposition of gaseous compound it is observed that a plot of logarithm of
its partial pressure versus time is linear. What is the order of decomposition?

6x
5. In the sodium chloride structure each Na+ is surrounded by next nearest Na+ ions. What is
2
the value of x?

Space for Rough work

Mathematics PART – III

SECTION – A

Straight Objective Type

This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D)
out of which ONLY ONE option is correct.

1. If [x3 + x2 + x + 1] = [x 3 + x2 + 1] + x, (where [.] denote the greatest integer function), then number
of solutions of the equation log |[x]| = 2 – |[x]| is
(A) 1 (B) 0
(C) 3 (D) 2

2. A is a set containing 7 elements and a subset A1 of A is chosen. Then set A is reconstructed by

replacing the elements of A1. Now another subset A2 of A is chosen. In this way 4 subsets A1, A2,
A3, A4 are formed. The number of ways of choosing A1, A2, A3, A4 so that these subsets are pair
wise disjoint, is
(A) 57 (B) 75
7
(C) 4 (D) 840

3. Let C0, C1, C2 ....., Cn be the binomial co-efficient in the expansion of (1 + x)n, n  N.
2C0 22 C1 23 C2 2n1Cn 1 1 S
If Sn    .....   , then the value of 7 is
2 6 12  n  1n  2  n  1 2 n  1 n  2  S6
7
(A) 7 (B)
6
7 7
(C) (D)
3 2

Space for rough work

z z
4. The number of complex numbers satisfying the conditions   1 , |z| = 1 and
z z
arg (z)  (0, 2) is
(A) 1 (B) 2
(C) 4 (D) 8

1  1 
5. tan  tan is equal to
2 2 4 4
 
(A) 4cot  cot  (B) cot  cot 
4 4
1  
(C) cot  cot  (D) cot  4 cot 
4 4 4

6. Equation of the parabola with latus rectum 2 2 units and axis as the line x – y = 0 and tangent at
the vertex as x + y = 1, is (origin lies outside parabola)
(A) x2 + y2 – 2xy + 4x + 4y + 4 = 0 (B) x2 + y2 – 2xy + 8x + 8y – 8 = 0
2 2
(C) x + y – 2xy – 4x – 4y + 4 = 0 (D) x2 + y2 – 2xy – 2x – 2y + 2 = 0

7. If the circles x2 + y2 + (3 + sin )x + (2 cos )y = 0 and x 2 + y2 + (2 cos )x + 2cy = 0 touch each
other then the maximum value of ‘c’ is
1
(A) (B) 1
2
3
(C) (D) 2
2

8. If f be a continuous function on [0, 1], differentiable in (0, 1) such that f(1) = 0, then exists some
c  (0, 1] such that
(A) cf(c) – f(c) = 0 (B) f(c) + cf(c) = 0
(C) f(c) – cf(c) = 0 (D) cf(c) + f(c) = 0

Space for rough work

Multiple Correct Choice Type

This section contains 4 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for its
answer, out which ONE OR MORE is/are correct

5x  1 if x  2
 x
9. Let f(x) = 
  5  1  t  dt if x  2

 0
then
(A) f(x) is continuous but not differentiable at x = 2
(B) f(x) is not continuous at x = 2
(C) f(x) is differentiable everywhere,  x  (2, )
(D) the right hand derivative of f(x) at x = 3 does not exist
       
10. Let a  ˆi  ˆj  kˆ , b  ˆi  ˆj  kˆ , a  c  b  a and a  c  1 then which of following is true
   8 2
(A)  a b c    (B)   
3 3

  7 1
(C)  a b c    (D)   
3 3

11. A right angle triangle ABC, right angle at A is inscribed in hyperbola xy = c2 (c > 0) such that
slope of BC is 2. If distance of point A from centre of xy = c2 is 10 , then which of the following
is/are correct for xy = c2
(A) the value of c is 2
(B) the value of c is 4
(C) the equation of normal at point A can be y = 2x  3 2
(D) the equation of normal at point A can be y = 3x + 8 2

12. For the equation |x2 – a| = cos–1 cos x to be inconsistent, the value of ‘a’ can be
1 1
(A) (B)
2 6
1 2
(C)  (D) 
2 3

Space for rough work

Comprehension Type

Paragraph for Question Nos. 13 to 14

Read the following write up carefully and answer the following questions:
x2 y 2
Consider three points A, B and C lying on 2  2  1 (a > b) such that origin is a point interior to the
a b
ABC. Lines through A, B, C are drawn perpendicular to major axis to cut circle x 2 + y2 = a2 at P, Q, R
respectively such that P, Q, R are on same side w.r.t major axis as that of A, B, C respectively

13. If PQR is equilateral then area of ABC is

3 3 3
(A) ab (B) ab
4 4
3 3
(C) ab (D) none of these
2

14. Let PQR is not necessarily equilateral, A and B lies in first quadrant such that slopes of OA and
OB are m 1 and m2 respectively, where O is the origin. If the area of sector OAB of ellipse is given
1
by ab then  is equal to
2
ab  m  m2  m  m2
(A) tan1 2 12 (B) tan1 1
b  a m1m2 1  m1m2
ab  m1  m2  ab  m1  m2 
(C) tan1 2 2
(D) tan1
a  b m1m2 a2  b2m1m2

Space for rough work

Paragraph for Question Nos. 15 to 16

Read the following write up carefully and answer the following questions:
Consider a ABC with AB = c, BC = a and CA = b. Internal and external bisectors of C intersect AB and
AB produced respectively at L and M, such that CL = CM. AD is a diameter of circum-circle of ABC. The
radius of circum-circle is R

15. The value of a2 + b2 is

(A) R2 (B) 2R2
(C) 3R2 (D) 4R2

16. Which of the following is incorrect?

(A) DC = a (B) BAC = CAD
 
(C) ADC = CBM (D) BAC =  , where  =  ACB
4 2

Paragraph for Question Nos. 17 to 18

Read the following write up carefully and answer the following questions:
 1
If the expression for the nth term of the infinite sequence 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, ….. is  n   ,
  
(where [.] denotes the greatest integer function) then

17. Let a = , b =  + 1 and c =  +  + 1 then number of numbers out of first 1000 natural numbers
which are divisible by a, b or c is
(A) 764 (B) 867
(C) 734 (D) none of these

18. Let a = , b =  + 1, c =  +  + 1 and d = 3 + 1. The number of divisors of the number

ac . bc . cb . db which are of the form 4n + 1, n  N is equal to
(A) 24 (B) 48
(C) 96 (D) none of these

Space for rough work

SECTION – C

Integer Answer Type

This section contains 5 questions. Each question, when worked out will result in one integer from 0 to 9
(both inclusive).

3
1. Let f : R  R defined as f(x) = x + x + 1, 1  x  2. The graph of y = g(x) is the reflection of graph
of y = f(x) through the line y = x. If the domain of g(x) is [a, b], then |a – b| is _____

2. In a paper of mathematics there are 5 questions, such that the sum of the marks is 30 and the
marks for any question is not less than 2 and not more than 8 (only integer marks are awarded). If
the number of ways in which the marks can be awarded is ABC, where A, B, C  N and A, B,
C  9. Then value of A + B – C is _____

2
3. If f: [0, 2]  R, f  x   x 3  2  2 x 3  1  x 3  10  6 x 3  1 then  f  x  dx is equal to _____
0

x y z
4. The plane    1 intersect x–axis, y–axis, z–axis at A, B, C respectively. If the distance
1 2 3
between origin and orthocentre of ABC is equal to k then the value of 7k is equal to _____

 4 sin2 x cos x  cos x  sin x 

5. The value of lim   is equal to _____
x 
3 sin x  cos x 
4

Space for rough work

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ANSWERS, HINTS & SOLUTIONS

FULL TEST– V
(Paper-1)
Q. No. PHYSICS CHEMISTRY MATHEMATICS
ALL INDIA TEST SERIES

1. A B B

2. A C A

3. D B C

4. B A D

5. A B C

6. B A C

7. A C B

8. C A D

A, C B, C A, C
9.
A, D B, C A, D
10.
A, B, C, D A, C
11. B, C
A, B, D A, B, C C, D
12.

13. C D B

14. C A A

15. B C D

16. D A D

17. A A C

18. C B B

1. 3 2 8

2. 2 4 4

3. 6 0 4

4. 6 1 6

5. 2 4 1

Physics PART – I
SECTION – A

m1v1  m 2 v 2
1. VCM   1m / s
m1  m2
At maximum extension both blocks will move in same direction with VCM now use energy y
conservation.

 M 2
2.  F  N   a1
2 12 1
F – N = Ma
N = Ma1 N
 M 2 N
N  1 a
2 12
Acceleration of hinge O 
  F
= a
  a1  1
2 2
 Nx = – F/4  a = 5F/4M and  = 9F/2M
    F
 ao   a    ˆi   ˆi
 2  M

9
3. y  y0 sin kx cos t and L
4

d KA
4.   T1  T2   equation of heat flow through the box.
dt L

5. Just after collision.

a  b  v
e 
ucos  u sin
 L2 
m2  v  ucos   a  b   m1   a2   v
 12 

1 2 2
6. L   constant
2 2C

7. Using COM;
u
vcos = u sin
v
 2v sin  v
and e   2 tan2  
ucos 

 2tan2 < 1
 1 
 tan < 1/2    tan1  
 2

8. Power factor = VRMS RMS cos 

np p
9. For maxima y  for minima y   2n  1
d 2d
y
10. y  A sin  t  kx   A  cos  t  kx 
t
C
11. Vmedium 

12. Isothermal PV  cons tan t adiabatic PV r  constant

13-14. h    eV0

15-16. E  mc 2

C  V0
17-18. f   f0
C  VS
SECTION – C

1. 3 2 km/hr

2. The centre of mass of system will not move horizontally.

dM
3. F  URel
dt

4. In steady state the current in branch containing capacitor will become zero.
  
5. F  q E  V  B 

Chemistry PART – II

SECTION – A
1  1 3
   
1. NH2OH  H  N2 O  2NH4  2H  3H3O

2. S

P
S
S
S P S P S
S
S S
P
Has 3 p - d P – S bond
Has 3 P = S bonds
Has 3 tetrahedral units of P
Has 6 P – S – P linkages.

3. KI  2KMnO4  H2O  KIO3  2KOH  MnO2

(nf = 6) (nf = 3)
Moles of KI × 6 = 3 × 0.2 × 10 × 10-3
Moles of KI = 10-3 = 1 milli mole = x

S  C  N  I2  H  SO42  HCN  I  H2O
To produce 10-3 moles I ,

Moles of S CN  6  10 3  1

10 3
Moles S CN   1.6  10 4 moles
6
4. O CH3 OH

 H  CH3OH
H2 O

OH
 Tautomerises
H 5 C6 C H
O O
C6H5 CHO

OH


A N2H4 / OH

OH
H 5C 6 C H

D
2 enantiomeric pairs possible for (D).

 Co  CN2  2KCl
CoCl2  2KCN 
5.
Buff coloured
Co  CN 2  4KCN 
 K 4 Co  CN 6 

o
6. Half cell Reaction E
L.H.S. H2  2H  2e Eoxo  0.00 V
R.H.S. Zn2  2e 
 Zn Eored  0.76 V
Net  zn2  H2 
 Zn  2H , Eocell  0.76 V
2
H  0.0591
K   2  log K
 Zn  n

o 0.0591
Ecell  Ecell  log K
n
 2
0.0591 log10 H 
0.46  0.76 
2 0.3
 6
H   4.60  10 M
 

HSO3 
 H  SO32
0.4 M 4.6  10 6 6.4  10 3
H  SO23  4.6  10 6  6.4  103
Ka         7.3  108
HSO3 0.4
pKa = 7.13

7. O O O O O
C
MeOH OMe OMe
O 
 PCl
3
  MeNH OMe
2
  NMe
OH Cl
C NHMe

O O O O O

1 1 1 
8. Shortest  of Lyman series    1 for H-atom  RH   2    1  R H
1 1  

9. 

(2) B  OH3  NaOH 
 NaBO2  Na B OH 4   H2O can be made to proceed forward by
adding cis-diol to stabilization by chelation.
(3) The solution turning milky on passing H2S through group II solution indicates the presence of
oxidizing agent where H2S is oxidised to 8O2.

10. (A) The heat of formation for NO2 is calculated as – 45 kJ mol-1.

(B) For reaction N2  2O2   2NO2 , H  310 kJ / mole. i.e. it is exothermic.
1 1
(C) N2  O2  NO, Ho  55 kJ / mol
2 2
1
(D) NO  O2  NO2 , Ho  100 kJ / mol, it is exothermic.
2

11. (A) No H-bonding can occur in chlorobenzene, so +I effect increases acidic strength.
(B) Due to both steric and polar effect has no ortho effect is observed in case of o-MeOC6H4NH2.
MeOC6H5NH2.
p-MeOC6H4NH2 > o-MeOC6H4NH2 > m-MeOC6H4NH2
pKa = 5.24 (pKa = 4.45) (pKa = 4.2)
N
N > NH2 > N
(C)
pKa = 5.25 pKa = 4.63 pKa = 1.3

NH
> (CH3 CH2) 2NH
(D)
Pyrrolidine
(pKa = 11.27) (pKa = 10.98)

V2
12. (A) is correct, S sys  nRn , V 2 > V 1.
V1
 S sys   ve
(B) PV = constant, as  is high, slope is higher.
(D) At boiling point, the process is at equilibrium  G  0 .

14. x = 6, y = 3, z = 2
Solution for the Q. No. 13 & 14.
IO3  6OH  Cl2  IO65  3H2O  2Cl

 IO56 
H
H5IO6
o o
100 C 200 C
 2H5IO6 
4H2O
2HIO 4  I2O5 .

17. V. D. = 28.75
Dd
  0.6
n  1 d



 2NO2
N2O 4
At equilibrium : 1  0.6  1.2
 0.4
PN2O 4 0.4

PNO2 1.2
If moles of N2O4 escaping out is x, then for NO2 it is (1 – x)
 x  0.4 46
   x  0.188
 1  x  1.2 92
XN2O4 escaping out   0.188, XNO2  escaping out   0.812

2
 1.2 
PNO2  1.6 
18. KP    2.2
PN2O4  0.4 
 
 1.6 
 KP does not change.

SECTION – C

1. Cl Cl
Cl
As As Cl

Cl Cl Cl
Cl

H
CrO  C H N
3 5 5 2
  B 
OH CH2Cl2
 Collins reagent 
O
ArO O
ArO O
O
O
CN CN
copolymerisation
3. nCH2  CH  CH  CH2  nCH2  CH   CH2  CH  CH  CH2  CH2  CH
n

Mathematics PART – III

SECTION – A

1. If [x3 + x2 + x + 1] = [x 3 + x2 + 1] + x, then x is integer

 log |[x]| = 2 – |[x]| has same solutions as log |x| = 2 – |x|, x is integer
 No integral solution  0

2. Each element has 5 choices, as it may be in any one of A1, A2, A3, A4 or may not be in any one
 The required number of ways = 57

3. (1 + x)n = C0 + C1x + C2x2 ..... Cnxn

1  x n 1  1 C1x 2 C2 x3 C x n 1
  C0 x   ..... n
n1 2 3 n 1
n2
1  x  
x

1 C x 2 C x3
 0  1 .....
Cn xn2

n  1n  2  n  1 n  1n  2  2  1 3  2 n  1n  2 
3n 2  1 2 C  22 C1  23 Cn  2n2
Put x = 2, we get   0  .....
n  1n  2  n  1 2  1 23 n  1n  2 
C0  2 C1  22 Cn  2n1 1 1 3n 2
   .....    
2 23 n  1n  2  n  1 2  n  1n  2  2 n  1n  2 
3n 2
 Sn 
2  n  1n  2 
S7 3 9  2  7  8 7
  
S 6 2  8  9  38 3

4. It is given that |z| = 1

z = cos  + i sin  (let),   (0, 2)
z z
Now,   1 gives 2 |cos 2| = 1
z z
1 1
 cos2  , 
2 2
 5 7 11  2 4  5
  , , , , , , , i.e. 8 values
6 6 6 6 3 3 3 3

5. As cot  – tan  = 2 cot 2

 tan  = cot  – 2 cot 2
1  1  1 
 tan  tan  cot  cot 
2 2 4 4 4 4

6. Latus rectum, 4a  2 2
1
 a
2
 1 1
Vertex =  , 
 2 2
Axis = x – y = 0
As origin lies outside the parabola
 Focus = (1, 1), Directrix = x + y = 0

2
xy 2 2
 Equation of parabola is   x  1   y  1
2
 x2 + y2 – 2xy – 4x – 4y + 4 = 0

7. Tangent at (0, 0) will be same

(3 + sin B)x + (2 cos )y = 0 and 2 cos  x + 2cy = 0 are same
2cos2 
 c  cmax = 1 where sin  = –1 and  = 0
3  sin 

8. Consider g(x) = xf(x)

g is continuous on [0, 1] and differentiable on (0, 1)
 f(1) = 0  g(0) = 0 = g(1)
By Rolles’ Theorem g(c) = 0 some for c  (0, 1)
 cf(c) + f(c) = 0

9. At x = 2, RHD = 5 + |1 – x| = 5 + 1 = 6 whereas LHD = 5

11. Let the coordinates of point A are (ct, c/t)

So, the slope of normal at A will be t2.
And normal will be parallel to BC.
So, t will be  2  c =  2.

12. ‘a’ must be negative and x2 – a = x should have no solution

1
Now, D < 0  1 + 4a < 0  a   y=x
4
0

ABC b
13.  
PQR a
b 3 3 2 3 3
 ABC   a  ab
a 4 4

am1 am2
14. Slope of OP and OQ will be and respectively
b b
a
m1  m2  ab m1  m2 
If POQ =  than tan  = b 2  2
a b  a2m1m2
1  2 m1m2
b
a2 1
Now, area of sector OPQ of circle x 2 + y2 = a2 =   a2 
2 2
b1 2  1 ab  m  m2 
Area of sector OAB = a     ab where  = tan1 2 12
a  2  2 b  a m1m2

15.-16.  LCM = 90º and CL = CM


CLM = CML =  A
4 
Let ACB = 2 4 L D
 3  
 LAC =   and LCB =  
4 B C
4 4
By sine Rule in ABC 
    3  4
a2 + b2 = 4R 2  sin2      sin2   
 4   4  M
    
a2 + b2 = 4R 2  sin2      cos2      = 4R2
  4   4 
Let ADC =   ABC =  –   CBM = 

   
4
 
 CAD =     
2 4
 BAC = CAD  BC = CD

k  k  1 k k  1
17.-18. We want to have an = k if n
2 2
k  k  1 1 k  k  1 1
 n is an integer, this is equivalent to  n 
2 8 2 8
1 1
 k2  k   2n  k 2  k 
4 4
1 1 1
 k   2n  k   k  2n   k  1
2 2 2
 1
Hence, an   2n     = 2 = 
 2

17. Now, a = 2, b = 3, c = 5
Let A = number of numbers which are divisible by 2
B = number of numbers which are divisible by 3
C = number of numbers which are divisible by 5
Required number = A + B + C – A  B – B  C – C  A + A  B  C
 1000  1000  1000  1000  1000  1000  1000 
=          734
 2   3   5   6   15   10   30 

18. a = 2, b = 3, c = 5, d = 7
Hence, the given number is 25 · 35 · 53 · 73
 4n + 1 is odd number therefore the factor 2 will not occur in divisor. 3 and 7 are of 4n + 3 form,
odd powers of 3 and 7 will be of 4n + 3 form and even powers will be 4n + 1 form
5 is 4n + 1 form and any power of 5 will be of 4n + 1 form
 Number of divisors of 4n + 1 type
= number of terms in the product (1 + 32 + 34)(1 + 5 + 52 + 53)(1 + 72)
2 5 2 3 3
+ number of terms in the product (3 + 3 + 3 )(1 + 5 + 5 + 5 )(7 + 7 ) = 48

SECTION – C

1. As g(x) is inverse of f(x)

 Domain of g(x) is range of f(x)
 a = 3, b = 11

2. Required number of ways = co-efficient of x30 in (x2 + x3 + x4 + x5 + x6 + x7 + x8)5

= co-efficient of x in 
 x 2 1  x7 
30 
 
 1 x 
 
20 7 5 –5
= co-efficient of x in (1 – x ) (1 – x)
24 17 10
= C20 – 5 C13 + 10 C6 = 826
 A = 8, B = 2, C = 6
A+B–C=4

2 2
3.  
x3  1  1   x3  1  3  = x3  1  1  x3  1  3

For x  [0, 2] 1  x3  1  3
 f(x) = 2
2
  f  x  dx  4
0

4. Orthocentre is foot of perpendicular drawn from origin on the plane

Paper 2

Time Allotted: 3 Hours Maximum Marks: 234

 Pl ea s e r ea d t h e i n s t r u c t i o n s c a r ef u ll y . Yo u a r e a l l o t t ed 5 m i n u t es
s p ec i f i c a ll y f o r t h i s p u r p o s e.
ALL INDIA TEST SERIES

 Yo u a r e n o t a l l o wed t o l ea v e t h e E xa m i n at i o n Ha l l b ef o r e t h e en d o f
t h e t es t .

INSTRUCTIONS

B. Filling of OMR Sheet

C. Marking Scheme For All Three Parts.

Name of the Candidate

Enrolment No.

Useful Data

PHYSICS
Acceleration due to gravity g = 10 m/s2

Planck constant h = 6.6 1034 J-s

Charge of electron e = 1.6  1019 C

Mass of electron me = 9.1  1031 kg

Permittivity of free space 0 = 8.85  1012 C2/N-m2

Density of water water = 103 kg/m3

Atmospheric pressure Pa = 105 N/m2

Gas constant R = 8.314 J K1 mol1

CHEMISTRY

Gas Constant R = 8.314 J K1 mol1

Atomic No: H=1, He = 2, Li=3, Be=4, B=5, C=6, N=7, O=8,

Physics PART – I

SECTION – A
Straight Objective Type
This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D)
out of which ONLY ONE option is correct.

1 from
1. A particle of mass 1 kg and charge C is projected
infinity
3
towards a non-conducting fixed spherical shell having 0.5mm
same charge uniformly distributed on its surface. Find
minimum initial velocity of projection required by the 1mm
particle, if it just grazes the shell :
2 2
(A) m/s (B) 2 m/s
3 3
2
(C) m/s (D) None of these
3

2. High-Quality camera lenses are coated to prevent reflection. A lens has an optical index of
refraction 1.72 and a coating with an optical index of refraction of 1.31. For near-normal
incidence, the minimum thickness of the coating to prevent reflection for wavelength of 5.3 × 10–
7
m is:
(A) 0.75 µm (B) 0.2 µm
(C) 0.1 µm (D) 1.75 µm

3. In the diagram shown, all the straight segments have resistance A

‘R’. The equivalent resistance between the upper dot (A) and lower
dot (B) is
3R
(A) (B) R
8
R 2R B
(C) (D)
2 5

Space for Rough work

4. Two rods each of length 6L carry charge Q are placed Q, 6L P

as shown in figure. Find magnitude of electric field at
5L
point P, which is at a distance of 5L from centre of 60°
each rod.
5L
1 3Q
(A) . 2
64 0 L Q, 6L

5Q
(B)
40 L2
1 Q
(C) .
20 L2
3Q
(D)
50 L2

5. A particle moves along x-axis following the relation px 2  qv 2  r, where p, q, r are positive
constants. x is co-ordinate of the particle and v is the instantaneous speed. Find the time interval
between two consecutive instants, when particle is at rest.
q q
(A) 2 (B) 
p p
p p
(C)  (D) 2
q q

 charge 
6. A charged particle of specific charge   ‘’ is released from origin at t = 0 with velocity
 mass 
 
v  v 0 (iˆ  ˆj) in uniform magnetic field B  B0 ˆi. Then co-ordinates of the particle at

t will be :
B0 
 v 2 v0 v   v 
(A)  0 , , 0  (B)   0 , 0, 0 
 2B   B B0    2B  
 0 0

 2v 0 v 0   v  2v 
(C)  0, ,  (D)  0 , 0,  0 
 B0  2B0    B0  B0  

Space for Rough work

7. A resistor ‘R’ is connected across a source of emf ‘E’ and unknown internal resistance.
Power of resistor is maximum when internal resistance is
(A) R (B) < R
(C) > R (D) zero

8. A child loves to watch as you fill a transparent plastic bottle with shampoo. Horizontal
cross-sections of the bottle are circles with varying diameters. You pour in shampoo with
3
constant volume flow rate 16.5 cm /s. At what rate is its level in the bottle rising at a point where
diameter of the bottle is 6.30 cm. Calculate keeping significant figures in view.
(A) 0.063 cm/s (B) 0.63/cm/s
(C) 0.529 cm/s (D) 0.5291 cm/s

Multiple Correct Choice Type

This section contains 4 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for its
answer, out which ONE OR MORE is/are correct.

9. A boy in the elevator with open roof shoots a bullet in vertical upward direction from a height of
1.5 above the floor of the elevator. The initial speed of the bullet with respect to elevator is 15
m/s. The bullet strikes the floor after 2 seconds. Then (Assuming g = 10 m/s2)
(A) Lift is moving with constant speed
(B) Lift is moving with upward acceleration of 5.75 m/s2
(C) Lift is moving with downward acceleration of 5.75 m/s2
(D) Distance travelled by bullet during its flight can not be calculated from the given data.

Space for Rough work

10. An idealized diesel engine operates in a cycle known as air P

standard diesel cycle as shown. Fuel is sprayed into the
cylinder at the point of maximum compression, B. B C Adiabotic
Combustion occurs during expansion BC. Process
(A) Heat is absorbed during process BC.
D
(B) Heat is released during process DA.
(C) Heat is absorbed during process CD. A
V
 1  T  TA 
(D) Efficiency of the cycle is 1   D 
   TC  TB 

11. Consider a non-conducting ring of radius ‘a’ in x-y plane. The y

two semi-circular portions have linear charge densities + 
and –  as shown. Then : – – –
+– – –
+
 
(A) Dipole moment of the ring is  2 3 ˆi  2ˆj  a2
+
+ 30° –
–
–
(B) Dipole moment of the ring is   2iˆ  2 3 ˆj   a 2
+ + x
+ –
+ –

(C) Electric field at the center of the ring is
20 a
 3 ˆi  ˆj  +
+
+ ++ +–
–


(D) Electric field at the center of the ring is
2 0 a

ˆi  3 ˆj 
12. Two plates, each of one side area ‘A’ are kept at a Q1 Q2
separation ‘d’. They have charge Q1 and Q2 and

uniform electric field E exist in the region as 
shown. Then E
Q  Q2
(A) Charge on surface (1) is 1  A 0 E
2
Q  Q2
(B) Field inside the gap is E  1
2 A 0 1 2 3 4
(C) Field outside the gap is E d
(D) Field on either side of the plates is identical

Space for Rough work

Comprehension Type

This section contains 3 paragraphs, each describing theory, experiments, data etc. Six questions relate
to the three paragraphs with two questions on each paragraph. Each question has only one correct
answer among the four given options (A), (B), (C) and (D).

Paragraph for Questions 13 & 14

Two astronauts, each having a mass ‘M’ are connected by a rope of length ‘d’ having negligible mass.
They are isolated in space, orbiting their center of mass at speed ‘v’. Treat astronauts as particles. By
d
pulling the rope, one astronaut shortens the distance between them to ' '.
2

13. Calculate work done by the astronaut in shortening the rope :

(A) 2Mv 2 (B) 3Mv 2
(C) Mv 2 (D) zero

14. Physical quantity that will remain conserved during the process of pulling in rope.
(A) Speed of astronauts
(B) Energy of the system
(C) Angular speed of astronauts about COM
(D) Angular Momentum

Space for Rough work

Paragraph for Questions 15 & 16

Ad
A parallel plate capacitor has plate area ‘A’ and plate separation ‘d’. A dielectric medium of volume
2
and dielectric constant ‘K’ is available. To use it in the capacitor, two alternative methods are available as
shown :
d x

(2) 2x
(1)

x d
d 2
2 d
 A
15. Capacitance in case (1) is X1 0 . Then x1 is
d
K 2K
(A) (B)
3 k 1
3K  1 2K
(C) (D)
2K K 1

0 A
16. Capacitance is case (2) is x 2 . Then x2 is
d
K 1 K K 1 1
(A)  (B) 
4 K 1 4 K
2K K K K 1
(C)  (D) 
4 3 3 2

Space for Rough work

Paragraph for Questions 17 & 18

In the set-up shown ‘E’ is unknown emf with internal resistance ‘r’. AB is a E r
uniform potentiometer wire, 11m long and resistance 1 m1. When AJ =
1m, current through galvanometer is 2.4 A when AJ = 10m, current J
6 A B
through galvanometer is A, in the opposite direction as in previous
11
case. Galvanometer has negligible resistance.
E0=12V 1
r0=0 

17. The value of E is

(A) 4 V (B) 6 V
(C) 8 V (D) 10 V

18. The vale of r is

(A) 1 (B) 2
(C) 2.5  (D) 4 
Space for Rough work

SECTION – C
Integer Answer Type

This section contains 5 questions. Each question, when worked out will result in one integer from 0 to 9
(both inclusive).

 
1. Ball A is at rest somewhere on a smooth billiards table   3  . 1 2
b 
–1
An identical ball B moving with velocity 2 ms , collides with a
1 
elastically. After sec both the balls simultaneously fall into
2
pockets marked 1 and 2. Find area of the table in m2.
3 4
b

2. A hydrogen atom in excited state n reaches the ground state by emitting three photons of
frequency in ratio 5 : 72 : 243. Wavelength corresponding to one of the transitions belongs to
visible region. Find the value of ‘n’.

3. A and B are thin lenses. 1, 2 and 3 are representative A 2 B 3

wave fronts of incident and refracted monochromatic
light with distances from the lenses as shown. The
radii of curvature of wave fronts 2 and 3 are 4m and
7m respectively. Find the focal length (modulus value)
1m 1m
of lens B in meter.
1
2m

4. A particle is taken to a height K times radius of Earth, above the surface of earth and projected at
an angle of 30° to the vertical. During its subsequent motion, it just grazes the surface of earth
15
with a velocity times the escape velocity from the surface of earth. Find the value of K.
4
(K < 9).

5. ABCD is a semi-circular wire carrying current in a region of uniform magnetic field.

Arc AB = Arc BC = Arc CD. The magnitude of the force on the wire is 8 N, while magnitude of the
force on arc AB alone is 9 N. The lines of action of these two forces include an angle of
2
cos1   . Find the magnitude of the force (is N) on arc CD alone.
3

Space for Rough work

Chemistry PART – II

SECTION – A

Straight Objective Type

This section contains 08 multiple choice questions. Each question has four choices (A), (B), (C) and (D)
out of which ONE OR MORE is/are correct.

1. In a rock salt crystal and fluorite structure, if atoms along the plane cutting the two opposite edges
are removed, then the new ratio of the cations in two respective unit cells and ratio of anions in
respective unit cells are:
(A) 5 : 6 / 10 : 1 (B) 6 : 5 / 5 : 8
(C) 12 : 5 / 5 : 8 (D) 3 : 4 / 5 : 4

2. Which of the following is reducing sugar?

(A) CH2OH O
CH2OH O
H
HO
OH
OH O
HO OH H
HO
(B) CH2OH O

HO H

OH
HO O
CH2OH
O
OH
CH2OH
OH
(C) HOH2C HOH2C
O O
HO O OH

HO OH HO OH
(D) HOH2C O HO
OH
H OH
HO
H CH2OH
OH O
HO

Space for Rough work

3. In compound (A) with molecular formula C8H8O, which responds to iodoform test?
NaCN  i MeMgBr H
 A  
HOEt
 B  
ii H ,H O
2
  C   D 
Major product 
The structure of (D) is:

(A) O (B) O
H

4. When copper is reacted with dil. HNO3 a colourless gas is liberated. Which of the following is true
for the gas?
(A) Has 11 valence e’s, where extra electron occupies an antibonding  *2p orbital.
(B) An acid anhydride.
(C) Paramagnetic in solid state.
(D) Has no resonance.

5. What is the number of five membered ring in the molecule of S4N4?

(A) 4 (B) 2
(C) 1 (D) 5

6. Consider the following complexes:

(I) [Cr(CO)x)] (II) [Cr(CO)x – 1 PF3]
If PF3 is better  accepter than CO, what will be the order of bond length of CO in complexes (I)
and (II):
(A) I > II (B) II > I
(C) I = II (D) cannot be compared

7. A compound (A) form a unstable pale blue colour solution in water. The solution decolourised Br2
water and an acidified solution of KMnO4. The possible compound (A) is:
(A) HNO2 (B) HNO3
(C) N2O3 (D) Both (A) and (C)

8. The 100 ml of 0.1 M CH3COOH (Ka = 10-5) is titrated with 0.1 M NaOH and pH is observed at
three stages, at 298 K.
st
1 step  50 ml of 0.1 M NaOH added.
nd
2 step  100 ml of 0.1 M NaOH added.
rd
3 step  120 ml of 0.1 M NaOH added.
Calculate the difference in pH between stages (I & II) and (II & III) stages.
(A) (pHII – pHI) = 3.84 & (pHIII – pHII) = 4.11 (B) (pHI – pHII) = 2 & (pHIII – pHII) = 5.2
(C) (pHI – pHII) = 3.84 & (pHIII – pHII) = 3.11 (D) (pHI – pHII) = 4.2 & (pHIII – pHII) = 3.11

Space for Rough work

Multiple Correct Choice Type

This section contains 4 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for its
answer, out which ONE OR MORE is/are correct.

9. OH
CH N
P2O5
CH CH    X

A
Which of the following are correct for compound (X)?
(A) (X) is aromatic and a heterocyclic compound.
(B) Compound (X) is more basic than aniline.
(C) The reaction that occur is stereospecific.
(D) The reaction do not follow a concerted mechanism and (X) is non-aromatic.

10. Which of the following the first pair has less parameter in the property mentioned:
(A) CH3NH2 or (CH3)2NH2+ (Angle H – N – H)
(B) PCl2F3 or PCl3F2 (Angle F – P – F)
(C) NO+ or NO (Bond length)
(D) CrO3 or Cr2O3 (In acidic nature)

11. Which of the following statements are incorrect?

(A) 1, 1, 1-trideutero – 2-propanol reacts with conc. H2SO4 at high temperature to give only one
alkene 3,3,3- trideuteropropene.
(B) Attack of chlorine radical on alkene is endothermic and attack of radical formed on HCl is
exothermic.
(C) Rearrangement will occur in forming the major product in reaction.
H HCl
CH2  E

H3C
(D) Cis form of 1,3-Dimethyl cyclohexane is more stable than its trans form with respect to atleast
one group at equatorial position.

12. During discharging of lead storage battery, which of the following is/are true?
(A) H2SO4 is produced. (B) H2O is consumed.
(C) PbSO4 is formed at both electrodes. (D) Density of electrolytic solution increased.

Space for Rough work

Comprehension Type

This section contains 3 paragraphs, each describing theory, experiments, data etc. Six questions relate
to the three paragraphs with two questions on each paragraph. Each question has only one correct
answer among the four given options (A), (B), (C) and (D).

Paragraph for Question Nos. 13 to 14

A substance ‘X’ dissociates by two parallel reactions with different Arrhenius constant (A1 & A2) and

different activation energies Ea1 & Ea2  as shown:
14 1 14 2 1
A1  210 s A 2 210 e s
X  g  
Ea 12 kcal/mol
 Y  g ; X  g 
Ea 9 kcal/mol.
 Z g
1 2

13. Temperature at which net activation energy for dissociation of X will be 10.5 kcal/mol?
(A) 200 (B) 750
(C) 800 (D) 1000

14. Which of the following statements are correct with respect to major product obtained in the above
parallel decay?
(A) ‘Y’ will be the major product at temperature 300 K.
(B) ‘Z’ will be the major product at temperature 800 K.
(C) Both statements are correct.
(D) None of the statement are correct.

Space for Rough work

Paragraph for Question Nos. 15 to 16

MeOK in KMnO4 conc. HNO3


chloroform  A    B    C

15. Identify the structure of (A) & (B):

(A) Cl
COCl
A B

COOH
(B) Cl Cl
COOH
A B

COOH
(C) CHCl2 HOOC COOH
A B

OMe HOOC Ome

(D) HOOC Cl
A B
Cl HOOC

16. How many isomers of (C) are possible and the major product is:

(A) HOOC Cl (B) COOH

3, HOOC Cl

HOOC NO2
3,

NO 2
(C) COCl (D) NO 2
4, Cl COOH

O2 N COOH 2,

COOH

Space for Rough work

Paragraph for Question Nos. 17 to 18


K2Cr2O7 /H
A metal sulphide A  B  C   Green solution  D   burns in air to form E 
 White   Colourless   Colourless solu.  Coloured
anhyd. CuSO4
E  B   D  colourless liquid  Blue
aq. NH3 excess of
C 
or NaOH
 precipitate 
reagent
clear solution.

17. Calculate the weight of compound (A) required to produce (B), so that (B) will completely react
with 10 ml of 0.2 M K2Cr2O7?
(A) 650 mg (B) 288 mg
(C) 583 mg (D) 458 mg

18. Compound (C) on reaction with NH3 or NaOH forms a precipitate and with excess of reagent
forms a clear solution of compound (Z), (Z) is?
(A) NaAlO2 (B) Na2ZnO2
(C) Cu(OH)2 (D) [Ag(NH3)2]+

Space for Rough work

SECTION – C

Integer Answer Type

This section contains 5 questions. Each question, when worked out will result in one integer from 0 to 9
(both inclusive).

1. How many are the intensive properties?

Density, pressure, emf, free energy, specific heat, heat capacity, specific conductance,
resistance, pH?

2. The sum of chiral centre and number of lone pair in compound (B) are:
Me
conc. H2 So4 HCN
Me  
  A    B 
OH OH

3. One mole of an ideal monoatomic gas is expanded irreversibly in two stage isothermal
expansion:
Stage 1. (8 bar, 4.0 litre, 300 K)
Stage 2. (2 bar, V1 litre, 300 K)
Stage 3. (1 bar, V2 litre, 300 K)
Calculate heat absorbed by the gas in the process (in kJ)

4. The sum of all sp3, sp2 & sp units present in solid PCl5, solid Pbr5 and solid N2O5 are:

5. How many diastereomeric pair are possible for compound [Co(en)2Cl2]+?

Space for Rough work

Mathematics PART – III

SECTION – A

Straight Objective Type

This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D)
out of which ONLY ONE option is correct.

1. Six different letters are placed in boxes of the given figure such that no row
is empty. (In each box only one letter can be placed). The number of such
arrangements is
(A) 2920 (B) 6480
(C) 8640 (D) 10800

2. Suppose b and c are real numbers randomly chosen in the interval [0, 1]. What is the probability
2
that the distance in complex plane between the two roots of the equation z + bz + c = 0 is not
greater than 1?
1 2
(A) (B)
3 3
1 3
(C) (D)
4 4

3. The equation of line which intersect each of two lines 2x + y – 1 = 0 = x – 2y + 3z and

x y z
3x – y + z + 2 = 0, 4x + 5y – 2z – 3 = 0 and is parallel to   is
1 2 3
(A) 4x + 7y – 6z – 3 = 0, 2x – 7y + 4z + 7 = 0 (B) 4x + 7y – 6z – 4 = 0, 2x – 7y + 4z + 2 = 0
(C) 4x + 7y – 6z – 2 = 0, 2x – 7y + 4z + 5 = 0 (D) none of these

1 1 1 1
4. If three positive distinct real number p, q, r satisfy the condition     0 . If q = 2,
p r pq rq
then minimum value of pr is
(A) 1 (B) 2
(C) 3 (D) 4

Space for rough work


1
5. The value of  dx is equal to
0 1  x 1  x 
17 2

 
(A) (B)
2 3
 
(C) (D)
4 6

6. If a curve is such that line joining origin to any point P(x, y) on the curve and line parallel to
y-axis through P are equally inclined to tangent to curve at P, then the differential equation of the
curve is
2 2
 dy  dy  dy  dy
(A) x    2y x (B)    2y x
dx
  dx  dx  dx
2 2
 dy  dy  dy  dy
(C) y    2y x (D) y    2y y
 dx  dx  dx  dx

7. Let f (x) = x3 + 2x2  x + 1, then which of the following statement(s) is/are correct
   
(A) sin1 sin       cos1 cos       6   (where  is real root of f(x) = 0 and [.] denotes the
G.I.F.)
(B) f (x) = 0 has three distinct real roots
(C) y = f (x) is increasing function
   
(D) sin1 sin       cos1 cos       5  2
(where  is real root of f(x) = 0 and [.] denotes the G.I.F.)

8. Which of the following functions is an injective (one–one) function in their respective domain?
(A) f(x) = 2x + sin 3x (B) f(x) = x · [x], (where [.] denotes the G.I.F)
x
2 1 2x  1
(C) f(x) = x (D) f(x) = x
4 1 4 1

Space for rough work

Multiple Correct Choice Type

This section contains 4 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for its
answer, out which ONE OR MORE is/are correct

9. Let point A be (1, 1) and equation of BC is (x + y – 1) + (2x – y + 4) = 0 and area of ABC is

maximum, then slope of the line AB and AC, may be equal to
(Given AB = AC and BC = 2 units)
2  5 2 5
(A) (B)
2 5 1 2 5 1
2 5 2 5
(C) (D)
2 5 1 2 5 1

tan x tan 2x tan 3x

10. The value of x which satisfy the equation tan 2x tan 3x tan x  0 is
tan 3x tan x tan 2x
13 13
(A) (B)
12 3

(C) (D) none of these
3

2 2

11. The co-ordinate of foci of ellipse

x  y y  x
  1 are
4 1
 3 3  3 3
(A)  ,   (B)   ,  
 2 2   2 2 
 3 3  3 3
(C)  ,  (D)   , 
 2 2  2 2

12. For different values of k, the circle x 2 + y2 + (8 + k)x + (8 + k)y + (16 + 12k) = 0, always passes
through two fixed points P and Q. For k = k1, then tangents at P and Q intersect at the origin.
Which of the following is/are correct?
(A) the mid-point of P and Q is (–6, –6) (B) the sum of ordinates of P and Q is –12
32 8
(C) k1 may be equal to  (D) k1 may be equal to
9 3

Space for rough work

Comprehension Type

This section contains 3 paragraphs, each describing theory, experiments, data etc. Six questions relate
to the three paragraphs with two questions on each paragraph. Each question has only one correct
answer among the four given options (A), (B), (C) and (D).

Paragraph for Question Nos. 13 to 14

 x 2  10x  8, x  2

Let continuous function f (x) = px 2  qx  r, 2  x  0 , where p  0. If a line ‘L’ touches the graph of
 2
 x  2x, x0
y = f (x) at 3 points, then

13. The line ‘L’ touches which of the following curve

(A) x2 + y2 = 1 (B) x2 + y2  2x  4y  9 = 0
2 2
(C) (y + 2) =  48 (x + 1) (D) (y + 2) = 13 (x + 1)

14. The value of p + q + r is equal to

(A) 5 2 (B) 5
(C) 6 (D) 7

Paragraph for Question Nos. 15 to 16

Read the following write up carefully and answer the following questions:
Consider some special type of matrices. A square matrix is called idempotent matrix if A2 = A
A matrix A is called nilpotent matrix if Ak = 0, for some k  N
A square matrix is called involutary matrix if A2 = I
2 3 5 1 3 4 0 1 1
Now consider the following matrices A  1 4 5 , B  1 3 4 , C  4 3 4
1 3 4 1 3 4 3 3 4

15. Which one of the following is a nilpotent matrix?

(A) A (B) B
(C) C (D) AC2

16. Which one of the following is not an idempotent matrix?

3 2 2 2
(A) A C (B) A C
2 2
(C) BC (D) C A

Space for rough work

Paragraph for Question Nos. 17 to 18

Read the following write up carefully and answer the following questions:
2 2
x + y  2ax  2y  8 = 0, where a is variable.

17. The equation represents a family of circle passing through two fixed point whose coordinates are
(A) (0, 2), (0, 2) (B) (0, 2), (0, 4)
(C) (4, 0), (2, 0) (D) none of these

18. Equation of a circle C of this family, tangent to which at these fixed points intersects on the line
2x + y + 5 = 0 is
(A) x2 + y2  2x  8y  8 = 0 (B) x2 + y2  2x + 6y  8 = 0
2 2 2 2
(C) x + y  2x + 8y  8 = 0 (D) x + y  6x  2y  8 = 0

SECTION – C

Integer Answer Type

This section contains 5 questions. Each question, when worked out will result in one integer from 0 to 9
(both inclusive).

1. n balls are arranged to form an equilateral triangle such that first row consists of one ball, second
row consists of two balls, third row consists of three balls and so on. 49 more balls are added to
form a square. If each side of the square contains 3 balls more than that of the equilateral
triangle, then sum of digits of n is _____

2. Let O, A(z1), B(z2), C(z3) are vertices of rhombus such that |z1| = |z2| = 4 and |z3| = 6 if
 3
arg(z2 – z1) = then arg(z3) is then k is equal to _____
3 k

  /2
cos x sin 2x 1 
3. If  dx  P and I   dx , then the value of P + 2I  is equal to then
2   4  2x 4 12
0  x  4 0

 is _____

4. The value of tan 1º tan 2º + tan 2º tan 3º + ..... + tan 88º tan 89º is equal to cot 2 1  n, where n is
a two digit number ab, then the value of b  a is _________

1 1 1 1 1 1 1 1
5. Let , , , , , , , are vertices of regular
a1  2i a2  2i a3  2i a4  2i a5  2i a6  2i a7  2i a8  2i
octagon. If the area of octagon is A, then the value of 8 2A is
(where ai  R for i = 1, 2, 3, 4, 5, 6, 7, 8) _______

Space for rough work

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ANSWERS, HINTS & SOLUTIONS

FULL TEST– V
(Paper-2)
Q. No. PHYSICS CHEMISTRY MATHEMATICS
ALL INDIA TEST SERIES

1. B B D

2. C A A

3. C A A

4. A A D

5. B A C

6. D A A

7. D D A

8. C C D
B, D A, B, C A, C
9.
A, B, D A, B, C B, C
10.
A, C A, B, C A, D
11.
A, B C, D A, B
12.

13. B A C

14. D D D

15. D D B

16. A A C

17. C C B

18. B B D

1. 3 7 6

2. 9 4 9

3. 2 4 3

4. 5 4 1

5. 7 2 2

Physics PART – I

SECTION – A

1. Apply conservation of energy and conservation of angular momentum between points at infinite
separation and on the surface of the sphere.

2. To prevent reflection  2µt
2
 5.3  10 7
t   0.1µm.
4 4  1.31
3. All the corners of the hexagonal faces will be at same potential. So wires joining the corners can
be removed. Then in the remaining circuit we will have six ‘3R’ resistors connected in parallel
between A and B.

4. E due to rod. Q, 6L
6L E
1  Q dx 1 Q
= .  2
= x  5L 
0
4 0  6L   8L  x  64 0 L2
Enet = 2 E cos 30°. dx

5. px 2  qv 2  r
dx dv p
 2px  2qv 0  a x
dt dt q
p
 Particle is executing S.H.M. with angular frequency  .
q

6. Particle will move in helical path having circular projection in y-z plane. In the given time it will
complete half revolution.
So, x = v0t, y = 0, z = – 2R

2
 E 
7. P  .R
Rr 
P will be maximum when r = 0

dV D2 dh
8.  .
dt 4 dt
dh 4  16.5
  0.529 cm / s
dt 3.14  (6.30)2

1
9. 1.5  15T  10  a  T 2 . Where T = 2 sec  a = 5.75 m/s2
2
As we can not find velocity of bullet with respect to ground, we can not calculate distance from
the given data.
10. QBC  nCP T  0
QDA  nCV T  0
W Q nCV  TD  TA  1  T  TA 
  1  released  1   1  D 
Qabsorbed Qabsorbed nCP  TC  TB    TC  TB 

 2a 
11. Dipole moment will be angle of 30° to ‘–x’ – axis of magnitude   .a  .  2   4 a 2 .
  
2
E will be 2. at angle of 30° with ‘+x’ – axis.
40 a

12. For EP = 0
q Q q Q2
E  1 
2A 0 2 A 0 2A 0
q Q1 – q Q1 + Q2 – q
Q  Q2
 q 1  0 AE
2
P
Q  Q2
E inside the gap = E  1
2 A 0
Q1  Q2 – (Q1 – q)
E on the left side of plates = E 
2 A 0
Q1  Q2
E on the right side of plates = E 
2 A 0

13. Apply conservation of Angular momentum about COM, to calculate final speed of the astronauts.
Work =  K.E. = 3 Mv2

14. Torque about COM is zero hence angular momentum remains conserved

 K 0 A  0 A
 d/ 2  d/ 2
15. D=  
k 0 A 0 A

d/2 d/ 2
16. K 0 A
4d

0 A
K 0 A d
d
0 A
4d

17-18. First case : E r

12 = 12i + 2.4
 i = 0.8 A 2.4 A
E = i + 2.4 (r + 1)
 E = 3.2 + 2.4 r (i  2.4)A 1 

i
12 V 11 

Second case : E r
60
12 = 12i1 –
11 6
 6  A
1 16  i' 11 A  10  11
 i  A  
11
6 i’
E  10i' 10  r 
11 12 V 2 
100 6r
 E  ……..(2)
11 11
Solving (1) and (2) E = 8V, r = 2

SECTION – C

1. v 2A t 2  v B2 t 2  b2 b
when identical balls collide obliquely and collision is elastic and
one of them at rest, they move mutually perpendicular to each
90° v
other after collision. vA B
From energy conservation
1 1 1
mu2  mv 2A  mv B2 A U=2m/s
2 2 2
B
so b2 = u2 t2 = 1
b=1m
 = 3m
Area = b  = 3m2.

2. Enm : Em2 : E21  5 : 72 : 243

Em2  Transition corresponding to visible region.
 1 1 1 1
En2 : E21  77 : 243   2  2  :  2  2   77 : 243  n = 9.
2 n  1 2 

3. For lens B according to the figure

u=+3
v= –6
1 1 1
   f = – 2 m.
f v u

4. Conservation of angular momentum 30° v0

15
mv 0 sin 30 (k  1)Re  m v e Re
4 KRe
conservation of M.E.
GMe m 1 GMe m 1  15 2 
  mv 20    m  ve 
(k  1)Re 2 Re 2  16 
solving we get k = 5 or 9. Re

5. Given :
  B C
Floop  i 1  B  8N 
3
  
FAB  i 2  B  9N  4
 1 2
FBC  Floop
2 A  D
Force on CD 1
      

FCD = i 4  B = i 1  2  3  B 
     
= i 1  B  i 2  B  i 3  B
    
= Floop  FAB  Floop = Floop  FAB
2 2

2
FCD = 42  92  2  4  9  = 7 N.
3

Chemistry PART – II

SECTION – A

1. On removing atoms along the plane cutting two opposite edges.

In new NaCl lattice;
+ –
No. of Na ion : No. of Cl atoms
+ 5
3Na atom : Cl atom.
2
In new fluorite structure:
No. of Ca2+ ion : No. of F – ion
5 2
Ca : 4F 
2
3
New cationic ratio  6:5
5
2
New anionic ratio = 5 : 8

3. O O
CN
CH3 NaCN CH3

HOEt
 
H /H2O
 
MeMgBr
 excess 

OH OH
CH3 O
H  C C CH3
C C H 
CH3 CH3
CH3

4. 3Cu  8HNO3  2NO  3Cu NO3  2  4H2O

NO  has 11 valence e– with an unpaired.
e– in  *2p orbital.
 is diamagnetic in solid state due to dimerisation.
 shows resonance.
5. S S

N
S 4N4 N  4 five membered ring
N
N

S
S

6. If PF3 is better  accepter than CO, than C – O bond order in II complex is more than I complex.
7. HNO2 pale blue in colour due to dissolution of N2O3.
Reaction with:
(A) Br2 water : HNO2  H2O  Br2   HNO3  2HBr
(B) KMnO4 : 2KMnO4  3H2SO4  5HNO2 
 K 2SO4  2MnSO4  3H2O  5HNO3

8. At stage I:
CH3COOH NaOH 
 CH3COONa
Left 5 meq 0 5 meq

So, its is a buffer solution, [Salt] = [Acid]

pH = pKa  pHI = 5
At statge II:
CH3COOH NaOH   CH3COONa
Left 0 0 10 meq in 200 ml.

So, c = 0.05
So, anionic hydrolysis takes place and pH is:
1
& pH = 7   pK a  log C 
2
1
pHII  7   5  log 0.05   8.84
2
Stage III:
CH3COOH NaOH  CH3COONa
Left 0 2 meq/220 ml 10 meq/220 ml

So, pH will be according to NaOH.

[NaOH] = [a × 10-3]
So, [OH–] = a × 10-3
pOH = 2.040
pHIII = 11.95.

9. Beckmann rearrangement:
OH
CH N CH
CH
P2O5
CH CH  
NH

C
O H

CH
CH
 H2O
 
N N
HO C
Isoquinoline  X  H
(B) is aromatic, more basic than aniline.

10. CH3NH2   CH3 2 NH2

PCl2F3  PCl3F2
180 
o

NO  NO  NO   NO

 B.O 3  B.O.2.5
6 3
Cr O3  Cr 2 O3  Acidic nature 

conc.H2SO4
11. (A) CD3  CH  OH   CH3   CD3  CH  CH2  CD2  CH  CH3
Major  Minor 

H H
CH2 H


CH3 No
 
(C) rearrangement willoccur

H3C H3C
(D) 1, 3  diequitorial  cis
1,3  axial and equatorial = 1 trans
So, groups are more stable at equatorial position.

12. Electrode reactions are:

Anodic reaction
Pb  s   HSO4  aq.  PbSO4  s   H  aq.  2e 
Cathodic reaction
 PbSO4  s   2H2O   
PbO2  s   HSO4  aq.  3H  aq.  2e 
 PbSO4  s   2H2O   
Net reaction: Pb  s   PbO2  s   2H2SO4  aq. 

15. Cl HOOC Cl
MeOK in KMnO4

chloroform



HOOC

16. NO 2
HOOC Cl
HOOC Cl HOOC Cl HOOC Cl
conc.

HNO3  
HOOC
HOOC NO 2 HOOC HOOC
Major  NO 2
Solution for the Q. No. 17 to 18.

ZnS H2SO 4  dil.  ZnSO 4  H2S 

 A C  B 
Eq. (2) 3H2S  K 2Cr2O7  4H2SO 4  K 2SO 4  Cr2  SO4 3  7H2O  3S
D 
2H2S B 
 D   O2  SO 2  
 2H2O  3S 
E   colourless liq. D 

CuSO 4 .5H2O
 Blue 
2NaOH
 Zn  OH2  
ZnSO4  2NaOH   Na2 ZnO 2  2H2O
 C  So luble
In Eq. (2) neq(H2S) = neq(K2Cr2O7)
Moles(H2S) × 2 = 0.2 × 10 × 6 = 6 milli moles
 wt. of ZnS required is  Co = (6 × 97.3) mg
= 583.8 mg.

SECTION – C

Me Me Me
2.
Me 
conc. H2So 4
 HCN
  Me
 Me
OH OH
O HO CN

1 chiral centre
3 lone pairs
Sum  1  3  4
3. q  U  w  0  w   w
[-2(16 – 4) + (-1 × (32 – 16)] = 4 kJ.
4. 
Solid PCl5  PCl4  PCl6 


 
sp3 sp3 d2
 
Solid PBr5  PBr4  Br 

sp3
 
Solid N2O5  NO 2  NO3 
 
sp sp2

5. Cl Cl Cl
Cl Cl
en Co en en Co Co en

Cl en
en
A  B  A

A & B
  diastereomers
A & C
B & C  enantiomers

Mathematics PART – III

SECTION – A

1. Six letters can be arranged in following ways: 1, 1, 1, 3 or 1, 1, 2, 2

(1, 1, 1, 3) selection of boxes can be done as 3C1  2
3 3
(1, 1, 2, 2) selection of boxes can be done as C1  C2
 Total such selections = 6 + 9 = 15
 The number of required arrangements = 15  6! = 10,800

2. Distance between two roots = b2  4ac

2
Required –1  b – 4ac  1
That means that the point P(b, c) lies on the intersection of the region in between two parabola
2 2
x = 4y – 1 and x = 4y + 1 and the square formed by x = 0, x = 1, y = 0 and y = 1
1
x2  1
 dx
0
4 1
Required probability = 
1 3

 2x  y  1    x  2y  3z   0
3. 
 3x  y  z  2     4x  5y  2z  3   0
x y z
Line is parallel to  
1 2 3
So, (2 + )  1 + (1 – 2)  2 + 3  3 = 0
2

3
(3 + 4)  1 + (–1 + 5)  2 + (1 – 2)  3 = 0
1

2

4. p, q, r will be in H.P.
 p + r = pr
pr
  pr
2
 pr  4

1
5. Put x 
t

6. 1    2
 P(x, y)
1  2 
2 
 
1  2  2  1  
2

22  1  2
2 1
2 tan  2  1 O(0, 0)
2

1   tan  2   tan 1

dy
2
dx x
2

 dy  y
1  
 dx 
2 2
dy  dy   dy  dy
2y  x    x  x    2y x
dx  dx  dx
  dx

7. f(2) = 3 and f (3) is negative

So, equation has one real root in between (3, 2)

2x  1 1
8. f(x)  x
 x
is a one-one function.
4 1 2 1

10. Either tan x + tan 2x + tan 3x = 0 or tan x = tan 2x = tan 3x

11. Major axis will be x + y = 0 and centre at (0, 0). Foci will be at distance 3 from centre on major
axis

12. Circle is x2 + y2 + 8x + 8y + 16 + k(x + y + 12) = 0

From origin chord of contact is T = 0
 (8 + k1)x + (8 + k1)y + 8(4 + 3k1)= 0
Which is same as x + y + 12 = 0
8  k1 8  4  3k1  16
   k1 
1 12 3
16
As PQ is chord to x2 + y2 + 8x + 8y + 16 +  x  y  12   0
3
From mid-point of PQ(h, k) chord is T = S1  (h, k) = (–6, –6)

13.-14. The equation of line L is y = 4x  1 and the value of p = 1, q = 6 and r = 0

15.-16. |A| = 0  A is singular

|B| = 0  B is singular
|C| = –1  C is non-singular
2 3 5
A 2  1 4 5  A  A is idempotent
1 3 4
0 0 0
2
B  0 0 0  B is nilpotent
0 0 0
1 0 0
2
C  0 1 0  I  C is involutary
0 0 1
Now, check the options

17. Equation of the given circle can be written as

(x2 + y2  2y  8)  2ax = 0
2 2
which represent the family of circle passing through points of intersection of x + y  2y  8 = 0
and 2ax = 0.
Now, 2ax = 0 ; x = 0 put in circle we get
y2  2y  8 = 0  (y  1)2 = 9
y=13
y = 2, 4
points can (0, 2), (0, 4)

18. Let the tangent at P and Q to a member of this family intersect at (h, k), then PQ is the chord of
contact of (h, k) and its equation is
hx + ky  a(x + h)  (y + k)  8 = 0
x(h  a) + y( k  1)  (ah + k + 8) = 0
Comparing this with equation x = 0 of PQ. We get
k = 1 and ah + k + 8 = 0
Since (h, k) lies on the given line 2x + y + 5 = 0
2h + 1 + 5 = 0
h=3
3a + 1 + 8 = 0
a=3
Hence the equation of the required member c of this family is x2 + y2  6x  2y  8 = 0

SECTION – C

1. Let the number of rows in triangle is k

 1 + 2 + 3 ..... k = n
k  k  1
 n
2
k  k  1 2
Also,  49   k  3 
2
 k2 + k + 98 = 2k2 + 12k + 18
 k2 + 11k – 80 = 0
k2 + 16k – 5k – 80 = 0
k=5
5  5  1
 Number of balls, n =  15
2

 
 cos x  sin x
3. P =     x  4dx , put x = 2t
 x  4 0 0
 /2
1 1 sin 2t
P=    dt
4 4 0
t2

sinrº sin r  1 º cos1º cos1º  sin (r  1)º r 

4. Tr   1 1 = 1 =   1
cos rº cos 1  r  º cosrº cos(r  1)º sin1º  cosrº cos(r  1)º 
= cot1º tan  r  1 º  tanrº  1  T1  T2  .....T88  cot1º (tan89º  tan1º )  88 = cot 2 (1º )  89

ar  2i
5. X  iy 
 ar 2  4
ar 2
X , Y
ar2 4 ar2 4
1
X2 + Y2 = Y
2
1
X2  Y 2  Y  0
2
y
So all points lies on circle x2 + y2 – 0
2
1
r
4
1  2   1 1 1 1
Area of regular octagon = 8   r 2  sin   = 8    
2  8  2 16 2 4 2

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HALF COURSE TEST – V
ALL INDIA INTEGRATED TEST SERIES
Time Allotted: 3 Hours Maximum Marks: 432
 Pl ea s e r ea d t h e i n s t r u c t i o n s c a r ef u l l y. Yo u a r e a l l o t t ed 5 m i n ut es
s p ec i f i c a ll y f o r t h i s p u r p o s e.
 Yo u a r e n o t a l l o wed t o l ea v e t h e E xa m i n at i o n Ha l l b ef o r e t h e en d o f
t h e t es t .

B. Filling of OMR Sheet

C. Marking Scheme For All Three Parts.

(i) Section-A (01 to 02 and 09 to 30) contains 24 multiple choice questions which have only one
correct answer. Each question carries +4 marks for correct answer and – 1 mark for wrong
answer.

Section-A (03 to 08) contains 6 multiple choice questions which have only one correct answer.
Each question carries +8 marks for correct answer and – 2 mark for wrong answer.

Name of the Candidate

Enrolment No.

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Useful Data

PHYSICS
Acceleration due to gravity g = 10 m/s2

Planck constant h = 6.6 1034 J-s

Charge of electron e = 1.6  1019 C

Mass of electron me = 9.1  1031 kg

2 2
Permittivity of free space 0 = 8.85  1012 C /N-m

Density of water water = 103 kg/m3

Atmospheric pressure Pa = 105 N/m2

Gas constant R = 8.314 J K1 mol1

CHEMISTRY

Gas Constant R = 8.314 J K1 mol1

Atomic No: H=1, He = 2, Li=3, Be=4, B=5, C=6, N=7, O=8,

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Physics PART – I

SECTION – A
(Single Correct Choice Type)
This section contains 30 multiple choice questions. Each question has four choice (A), (B), (C) and (D) out
of which ONLY ONE is correct.
1. Kepler's third law of planetary motion when stated for circular orbits, has the form
(v = orbital velocity of a planet and R = distance from the sun)
(A) v  R (B) v  1/R
(C) v 2  1/R (D) none of the above

2. An air bubble of radius r rises steadily through a liquid of density  at

the rate of v. Neglecting density of air, the coefficient of viscosity of

liquid is
2 r 2 g 1 r 2 g
(A) (B) v
9 v 3 v
1 r 2 g
(C) (D) None of these
9 v
0
3. A smooth wedge of mass m and angle of inclination 60 rests
4k
unattached between two springs of spring constant k and 4k, on
a smooth horizontal plane, both springs in the unextended k
position. The time period of small oscillations of the wedge
(assuming that the springs are constrained to get compressed 0
60
along their length) equals
 1 m  1  m
(A)   1   (B)   1  
 2 k  3 k
m
(C) 2 (D) none of the above
k

4. Two waveforms, 1 (x, t) = A sin(kx - t) and 2 (x, t) =A sin(kx + t + /3), travelling along the x-axis,
2
are superposed. The position of the nodes is given by (k = ,  = 2f)

  1
(A) x n = (n + 1/6) (B) x n =  n  
2  62

(C) x n = n (D) none of these
2
Space for rough work

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5. A uniform ring of mass m and radius r is hinged at point O so that it can rotate about
horizontal axis passing through O in vertical plane. If the system is released from
O
rest, then the initial acceleration of the ring is [Ring is in the vertical plane.]
(A) g (B) g/2
(C) g/4 (D) g/8

6. A sphere of mass M and radius b has a concentric cavity of radius a

as shown in figure. The graph showing variation of gravitational
potential V with distance r from the center of sphere is
a
b

(A) V (B) V

b a b
r r

7. Oil enters the bend of a pipe in the horizontal plane with v2

–1 3 –2
velocity 4 ms and pressure 280 × 10 Nm as shown A1
in the figure. The pressure of oil at the point Q is 86. Q A2 =
4
Find the value of . (KNm–2).
(Take specific gravity of oil as 0.9 and sin 37º = 0.6) 37º
(A) 1 (B) 2 A1=0.2m2
(C) 3 (D) 4
v1=4ms–1 P

Space for rough work

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8. In the given figure an impulse J is given to the block of mass m in the

downward direction. As a result of the impulse
J
(A) both the blocks start moving with speed in opposite directions J
3m
m
J
(B) both the blocks start moving with speed in opposite directions 2m
m
(C) the centre of mass of the system is moving downwards
(D) the centre of mass of the system is not moving

9. A rod of mass m and length 2r is attached to the centre of a semi-circular

lamina of equal mass m and radius r, such that it remains in the plane of the O
lamina. The moment of inertia of the system, about an axis passing through
m
the centre O, perpendicular to the plane of the lamina equals m
19 2 5 2
(A) mr (B) mr
6 6
11
(C) mr2 (D) none of the above
6
10. A small ball of density  plunges into a liquid of density 2, after falling freely from a height h above its
surface. The motion of the particle will be (assuming no loss of energy and no splashing of liquid)
2h 2h
(A) simple harmonic with a period of 4 (B) simple harmonic with a period 2
g g
2h 2h
(C) periodic with a time period 2 (D) periodic with a time period 4
g g

11. An ideal gas is taken through a process where dQ = dU - dW. The specific heat of the ideal gas in
this process is (dQ = heat supplied to the gas, dU = change in internal energy, dW = work done by
gas)
3
(A) Cv (B) Cv + R
2
(C) Cv - R (D) the process is impossible
Space for rough work

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12. If the three point masses are released simultaneously to move under m
influence of mutual gravitational attraction, speed of each particle
 a 
when each one of them has moved a distance  
2 3 a
a
6 3Gm 2Gm
(A) (B)
2  3  a a

Gm
(C) (D) None of these m m
a a

13. If the pulley 'P' as shown in the given arrangement is

m P
massless and there is no friction, force exerted by the
pulley 'P' on the clamp would be
mg
(A) 2 mg (B)
2
mg m
(C) 2mg (D)
2

14. A block of mass M is performing SHM with amplitude A on a smooth horizontal surface. At the
extreme position a small block of mass m falls vertically and sticks to M. New amplitude of oscillation
will be
M
(A) A (B) A
Mm
M Mm
(C) A (D) A
Mm m

15. A block can slide down the incline plane with constant velocity. If it is u
projected with a velocity ‘u’ up the inclined plane at t = 0. The displacement
u
of particle after a time t = from O will be
g sin  
2 2 O
u u
(A) (B)
4gcos  2g sin 
u2
(C) zero (D)
4g sin 

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16. Consider the system shown when pulleys and strings are
A
massless. If the system is released from rest, then centre of
mass of the system (3m, 2m and m)
(A) is always at rest
(B) accelerates upwards 3m
(C) accelerates downwards. B
(D) has acceleration depending on the value of m.
m
2m

17. A solid sphere of mass M and radius R is placed vertically on a rough M

horizontal surface. If a horizontal force F is applied at a distance R/6 below R

the centre line parallel to the diameter of the sphere, the direction of frictional
(R/6)
force will be F
(A) forward (B) backward
(C) depends upon value of F (D) insufficient information.

18. A cabin is moving vertically upwards with an acceleration a. A massless string is

a
attached to the ceiling of the cabin on which a bead of mass m is sliding m
a
downwards w.r.t. string with an acceleration a. The tension in the string is
(A) m(g + 2a) (B) m (g + a)
(C) mg (D) m (g – a)

19. A ball of mass m strikes the inclined face of the wedge m

perpendicularly with speed v 0. The wedge is at rest on a 0
X
90 Y
rough horizontal surface before collision. The conservation Y
of momentum is applicable for 
M X
(i) m as system, along y’ 
(ii) M as system, along y’
(iii) (M + m) as system, along x
(iv) (M + m) as system, along y
which of the following is correct ?
(A) (i) only (B) (i) + (ii) only
(C) (iii) only (D) (iii) + (iv) only.

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2
20. A mass m is performing linear simple harmonic motion, then graph for square of acceleration f and
2
corresponding square of linear velocity v is
(A) (B)

v2  v2 

f2 f2
(C) (D)

v2  v2 

f2 f2

21. One mole of a monoatomic gas undergoes the processes

1 Isothermal
1 – 2 and 2 – 3 as shown. The corresponding graph for
the above process for 2 moles of the gas is P
3 2

V
(A) (B)
2
2
V 1 V
3 1 3

T T
(C) (D)
1
1 2
V V
3 2 3

T T

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22. If 1 and 2 (> 1) are the lengths of an air column for the first and second resonance when a tunning
fork of frequency n is sounded on a resonance tube, then the minimum distance of the anti-node from
the top end of the resonance tube is
(A) 2 (2 - 1) (B) ½ (21 - 2)
 2  3 1  2  1
(C) (D)
2 2
23. Two different masses m and 2m of same ideal gas are heated separately in vessels of equal volume.
The T-P curve for mass 2m makes angle  with T-axis and that for mass m makes angle  with T-axis
then
(A) tan  = tan  (B) tan  = 2 tan
(C) tan  = 2 tan (D) None of these

24. A viscous liquid flows through a uniform tube. As we move from the axis of the tube towards its wall,
the speed of the liquid relative to the tube
(A) increases from zero to maximum value.
(B) decreases from maximum value to zero
(C) remains same throughout the cross section.
(D) increases from zero to a maximum value on one side and decreases from a maximum value to
zero on the other side.
1 
25. The displacement y of a particle executing a certain periodic motion is given by y  4 cos2  t 
2 
sin (1000t). This expression may be considered to be the superposition of n independent harmonic
motions. Then, n is equal to
(A) 2 (B) 3
(C) 4 (D) 5

26. The escape velocity for a planet is ve. A tunnel is dug along a diameter of the planet and a small body
is dropped into it from the surface. When the body reaches the centre of the planet, its speed will be
(A) v e (B) Ve 2
v
(C) e (D) zero
2
27. In a resonance tube, closed at one end by a smooth moving piston and the other end open, exhibits
the first three resonance lengths, L1, L2 and L3 for the same tuning fork. Then they are related by
5
(A) L3 = 2L2 = 4 L1 (B) L3 = L2 = 5 L1
3
(C) (L3 - L2) = (L2 - L1) (D) (L3 - L2) = 2(L2 - L1)
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28. A syringe of diameter D = 8 mm and having a nozzle of V=0.25 m/s

diameter d = 2 mm is placed horizontally at a height of 1.25
m as shown in the figure. An incompressible and non-
h=1.25 m
viscous liquid is filled in syringe and the piston is moved at
speed of 0.25 m/s. Find the range of liquid jet on the ground.
(A) 1 m (B) 4 m
(C) 3 m (D) 2 m

29. A pendulum of length  = 1 m having a bob of mass m = 1 kg is hanging

v
from a rigid support. If the bob is projected horizontally with a velocity

v0 = 35 m/s. The tension in the string is 6k Newton when angle made by
v0
string is 60 from vertical as shown. Find the value of k.
(A) 6 (B) 3
(C) 5 (D) 1

30. A particle of mass m moves towards a smooth vertical wall with a speed u (relative to the ground)
and collides elastically with the wall; the wall moving towards the particle with a speed 2u (also
relative to the ground). Assuming that the wall is extremely massive, the impulse delivered to the
particle equals
(A) 5 mu (B) 6 mu
(C) 4 mu (D) none of these.
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Chemistry PART – II

SECTION – A
(Single Correct Choice Type)
This section contains 30 multiple choice questions. Each question has four choice (A), (B), (C) and (D) out
of which ONLY ONE is correct.
2
1. If pKa = 4 and Ka = Cx then van’t Hoff factor for weak monobasic acid when C = 0.01 M is:
(A) 1.01 (B) 1.02
(C) 1.10 (D) 1.20

2. 10 g CaCO3 is completely decomposed to X and CaO. X is passed into an aqueous solution

containing one mole of sodium carbonate. What is the number of moles of sodium bicarbonate
formed?
(A) 0.2 (B) 0.1
(C) 0.01 (D) 10
rd
3. The average molar mass of air is 28.8. An open vessel at 27°C is heated upto t°C until 1/3 of the air
measured at final temperature escapes out. The rms velocity of air molecules at t°C is
(A) 340 m sec1 (B) 420 m sec1
1
(C) 515 m sec (D) 588 m sec1

4. Two solid compounds ‘X’ and ‘Y’ dissociates at certain temperature as follows :

X(s)  –3
 A(g) + 2B(g) Kp1 = 9 x 10 atm
3


Y(s)   –3
 2B(g) + C(g) Kp2 = 4.5 x 10 atm .
3

The total pressure(in atm) of gases over a mixture of ‘X’ and ‘Y’ is
(A) 1.5 (B) 0.45
(C) 0.6 (D) 0.9
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5. Consider the below given intermediates

CH2 CH2
CH2 H
C H
H H
C H
C H
H H
H
(i) (ii) (iii)

The stability order of above intermediate are

(A) i > ii > ii (B) ii > i > iii
(C) iii > i > ii (D) iii = i > ii
4 –1
6. At 550 K, the Kc for the following reaction is 10 mol L.
X(g) + Y(g) Z(g)
At equilibrium, it was observed that
1 1
[X] = [ Y ]  [ Z]
2 2
What is the value of [Z] at equilibrium?
(A) 2 × 10–4 mol L–1. (B) 10–4 mol L–1.
4 –1 4 –1
(C) 2 × 10 mol L . (D) 10 mol L .

7. The predominant product formed when 3methyl2pentene reacts with chlorine water is
Cl Cl
(A) CH3CH2CCH(OH)CH3 (B) CH3CH2C CH(OH)CH3
CH3 CH3 Cl
OH CH3
(C) CH3CH2C CHCH3 (D) CH3C CHCH3
CH3 Cl CH3 OH
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8. The major product expected from the monobromination of phenyl benzoate is

Br Br

(A) —COO—C6H5 (B) C6H5COO—

(C) Br— —COO—C6H5 (D) C6H5COO— —Br

9. Which of the following conformer of nbutane is achiral?

CH3 H3C H
H CH3
(A) (B)
H H H H
H CH3
H
CH3 CH3
H H H3C H
(C) (D)
H H H H
CH3 H

10. The correct order of basic strength the among the following is
(A) Li2O < Na2O < K2O < Rb2O (B) Rb2O > Na2O = K2O > Li2O
(C) Li2O > Na2O > K2O > Rb2O (D) Na2O > Li2O > K2O > Rb2O

11. A hydrogen like ion in excited state can emit 3 spectral lines and the minimum energy emitted in one
of the transition is 7.5 eV. The species and the excited state ion belongs to, is
(A) H and 2nd (B) He+ and 2nd
2+ nd
(C) Li and 2 (D) Can’t be predicted
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12. Which of the following is the correct order of basic nature of the compounds?
OH OH OH OH

NO 2 O Cl
H3C
(I) (II) (III) (IV)

(A) I > II > III > IV (B) III > I > IV > II
(C) IV > III > I > II (D) III > IV > I > II

13. Arrange the following in order of decreasing dipole moment:

H 3C NO2 , Cl Cl , H 3C Cl

(I) (II) (III)

(A) I > II > III (B) III > I > II
(C) I > III > II (D) III > II > I

14. In the case of a first order reaction, the time required for 93.75% of reaction to take place is
x times that required for half of the reaction. The value of ‘x’ is
(A) 1 (B) 2
(C) 3 (D) 4

15. In the following groups

OAc OMe OSO2Me OSO2CF3
(1) (2) (3) (4)
the order of leaving group ability is
(A) 1 > 2 > 3 > 4 (B) 4 > 3 > 1 > 2
(C) 3 > 2 > 1 > 4 (D) 2 > 3 > 4 > 1
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–5
16. If 10 % reactant molecules are crossing over the barrier in transition state at 298 K, then the
activation energy is
(A) 39.94 kJ (B) 49.94 kJ
(C) 79.94 kJ (D) 97.97 kJ

17. CH3

(NH4 )2 S (i)NaNO 2 HCl / low temp

  A 
(ii) CuCl / HCl
B

O 2N NO2
CH3 CH3

(A) (B)

Cl NH2 H2N NO2

CH3

(C) (D) None of the above

Cl NO 2

18. Which among the following compounds will give nucleophilic addition reaction?
(A) carbonyl compounds (B) nitriles
(C) isocyanides (D) all of these

19. The pair of molecule given below represents

CH3 CH3
H HO

CH3
HO H
CH3
(A) Enentiomers (B) Diastereomers
(C) Regiomers (D) None of these
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20. Br  Zn O3
(A) 
 (B)
 (CH3 )2 S
Br
The product (A) and (B) in above sequence of reactions are
(A) (A)  H2C C C CH2 (B)  HCHO
COOH
(B) (A)  (B) 
COOH
CHO
(C) (A)  (B) 
CHO
OH

(D) (A)  (B)  S

21. The hybridization of central atom in N3 and ICl4 are, respectively,
3 2 3 2
(A) sp d , sp (B) sp, sp d
2 3 2
(C) sp , sp d (D) sp , sp3d2
3


22. The correct order of increasing N – O bond length in NO2 , NO 2 and NO3 is

(A) NO3  NO2  NO2 (B) NO 2  NO2  NO3
 
(C) NO2  NO3  NO2 (D) NO2  NO2  NO3

23. Atoms 7A, 8B and 9C17 are such that 8B is an isobar of 7A and atom 9C17 is isotone to B from above
information. Mass number of A and B are ………………….. respectively.
(A) 16 and 17 (B) 16 and 16
(C) 17 and 16 (D) 15 and 16

24. Six moles of Cl2 undergo a loss and gain of 10 mole of electrons to form two oxidation states of Cl.
The difference between the maximum and minimum oxidation states of Cl in this reaction is
(A) 4 (B) 8
(C) 6 (D) 3
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+
25. 50 ml of 0.1 M solution of protonated form of alanine (H2A , pKa1 = 2.34, pKa2 = 9.70) is treated with
50 ml of 0.01 NaOH. The pH of the solution will be
(A) 5.2 (B) 7.02
(C) 6.02 (D) 4.6
o
26. A mixture of NH3(g) and N2H4(g) is placed in a sealed container at 27 C, 0.5 atm. After complete
decomposition of NH3 and N2H4 at 927oC, the total pressure reached to 4.5 atm. The percentage of
NH3 (by mole) is
(A) 75% (B) 25%
(C) 40% (D) 60%

27. When one mol of an ideal gas is compressed to half of the initial volume and simultaneously heated
to twice of its initial temperature. The change in enthalpy (S) is
(A) Cv ln2 (B) Cp ln2
(C) R ln2 (D) (Cv-R) ln2

28. Which of the following compounds is resolvable?

H CO NH CH3 CH3 CO NH CH3

(A) (B)
CH3 NH CO H H NH CO H
COOH O 2N
H3C C2 H5

29. Borazene, B3N3H6, is isoelectronic and isostructural with benzene. Which of the following
statement(s) is/are true about borazene?
(i) Borazene is aromatic.
(ii) There are four isotopic disubstituted borazene molecules, B3N3H4X2.
(iii) Borazene is more reactive towards addition reactions than benzene.
(A) Only (i) (B) (i) and (ii)
(C) (i) and (iii) (D) (i), (ii) and (iii)

30. Structure of Na2B 4 O7 .10H2 O contains:

(A) Two triangular and two tetrahedral units (B) Three triangular and one tetrahedral units
(C) All tetrahedral units (D) All triangular units.
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Mathematics PART – III

SECTION – A

Straight Objective Type

This section contains 30 multiple choice questions numbered 1 to 30. Each question has 4 choices (A), (B),
(C) and (D), out of which ONLY ONE is correct.

x2 y2
1. If e is the eccentricity of the hyperbola   1 and  is angle between the asymptotes, then
a2 b2

cos is equal to
2
1 e 1
(A) (B) 1
2 e
1
(C) (D) none of these
e

2. The sum of the abscissas of all the points on the line x + y = 4 that lie at a distance from the line
4x + 3y – 10 = 0 is
(A) 4 (B) –4
(C) 2 (D) –3

3. If (x2 + x + 2)2 – (a – 3)(x2 + x + 1)(x2 + x + 2) + (a – 4)(x2 + x + 1)2 = 0 has atleast one root then
complete set of value of a is
 7 7 
(A)  1,  (B)  , 5 
 3 3 
3   19 
(C)  , 1 (D)  5,
7   3 

4. Slope of the line za  za  b  0 , b  R is

Re  a  Im  a 
(A)  (B) 
Im  a  Re  a 
Re  a  Im  a 
(C) (D)
Im  a  Re  a 

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5. If f : {1, 2, 3, 4, 5}  {x, y, z, t}, total number of onto functions ‘f’ is equal to

(A) 242 (B) 245
(C) 1024 (D) 240

2 2 2
6. Locus of mid–point of chords of circle x + y + 2gx + 2fy + c = 0 that subtends angle at the centre
3
of the circle is
(A) 4x2 + 4y2 + 8gx + 8fy + 3g2 + 3f 2 + c = 0 (B) 4x2 + 4y2 – 8gx – 8fy + 3g2 + 3f 2 + c = 0
(C) 4x2 + 4y2 + 8gx – 8fy + 3g2 + 3f 2 + c = 0 (D) 4x2 + 4y2 – 8gx + 8fy + 3g2 + 3f 2 + c = 0

x2 y2
7. If the normals to the ellipse 2
  1 at the extremities of the chords and px + qy = 1 are
a b2
concurrent, then
2 2 2 2
(A) a  p = bm q = –1 (B) a  p = b mq = –1
(C) a  p = bmq = 1 (D) a  p2 = mbq2 = 1
2 + 2
8. Let f(x) = ax + bx + c where a  R and b – 4ac < 0. Area bounded by f(x), x–axis and the line x = 0,
x = 1 is equal to
1 1
(A) 3f 1  f  1  2f  0  (B) 5f 1  f  1  8f  0 
6 12
1 1
(C) 3f 1  f  1  2f  0  (D) 5f 1  f  1  8f  0 
6 12

n
3
9. If sin x sin 3x =  a cos rx  ,  x  R where, a
r 0
r 0 a1 a2 ….. an are constants and an  0, then

(A) n = 5 (B) n = 6
(C) n = 7 (D) none of these

18
2
10. The numerical value of  sin
r 1
5r  º is equal to

19
(A) 9 (B)
2
21 17
(C) (D)
2 2

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    tan x  1  3 sin2x 
11. If x    ,  , then the value of tan1    tan   is
 2 2  4   5  3 cos 2x 
x
(A) (B) 2x
2
(C) 3x (D) x

12. If a, b, c, d are positive numbers such that a + b + c + d = 2, then M = (a + b)(c + d) then M satisfies
the relation
(A) 0 < M  1 (B) 1  M  2
(C) 2  M  2 (D) 3  M  4
2 2 2 2
13. If 1 + 2 + 3 + ….. + 2003 = (2003)(4007)(334) and (1)(2003) + (2)(2002) + (3)(2001) + ….. +
(2003)(1) = (2003)(334)(x), then the value of x is
(A) 2005 (B) 2004
(C) 2003 (D) 2001

n
n
14. In ABC, the value of the expression  Cr ar bnr cos rB   n  r  is equal to
r 0
(A) cn (B) 0
(C) an (D) bn

15. Total number of ways, in which 22 different books can be given to 5 students, so that two students
get 5 books each and all the remaining students get 4 books each, is equal to
22! 22!
(A) 3
(B) 2

2!3!5! 4!   
3! 2! 5!
22!
(C) (D) none of these
3!2!5!4!

 z  1 
16. In a complex number z = x + iy such that   , z  –1 whose argument is equal to , then relation
 z  1 4
of variables is
2 2 2 2
(A) x – y = 1 (B) x + y – 2y = 1
2 2
(C) x – y + 2y + 1 = 0 (D) none of these

Space for rough work

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2 2 4 8
17. If 1, ,  are cubes roots of unity of complex number, then (1 + )(1 +  )(1 +  )(1 +  ) ….. 2n
term is
(A) 0 (B) –1
(C) 1 (D) 2

 
18. If ii  e 4n1  / 2 and the value of ii are in G.P. then the common ratio is
–2 2n
(A) e (B) e
(C) e (D) e–1

19. The value of cos  tan1 sin  cot 1 x  is

x2  1 x2  2
(A) (B)
x2  2 x2  1
x2 x 1
(C) 2
(D)
x 2 x2  2

 1 x   3x  x 3 
20. If f  x   log   , and g  x     then f[g(x)] is equal to
1 x   1  3x 2 
(A) –f(x) (B) 3f(x)
(C) [f(x)]3 (D) [f(x)]2

21. If log (1 + x + x 2 + x3 + …..) = A log (1 – x), and A = cos , then the value of  is

(A) 1 (B)
2
(C)  (D) 0

22. The complex number sin x + i cos 2x and cos x – i sin 2x are conjugate to each other for
 1
(A) x = n (B) x   n   
 2
(C) x = 0 (D) no value of x

7  3 5
23. If z + 1 = 0 then cos cos cos is
7 7 7
1 1
(A) (B) 
8 8
1 1
(C) (D)
2 2 2

Space for rough work

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24. An ordinary cube has 4 blank faces, one face mark 2 and another marked 3, and then probability of
obtaining 12 in 5 throws is
5 5
(A) (B)
2592 1944
5 5
(C) (D)
1296 648

25. Negation of “2 + 3 = 5” and “8 < 10” is

(A) 2 + 5  and < 10 (B) 2 + 3 = 5 and 8 < 10
(C) 2 + 3  5 or 8 < 10 (D) none of these

26. If (a, 0), (0, b) and (a, b) are the vertices of  circumscribed by a circle then radius of circle is
(A) a2 + b2 (B) a2 – b2
1 2
(C) a  b2 (D) a2  b2
2
2 2
27. The locus of mid–point of the chord of circle x + y = y such that segment intersected by the chord on
curve x2 – 2x – 2y = 0 subtends right angles at the origin is
(A) x 2 + y2 – 2x – 2y = 0 (B) x 2 + y2 + 2x + 2y = 0
2 2
(C) x + y = 0 (D) none of these

28. If C0, C1, C2 ….. Cn are the combinational coefficients in the expansion of (1 + x)n, n  N, then
C0 + 3C1 + 5C2 + ….. + (2n + 1)Cn is equal to
n n–1
(A) (n + 1) 2 (B) (n + 2) 2
n –1
(C) n 2 (D) none of these

29. Let a, b, c are the sides of a triangle. No two of them are equal and   R. If the roots of the equation
x2 + 2(a + b + c)x +3(ab + bc + ca) = 0 are real then
4 5
(A)   (B)  
3 3
 1 5  4 5
(C)    ,  (D)    , 
3 3 3 3
st
30. The sum of 1 nth terms of an A.P. is given by Sn = (1 + 2Tn)(1 – Tn) where Tn is nth term of series,
a b
then Tn2  , a, b  N, then (a + b) is
4
(A) 4 (B) 5
(C) 6 (D) 7

Space for rough work

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ALL INDIA INTEGRATED TEST SERIES

ANSWERS, HINTS & SOLUTIONS
HALF COURSE TEST–V
(Main)

Q. No. PHYSICS CHEMISTRY MATHEMATICS

1. C C C
2. A A B
3. B D D
4. B B A
5. B C D
qualified in JEE (Advanced), 2014.

6. B A A
7. B C B
8. A D D
9. C C B
10. D A B
11. A B D
12. B B A
13. B C A
14. A D A
15. D D A
16. C A B
17. B C C
18. C D A
19. A D A
20. B C B
21. D B C
22. C D D
23. B B B
24. B C C
25. B C D
26. B A C
27. C D A
28. D B A
29. C D A
30. B A C

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Physics PART – I

SECTION – A

42 3
1. Kepler's IIIrd law may be written as T2 = R
GM
2R 1
v= , v2 
T R
4 3
2. Buoyant force = Weight of liquid displaced = r g
3
Viscous force = Stoke’s drag force = 6 rv
4
 6rv = r 3 g
3
3. The oscillations in one direction are governed by the equation,
m x = –kx
in the other by,
mx  4kx sin2 600

4. Superposing them gives 2A sin (kx + /6) cos (t + /6)

Nodes are given by kx + /6 = n
 mgr g
5. =  
I 2mr 2 2r
g
a = r = .
2
6. Gravitational potential inside a shell = constant

7. Let A1 and A2 be the cross–sectional area of the pipe at points P and Q respectively.
Let v 1 and v 2 be the velocities of oil at the points P and Q respectively.
By conservation of mass,
Q = A1v 1 = A2v 2
A 
 v 2   1  (v1 )  4v1
 A2 
Applying Bernoullis’s equation between points P and Q, we have
1 1
P1  v12  P2  v 22
2 2
1
P2  P1  (v12  v 22 )
2
1
 P1   0.9[16  256]  103
2
 280  103  ( 108)  103
P2 = 172 × 103 Nm–2
P2 = 172 (KN m–2).

8. By conservation of momentum
J = mV + 2 mV
V= J/3m.
1 2 1
9. I = mr + m(2r)2
2 3

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10. Within the liquid, the acceleration is g upward, outside it is g downward.

11. dQ = dU - dW, again dQ = dU + dW

dW = pdV = 0

12. Conservation of total mechanical energy

13. mg  T = ma T
N

mg mg 2
T = ma  T = or N =
2 2
T

14. At extreme position velocity of M will be zero.

u
15. At the highest point velocity v = 0 after t = and it will not return to point O.
2gsin 
u2
So distance covered before coming to rest =
4gsin
16. Tension in AB, 2T < 6 mg
 Net downward force
 C.O.M. has a downward acceleration.

17. The contact point has a tendency to slide in forward direction, so direction of friction will be in
backward direction.

18. ma + mg – T = ma  T = mg

19. Normal force by the surface and frictional force an impulsive in nature.
Hence COM is applicable for m only along common tangent.

20. For y = a sin t, v = a cos t

2
 Acceleration f = - a  sin t
f2
2
 v 2  a 2 2


21. 1 – 2 is isothermal with volume increasing

2 – 3 is isobaric
 v/T = k
 v = kT
Change in number of moles of the gas does not affect the nature of graph.

22. 1 + x = /4 x
3 1
2 + x =
4
 2  3 1 2
 x=
2

23. PV = nRT
m
PV = RT  slope is directly proportional to the mass of the gas.
M
30. Use the definition of an elastic collision.

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Chemistry PART – II

SECTION – A

1. Ka = Cx2
10–4 = 10–2 × x 2
x = 0.1
i = 1 + x = 1.1


2. CaCO 3  CaO  CO 2 ( X)
So, X is CO 2
Na 2 CO3  H 2 O  CO 2  2 NaHCO3
10
Moles of CaCO 3   moles of CO2
100
Mole of Na 2 CO 3  1mole
So, moles of NaHCO 3 formed according to the moles of CO 2 , which is limiting reagent so moles of
NaHCO 3 formed will be (0.2) mole.

3. Initial temperature = 300 K at

Let new temperature corresponding to t°C is T K.
Let number of moles present at T K = n
Moles of air escaped at T K = n/3
n 4n
So, number of moles at 300 K = n + =
3 3
Volume and pressure are kept constant
n1 T
 = 2
n2 T1
4n
 300 = nT
3
 T = 400 K
3RT 3  8.314  400
Now r.m.s velocity = = = 588.6 m sec1.
3
M 28.8  10

[ Z]
6. Kc 
[ X][Y]
1 1
[X ]  [Y ]  [ Z]  a (let)
2 2
 [Z] = 2a, [Y] = 2a, [X] = a
2a
 104 =
a  2a
–4
a = 10
 [Z] = 2a = 2 × 10–4 mol L1.

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..
8. In phenyl benzoate C–O
.. the 1st ring gets deactivated, while 2nd ring gets

(1) O (2)
activated due to lone pairs on oxygen atom. So it can give ortho or para bromo product. But the ortho
position is sterically hindered, so it will give para compound as a major product.

9. Staggered conformer of nbutane is achiral because it has a plane of symmetry as well as a centre of
symmetry.

10. Basic strength of the oxides of alkali metals increases down the group.

0.0693
14. t1/ 2 
k1
... (1)
2.303 100
Also, t93.75  log
k1 100  93.75
2.303 100 2.303 4  2.303  log2 4  0.693
 log  log24    4 t1/ 2
k1 6.25 k1 k1 k1
Hence, (D).

15. Weak base groups are the best leaving groups also stronger is an acid weaker its conjugate base.
The CH3OH is the weakest base among the given group.

105
16. eEa / RT 
100
E 105
 a  n
RT 100
 Ea  39.94 kJ

20. In the presence of Zn, only Br2 can be removed, not as HBr
In the presence of (CH3)2S
COOH

COOH
can not be formed (reductive ozonolysis)

21.
Cl Cl
  
I
N  N  N sp ; sp3 d2
Cl Cl

23. Given 8B and 7A are isobars and thus should have same mass number.
m m
8B and 7A :
Now, also given 9C17 is isotone to 8Bm.
 Number of neutrons in both should be same.
or, 17 – 9 = m – 8
m = 16
Mass number of A and B are 16.

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24. 6Cl2  10e  2Cl5   10Cl

Hence, (C).

26. Let, NH3 = x mole

N2H4 = y mole
2NH3  N2  3H2
x 0 0
x 3x
0
2 2
N2H4  N2  2H2
y 0 0
0 y 2y
Initial total mole = x + y
Final total mole = 2x + 3y
2x  3y 4.5

xy (0.5)  (4)
 x = 3y  %x = 75

 T2  V2
27. S = Cv ln    R ln
 T1  V1
= Cvln2 + R  ln1/2
= (CvR) ln2

28. (B) don’t have any symmetry.

30. OH

O B O
HO B O B OH
O B O

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Mathematics PART – III

SECTION – A
x2 y2 b
1. Angle between the asymptotes of 2
 2
 1 is 2 tan1   (formula)
a b a
x2 y2
Given  is angle between asymptotes of hyperbola  1 , and e be its eccentricity
2

a b2
b   b  b
According to question  = 2 tan1     tan1  tan 
a 2 a 2 a
 b2
1  tan2 1 2 2 2 2 2
cos   2 = a  a  b  2cos2   1  a  b
1  tan2
 b 2 a 2  b2 2 a2  b2
1 2
2 a
 a2  b 2 a2  b2  a2  b 2 2a 2
2cos2  2  1  
2 a  b2 a2  b2 a 2  b2
 1 1 1
cos   
2 b 2
e 2 e
1 2
a

2. Any point on the line x + y = 4 can be taken as (x1, 4 – x 1) as it is a limit distance from the line
4x1  3  4  x1   10
4x + 3y – 10 = 0 we get  1  |x 1 + 2| = 5
25
 x1 + 2 = 5, x1 = 3 or – 7
 The required sum = 3 + (–7) = –4

3. Let A = x 2 + x + 1
3 
 A   , 
4 
 (A + 1)2 – (a – 3)A(A + 1) + (a – 4)A2 = 0
 A2 + 2A + 1 – (a – 3)(A2 + A) + (a – 4)A2 = 0
 A(2 – a + 3) + 1 = 0
 A(5 – a) = –1
1
 A
a5
1 3
 
a5 4
19  3a
 0
a5
 19 
 a   5,
 3 

4. z = x + iy then line becomes (x + iy) a + (x – iy)a + b = 0

 x(a + a ) + y( a – a)i + b = 0
a  a  2Re  a  Re  a 
 Slope =  
i a  a  i  2iIm  a  Im  a 

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5
5. Total function = 4
4 5
Total functions when any one element is the left out = C1 3
4 5
Total functions when any two elements are left out = C2 2
Total functions when any three of elements are left out = 4C3 15
5 4 5 4 5 4
Thus the total number of onto functions = 4 – ( C 1 3 + C2 2 – C3) = 240

6. Let D(x, y) be the mid–point of chord AB, we have B

2 
BPA  , DPA  D
3 3
A

 PD  PA cos P(–g,–f)
3
2 2
 4PD = PA
 4[(x + g)2 + (y + f)2] = g2 + f2 – c
2 2 2 2
 4x + 4y + 8gx + 8fy + 3g + 3f + c = 0

a 2  x  x1  b2  y  y1 
7. Equation of normals at any point P(x1, y1) on the 
x1 y1
2 2
If it passes through Q(h, k) then a y1(h – x1) = b x 1(k – y1) thus foot of normals drawn from Q(h, k) to
the ellipse will lie on the hyperbola, whose equation will be
2 2 2 2
b xk – a hy + (a – b )xy = 0 ….. (1)
In the given question if the normals are concurrent at Q(h, k) then
x2 y2
  1    x  my  1 px  qy  1  0 ….. (2)
a2 b2
Equation (2) will represent hyperbola given by (1) by comparing the coefficients a2  p = b2mq = 1

8. f(x) > 0,  x  R
1

 ax  bx  c  dx
2
 Required area A 
0
a b 1
A   c   2a  3b  6c 
3 2 6
Now, f(0) = c, f(1) = a + b + c
f(–1) = a – b + c
1 1
a  f 1  f  1  f  0  and b  f 1  f  1
2 2
1
= 5f 1  f  1  8f  0 
12

1 2
9. sin3x sin 3x = sin x  cos 2x  cos 4x 
2
1 1
=  1  cos 2x   cos 2x  cos 4x 
2 2
1
=  cos 2x  cos 4x  cos2 2x  cos2x cos 4x 
4
1 1  cos 4x 1  
= cos 2x  cos 4x   cos6x  cos 2x  
4 2 2 
1 3 3 1 1
=  cos 2x  cos 4x  cos 6x  
4 2 2 2 2
n=6

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18 18 18
10. I=  sin2  5r  º =  sin2  5r  º =  sin 2
5 18  r  º
r 1 r 0 r 0
18 18
I=  sin2  90  5r  º =  cos 2  5r  º
r 0 r 0
18
2I =   sin
r 0
2
5r  cos2 5r   19

19
I=
2

 tan x  1  3 sin2x 
11. tan1    tan  
 4   5  3 cos 2x 
 tan x  1  6 tan x  1  tan x  1  3 tan x 
= tan1    tan    tan  4   tan  
 4   8  2 tan2 x   4  tan2 x 
 16 tan x  tan3 x 
= tan1  1
  tan  tan x   x
 16  tan2 x 

12. a+b+c+d=2
AM  GM
 a  b    c  d
 [(a + b)(c + d)]1/2
2
1  [(a + b)(c + d)]1/2  (a + b)(c + d)  M
 (a + b)(c + d) > 0
So 0 < M  1
2 2 2 2
13. If 1 + 2 + 3 + ….. + 2003 = (2003)(4007)(334)
(1)(2003) + (2)(2002) + 3(2001) + ….. + (2003)(1) = (2003)(334)(x)
2003
  r  2003  r  1   2003 334   x 
r 1
 2004(r) – (r2) = (2003)(334)(x)
 2003  2004 
 2004   2003  4007  334 
2
 (2003)(334)x  x = 2005

n  n n 
14. 
r 0
n
Cr ar bnr cos rB   n  r  A = Re 

 r 0

   
Cr ar bnr ei rB nr A 

 n n r n r  n
= Re  Cr  aeiB   be iA   = Re aeiB  be iA 

 r 0 
n
= Re(a cos B + ia sin B + b cos A – ib sin A)
n n
= Re(a cos B + b cos A) = c

15. It is same as dividing 22 books in 5 packets such that two packets have 5 books, each and remaining
three have four books each and then the distributing these packets thus total ways fo distributing
22!  5! = 22!
these books = 2 3 3
 
 5! 4! 2!3! 2!3!5!  4! 

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 z  1   x  iy  1    x  1  iy    x  1  iy  x2  y2  1 2iy
16.           
   
 z  1   x  iy  1   x  1  iy   x  1  iy   x  1  y 2 2  x  12  y 2
 z  1 
Now, agr  
 z  1 4
 2y  x  12  y 2  
 tan1   
2
  x  1  y
2
 x 2  y 2  1  4
2 2
 x + y – 2y = 1

17. (1 + )(1 + 2)(1 + 4)(1 + 8) ….. 2n term (  3 = 6 = 1)

2 3 6 2 2 2
= (1 + )(1 +  )(1 +  )(1 +   ) ….. = [(1 + )(1 + ) ….. n term][1 +  )(1 +  ) ….. n term]
n 2 n 2 n n
= (1 + ) (1 +  ) = [(1 + )(1 +  )] = 1 = 1

 1  1 
i 2ni  logii tan 1  
18. ii  eilogi  ei2ni logi  e  2  0 

  
i 2ni i    4n 1
= e 2 e 2 (where n = 0, 1, 2, 3 …..)
 5 9
  
Put n = 0, 1, 2, 3, ….. then ii  e 2 , e 2 , e 2 ….. are in G.P.
–2
 Common ratio = e

19. cos  tan1 sin  cot 1 x 

   1    1 1 
= cos  tan1 sin  sin1   cos  tan
2  
   1  x    1  x2 
 1  x2  x2  1
= coscos 1  
 2  x2  2
 2x 

3
 3x  x3  1 x   1 x 
20. f[g(x)] = f  2  = log    3log    3f  x 
 1  3x   1 x   1 x 

2 –1
21. log (1 + x + x + …..) = log (1 – x) = –log (1 – x)
 A = –1
 cos  = –1
=

22. No value of x

23. z7 + 1 = 0
 z7 = –ei( + 2k) (where k = 0, 1, 2, ….. 6)
 3 5
i i i
z = –1, e 7 , e 7 , e 7

   3   3   5   5 
i  i   i i i i
z7  1   z   1   z  e 7   z  e 7   z  e 7   z  e 7   z  e 7   z  e 7 
Put z2 + 1 = 0 z =  i
 3 5
8cos cos cos 1
7 7 7
 3 5 1
 cos cos cos 
7 7 7 8

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2 2
26. x + y + 2gn + 2fg + c = 0 is the equation of circle
 (a, 0), (0, b) and (a, b) passes through the circle
 a2 + 2ag + c = 0
b2 + 2bf + c = 0
2 2
a + b + 2ag + 2bf + c = 0
a b
g , f  ,c=0
2 2
1 2
Radius = a  b2
2

27. Equation of chord T = S1

2 2
 xh + ky = h + k ….. (1)
 xh + hk – (h + k2) = 0
2 (0, 0)
x2 – 2x – 2y = 0 2 2
2
 x – 2x – 1 – 2y – 1= 0
 hx  ky   hx  ky  M(h, k)
 x 2  2x  2 2
 2y  2 0
h  k  h  k2 
 h2 + k2 – 2h – 2k = 0
2 2
 x + y – 2x – 2y = 0

28. (1 + x)n = C0 + C1x + C2x 2 + C3x3 …..

C0 + 3C1 + 5C2 + ….. (2n + 1)Cn = (C0 + C1 + C2 + ….. + Cn) + 2(C1 + 2C2 + 3C3 + ….. + nCn)
n n n
= 2 + n 2  (n + 1) 2

29. D>0
2
 (a + b + c) – 3(ab + bc + ca) > 0
 a2  b2  c 2 
 3 – 4 < 0,   2
 ab  bc  ca 
4
 
3

30. Sn = (1 + 2Tn)(1 – Tn)

Put n = 1 then T1 = (1 + 2T1)(1 – T1)
1
 T1 
2
1 1 4 2 a b
Put n = 2 we get T22    
2 2 2 4 4
 (a + b) = 6

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FIITJEE JEE (Main), 2015
2521 FIITJEE Students from Classroom / Integrated School Programs & 3579 FIITJEE Students from All Programs have qualified in JEE (Advanced), 2014.
HALF COURSE TEST –VI
ALL INDIA INTEGRATED TEST SERIES
Time Allotted: 3 Hours Maximum Marks: 432
 Pl ea s e r ea d t h e i n s t r u c t i o n s c a r ef u l l y. Yo u a r e a l l o t t ed 5 m i n ut es
s p ec i f i c a ll y f o r t h i s p u r p o s e.
 Yo u a r e n o t a l l o wed t o l ea v e t h e E xa m i n at i o n Ha l l b ef o r e t h e en d o f
t h e t es t .

B. Filling of OMR Sheet

C. Marking Scheme For All Three Parts.

(i) Section-A (01 to 03 and 10 to 12) contains 6 multiple choice questions which have only one
correct answer. Each question carries +8 marks for correct answer and – 2 mark for wrong
answer.

Section-A (04 to 09 and 13 to 30) contains 24 multiple choice questions which have only one
correct answer. Each question carries +4 marks for correct answer and – 1 mark for wrong
answer.

Name of the Candidate

Enrolment No.

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Useful Data

PHYSICS
Acceleration due to gravity g = 10 m/s2

Planck constant h = 6.6 1034 J-s

Charge of electron e = 1.6  1019 C

Mass of electron me = 9.1  1031 kg

Permittivity of free space 0 = 8.85  1012 C2/N-m2

Density of water water = 103 kg/m3

Atmospheric pressure Pa = 105 N/m2

Gas constant R = 8.314 J K1 mol1

CHEMISTRY

Gas Constant R = 8.314 J K1 mol1

Atomic No: H=1, He = 2, Li=3, Be=4, B=5, C=6, N=7, O=8,

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Physics PART – I

SECTION – A
Single Correct Choice Type

This section contains 30 multiple choice questions. Each question has four choices (A), (B), (C) and (D)
out of which ONLY ONE is correct.

1. A cylinder released from the top of an inwclined plane of inclination  and having coefficient of
friction  rolls without sliding and reaches the bottom with speed v r. Another cylinder, released
from the top of the same plane rolls with sliding till it reaches bottom with speed v s. Then
tan 
(given that   )
3
(A) Vs > Vr (B) Vr > Vs
(C) Vr = Vs (D) none of these

2. A uniform rod of length  is suspended from one end such that it is free to rotate about an axis
passing through that end and perpendicular to the rod. A small sphere of same mass as that of
rod strikes the other end of the rod and sticks to it. The minimum velocity of the sphere for which
the rod can make full rotations is
(A) 8g (B) 4g
(C) 5g (D) none of these

3. From A ring of mass M and radius R rotating with angular velocity  about 
it’s diameter (being vertical) two beads of mass m are released from rest
from the upper end of the diameter. The angular velocity of the ring when R
the two bead’s meet at the lower end is
M
(A)  (B) 2
(C) M/M+2m (D) none of these

4. The moment of inertia of a uniform disc with centre C and radius R, having
circular cavity of diameter CB = R about an axis passing through A, B and C and A
C
B
perpendicular to the plane of disc is IA, IB and IC respectively. Then
(A) IC > IB > IA (B) IB > IC > IA
(C) IA > IC > IB (D) IB > IA > IC
Space for rough work

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5. If moment of inertia of a rod of mass m and length  about an axis through it’s end and
perpendicular to the rod is I, then the moment of inertia about the same axis if the rod is bent at
90 at it’s mid point is
5 5
(A) I (B) I
4 8
13
(C) I (D) none of these
8

6. The system shown in figure is released from rest, if block B is smooth

stopped for a moment after 1 sec from the start, the time after m A
which the string will again become taut is
(A) 1 sec (B) 2 sec
(C) 3 sec (D) none of these
m B

7. The system shown in figure is released from rest. The ratio of

velocities of block and wedge after 2 sec from the start of motion is
(Assuming all the surface to be frictionless) m M
(A) 1/2 (B) 1 =60
(C) 2 (D) none of these

8. In the figure shown pulley string and spring are massless and the system is in
equilibrium. If the string connected to the ground is cut, the instantaneous
acceleration of m1 mass is k1
(A) g/2 (B) g/3 k2
m1g m2  m1  g
(C) (D) m1
m2 m1 m2

9. A particle is observed from two frames S1 and S2. The graph of relative velocity of

Rel. velocity
S1 with respect to S2 is shown in figure. Let F1 and F2 be the pseudo forces on the
particle when seen from S1 and S2 respectively. Which of the following is
impossible? O
(A) F1 = 0, F2  0 (B) F1  0, F2 = 0 Time
(C) F1  0, F2  0 (D) F1 = 0, F2 = 0
Space for rough work

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10. In the figure shown neglect the masses of the pulley and also neglect friction. Then P2
(A) tension in the string connected to block is mg/2
(B) Acceleration of pulley P1 is g upwards P1
(C) Acceleration of pulley P3 is g/2 downwards P3
(D) Acceleration of block is g downwards.
m

11. A block of mass m is given velocity u over a long plank of mass M as Rough
shown in figure. The velocity of the block after a long time is smooth M u
m
mu mu
(A) (B)
M Mm
u Mu
(C) (D)
2 Mm
12. The end B of the chain of mass per unit length  and length  is released from rest as B
A
shown in figure. The force at the hinge when the end B is at /4 from the ceiling is
3 gL
(A) gL (B)
4 4
4
(C) gL (D) none of these
3
13. Choose the correct statement.
(A) Work done by friction force is always negative.
(B) Static friction is always greater than kinetic friction
(C) When the body slides on a surface the value of friction is equal to limiting friction.
(D) The coefficient of friction between two bodies does not depend upon the area of contact.
Space for rough work

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14. For a body moving on a circular path, the angular velocity, linear velocity, angular acceleration,
   
centripetal acceleration and tangential acceleration of the particle at any instant are ,v, , ac and

a t respectively. Which of the following relation is incorrect.
   
(A)   v (B) a t  ac
   
(C)   at (D)   

15. A particle is projected from corner P of an elevator moving downwards S R

with acceleration a towards diagonally opposite corner R. Then
(A) particle will hit R only when a = g. P Q
(B) particle will hit the wall QR if a > g
(C) particle will hit the roof RS if a < g.
(D) data insufficient

v
16. For the given velocity time curve, displacement (m/s)
is zero at t = 0 the corresponding displacement 10

time curve is
0 2 4 6
t(s)
10

(A) s(m) (B) s(m)

20 20
10 10

2 4 6 t(s) 0 2 4 6 t(s)

(C) s(m) (D) s(m)

20 20
10 10

2 4 6 t(s) 0 2 4 6 t(s)

Space for rough work

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17. A block of mass 1 kg is pulled along the curve path ACB by a =0.2
tangential force as shown in figure. The work done by the friction force C
when the block moves from A to B is
(A) 5 J (B) 10 J A X=10m B
(C) 20 J (D) none of these

18. An inclined plane of height 10 m and angle of inclination 30 with the horizontal is having a
groove at an angle of 45 with the line of greatest slope. A particle is released from rest from the
top of incline plane in the groove, the time taken by the particle to reach the bottom of the groove
2
is (g = 10 m/s )
(A) 1 sec (B) 2 sec
(C) 4 sec (D) none of these

19. A light rod of length L, is hanging from on the vertical wall of a vehicle moving with acceleration
3 g having a small mass attached at it’s one end is free to rotate about an axis passing through
the other end. The minimum velocity given to the mass at it’s equilibrium position so that it can
complete vertical circular motion is
(A) 5gL (B) 4gL
(C) 8gL (D) none of these

20. A particle moves with a speed v = |t  3| m/s along a straight line, where t is time in seconds.
Distance travelled by the particle during first 5 seconds is equal to:
(A) 1.5 m (B) 2m
(C) 6.5 m (D) none of these

21. Two particles A and B starts from rest and move for equal time on a straight line. Particle A has
an acceleration of z m/s2 for the first half of the total time and 2zm/s2 for the second half. The
particle B has an acceleration of 2z m/s2 for the first half of the total time and z m/s2 for the
second half. Which particle has covered larger distance?
(A) A has covered the larger distance (B) B has covered the larger distance
(C) both have covered the same distance (D) none of these
Space for rough work

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22. A uniform circular ring of mass m and radius R is rotating in a gravity free space with angular
velocity  about its own axis. Tension in the ring is
mR2 mR2
(A) (B)
2 4
(C) zero (D) mR2

23. The displacement time graph for a particle in motion is shown in figure. X(m)
C
For which portion of the graph the force acting on the particle is opposite
to the direction of velocity B D
(A) AB (B) OA A
(C) BC (D) CD
0 t(s)

24. A cylinder of mass m and radius R is given velocity V0 over a smooth

V0
plank of same mass having velocity V0/2. The velocity of the V0/2
cylinder when it starts rolling without slipping is
V 7V0
(A) 0 (B)
8 8
V0 3V0
(C) (D)
4 4

25. In the previous question if coefficient of friction between the plank and cylinder is , the work
done by friction in the motion is
1 1
(A)  mv 20 (B)  mg
16 32
1
(C)  mv 20 (D) zero
32
Space for rough work

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26. A small sphere and a big sphere are released from rest with a very small O
gap from height h as shown in figure. The mass of bigger sphere is very
large as compared to mass of smaller sphere the height from the point
of collision of smaller sphere with the bigger sphere to which the smaller h
sphere will rise if all the collision is elastic
(A) 2h (B) 4h
(C) 6h (D) 9h

27. A particle is projected from a point at an angle with the horizontal at, t = 0. At an instant t, if p is
magnitude of linear momentum, x is horizontal displacement, y is vertical displacement E is
kinetic energy of the particle, then which of the following graph is correct.
E E

(A) (B)

O y O y
E E

O y O y

28. A particle on the top of a fixed sphere of radius R is given horizontal velocity 2gR , the angle
measured from vertical at which it will leave contact with the sphere is
(A) zero (B) cos 1  2 / 3 
(C) 30 (D) 45

29. Work will be done by a force on an object, if

(A) the object is stationary but the point of application of the force moves.
(B) the force is always perpendicular to its velocity
(C) the object moves in such a way that the force is perpendicular to its displacement.
(D) none of these

30. A block is suspended by an ideal spring constant K. If the block is pulled down by constant force
F and if maximum displacement of block from it’s initial position of rest is z, then
(A) z = F/K
(B) z = 2F/K
(C) work done by force F is equal to 2Fz.
1
(D) increase in potential energy of the spring is Kz2
2
Space for rough work

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Chemistry PART – II

SECTION – A

Straight Objective Type

This section contains 30 multiple choice questions. Each question has four choices (A), (B), (C) and (D)
out of which ONLY ONE is correct.
1. How many substituted products are possible on monochlorination of isohexane (including
isomers)?
(A) 5 (B) 6
(C) 7 (D) 8

2. For the reaction: N2 g  3H2 g 

 2NH3 g equilibrium constant is


2 2

(A)
 NH3
V (B)
  NH3
V2
3 3
    
N2 H2     
N2 H2

2 2

3. The rate constant for the reaction: 2N2O 5   4NO2  O2 is 6  105 s1. If the rate is
2.40  105 mol L1s1, then the concentration of N2O5 (in mol L1) is
(A) 0.8 (B) 0.4
(C) 0.2 (D) 0.1

4. Dolomite is mineral whose formula is

(A) CaCO3 (B) CaCO3.MgCO3
(C) MgCO3 (D) CaSO4.2H2O

5. Which of the following thermodynamic relation is correct for ideal gas?

 V   P 
(A)   0 (B)   0
 T P  T  V
 E   V 
(C)   0 (D)   0
 V  T  T P
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6. How many bonds are there in naphthalene?

(A) 18, 4 (B) 18 , 5
(C) 19 , 5 (D) 19 , 4

7. Which of the following is correct relation between mole fraction of solute into molarity of solution?

(A) M 
x2
(B) M 
x M
1 1
 x 2M2 
M2  X1M1  x 2M2  x 2M2
x2 1  m 
(C) M  (D) M 
x M
1 1
 x 2M2  mM2

8. Which of the following expression is correct for rate of diffusion of 2 gases?

(A) rH2  rD2 (B) rCO2  rNO 2
(C) rN2  rCO2 (D) rCO2  rN2 O
6 3 3
9. Ksp of a substance MX2 is 4  10 mol L . Molecular mass of the substance is 200 its solubility
would be
(A) 0.2 g/L1 (B) 2 g/L
(C) 20 g/L (D) 200 g/L

10. Which of the following represents the given mole of hybridization sp – sp – sp2 – sp2 – sp3 from
left to right?
(A) CH3 – CH3 – CH2 = CH – NH2 (B) CH  CH – CH2 – CH2 – NH2
(C) CH  CH – CH2 – NH – CH3 (D) CH  CH – N = N – CH3

11. The hybridization of diborane is

(A) sp (B) sp2
(C) sp3 (D) sp3d

12. Which of the following ion has the maximum value of magnetic moment?
(A) Mn2+ (B) Fe2+
2+
(C) Co (D) Ni2+
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13. 2 moles of SO2 and 1 mole of O2 are allowed to react in a container of capacity 2 litre and
following equilibrium is allowed to establish: 2SO2 g  O2 g
 2SO3 g


The equilibrium mixture requires 200 ml of 0.1 M acidified K2Cr2O7 solution. Hence Kc for the
above equilibrium is approximately
(A) 6.96  102 (B) 6.96  103
4
(C) 6.96  10 (D) 6.96  105

14. Which of the following radial probability curve is correct for a 3p – orbital? (Assume that all graphs
have been plotted for the same parameters. In all these curves principal quantum number < 4)
(A) (B)

Radial probability
Radial probability

density (A)
density (A)

2
2 3 5 10
Radial distabnce (r) (A)
Radial distabnce (r) (A)
(C) (D)

Radial probability
Radial probability

density (A)
density (A)

2 3 6 10 1.5
Radial distabnce (r) (A) Radial distabnce (r) (A)

15. Which of the following species is incorrect method with its shape?
(A) ClF3   T  shaped (B) SO32   tetrahedral
(C) C2 O42  
 planar (D) I3 
 linear

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16. Identify the incorrect statement (s) among the following:

(A) if spin quantum number curve having 3 possible values then 4th period of periodic table
would have maximum of 27 elements.
(B) platinum (Pt) is one of the typical transition metal.
(C) total number of f – electrons in Gd (at. No. 64) is 8.
(D) atomic radius of Ga is nearly equal to that of Al.

17. To prepare a buffer of pH 8.26 amount of (NH4)2SO4 to be added into 500 ml of 0.01 M NH4OH
solution [pka(NH4+) = 9.26] is
(A) 0.05 mole (B) 0.025 mole
(C) 0.10 mole (D) 0.005 mole

18. Consider the following equilibria:

 

CH3 COOH  H2 O   H3 O  CH3 COO
 


H2 SO 4  CH3 COOH 
 CH3 COOH2  HSO4

 H2NO3   HSO4



H2 SO 4  HNO3 
 


NH3  H2 O 
 NH4  OH
and, now select the incorrect statement among the following
(A) H2SO4 is a stronger acid than HNO3
(B) NH4+ is a weaker acid than H3O+
(C) acetic acid is stronger acid than H2O
(D) H2O is a weaker acid than H2SO4 and HNO3 both

19.
HBr
H3C CH CH 
peroxide
 A  major product 

Hence the major product (A) in the reaction is

(A) (B)
Br CH2 CH CH H3C CH CH Br

Br Br

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20. An inorganic compound (X) on hydrolysis produces a gas which on treatment with sodium
followed by its reaction with ethyl chloride gives another compound (Y). Compound (Y) on heating
with P – 2 catalyst gives (2Z) – pent – 2 – ene as major product. Hence the inorganic compound
(X) is
(A) Tl4C (B) BaC2
(C) SiC (D) Mg2C3

21. Which of the following metal reacts most vigorously with water?
(A) Li (B) Na
(C) K (D) Be

22. Boron trichloride on heating with ammonium chloride at 140C gives a compound (A) which on
reduction with sodium borohydride followed by hydrolysis gives another compound (B). Hence
compound (A) and (B) are respectively.
(A) inorganic graphite & ammonia (B) inorganic benzene & orthoboric acid
(C) trichloroborazine & orthoboric acid (D) diborane & orthoboric acid

23. (S) – 2 – chloropentane on reaction with Cl2 (g) in the presence of U.V. light, gives ‘N’ number of
optically active dichloro derivatives. Hence the value of N is
(A) 5 (B) 6
(C) 7 (D) 4

24. Which of the following statement regarding keto - enol forms of acetone and ethyl actoacetate is
correct?
(A) both are more stable in enol form than in keto form
(B) both are more stable in keto form than in enol form
(C) acetone is more stable in keto form whereas enol form is more stable in ethyl acetoacetate
(D) acetone is more stable in enol form whereas keto form is more stable in ethyl actacetate

25. 3 mole of KIO3, 4 mole of KI and 5 mole of HCl are allowed to react according to the equation
given below:
KIO3  2KI  6HCl  3ICl  3KCl  3H2 O
The number of moles of products are
(A) 27 (B) 18
(C) 7.5 (D) 9.5
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26. Which of the following statement is incorrect for the given graph?

Gas A
Fraction of Gas B (temp=T)
molecules

Velocity
(A) The molecular weight of gas A will be greater than gas B
(B) The average kinetic energy of both the gases will be similar at a particular temperature
(C) The average velocity of particles of gas A will always be less than of particles of gas B at
temperature T.
(D) The velocity of any particle of gas A can not be greater than velocity of any particle of gas B
at same temperature.

27. A water sample is found to have 384 ppm of SO 24 . What will be the degree of hardness of water
sample?
(A) 200 ppm (B) 300 ppm
(C) 400 ppm (D) 100 ppm

28. How many structural isomers are possible with molecular formula C7H16?
(A) 6 (B) 8
(C) 10 (D) 12

29. Which of the following is mismatched?

(A) LiCO3  least stable carbonate among alkali metal carbonates.
(B) LiF  almost insoluble in water due to its high covalent character.
(C) Ba(OH)2  most soluble hydroxide among alkaline earth metal hydroxide.
(D) BeSO4  more soluble than BaSO4 in water.

30. O

H /H2 O
' A ' 

H /
 B  O3
 C  
 H
Zn
Major 

O
(A) and (C) are isomers.
Which of the following can be structure of (A)?
(A) (B)

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16

Mathematics PART – III

SECTION – A

Straight Objective Type

This section contains 30 multiple choice questions. Each question has four choices (A), (B), (C) and (D)
out of which ONLY ONE is correct.

1. If the circle x 2 + y2 + 2gx + 2fy + c = 0 is touched by y = x at P such that OP = 6 2 , where O is

the origin, then the value of c is
(A) 36 (B) 144
(C) 72 (D) none of these

2. If  is the nth root of unity, then 1 + 2 + 32 + ..... to n terms is equal to

n n
(A)  2
(B) 
1    1 
2n 2n
(C)  (D) 
1  1   2

3. If (3 + x2008 + x2009)2010 = a0 + a1x + a2x2 + ..... + anxn, then the value of

1 1 1 1
a0  a1  a2  a3  a 4  a5  a6  ..... is
2 2 2 2
(A) 32010 (B) 1
(C) 22010 (D) none of these

4. Total number of words that can be formed using all letters of the word ‘BRIJESH’ that neither
begins with ‘I’ nor ends with ‘B’ is equal to
(A) 3720 (B) 4920
(C) 3600 (D) 4800

5. If cos2 A + cos2 B + cos2 C = 1, then ABC is

(A) equilateral (B) isosceles
(C) right angled (D) none of these

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 C
6. In triangle ABC, A = , then tan   is equal to
2 2
ac ab
(A) (B)
2b 2c
ac ab
(C) (D)
b c
–1 –1
7. For the equation cos x + cos 2x +  = 0, the number of real solution is
(A) 1 (B) 2
(C) 0 (D) 

8. Let A  (3, –4), B  (1, 2), let P  (2k – 1, 2k + 1) be a variable point such that PA + PB is
minimum. Then k is
7
(A) (B) 0
9
7
(C) (D) none of these
8

9. If x, 2y, 3z are in A.P., where the distinct numbers x, y, z are in G.P., then the common ratio of
the G.P. is
1
(A) 3 (B)
3
1
(C) 2 (D)
2

10. If |z2 – 3| = 3|z| then the maximum value of |z| is

3  21
(A) 1 (B)
2
21  3
(C) (D) none of these
2

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11. Total number of values of a so that x 2 – x – a = 0 has integral roots, where a  N and 6  a  100,
is equal to
(A) 2 (B) 4
(C) 6 (D) 8

12. If normals are drawn from a point P(h, k) to the parabola y2 = 4ax, then the sum of the intercepts
which the normals cut off from the axis of the parabola is
(A) h + a (B) 3(h + a)
(C) 2(h + a) (D) none of these

x2 y2 5
13. The eccentric angle of a point on the ellipse   1 at a distance of units from the focus
4 3 4
on the positive x–axis, is
3 3
(A) cos1   (B)   cos 1  
4 4
3
(C)   cos1   (D) none of these
4

14. The eccentricity of the conic represented by x 2 – y2 – 4x + 4y + 16 = 0 is

(A) 1 (B) 2
1
(C) 2 (D)
2

15. If centre of a regular hexagon is at origin and one of the vertices on Argand diagram is 1 + 2i,
then its perimeter is
(A) 2 5 (B) 6 2
(C) 4 5 (D) 6 5

16. The number of even divisors of the number N = 12600 is

(A) 72 (B) 54
(C) 18 (D) none of these

17. If x, y, z are in G.P. and ax = by = cz, then

(A) logb a = loga c (B) logc b = loga c
(C) logb a = logc b (D) none of these

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2 3 10
18. Consider the ten numbers ar, ar , ar , ....., ar . If their sum is 18 and the sum of their reciprocals
is 6 then the product of these ten numbers, is
(A) 81 (B) 243
(C) 343 (D) 324
2
19. The value of m for which one of the roots of x – 3x + 2m = 0 is double of one of the roots of
x2 – x + m = 0 is
(A) –2 (B) 1
(C) 2 (D) none of these
2 2
20. x – xy + y – 4x – 4y + 16 = 0 represents
(A) a point (B) a circle
(C) a pair of straight lines (D) none of these

1  2log3 2 2
21. The value of 2
  log6 2  is
1  log3 2 
(A) 2 (B) 3
(C) 4 (D) 1

sin3  sin5  sin7  sin9

22. is equal to
cos 3  cos5   cos 7   cos 9
(A) tan 3 (B) cot 3
(C) tan 6 (D) cot 6

 r  r 
23. If in a triangle,  1  1  1  1   2 , then the triangle is
 r2  r3 
(A) right angled (B) isosceles
(C) equilateral (D) none of these

x xy
24. tan1    tan1   is
y xy
 
(A) (B)
2 3
  3
(C) (D) or 
4 4 4

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25. The circle x2 + y2 + 2x = 0,   R, touches the parabola y2 = 4x externally. Then

(A)  > 0 (B)  < 0
(C)  > 1 (D) none of these

26. In a triangle ABC, A  (, ), B  (2, 3) and C  (1, 3) and point A lies on line y = 2x + 3 where
  I. Area of ABC, , is such that [] = 5. Possible coordinates of A are (where [.] denotes the
greatest integer function)
(A) (2, 3) (B) (5, 13)
(C) (–5, –7) (D) (–3, –5)
3 2 1985 100
27. Sum of common roots of the equations z + 2z + 2z + 1 = 0 and z +z + 1 = 0 is
(A) –1 (B) 1
(C) 0 (D) none of these

28. The last two digits of the number 3400 are

(A) 81 (B) 43
(C) 29 (D) 01

29. If tn denotes the nth term of the series 2 + 3 + 6 + 11 + 18 + ….. then t50 is
(A) 492 – 1 (B) 492
2
(C) 50 + 1 (D) 492 + 2

30. Number of ways in which Rs. 18 can be distributed amongst four persons such that no body
receives less than Rs. 3 is
(A) 126 (B) 84
(C) 4! (D) none of these

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(Advanced), 2014: 2521 FIITJEE Students from Classroom / Integrated School Programs & 3579 FIITJEE Students from All Programs have
JEE(Main), 2015
ALL INDIA INTEGRATED TEST SERIES ANSWERS, HINTS & SOLUTIONS

HALF COURSE TEST–VI

(Main)

Q. No. PHYSICS CHEMISTRY MATHEMATICS

1. B D C
2. A B B
3. A B C
4. D B A
5. B C C
qualified in JEE (Advanced), 2014.

6. A C D
7. B C C
8. D D C
9. D B B
10. D D B
11. B C D
12. C A C
13. D C A
14. D C B
15. A B D
16. B C B
17. C B C
18. C B B
19. C C A
20. C D A
21. B C D
22. A C C
23. C A A
24. B C C
25. C C A
26. D D B
27. D C A
28. A B D
29. D B D
30. B C B

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2

Physics PART – I
SECTION – A
1. In sliding maximum friction will act where as in rolling static friction will act.
2. Conservation of angular momentum
 m 2 
mv    m 2   (1)
 3 
conservation of energy
 1  m 2  
mg  mg    m 2  2  mg  mg (2)
2 2 3  2
 v  8g
3. Apply conservation of angular momentum.
4. Cheek the distribution of mass about the three axis.
 1  m   2 m  52 2
2
m   1  m    5
5. I , I'               I
3 12  2  2  2  16   3  2  2  8

6. acceleration a = g/2
g g
velocity after 1 sec v   1  m/s
2 2
g 1
the string will given become taut when t  gt 2
2 2
 t = 1 sec.
7. Vm /M  VM (1)
2 2 2
V V
m m /M  V  2Vm /M VM cos120
M

 Vm / VM  1

spring force  mg m2 g  m1g

8. a 
m1 m1
9. From the graph it can be concluded that both S1 & S2 are accelerated frame or either S1 or S2 is
accelerated.

10. The tension in the string is zero and block is falling freely.

11. Conservation of linear momentum

mu
mu = (M + m)V  V 
Mm
mu
12. Using conservation of energy and F 
Mm
14. Angular velocity and angular acceleration are parallel or antiparallel in a circular motion.

16. Displacement time curve will be a parabola.

  x dx N
17. Work done by friction   F  ds   mgcos  
0 cos 
f
0
= mg x = 20 J
mg dx ds dy
cos   
ds
dx

1
18.   2 2h   gsin30 cos 45  t 2 2h h=10 m
2
45
t = 4 sec. 30

1    3
19. Conservation of energy mv 2  mg  mg  m 3  2 
2 2 2  2 

v  8g 60

3 5

20. s     3  t  dt    t  3  dt   6.5m
0 3 
d
22. 2T sin  dmR2
2 T T
m d
T R2
2
23. OA  acceleration is positive
AB  acceleration zero
BCD  velocity is decreasing i.e. acceleration is negative
24. for cylinder Vc = V0  gt (1)
2g
=0+ t (2)
R
V
for plank VP  0  gt (3)
2
VP  VC  R (4)
7
 VC  V0
8
25. Apply work energy theorem to calculate work done by friction

26. conservation of momentum

M 2gh  m 2gh =MV1 + mV2 ...(1)
V1  V2
1  ...(2)
2 2gh
V2  3 2gh
V22
h'   9h
2g
mv 2
28. The particle will leave contact at the same point if mg  N 
R
i.e N = 0; for v  gR

30. apply work-energy theorem

2 2
 1  mg  1  mg  
mgz   K   z  K    Fz  0
 2  K  2  K  
 z = 2F/K.

Chemistry PART – II
SECTION – A
1. CH3
H3C CH CH2 CH2 CH3
1 2 3 4 5
Iso hexane has 5 type of H for substitution in which C–1, C–3 & C–4 will form optically active
compound on mono chlorination.

2. 
 2NH3
N2  3H2 
2 2

Kc 
NH3

/V 
  NH3
V2
3 3
  / V   / V      
N2 H2 N2 H2

Rate 2.4  10 5
3. N O  
2 5
K

6  10 5
 0.4 mol L1

6. H H
H H

H H
H H

n2
7. x2 
n1  n2
Mass of solution = n1M1 + n2M2

Volume of solution =
n M 1 1
 n2M2 

n2 n2 x2
M  
V  n1M1  n2M2   x1M1  x 2M2
1
8. r
M
MCO 2
 MN2O


 M2  2X
MX2 
9.
s 2s
6 3
Ksp = 4  10 = 4s
s = 102 mol/lit
s = 200  102 g/L = 2g/lit

12. Mn2+ = 3d54s0

Fe2+ = 3d64s0
Co2+ = 3d74s0
2+ 8 0
Ni = 3d 4s

2SO 2 g  O2 g

 2SO3 g


13. Initial mol. 2 1 0
2  2x 1 x2x
Eq. conc.
2 2 2
K2Cr2O7 oxides, SO2 remaining of equilibrium, Hence
Equivalent of SO2 left = equivalents of K2Cr2O7 used
(2 – 2x) × 2 = (0.2 × 0.1 ×6)
This gives, x  0.97
2
 2x 
 
Now, K   2   6.96  104
2
 2  2x   1  x 
 2   2 
   
Hence (C) is correct.

14. Number of radial node in 3p orbital = 3 – 1 – 1 = 1 is its radial probability curve will meet the
x – axis only at one point and this is the case in curve (A) and (C) is either for 3p or 2s orbital.
Also, as the shell number increases, the distance of maximum probability increases from the
nucleus. Hence (C) is correct.

15. In, SO32  , S has 3 bond pairs and one lone pair and hence, it is pyramidal in shape not tetrahedral.
Hence (B) is correct.

16. (A). Three possible spin quantum number means an orbital can have at the most three electrons.
So 4th period involves,
3s,3d and 4p subshells.
So, maximum of electrons (and hence elements) = 4s33d154p9 = 3 + 15 + 9 = 27
(B). All elements of VIIIth group (i.e. 8, 9, 10 groups) are called typical transition metals. Pt also
belongs to 10th group. So it is a typical transition metal.
(C). E.C. of Gd (64) = [Xe]4f 75d16s2
So number of f – electrons = 7
(D). Due to ineffective shielding by 3d – electrons in Ga, size of Ga is smaller (almost equal) to
that of Al.
Hence (C) is correct.

17. A mixture of NH4OH and (NH4)2SO4 is a basic buffer.

pOH  pkb  log


NH4 
NH4OH
K
p b (NH4OH) = 14 – 9.26 = 4.74
pOH = 14 – 8.26 = 5.74
Let, moles of (NH4)2SO4 to be added = x
2x
5.74  4.74  log
5  10 3
 x = 0.025 mole
Hence (B) is correct.

18. The following equilibrium shows that

 

NH3  H2 O  
 NH4  OH
So, NH4+ is a stronger acid than H3O+.
Hence (B) is correct.

19. It is a case of free radical addition to C = C bond.

Br 0 o o
H3C CH CH   H3C CH CH H3C CH CH
Br Br
(more stable)

H3C CH 2 CH

Br
major product (A)

20. H2 O
Mg2 C3   CH3  C  C  H   i Na
 ii CH3 CH2  Cl  H3 C  C  C  CH2  CH3
 X  Y

P 2 - catalyst

Me Et
C C
H H
( 2Z ) - pent - 2 - ene
Hence (D) is correct.

21. Since melting point of K is least among all, so heat released in the reaction immediately melts its
surface and catches fire.
Hence (C) is correct.
0
140 C NaBH4 H2 O
22. BCl3  3NH4 Cl  B3N 3H3 Cl3 
 reduction 
 BNH3 3 6   NH3  H3BO3  H2
(A )
 trichloroborazine   Borazine or
Inorganic benzene  B

Hence (C) is correct.

23. CH3 H 2C Cl CH3 CH3 CH3 CH3

Cl2
H Cl  H
 hv
Cl Cl Cl H Cl H Cl H Cl
H 2C CH 2 CH3 H 2C CH2 CH3 H 2C CH2 CH3 CH Cl CH 2 CH2
(S)-2-chloropentane (optically active (1) (optically inactive) H 2C CH3 CH Cl CH2
(optically CH3 H2 C Cl
active (2)
(one optically (optically
active and one active (1)
optically
inactive)

Hence (A) is correct.

24. Stability due to H – boding is present in ethyl acetoacetate.

O O H
O O
H3C C CH2 C OEt
C C
H3C CH2 OEt

25. The limiting reagent here is HCl. So calculating with the limiting reagent, the total number of
produces is 7.5.

26. It may be possible that few particles of gas A have higher velocities than that of few particles of
gas B.

27. 96 ppm of SO24  = 100 ppm of CaCO3 so 384 ppm of SO24  = 400 ppm of CaCO3.

28.

29. LiF is insoluble in water but not because of its covalent character rather its high lattice energy
which is not easily compensated by its hydration energy.

30.
  O3
H /H2 O H /
  OH  
Zn

 O
O
H
(Major)

Mathematics PART – III

SECTION – A

1. Here, P  (6, 6)
It lies on the circle, thus 72 + 12(g + f) + c = 0 ..... (1)
Since y = x touches the circle, hence (g + f)2 = 2c ..... (2)
Solving (1) and (2), c = 72
2 n–1
2. Let, S = 1 + 2 + 3 + ..... + n
 S =  + 22 + 33 + ..... + (n – 1)n – 1 + nn
On subtracting, S(1 – ) = 1 + [ + 2 + ..... + n – 1] – nn
n
 S
1 

3. Put x = , 2 ..... (1)

2010 2
2 = a0 + a1 + a2 + ..... ..... (2)
22010 = a0 + a12 + a2 + .....
Adding (1) and (2) we get, 2  22010 = 2a0 – a1 – a2 + 2a3 – a4 – a5 + 2a6 – .....
1 1 1 1
 22010 = a0  a1  a2  a3  a 4  a5  a6  .....
2 2 2 2

4. Total number of required words = 7! – 6! – 6! + 5!

5. cos2 A + cos2 B – (1 – cos2 C) = 0

cos2 A + cos2 B – sin2 C = 0
cos2 A + cos(B + C) cos(B – C) = 0
 2 cos A cos B cos C = 0
Hence, either A or B or C is 90º

6. Here, a2 = b2 + c2
C
2 tan
c 2
sinC  =
a 2 C
1  tan
2
C C
 c tan2  2a tan  c  0
2 2
C ab
 tan 
2 c
C ab ab 
 tan  ( for , C )
2 c c 2

7. cos–1 x + cos–1 2x = –, which is not possible

8. PA + PB is minimum if PA + PB = AB
3 4 1
 1 2 10
2k  1 2k  1 1
7
 k
8

9. Here, y = xr, z = xr2

Also, x, 2y and 3z are in A.P.
Hence, 4y = x + 3z
2
 4xr = x + 3xr
2
 3r – 4r + 1 = 0
1
 r
3

10. |z2 – 3|  |z|2 – 3

3|z|  |z|2 – 3
|z|2 – 3|z| – 3  0
3  21
0 < |z| 
2

11. x2 – x – a = 0, D = 1 + 4a = odd
D must be perfect square of some odd integer
Let D = (2 + 1)2
 1 + 4a = 1 + 42 + 4
 a = ( + 1)
Now, a  [6, 100]
 a = 6, 12, 20, 30, 42, 56, 72, 90

12. Equation of normal y = mx – 2am – am3

Put y = 0, we get x 1 = 2a + am12
x 2 = 2a + am22
x 3 = 2a + am23
The normal passes through (h, k)
 am3 + (2a – h)m + k = 0
 m1 + m2 + m3 = 0
2a  h
m1m2 + m2m3 + m3m1 =
a
2 2a  h 

 m12  m22  m32  
a
 x1 + x2 + x3 = 6a – 2(2a – h) = 2(h + a)

13. Any point on the ellipse is (2 cos , 3 sin )

The focus on the positive x-axis is (1, 0)
2 2 25
Given that (2 cos  – 1) + 3 sin  =
16
3
 cos  
4

 x  2 2  y  2 2
14. Given curve is   1
42 42
So, a = b = 4
 e 2

15. Let the vertices be z0, z1, z2, z3, z4, z5 w.r.t. centre O at origin and z0  5
Now, OA2A3 is equilateral

 A 2 A3  5
 Perimeter = 6 5

16. N = 12600 = 23325271

Number of even divisors = (3)(2 + 1)(2 + 1)(1 + 1) = 54
2
17. Here, y = xz ..... (1)
x y z
Also, a = b = c =  (Say)
log  log  log 
 x , y , z
loga logb logc
Putting these values in (1), we get logb a = logc b

ar  r10  1
18. Given  18 ..... (1)
r 1
1 1 
1  
ar  r10 
Also, 6
1
1
r
1 ar r10  1
 2 11
6 ..... (2)
a r r 1
1
From equation (1) and (2), we get 2 11
 18  6
ar
 a2r11 = 3
Now, P = a10r53 = 35 = 243

19. Let  be the root of x 2 – x + m = 0 and 2 be the root of x2 – 3x + 2m = 0

Then, 2 –  + m = 0 and 42 – 6 + 2m = 0
Eliminating , m2 = –2m  m = 0, m = –2

20. Given equation is, x2 – (y + 4)x + y2 – 4y + 16 = 0

Since x is real, so, D  0
 (y + 4)2 – 4(y2 – 4y + 16)  0
 (y – 4)2  0
y=4
Since equation is symmetric in x and y
Hence, x = 4 only

2
21.
1  2log3 2 log3 2 
2
 2
1
1  log3 2  1  log3 2 
 sin3  sin9    sin5  sin7 
22.
 cos 3  cos 9   cos 5  cos 7  
2 sin6  cos 3  cos  
=  tan6
2cos 6  cos 3  cos  

  s  b    s  c  
23. 1   1  2
 s  a  sa 
2 2 2
a =b +c
Hence, triangle is right angled

x xy
24. tan1    tan1  
y xy
 y
 x   1   
= tan1    tan1   x  
y  1  y  
  
 x 
x  y
= tan1     tan1 1  tan1 
y  x
x y    
= tan1  tan1    
y x 4 2 4 4

25. The graph shows  > 0 y

2
y = 4x
(–, 0)
x x
(0, 0)

y

26. A  (, 2 + 3), BC = 1 unit. Equation of BC is y – 3 = 0

Distance of A from BC is p = |2 + 3 – 3|
Area of ABC =  = ||; 5   < 6
 5  || < 6

27. We have, z3 + 2z2 + 2z + 1 = 0

 z = –1, , 2
Since z = –1 does not satisfy z1985 + z100 + 1 = 0
While z = , 2 satisfy it
Hence, sum =  + 2 = –1

28. 3400 = (81)100 = (1 + 80)100

= 100C0 + 100C180 + ….. + 100C10080100
 Last two digits are 01

29. 2 + 3 + 6 + 11 + 18 + ….. = (02 + 2) + (12 + 2) + (22 + 2) + (32 + 2) + …..

Hence, t50 = 492 + 2

30. Distribute Rs. 3 to each person

Then, we have to find non–negative integral solution of y1 + y2 + y3 + y4 = 6
6+4–1 9
Which is given by C4 –1 = C3 = 84

Time Allotted: 3 Hours Maximum Marks: 243

 Pl ea s e r ea d t h e i n s t r u c t i o n s c a r ef u ll y . Yo u a r e a l l o t t ed 5 m i n u t es
s p ec i f i c a ll y f o r t h i s p u r p o s e.
 Yo u a r e n o t a l l o wed t o l ea v e t h e E xa m i n at i o n Ha l l b ef o r e t h e en d o f
t h e t es t .

INSTRUCTIONS
A. General Instructions
1. Attempt ALL the questions. Answers have to be marked on the OMR sheets.
2. This question paper contains Three Parts.
3. Part-I is Physics, Part-II is Chemistry and Part-III is Mathematics.
4. Each part is further divided into two sections: Section-A & Section-B
5. Rough spaces are provided for rough work inside the question paper. No additional sheets will be
provided for rough work.
6. Blank Papers, clip boards, log tables, slide rule, calculator, cellular phones, pagers and electronic
devices, in any form, are not allowed.
B. Filling of OMR Sheet
1. Ensure matching of OMR sheet with the Question paper before you start marking your answers
on OMR sheet.
2. On the OMR sheet, darken the appropriate bubble with black pen for each character of your
Enrolment No. and write your Name, Test Centre and other details at the designated places.
3. OMR sheet contains alphabets, numerals & special characters for marking answers.
C. Marking Scheme For All Three Parts.
(i) Section-A (01 to 09) contains 09 multiple choice questions which have only one correct answer.
Each question carries +3 marks for correct answer and – 1 mark for wrong answer.
Section-A (10 – 13) contains 4 Assertion-Reasoning (multiple choice questions) which have
only one correct answer. Each question carries +3 marks for correct answer and
– 1 mark for wrong answer.
Section-A (14 – 19) contains 2 paragraphs. Based upon paragraph, 3 multiple choice questions
have to be answered. Each question has only one correct answer and carries +4 marks for
correct answer and – 1 mark for wrong answer.
(ii) Section-B (1 – 03) contains 3 Matrix Match Type (4 × 4 Matrix) questions containing statements
given in 2 columns. Statements in the first column have to be matched with statements in the
second column. Each question carries +6 marks for all correct answer. There is no negative
marking.

Name of the Candidate

Enrolment No.

Useful Data

PHYSICS
2
Acceleration due to gravity g = 10 m/s

Planck constant h = 6.6 1034 J-s

Charge of electron e = 1.6  1019 C

Mass of electron me = 9.1  1031 kg

Permittivity of free space 0 = 8.85  1012 C2/N-m2

Density of water water = 103 kg/m3

Atmospheric pressure Pa = 105 N/m2

Gas constant R = 8.314 J K1 mol1

CHEMISTRY

Gas Constant R = 8.314 J K1 mol1

Atomic No: H=1, He = 2, Li=3, Be=4, B=5, C=6, N=7, O=8,

Physics PART – I

SECTION – A
Single Correct Choice Type

This section contains 9 multiple choice questions. Each question has four choices (A), (B), (C) and (D)
out of which ONLY ONE is correct.

1. A particle moves over the sides of an equilateral triangle of side  with B

constant speed v as shown in figure. The magnitude of average
acceleration as it moves from A to C is  
v2 3 v2
(A) (B)
 2  A  C

3v 2 v2
(C) (D)
 2

2. In the arrangement shown in the figure pulley is light and smooth.

The extension in the spring is (g = 10 m/s2)
(A) 1.33 cm (B) 1 cm K = 1000 N/m
(C) 1.67cm (D) 2 cm

1 kg
2 kg

3. A block of mass m is gently placed over a massive plank moving horizontally over a smooth
surface with velocity 6 m/s. The coefficient of friction between the block and plank is 0.2. The
distance travelled by the block till it stop with respect to the plank is (g = 10 m/s2)
(A) 4 m (B) 6 m
(C) 9 m (D) 12 m

4. A particle is given velocity 5 m/s on a fixed large inclined surface as

shown in the figure. The radius of curvature of the path of the particle 5 m/s
after 1 sec from the start is about(g = 10 m/s2) 90
(A) 7 m (B) 14 m
(C) 21 m (D) 28 m 30

Space for rough work

5. A liquid drop at temperature t, isolated from its surrounding breaks into a number of droplets the
temperature of the droplets will be
(A) equal to t (B) greater than t
(C) less than t (D) none of these

6. In a mixture of gases, the average number of degree of freedom per molecule is 6. The rms
speed of the molecules of the gas is C. The velocity of sound in the gas is
C 3C
(A) (B)
2 4
2C C
(C) (D)
3 3

7. A wall of width  and cross sectional area A is having variable coefficient of thermal conductivity
given by k = k0 + x (where k0 and  are positive constant) and x is measured from outer surface
of wall. If the temperature of surroundings is T0C and the temperature of the room is maintained
at Tr C(T0 > Tr). The temperature of the wall at x = /2 is (given /K0 = 1 S.I. unit, T0 = 45C, and
Tr = 20C)
(A) 20C (B) 25C
(C) 28C (D) 30.37C

8. A particle of mass 2 kg is moving with a uniform speed of 3 2 m/s in XY plane along the line y =
(x + 4)m. The magnitude of the angular momentum about origin is
(A) zero (B) 24 kgm 2/sec
2
(C) 30 kgm /sec (D) 35 kgm 2/sec

9. A bob of mass m is in equilibrium with the help of two inextensible P

A C
string connected to fixed support. The bob is slightly displaced
perpendicular to the plane of figure and released. The time period of  
oscillation of bob is B
 d d
(A) 2 (B) 2
g g

d ( 42  d2 )
(C) 2 (D) 2
g 2g

Space for rough work

Assertion - Reason Type

This section contains 4 questions numbered 10 to 13. Each question contains STATEMENT  1
(Assertion) and STATEMENT -2 (Reason). Each question has 4 choices (A), (B), (C) and (D) out of which
ONLY ONE is correct.

10. STATEMENT-1: A ball is projected upwards with such a velocity so

as to reach to a height H. Collision of ball with ground is elastic. h<H
There is a roof at height h < H. Whatever may be the value of h and
e (the coefficient of restitution between the roof and ball), the ball
will definitely collide with the roof for second time provided e  0.

because
STATEMENT-2: Magnitude of velocity will not change by the collision with ground.
(A) Statement-1 is True, Statement-2 is True; Statement -2 is a correct explanation for
Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement -2 is NOT a correct explanation for
Statement-1.
(C) Statement -1 is True, Statement-2 is False.
(D) Statement -1 is False, Statement-2 is True.

11. STATEMENT-1: Radius of circular orbit of a satellite is made two times, then its areal velocity will
also becomes two times.

because
dA L mvr
STATEMENT-2: Areal velocity is given as   .
dt 2m 2m
(A) Statement-1 is True, Statement-2 is True; Statement -2 is a correct explanation for
Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement -2 is NOT a correct explanation for
Statement-1.
(C) Statement -1 is True, Statement-2 is False.
(D) Statement -1 is False, Statement-2 is True.
Space for rough work

12. STATEMENT-1: If a particle is projected from the ground with some velocity (not very large) at an
angle with horizontal, path of the particle is assumed parabolic, although it is not exactly
parabolic.

because
STATEMENT-2: Magnitude of acceleration due to gravity at all places is assumed to be the
constant, although it is not so.
(A) Statement-1 is True, Statement-2 is True; Statement -2 is a correct explanation for
Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement -2 is NOT a correct explanation for
Statement-1.
(C) Statement -1 is True, Statement-2 is False.
(D) Statement -1 is False, Statement-2 is True.

13. STATEMENT-1: The two bodies of masses M and m (M > m) are allowed to fall from the same
height. If the air resistance for each be the same then both the bodies will reach the earth
simultaneously.

because
STATEMENT-2: For same air resistance, acceleration of both the bodies will be different.
(A) Statement-1 is True, Statement-2 is True; Statement -2 is a correct explanation for
Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement -2 is NOT a correct explanation for
Statement-1.
(C) Statement -1 is True, Statement-2 is False.
(D) Statement -1 is False, Statement-2 is True.
Space for rough work

Linked Comprehension Type

This section contains 2 paragraphs. Based upon each paragraph, 3 multiple choice questions have to be
answered. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

Paragraph for Question Nos. 14 to 16

The air column in a pipe closed at one end is made to vibrate in its third over tone by tuning fork of
frequency 220 Hz. The speed of sound in air is 330 m/sec. End correction may be neglected. Let P0
denote the mean pressure at any point in the pipe and P0 the maximum amplitude of pressure vibration.

14. Find the length of the air column

(A) 3.2 m (B) 2.625 m
(C) 4.23 m (D) 1.16 m

15. What is the amplitude of pressure variation at the middle of the column
P0 3P0
(A) (B)
2 2
P0
(C) P0 (D)
2

16. What is the maximum pressure at the open end of the pipe?
(A) P0 (B) P0  P0
P
(C) P0  0 (D) none of these
2

Space for rough work

Paragraph for Question Nos. 17 to 19

P
Rolling motion can be considered as superposition of pure translation
and pure rotation about centre of mass of a body of uniform density. C1 C2

Rolling without slipping is a special case of rolling, in which the

P Q
distance travelled by centre of mass of body of uniform density is
t = 0 sec t = t sec
equal to the distance travelled by point of contact on body of uniform
density [say P] in a given time interval.

If C1C2 = PQ  Rolling without slipping, and if C1C2  PQ  Rolling with slipping

Hence if velocity and tangential acceleration of point P with respect to 0

surface must be zero but point P has a total acceleration in case of rolling A
without slipping. The rolling without slipping can also be viewed as a pure
rotation about an axis passing through the point of contact and
perpendicular to plane of motion, called “Instantaneous axis of rotation 
B
A disc having radius R and mass M is placed gently on rough inclined surface as shown in the figure.
Disc rolls down and comes to B in time t after placing at A. If AB is  and 0 is the angular velocity of disc
in clockwise direction just before placing it at A (Assume that the friction is sufficient to provide pure rolling
to the disc for the given length of inclined plane).

17. A body performing rolling with slipping then

(A) the radial velocity of point of contact of body relative surface is zero
(B) the tangential velocity of point of contact of body relative to surface is zero
(C) the net velocity of point of contact of body relative to surface is zero
(D) none of the above

18. The nature of rolling of disc during motion from A to B is

(A) Rolling with slipping (B) rolling without slipping
(C) some time (A) sometime (B) (D) can’t say

19. Work done by frictional force during time interval t is

(A) zero (B) positive
(C) Negative (D) none of the above
Space for rough work

SECTION – B
(Matching List Type)

This section contains 3 multiple choice questions. Each question has matching Column(s). The codes
for the Column(s) have choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

1. Match the processes in column I with the possibilities in column II, where Q and W are heat
change and work done respectively.
Column A Column B
(A) Isobaric (p) Q > O
(B) Isothermal (q) Q < O
(C) Isochoric (r) W>O
(D) Adiabatic (s) W<O

Space for rough work

2. If the position vector of a particle at point P moving in space and distance traveled from a fixed

point on the path are given by r and s respectively. Three vectors are defined as follows

 dr  dN   
N , T R and B  N  T
ds ds

where R is the radius of curvature at point P and r is a non-zero vector. Now match the following
on the basis of above concept
Column A Column B

(A) T (p) Unit vector

(B) N (q) Zero
 
(C) N  T (r) v̂ (unit vector along the direction of velocity)
  
(D) B va  v aT 
  , where a and aT are acceleration and magnitude of
(s) v a
tangential acceleration of particle P.

3. Match the following.

Column – A Column – B
(A) Equation of continuity (Av = constant) (p) non-viscous flow
(B) Bernoulli’s equation (q) steady flow
(C) Velocity at a given point is constant with time. (r) incompressible flow
(D) Ideal flow (s) irrotational flow

Space for rough work

Chemistry PART - II

SECTION – A
Single Correct Choice Type

This section contains 9 multiple choice questions. Each question has four choices (A), (B), (C) and (D)
out of which ONLY ONE is correct.

1. Lattice energy of chloride of alkali metals is in order:

(A) LiCl > NaCl > KCl > RbCl > CsCl (B) LiCl < NaCl < KCl < RbCl < CsCl
(C) NaCl < KCl < LiCl < RbCl > CsCl (D) NaCl < KCl < RbCl < CsCl > LiCl

2. A ball of mass 200 g is moving with a velocity of 10 m/sec. if the error is the measurement of
velocity is 0.1%, then the uncertainty in its position is
(A) 3.3  1031 m (B) 3.3  1027 m
25
(C) 5.3  10 m (D) 2.6  1032 m

3. Fluorine reacts with uranium to form UF6:

U  s  + 3F2  g   UF6  g 
How many fluorine molecules are required to produce 2 mg of UF6 from an excess of uranium?
(The molar mass of UF6 is 352 g/mol)
(A) 3.4  1018 (B) 1  1019
19
(C) 2  10 (D) 3.4  1021

4. In the disproportion reaction:

H3PO2  PH3 + H3PO3
The equivalent mass of H3PO2 is (m = molecular mass of H3PO2)
(A) 3m (B) 3m
4 5
(C) 5m (D) m
6 4
Space for rough work

5. For the reaction: N g + 3H g 

 2NH3 g; K p = 6.1×10-5 at 25C. G for the reaction will

 
2 2
be (log1061 = 1.78)
(A) 24 kJ (B) 24 kJ
(C) 7.3 kJ (D) 2 kJ

6. A diatomic molecule has a dipole moment of 1.2 D. If the bond length is 1.0  108 cm, what
fraction of charge does exist on each atom?
(A) 0.1 (B) 0.2
(C) 0.25 (D) 0.3

7. Which of the following has longest C – O bond?

(A) O (B) O

CH2

8. The given pair is

Br Br
H and C2H5
H 5C 2 CH3 H CH3
(A) enantiomers (B) homomers
(C) constitutional isomers (D) diastereomers

9. Which of the following species will act only as electrophile and not as nucleophile?

H3C B CH2 CH2 NH2

(A) (B)
NH CH3

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Assertion - Reason Type

10. STATEMENT-1: In Lother Meyer Curve the alkali elements occupy maxima of the curve.
and
STATEMENT-2: Alkali elements have largest volume in a period.
(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for
Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for
Statement-1.
(C) Statement-1 is True, Statement -2 is False.
(D) Statement-1 is False, Statement-2 is True.

11. STATEMENT-1: Cyclopropane has greater ring strain than cyclobutane ,yet this is readily formed.
and
STATEMENT-2: Formation of four membered ring involves larger decrease in entropy than that of
three membered ring.
(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for
Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for
Statement-1.
(C) Statement-1 is True, Statement -2 is False.
(D) Statement-1 is False, Statement-2 is True.
Space for rough work

12. STATEMENT-1: pH of a neutral solution is always equal to pk w / 2 at any temperature.

and
STATEMENT-2: A neutral solution at any temperature will have same concentration of H  and
OH  ion in the expression for Kw.
(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for
Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for
Statement-1.
(C) Statement-1 is True, Statement -2 is False.
(D) Statement-1 is False, Statement-2 is True.

13. STATEMENT-1: 0.74g of Ca  OH2 is required for 10 litre of water to remove temporary hardness
of 100 ppm due to Mg(HCO3)2 (atomic mass: Ca = 40, O = 16, H = 1,Mg = 24, C = 12).
and
STATEMENT-2: Hardness of water is caused due to calcium and magnesium salts of
Cl ,SO24 and HCO3 .
(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for
Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for
Statement-1.
(C) Statement-1 is True, Statement -2 is False.
(D) Statement-1 is False, Statement-2 is True.
Space for rough work

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Linked Comprehension Type

This section contains 2 paragraphs. Based upon each paragraph, 3 multiple choice questions have to be
answered. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

Paragraph for Question Nos. 14 to 16

Alkene undergo free radical substitution reaction as it is carried out in the presence of sunlight.
Experimentally it is found that relative rate of abstraction of hydrogen atoms from 1, 2 and 3 carbon
atoms are in the ratio 1: 3.8 : 5 during the monobromination while this ratio is 1 ; 82 : 1600 during the
monobromination. Alkyl halide undergoes nucleophilic substituton reaction which may be unimolecular or
bimolecular depending upon the nature of substrate alkyl halide. Former reaction proceeds via the
carbocation intermediate while latter one proceeds via transition state.

Cl2 / 
14. CH2  CH  CH2  CH3    A   B 
In the given equation (A) and (B) are
(A) H2C CH CH CH3 and H2C CH CH2 CH2
Cl Cl
(B H2C CH CH CH3 and H2C CH CH CH3
Cl Cl
(C) H2C CH CH2 CH3 and HC CH CH2 CH3
Cl Cl
(D) H2C C CH2 CH3 and H2C CH CH CH3
Cl Cl

15. Predict the product ratio during the monochlorination of 2 – methylbutane?

(A) 9 : 5 : 3 (B) 9 : 5 : 3.8 : 3
(C) 9 : 7.6 : 5 (D) 6 : 3 : 7.6 : 5

16. % isomer of products corresponding to 1 and 3 C – atoms obtained during the

monobromination of 2 – methylpropane are
(A) 0.5% and 99.5% (B) 80% and 20%
(C) 75% and 25% (D) 15% and 85%
Space for rough work

Paragraph for Question Nos. 17 to 19

Werner Heisenberg stated the uncertainty principle which is the consequence of dual behaviour of matter
and radiation. It states that if is impossible to determine simultaneously, the exact position and exact
momentum (or velocity) of an electron.
Mathematically, it can be given as the equation:
h
x  p 
4
Where x is the uncertainty in position and px (or Vx) is the uncertainty in momentum (or velocity) of the
particle.
If the position of the electron known with high degree of accuracy (x is small), then the velocity of the
electron will be uncertain [(Vx) is large]. On the other hand, if the velocity of the electron is known
precisely [(Vx) is small], then the position of the electron will be uncertain (x will be large). Thus if we
carry out some physical measurements on the electron’s position or velocity, the outcome will always
depict a fuzzy or blue picture.

17. An electron moves around the nucleus in a circle of radius r. Assuming that the uncertainty of the
momentum of electron is of the same order as momentum itself, the momentum of the electron
would be
h
(A) (B) 2hr
4r
h
(C) (D) 4hr
2r

18. The uncertainty in the velocity of a cricket ball of mass 100 g, when uncertainty in its position is of
the order of 1A, would be
(A)  5.27  10 24 ms 1 (B)  2.35  10 23 ms 1
 22 1
(C)  3.16  10 ms (D)  8.51  10 24 ms 1

19. If one tries to find the exact location of an electron to an uncertainty of only 108 m, then
uncertainty in velocity, i.e. V would be
(A) 104 ms1 (B) 104 ms1
8
(C) 10 ms 1
(D) 108 ms1
Space for rough work

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SECTION – B
(Matching List Type)

This section contains 3 multiple choice questions. Each question has matching Column(s). The codes
for the Column(s) have choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

1. Match the electronic configuration listed in Column – II with the descriptions listed in Column – I:
Column – I Column – II
(A) Violation of Aufbau’s rule (p)

(B) Violation of Pauli’s exclusion (q)

principle
(C) Violation of Hund’s rule (r)

(D) Violation of both Pauli’s and (s)

Hund’s rules

2. Match the following Column – I with Column – II:

Column – I Column – II
(A) C2 (p) 2pZ (2px = 2py)
(B) O2 (q) Paramagnetic
(C) NO+ (r) Triple bond
(D) N2 (s) Diamagnetic

3. Match the following Column – I with Column – II:

Column – I Column – II
(A) H3C CH CH2  H CH2 CH CH2 (p) I effect
(B) (q) R effect
H2C CH Cl  CH2 CH Cl
(C) O O (r) +R effect
H2C CH N  H2C CH N
O O
(D)   (s) Hyper conjugation effect
CH3 
 CN

Space for rough work

Mathematics PART – III

SECTION – A
Single Correct Choice Type

This section contains 9 multiple choice questions. Each question has four choices (A), (B), (C) and (D)
out of which ONLY ONE is correct.

1. Consider the sequence 1, 2, 2, 4, 4, 4, 4, 8, 8, 8, 8, 8, 8, 8, 8, ….. then 1025th term will be

(A) 29 (B) 211
10 12
(C) 2 (D) 2

2. A student is allowed to select at most n books from a collection of (2n + 1) books. If the total
number of ways in which he can select at least one book is 63, then the value of n is
(A) 2 (B) 3
(C) 4 (D) 5

3. If |z – 1|  2 and |z – 1 – 2| = a (where  is cube root of unity) then complete set of values of a
is
1 3
(A) 0  a  2 (B)  a 
2 2
3 1 1 3
(C)  a  (D) 0  a  4
2 2 2 2

4. In triangle ABC, a = 5, b = 4 and c = 3. G is the centroid of the triangle. Circumradius of triangle

GAB is equal to
5
(A) 2 13 (B) 13
12
5 3
(C) 13 (D) 13
3 2

Space for rough work

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5. If x  [–1, 0), then cos–1(2x2 – 1) – 2 sin–1 x is equal to


(A)  (B) 
2
3
(C) (D) –2
2

6. The range of values of  for which the line 2y = gx +  is a normal to the circle
x2 + y2 + 2gx + 2gy – 2 = 0 for all values of g is
(A) [1, ) (B) [–1, )
(C) (0, 1) (D) (–, 1]
2
7. If y = m1x + c and y = m 2x + c are two tangents to the parabola y + 4a(x + a) = 0, then
(A) m1 + m2 = 0 (B) 1 + m1 + m2 = 0
(C) m1m2 – 1 = 0 (D) 1 + m1m2 = 0

8. (x – 1)(y – 2) = 5 and (x – 1)2 + (y + 2)2 = r2 intersect at four points A, B, C, D and if centroid of

ABC lies on line y = 3x – 4, then locus of D is
(A) y = 3x (B) x2 + y2 + 3x + 1 = 0
(C) 3y = x + 1 (D) y = 3x + 1

9. If roots of an equation xn – 1 = 0 are 1, a1, a2, ….., an – 1, then the value of

(1 – a1)(1 – a2)(1 – a3) ….. (1 – an – 1) will be
(A) n (B) n2
n
(C) n (D) 0

Assertion - Reason Type

10. STATEMENT 1: 1111 …... 91 times is composite number

STATEMENT 2: 91 is composite number
(A) Both the statements are true and Statement 2 is correct explanation of Statement 1
(B) Both the Statements are true and Statement 2 is not the correct explanation of Statement 1
(C) Statement 1 is true and Statement 2 is false
(D) Statement 1 is false and Statement 2 is true
Space for rough work

11. STATEMENT 1: If n is an odd integer greater than 3 but not a multiple of 3, then (x + 1)n – xn – 1
3 2
is divisible by x + x + x
STATEMENT 2: If n is an odd integer greater than 3 but not a multiple of 3,
n 2n
we have 1 +  +  = 3
(A) Both the statements are true and Statement 2 is correct explanation of Statement 1
(B) Both the Statements are true and Statement 2 is not the correct explanation of Statement 1
(C) Statement 1 is true and Statement 2 is false
(D) Statement 1 is false and Statement 2 is true

1
12. STATEMENT 1: The minimum value of 27cos 2x 81sin 2x is
243
STATEMENT 2: The minimum value of a cos  + b sin  is  a2  b2
(A) Both the statements are true and Statement 2 is correct explanation of Statement 1
(B) Both the Statements are true and Statement 2 is not the correct explanation of Statement 1
(C) Statement 1 is true and Statement 2 is false
(D) Statement 1 is false and Statement 2 is true

2 2
13. STATEMENT 1: The locus of a moving point satisfying  x  2   y2   x  2   y 2  4 is
ellipse
STATEMENT 2: Distance between (–2, 0) and (2, 0) is 4
(A) Both the statements are true and Statement 2 is correct explanation of Statement 1
(B) Both the Statements are true and Statement 2 is not the correct explanation of Statement 1
(C) Statement 1 is true and Statement 2 is false
(D) Statement 1 is false and Statement 2 is true

Space for rough work

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Linked Comprehension Type

This section contains 2 paragraphs. Based upon each paragraph, 3 multiple choice questions have to be
answered. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

Paragraph for Question Nos. 14 to 16

Read the following write up carefully and answer the following questions:
 2 2
A(1, 3) and C   ,   are the vertices of a triangle ABC and the equation of the internal angle bisector
 5 5
of ABC is x + y = 2

14. Equation of side BC is

(A) 7x + 3y – 4 = 0 (B) 7x + 3y + 4 = 0
(C) 7x – 3y + 4 = 0 (D) 7x – 3y – 4 = 0

15. Co-ordinates of vertex B are

 3 17   17 3 
(A)  ,  (B)  , 
 10 10   10 10 
 5 9
(C)   ,  (D) (1, 1)
 2 2

16. Equation of side AB is

(A) 3x + 7y = 24 (B) 3x + 7y + 24 = 0
(C) 13x + 7y + 8 = 0 (D) 13x – 7y + 8 = 0

Space for rough work

Paragraph for Question Nos. 17 to 19

Read the following write up carefully and answer the following questions:
Five balls are to be placed in three boxes. Each box should hold all the five balls so that no box remains
empty

17. Number of ways if balls are different but boxes are identical is
(A) 30 (B) 25
(C) 21 (D) 35

18. Number of ways if balls and boxes are identical is

(A) 3 (B) 1
(C) 2 (D) none of these

19. Number of ways if balls as well as boxes are identical but boxes are kept in a row is
(A) 10 (B) 15
(C) 20 (D) 6

SECTION – B
(Matching List Type)

This section contains 3 multiple choice questions. Each question has matching Column(s). The codes
for the Column(s) have choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

1. Match the following Column–I with Column–II

Column – I Column – II
(A) If a, b, c are in G.P., then loga 10, logb 10, logc 10 are in (p) A.P.
a  be x b  cex c  de x
(B) If   , then a, b, c, d are in (q) H.P.
a  be x b  ce x c  de x
(C) If a, b, c are in A.P., a, x, b are in G.P. and b, y, c are in
(r) G.P.
G.P., then x2, b2, y2 are in
(D) If x, y, z are in G.P., ax = by = cz, then log a, log b, log c are
(s) None of these
in
Space for rough work

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2. Match the following Column–I with Column–II

Column – I Column – II
A bc
(A) cot  (p) always right angled
2 a
A B
(B) a tan A + b tan B = (a + b) tan   (q) always isosceles
 2 
(C) a cos A = b cos B (r) may be right angled
sinB
(D) cos A = (s) may be right angled isosceles
2 sinC

3. Match the following Column–I with Column–II

Column – I Column – II
(A) If two circles x2 + y2 + 2a1x + b = 0 and x2 + y2 + 2a2x + b = 0
(p) (2, 2, 2)
touch each other then triplet (a1, a2, b) can be
(B) If two circles x2 + y2 + 2a1x + b = 0 and x2 + y2 + 2a2y + b = 0  1
(q)  1, 1, 
touch each other then triplet (a1, a2, b) can be  2
(C) If the straight line a1x – by + b2 = 0 touches the circle
(r) (2, 1, 0)
x2 + y2 = a2x + by, then triplet (a1, a2, b) can be
(D) If the line 3x + 4y – 4 = 0 touches the circle  3
(s)  1, 1, 
(x – a1)2 + (y – a2)2 = b2, then triplet (a1, a2, b) can be  5

Space for rough work

sw
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ALL INDIA INTEGRATED TEST SERIES

ANSWERS, HINTS & SOLUTIONS
HALF COURSE TEST –VI
(Paper-1)

Q. No. PHYSICS CHEMISTRY MATHEMATICS

1. B A C
2. A D B
3. C B D
4. B A B
5. C A B
C C B
qualified in JEE (Advanced), 2014.

6.
7. D B D
8. B A A
9. D D A
10. A A B
11. D A C
12. A A A
13. D B D
14. B B B
15. A D C
16. A A A
17. A A B
18. C A C
19. C B D
(A)  (p), (q), (r), (s) (A)  (s) (A)  (q)
(B)  (p), (q), (r), (s) (B)  (p, r) (B)  (r)
1. (C)  (p), (q) (C)  (p, q) (C)  (p)
(D)  (r), (s) (D)  (p) (D)  (r)
(A)  (p), (s) (A)  (s) (A)  (p, r)
(B)  (p), (r) (B)  (p, q) (B)  (q, r)
2.
(C)  (q) (C)  (p, r, s) (C)  (r, s)
(D)  (p) (D)  (r, s) (D)  (q, r, s)
(A)  (q) (A)  (s) (A)  (r)
(B)  (p), (q), (r), (s) (B)  (r) (B)  (p, q)
3.
(C)  (q) (C)  (q) (C)  (q, r)
(D)  (p), (q), (r), (s) (D)  (p) (D)  (p, s)

Physics PART – I
SECTION – A
 
| v 2  v1 | 3v 3v 2
1. a = =
t (2 / v) 2

2. 2g  T = 2a …(i)
Tg=a …(ii)
T = kx …(iii)
x = 1.33 cm

3. v 2  u2  2as
36 = 0 + 2  2  S
S=9m

4. Velocity after 1 sec = 5 2

(5 2)2
Radius of curvature =  10 2  14 m
5/ 2

Chemistry PART – II
SECTION – A

9. The compounds given in options (A) & (C) has no electron deficiency. The compound (B) has
both electrophilic and nucleophilic site. While the compound (D) has got only electrophilic site.

10. Lother Meyer Curve is plotted between atomic volume and atomic mass. More is the atomic
volume, higher is the position occupied.

11. Decrease in entropy during formation of four membered ring is higher than that of three
membered ring. That’s why formation of four membered ring is thermodynamically less feasible.

12. K w  H  OH 

2

For neutral solution H   K w since H   OH  

2log H    logK w  pK w
pKw
pH 
2

13. Mg HCO3 2  Ca  OH2  CaCO3  2H2O  MgCO3

100ppm  106 g of water has 100g of CaCO3
100
10 l of H2O  104 g of water has  104 g of CaCO3 = 1g of CaCO3
106
1
Moles of CaCO3 =  moles of Ca(OH)2
100
1
 Mass of Ca(OH)2 =  74  0.74g
100
14. At high temperature, there is allyl substitution.

15. H3C CH CH2 CH3

CH3
It has four types of replaceable H – atoms hence monochloro derivatives will be four. The product
ratio will be 6 : 3 : 7.6 : 5. It is total number of hydrogen atoms multiplied by relative reactivity.

16. H3C CH CH3

CH3
1 = 9  1 = 9
3 = 1  1600 = 1600
Thus the product ratio is 9 : 1600 and hence
9
% of 1   100  0.5%
1609
1600
% of 3   100  99.5%
1609

17. x = r
h h h
x.p   p  
4 4 .x 4 r

h
p  p  p 
4r

h 6.626  1034
18. x  m V  or V  m sec 1
4 4  3.14  0.1  1010
 5.27  10 24 m sec 1

h 6.626  1034
19. V.x   = 104 m2s1
4 m 4  3.14  9.11  1031
104
 V  8  104 ms1
10

Mathematics PART – III

SECTION – A
1. Let the 1025th term falls in the nth group.
n–1 n
Then, 1 + 2 + 4 + ….. + 2 < 1025  1 + 2 + 4 + ….. + 2
 2n – 1 < 1026  2n + 1
 n = 10
 1025th terms is 210
2n + 1 2n + 1 2n + 1
2. Here, C1 + C2 + ….. + Cn = 63
n=3
3. |z – 1 – 2| = a
 |z + 1| = a  |z – 1 + 2| = a
 |z – 1| + 2  a  0  a  4
1 1
4. AG  2b2  2c 2  a2 ; BG  2a2  2c 2  b2
3 3
a 1 2 2 2 2
 AG  ; BG  b  4c 2 as a = b + c
3 3
5 1 2
 AG  ; BG  16  36  13
3 3 3
1
Also, AB = c = 3 and GAB = ABC = 2
3
If R1 is the circumradius of triangle GAB, then
 AG  BG   AB  5 2 1 5 13
R1 =   13  3   units
4 GAB 3 3 42 12
5. cos–1 (2x2 – 1) = 2 – 2 cos–1 x (as x < 0)
 cos–1 (2x2 – 1) – 2 sin–1 x = 2 – 2 cos–1 x – 2 sin–1 x

= 2 – 2 (cos–1 x + sin–1 x) = 2  2    
 2
6. The line 2y = gx +  should pass through (–g, –g)
So –2g = –g2 +    = g2 – 2g = (g – 1)2 – 1  –1
7. Tangents y = m 1x + c and y = m 2x + c intersect at (0, c) which lies on the directrix of the given
parabola
Hence, tangents are perpendicular for which m 1m2 = –1
8. If (xi, yi) is the point of intersection of given curves
4 4

i 1
xi
1 1
y
i 1
i
Then  and 0
4 2 4
3 3

i 1
xi
4  x4 i 1
yi y
Now,  and  4
3 3 3 4
 3 3 


xi 
yi 

Centroid  i1 , i1  lies on the line y = 3x – 4
 3 3 
y 3  4  x4 
Hence,  4  4
3 3
 y4 = 3x4
Hence, the locus of D is y = 3x

9. Clearly, xn – 1 = (x – 1)(x – a1)(x – a2) ….. (x – an – 1)

xn  1
   x  a1  x  a2  .....  x  an 1 
x 1
 1 + x + x2 + ….. + xn – 1 = (x – a1)(x – a2) ….. (x – an – 1)
 n = (1 – a1)(1 – a2) ….. (1 – an – 1) (Putting x = 1)

10. x = 1111 ….. 91 times

1091  1
= 1 + 10 + 102 + ….. + 1090 =
10  1
1091  1 1013  1
= 
1013  1 10  1
= (1 + 1013 + 1026 + ….. + 1078)  (1 + 10 + 102 + ….. + 1012)
= Composite number

11. x3 + x2 + x = x(x2 + x + 1) = x(x – )(x – 2)

n n 3 2 2
Now f(x) = (x + 1) – x – 1 is divisible by x + x + x, if f(0) = 0, f() = 0 and f( ) = 0
n n
Now, f(0) = (0 + 1) – 0 – 1 = 0
f() = ( + 1)n – n – 1 = (–2)n – n – 1 = 0
Similarly, f(2) = 0

12. Let y  27cos 2x  81sin 2x  33 cos 2x  4 sin 2x

Now,  32  42  3cos 2x  4 sin2x  32  42
 3 5  33 cos 2x  4 sin 2x  35

13. Locus of P is line segment between (–2, 0) and (2, 0)

14.-16. Image of A(1, 3) in line x + y = 2 is A(1, 3)

 2  2 2  2  x+y=2
1  , 3   (–1, 1)
 2 2 
 2 2
So line BC passes through (–1, 1) and   ,   (–1, 1)  2 2
 5 5 B C ,  
 2   5 5
   1
The equation of line BC is y  1   5  (x + 1)
 2 
   1
 5 
 7x + 3y + 4 = 0
 5 9
Vertex B is point of intersection of 7x + 3y + 4 = 0 and x + y = 2, i.e., B    , 
 2 2
 9 
3  
Line AB is y  3   2   x  1  3x + 7y = 24
 5
1  
 2

17.-19. If no box remains empty, we can have (1, 1, 3) or (1, 2, 2) distribution pattern
5! 5!
17. Here, number of ways =   25
1! 2! 3!  2! 2 1! 2!

18. Here, we have only two ways (1, 1, 3) and (1, 2, 2)

Hence, number of ways is 2

19. When boxes are kept in a row, they will be treated as different. In this case, the number of ways
5–1 4
will be C3 – 1 = C2 = 6
SECTION – B
2
1. (A) a, b, c are in G.P. Hence, b = ac
 2 log10 b = log10 a + log10 c
2 1 1
  
logb 10 loga 10 logc 10
2 1 1
  
y x z
Hence, x, y, z are in H.P.
a  be x b  ce x c  de x
(B)  
a  be x b  cex c  de x
2a 2b 2c
 x
1 x
1 1
a  be b  ce c  de x
a  be x b  ce x c  de x
  
a b c
 a, b, c, d are in G.P.
(C) Given, 2b = a + c, x 2 = ab, y2 = bc
Now, x2 + y2 = b(a + c) = 2b2
 x2 + y2 = 2b2
Hence, x2, b2, y2 are in A.P.
(D) x log a = y log b = z log c = k (say)
Also, y2 = xz
k2 k2
 
logb 2 logalogc
 log a, log b, log c are in G.P.

A bc
2. (A) cot 
2 2
A BC BC
cos   2sin  cos
 2  sinB  sinC 2 2
  
A sin A A A
sin   2sin cos
2 2 2
A B C
 cos  cos
2 2

A = B – C  A + B + C = 2B =   B =
2
A B
(B) In ABC, a tan A + b tan B = (a + b) tan  
 2 
  A  B    A B 
 a  tan A  tan     b  tan    tanB 
  2     2  
 A B  A B 
a sin  A   b sin   B
 2    2 
 A B  A B 
cos A cos   cos   cosB
 2   2 

 A B  A B
2R sin A sin   2R sinB sin  
  2   2 
cos A cosB
 A  B 
 sin   tan A  tanB   0
 2 
A=B
(C) a cos A = b cos B
2R sin A cos A = 2R sin B cos B
 sin 2A = sin 2B
 2A = 2B or 2A =  – 2B

 A = B or A + B =
2
sinB
(D) cos A 
2sinC
b2  c 2  a2 b
 
2bc 2c
 b2 + c2 – a2 = b2
 c2 = a2  c = a   is isosceles

3. (A) Radical axis of x2 + y2 + 2a1x + b = 0 and x2 + y2 + 2a2x + b = 0 is (a1 – a2)x = 0 or x = 0

It must touch both the circles
Solving it with one of the circles we get y2 + b = 0
b0
(B) Radical axis of given circle is a1x – a2y = 0
a 
Solving it with one of the circles, we have x 2   1  x 2  2a1x  b  0
 a2 
This must have equal roots
 a2 
Hence, 4a12  4b 1  12   0
 a2 
 a2 
 a12  b  1  12   0
 a 
 2 
Options (p) and (q) satisfy this
(C) If the straight line a1x – by + b2 = 0 touches the circle x2 + y2 = a2x + by
a b
a1  2  b   b2
2 2 a22 b2
  
a2  b2 4 4
1

 a12a22  2b2a1a2  b4  a12a22  a12b2  a22b2  b 4

 b2 = 0 or 2a1a2 = a12  a22
 Options (q) and (r) satisfy this
(D) Line 3x + 4y – 4 = 0 touches the circle (x – a1)2 + (y – a2)2 = b2
3a1  4a2  4
 b
5

Time Allotted: 3 Hours Maximum Marks: 243

INSTRUCTIONS
A. General Instructions
1. Attempt ALL the questions. Answers have to be marked on the OMR sheets.
2. This question paper contains Three Parts.
3. Part-I is Physics, Part-II is Chemistry and Part-III is Mathematics.
4. Each part is further divided into two sections: Section-A & Section-B
5. Rough spaces are provided for rough work inside the question paper. No additional sheets will be
provided for rough work.
6. Blank Papers, clip boards, log tables, slide rule, calculator, cellular phones, pagers and electronic
devices, in any form, are not allowed.
B. Filling of OMR Sheet
1. Ensure matching of OMR sheet with the Question paper before you start marking your answers
on OMR sheet.
2. On the OMR sheet, darken the appropriate bubble with black pen for each character of your
Enrolment No. and write your Name, Test Centre and other details at the designated places.
3. OMR sheet contains alphabets, numerals & special characters for marking answers.
C. Marking Scheme For All Three Parts.
(i) Section-A (01 to 09) contains 09 multiple choice questions which have only one correct answer.
Each question carries +3 marks for correct answer and – 1 mark for wrong answer.
Section-A (10 – 13) contains 4 Assertion-Reasoning (multiple choice questions) which have
only one correct answer. Each question carries +3 marks for correct answer and
– 1 mark for wrong answer.
Section-A (14 – 19) contains 2 paragraphs. Based upon paragraph, 3 multiple choice questions
have to be answered. Each question has only one correct answer and carries +4 marks for
correct answer and – 1 mark for wrong answer.
(ii) Section-B (1 – 03) contains 3 Matrix Match Type (4 × 4 Matrix) questions containing statements
given in 2 columns. Statements in the first column have to be matched with statements in the
second column. Each question carries +6 marks for all correct answer. There is no negative
marking.

Name of the Candidate

Enrolment No.

Useful Data

PHYSICS
2
Acceleration due to gravity g = 10 m/s

Planck constant h = 6.6 1034 J-s

Charge of electron e = 1.6  1019 C

Mass of electron me = 9.1  1031 kg

Permittivity of free space 0 = 8.85  1012 C2/N-m2

Density of water water = 103 kg/m3

Atmospheric pressure Pa = 105 N/m2

Gas constant R = 8.314 J K1 mol1

CHEMISTRY

Gas Constant R = 8.314 J K1 mol1

Atomic No: H=1, He = 2, Li=3, Be=4, B=5, C=6, N=7, O=8,

Physics PART – I

SECTION – A
Single Correct Choice Type

This section contains 9 multiple choice questions. Each question has four choices (A), (B), (C) and (D)
out of which ONLY ONE is correct.

x2 t2
1. A particle moves along the positive branch of the curve y  with x governed by x = where
2 2
x and y are in meter and t in second. At t = 2 the acceleration of the particle is
(A) ˆi  6ˆj (B) ˆi  6ˆj
(C) 2iˆ  4ˆj (D) 3iˆ  6ˆj

2. A block of mass ‘m’ is placed on a frictionless horizontal table and attached to

a string passing through a small hole in the surface. Initially the mass moves
in a circle of radius r0 with a speed v 0 and the string is held by a person. The m
person pulls the string slowly to decrease the radius of the circle to r. The
tension in the string depends on r as :
(A) r–3 (B) r3
(C) r (D) r–1

3. Two identical rods are joined to form a ‘X’. The smaller angle between the rod is . The moment
of inertia of the system about an axis passing through the point of intersection of the rods and
perpendicular to their plane is proportional to
(A)  (B) sin2 
2
(C) cos  (D) independent of 

4. Three identical solid spheres move down on three fixed inclined planes A, B, and C all of the
same dimensions A is without friction, the friction between B and sphere is sufficient to cause
rolling without slipping, the friction between C and sphere cause rolling with slipping. The kinetic
energies of sphere at the bottom of the inclines A, B and C are EA, EB and EC respectively, then
(A) EA  EB  EC (B) EA  EB  EC
(C) EA  EB  EC (D) EA  EB  EC

Space for rough work

5. A particle of mass m is at a distance 2R from the centre of a M

thin shell of mass M and having radius R as shown in figure.
The gravitational field at the centre of shell is O m

GM
(A) zero (B) 2
R
2R
G(M  m) Gm
(C) (D)
4R2 4R2
2
6. An organ pipe of cross sectional area 100 cm resonates with a tuning fork of frequency 1000 Hz
in fundamental tone. The minimum volume of water to be drain out so that the pipe again
resonate with the same tuning fork is (take velocity of wave = 320 m/s)
(A) 800 cm3 (B) 1200 cm3
3
(C) 1600 cm (D) 2000 cm3
2
7. An elevator car starts descending with constant acceleration 2 m/s , 2 second after the start a ball
is thrown horizontally with 3 m/s with respect to elevator from point P, then the magnitude of
displacement of the ball w.r.t. the point P after 1 sec is
(A) 5 m (B) 10 m
(C) 3 m (D) 6 m

8. The standing wave on a 5 m long string clamped at both ends is represented by the equation y =
4 sin 6.28 x cos 6.28 t, where y is in cm. The phase difference between two points at x = 1.51 m
and x = 2.75 m is
(A)  (B) zero
(C) /2 (D) 3/2

9. An ideal diatomic gas undergoes a process in which the heat absorbed by the gas is twice the
increment of its internal energy. Then the polytropic exponent of the process is
(A) 2 (B) 3/5
(C) 2/5 (D) 5/2

Space for rough work

Assertion - Reason Type

10. STATEMENT-1: A force F1 is applied on the 1 1

m1 m1
F1 F2
lower block in case (1) due to which only m2 m2
lower block moves with constant velocity. Case (1) Case (2)

A force F2 is applied on the lower block in case (2) due to which both the block moves with
constant velocity. F1 and F2 will be equal. (Given that nature of surfaces is same for both the
cases)

because
STATEMENT-2: Frictional force between ground and m2 will be same for both the case.
(A) Statement-1 is True, Statement-2 is True; Statement -2 is a correct explanation for
Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement -2 is NOT a correct explanation for
Statement-1.
(C) Statement -1 is True, Statement-2 is False.
(D) Statement -1 is False, Statement-2 is True.

11. STATEMENT-1: When one object collides with another object, the impulse during deformation
and restitution will be in same direction.

because
STATEMENT-2: Due to this impulse the objects first deform and due to the same impulse they
again try to regain its original shape.
(A) Statement-1 is True, Statement-2 is True; Statement -2 is a correct explanation for
Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement -2 is NOT a correct explanation for
Statement-1.
(C) Statement -1 is True, Statement-2 is False.
(D) Statement -1 is False, Statement-2 is True.
Space for rough work

12. STATEMENT-1: Rolling without slipping can not be possible in absence of friction.

because
STATEMENT-2: During rolling without slipping, energy can remain conserved in absence of
external applied force.
(A) Statement-1 is True, Statement-2 is True; Statement -2 is a correct explanation for
Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement -2 is NOT a correct explanation for
Statement-1.
(C) Statement -1 is True, Statement-2 is False.
(D) Statement -1 is False, Statement-2 is True.

13. STATEMENT-1: In absence of air friction, it is claimed that all objects fall with the same
acceleration although, a heavier object is pulled towards the earth with more force than a lighter
object.
because
STATEMENT-2: Net external force is always equal to rate of change of linear momentum.
(A) Statement-1 is True, Statement-2 is True; Statement -2 is a correct explanation for
Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement -2 is NOT a correct explanation for
Statement-1.
(C) Statement -1 is True, Statement-2 is False.
(D) Statement -1 is False, Statement-2 is True.
Space for rough work

Linked Comprehension Type

This section contains 2 paragraphs. Based upon each paragraph, 3 multiple choice questions have to be
answered. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

Paragraph for Question Nos. 14 to 16

If a point P moves in plane along a given curve y = f(x), the angular y = f(x)
velocity of point P about a fixed point O in the plane is the rate of Q
R
change of the angle that OP line makes with a fixed direction 
M
P
OX  line in the plane 90

Let OP = r at t = t sec N
PM = rd = PQ sin, But if d is very small then. PQ  PR  ds (arc d 90
length) p
 rd = ds sin   reference line x
O

d 1 ds v sin  Magnitudeof component of velocity of point perpendicular to radius vector

  sin   
dt r dt r Magnitudeof radius vector
d vp
 
dt r 2
Read the above passage carefully and answer the following questions
d
14. r2 represents
dt
(A) rate at which radius vector sweeps out area
(B) angular momentum
(C) moment of velocity about origin
(D) rate of increase of sectional area as P moves along curve

15. If two particles A and B are having speed 103 m/s and 20 m/s 103 m/s
20 m/s
at a particular instant as shown in the figure, then the angular
velocity of A with respect to B at the same instant is 30 60
A 5m B

(A) 1 rad/s clockwise (B) 1 rad/s anticlockwise

(C) 2 rad/s clockwise (D) 2 rad/s anticlockwise
16. If point P moves on parabolic path y2 = 4(x + 1), where x and y are in meter with constant speed 2
m/s. Its angular velocity about focus at an instant when it makes angle 60 at focus with x-axis is
[all angles are measured in anticlockwise direction with positive x-axis]
(A) 0.25 rad/s (B) 0.50 rad/s
(C) 0.12 rad/s (D) none of the above

Space for rough work

Paragraph for Question Nos. 17 to 19

2
A cubical box of side 2 metre contains oxygen gas (atomic weight 32) at a pressure of 100N/m . During
an observation time of one second, an atom travelling with the root-mean square speed parallel to one of
the edges of the cube was found to make 500 hits with a particular wall, without any collision with the
25
other atoms. ( R  J/mol–K and K  1.38  1023 J/K)
3

17. What is the temperature of the gas

(A) 500 K (B) 4520 K
(C) 5120 K (D) 3600 K

18. What is the average translational kinetic energy per atom

(A) 106  1021 J (B) 96  1021 J
21
(C) 116  10 J (D) 112  1021 J

19. Find the total mass of oxygen gas in the box.

(A) 30  104 kg (B) 2.4  104 kg
(C) 6  104 kg (D) 15  104 kg

Space for rough work

SECTION – B
(Matching List Type)

This section contains 3 multiple choice questions. Each question has matching Column(s). The codes
for the Column(s) have choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

t3
1. A particle is moving according to the displacement time relation x  3t 2  (where x is in meters
2
and t is in seconds). Match the condition of column I with time interval and instant of column II
Column A Column B
(A) Velocity and acceleration will be in same
(p) At t = 0 and t = 6 sec
direction
(B) particle will be at origin (q) 0 < t < 2 sec
(C) particle will retard (r) at t = 0 and t = 4 sec
(D) velocity is zero (s) 2<t<4

Space for rough work

2. A bob is attached to a string of length  whose other end is fixed and is given horizontal velocity
at the lowest point of the circle so that the bob moves in a vertical plane. Match the velocity given
at the lowest point of circle in column I with tension and velocity at the highest point of the circle
corresponding to velocity of column I of column II
Column A Column B
(A) 2g (p) T = mg, v > g

(B) 6g (q) T > 2mg and v > 3g

(C) 8g (r) T = 0 and v  g

(D) 5g (s) T=0

3. Match the following.

Column – A Column – B
(A) Ideal gas (p) Maximum emissivity
(B) Black body (q) Maximum absorptive power
(C) Wien’s displacement law (r) An ideal concept
(D) A red object having 0 K temperature. (s) Dominant wave in composite radiations.

Space for rough work

Chemistry PART - II

SECTION – A
Single Correct Choice Type

This section contains 9 multiple choice questions. Each question has four choices (A), (B), (C) and (D)
out of which ONLY ONE is correct.

1. Ag(CN)2 is stable while AgCl2 is unstable because

(A) CN– is stronger than Cl–
(B) Ag+ is soft acid, CN– is soft base while Cl– is hard base
(C) both are equally stable
(D) None of the above

2. Glycerol on oxidation with Fenton’s reagent produces

(A) glyceraldehyde (B) dihydroxyacetone
(C) tartaric acid (D) glyceraldehyde and dihydroxyacetone

3. Consider following reaction,

Zn  2H  Zn2   H2
half-life period is independent of concentration of Zn at constant pH. At constant Zn
concentration, half-life is 10 minutes at pH = 2 and half-life is 100 minutes at pH = 3. Hence, rate
law is
(A) k [Zn] [H+] (B) k [Zn] [H+]2
0 +
(C) k [Zn] [H ] (D) k [Zn]0 [H+]2
Space for rough work

4.
Br / Fe
COOPh  2
cold and dark
P (a monobromo derivative), where P is

(A) OOC - Ph (B) COOPh

Br
Br
(C) OOC - Ph (D) COOPh

5. Which of the following acid contains one S – S bond?

(A) H2S2O7 (B) H2S2O8
(C) H2S2O6 (D) none of these

6. Which of the following plots represent the behaviour of an ideal binary liquid solution?
1
(A) Plot of PTotal vs YA is linear (B) Plot of vs YB is linear
PTotal
(C) Plot of PTotal vs YB is linear (D) None of the above

RT RT
7. The value of for a gas at critical condition is how many times of at normal condition?
PV PV
8 3
(A) (B)
3 8
1
(C) 3 (D)
3

Space for rough work

8. In a closed vessel, ozone transforms into oxygen through two steps as mentioned below:

O3  
 O2  O (fast)
O  O3  2O2 (slow)
If half of the oxygen produced is removed from the vessel without disturbing the temperature,
then the rate of transformation of ozone into oxygen will be
(A) unaffected (B) half the initial value
(C) double the initial rate (D) None of the above

9. The stoichiometric equation for the oxidation of bromide ions by hydrogen peroxide in acidic
solution is
2Br   H2 O2  2H  Br2  2H2 O
r  k [H2O2 ] [H ] [Br  ]
If, by the addition of water, the total volume were doubled, what would be the effect on the rate of
–
disappearance of Br and rate of reaction?
– 1 1
(A) Rate of disappearance of Br becomes times and rate of reaction times
8 4
1 1
(B) Rate of disappearance of Br– becomes times and rate of reaction times
4 8
1
(C) Both rate of disappearance of Br– and rate of reaction becomes times
8
(D) Rate of disappearance of Br– becomes 4 times and rate of reaction 8 times
Space for rough work

Assertion - Reason Type

10. STATEMENT-1: The pH of pure water is less than 7 at 60oC

and
STATEMENT-2: As the temperature increases, pure water becomes slightly acidic
(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for
Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for
Statement-1.
(C) Statement-1 is True, Statement -2 is False.
(D) Statement-1 is False, Statement-2 is True.

11. STATEMENT-1: CO2 is a gas and SiO2 is a solid at room temperature.

and
STATEMENT-2: van der Waals’ forces are far weaker than ordinary covalent bond.
(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for
Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for
Statement-1.
(C) Statement-1 is True, Statement -2 is False.
(D) Statement-1 is False, Statement-2 is True.
Space for rough work

12. STATEMENT-1: Acidic strength boron halide is as BF3 < BCl3 < BBr3 < BI3
and
STATEMENT-2: Fluorine is more electronegative & creates greater electron deficiency on boron.
(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for
Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for
Statement-1.
(C) Statement-1 is True, Statement -2 is False.
(D) Statement-1 is False, Statement-2 is True.

3PV
13. STATEMENT-1: R.M.S velocity of a gas sample can be expressed as Crms 
M
and
STATEMENT-2: The R.M.S velocity of a gas sample decreases if its volume is decreased.
(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for
Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for
Statement-1.
(C) Statement-1 is True, Statement -2 is False.
(D) Statement-1 is False, Statement-2 is True.
Space for rough work

Linked Comprehension Type

This section contains 2 paragraphs. Based upon each paragraph, 3 multiple choice questions have to be
answered. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

Paragraph for Question Nos. 14 to 16

Parallel or competing reactions–

The reactions in which a substance reacts or decomposes in more than one B
k1
manner are called parallel or side reactions. This type of kinetics can be
observed in radioactive conversions where an element loses ,  particles
simultaneously to form new products at the same time through different A
paths. k2
d[A]
Then,   (K 1  K 2 ) [A]  K av. [A] C
dt
K1 = Fractional yield of B × Kav.
K2 = Fractional yield of C × Kav.
If K1 >> K2, then
A  B will become main reaction
A  C will become side reaction
Read the above paragraph carefully and answer the questions given below it:

14. A substance undergoes first order decomposition. The decomposition follows two parallel first
order reactions as–
k1 B
K1  1.26  10 4 sec 1
A K 2  3.8  10 5 sec 1
k2

C
The percentage distribution of B and C are–
(A) 75% B and 25% C (B) 80% B and 20% C
(C) 76.83% B and 23.17% C (D) 90% B and 10% C
Space for rough work

15. For a first order decomposition–

k A1
1

Kn
A n  A
k2

A2
Overall K will be given by–
(K1  K 2  .........  K n )
(A) K = K1 + K2 + ………… + Kn (B) K 
n
(C) K  K 1  K 2  K 3 ............  K n (D) None

16. For a reaction,

k1 2B

A
k2

3C
Follow parallel kinetics, the ratio of [B] and [C] at any instant can be given by,
K1 K2
(A) (B)
K2 K1
2K 1 3K 1
(C) (D)
3K 2 2K 2

Space for rough work

Paragraph for Question Nos. 17 to 19

The cis and trans isomers of alkene do not have the same stability. The stability can be measured by
hydrogenation and combustion. The reaction of alkene with hydrogen is endothermic and the enthalpy
change in the reaction is called heat of hydrogenation.

Pt
 H2   H  120 kJ / mol

H H
In all the isomers of 2–butene, the product is same but different amount of heat is evolved. In each
reaction, it must be related with different relative stability. 1–butene evolves greatest amount of energy
and trans 2–butene evolves least amount of energy. So, 1–butene must have greatest energy and it is
least stable whereas trans 2–butene must have lowest energy and it is more stable. cis 2–butene has
intermediate energy in relation to above.
Read the above paragraph carefully and answer the questions given below:

17. Which alkene has higher heat of hydrogenation?

H3 C CH3
CH3 CH3
CH3

, , ,
CH3 CH3 CH3
I II III IV
(A) I (B) II
(C) III (D) IV
Space for rough work

18. Which alkene is most stable?

(A) CH3 (B) H3 C CH3

H3 C

CH3 H H
(C) H3 C H (D) CH2
H3 C

H CH3

19. For the reaction,

conc. H2 SO 4
 
OH
the major product is
(A) (B)
CH2 CH3

Space for rough work

SECTION – B
(Matching List Type)

This section contains 3 multiple choice questions. Each question has matching Column(s). The codes
for the Column(s) have choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

1. Match Column–I and Column –II.

ColumnI ColumnII
(A) The volume of mixture has no effect on (p) Common ion effect
the equilibrium constant.
(B) Increasing the pressure moves the (q) pH = 4
equilibrium to the left.
(C) The solution has hydronium ion (r) There is an increase in the number
concentration of 0.0001 mol/litre. of moles from products to reactants.
(D) The addition of NaOH to Ca(OH)2 (s) There is no change in number of
solution precipitates Ca(OH)2. moles.
Space for rough work

2. Match the following:

Column – I Column – II
(A) OH (p) OH
K 2 Cr2 O7

H 
 O 2N NO2
 

NO2
(picric acid)
(B) OH (q) OH
Br2 water
 Br

(ortho-bromophenol)
(C) OH (r) OH
HNO3


conc. H2 SO 4 Br Br

Br
(tribromophenol)
(D) OH (s) O
Br2

Cs2   


O
(para-benzoquinone)

3. Match the following:

Column – I Column – II
(Thermo chemical equations) (Characteristics)
(A)  CuSO 4 .5H2O  aq 
CuSO 4 .5H2 O  Excess H2O  2.9 kcal  (p) Enthalpy change H  0
(B) 2H2  g  O2  g  
 2H2O     115.0 kcal (q) Enthalpy of combustion
(C) H2 SO 4  aq   NaOH  aq 
 Na2 SO 4  aq   H2 O  57.1 kJ (r) Enthalpy of solvation
(D) CH4  g   2O2  g   2H2O  g   CO2  g   803.3 kJ (s) Enthalpy of neutralization

Space for rough work

Mathematics PART – III

SECTION – A
Single Correct Choice Type

This section contains 9 multiple choice questions. Each question has four choices (A), (B), (C) and (D)
out of which ONLY ONE is correct.

2 n
1. Maximum sum of coefficient in the expansion of (1 – x sin  + x ) is
n
(A) 1 (B) 2
n
(C) 3 (D) 0

2. The number of integral solutions of x + y + z = 0 with x  – 5, y  –5, z  –5 is

(A) 134 (B) 136
(C) 138 (D) 140


3. Double ordinate AB of the parabola y2 = 4ax subtends an angle at the focus of the parabola,
2
then tangents drawn to parabola at A and B will intersect at
(A) (–4a, 0) (B) (–2a, 0)
(C) (–3a, 0) (D) none of these

x2 y2
4. The tangent at a point P on the hyperbola  1 meets one of the directrix in F. If PF

a2 b2
subtends an angle  at the corresponding focus, then  equals
 
(A) (B)
4 2
3
(C) (D) 
4

Space for rough work

 1 2  11 4 
5. If in triangle ABC, A  (1, 10), circumcentre    ,  and orthocentre   ,  , then the
 3 3  4 3
coordinates of mid–point of side opposite to A is
 11 
(A)  1,   (B) (1, 5)
 3
(C) (1, –3) (D) (1, 6)

n
6. cos3 x sin 2x =  a sin rx   x  R, then
r
x 0
1 1
(A) n = 5, a1  (B) n = 5, a1 
2 4
1 1
(C) n = 5, a2  (D) n = 5, a2 
8 4

7. If the roots of the quadratic equation (4p – p2 – 5)x2 – (2p – 1)x + 3p = 0 lie on either side of unity,
then the number of integral values of p is
(A) 1 (B) 2
(C) 3 (D) 4

8. If z     3   i 5   2 , then the locus of z is

(A) ellipse (B) semicircle
(C) parabola (D) straight line

1 1 1 1 1  1 1 1
9. If 1      .....  , then value of    ..... is
3 5 7 9 11 4 1 3 5  7 9  11
 
(A) (B)
8 6
 
(C) (D)
4 36

Space for rough work

Assertion - Reason Type

–1
10. STATEMENT 1: Principal value of cos (cos 30) is 30 – 9
STATEMENT 2: 30 – 9  [0, ]
(A) Both the statements are true and Statement 2 is correct explanation of Statement 1.
(B) Both the Statements are true and Statement 2 is not the correct explanation of Statement 1.
(C) Statement 1 is true and Statement 2 is false.
(D) Statement 1 is false and Statement 2 is true.

11. STATEMENT 1: If there is exactly one point on the line 3x + 4y + 5 5 = 0, from which
x2
perpendicular tangents can be drawn to the ellipse 2
 y 2  1 (a > 1), then the eccentricity of the
a
1
ellipse is
3
STATEMENT 2: For the condition given in statement 1, given line must touch the circle
x2 + y2 = a2 + 1
(A) Both the statements are true and Statement 2 is correct explanation of Statement 1.
(B) Both the Statements are true and Statement 2 is not the correct explanation of Statement 1.
(C) Statement 1 is true and Statement 2 is false.
(D) Statement 1 is false and Statement 2 is true.

12. STATEMENT 1: If the point (2a – 5, a2) is on the same side of the line x + y – 3 = 0 as that of the
origin, then a  (2, 4)
STATEMENT 2: The points (x1, y1) and (x2, y2) lie on the same or opposite sides of the line
ax + by + c = 0, as ax 1 + by1 + c and ax2 + by2 + c have the same or opposite signs.
(A) Both the statements are true and Statement 2 is correct explanation of Statement 1.
(B) Both the Statements are true and Statement 2 is not the correct explanation of Statement 1.
(C) Statement 1 is true and Statement 2 is false.
(D) Statement 1 is false and Statement 2 is true.

13. STATEMENT 1: If px 2 + qx + r = 0 is a quadratic equation (p, q, r  R) such that its roots are , 
and p + q + r < 0, p – q + r < 0 and r > 0, then [] + [] = –1, (where [.] denotes the greatest
integer function)
STATEMENT 2: If for any two real numbers a and b, function f(x) is such that f(a) f(b) < 0,
 f(x) has at least one real root lying in (a, b)
(A) Both the statements are true and Statement 2 is correct explanation of Statement 1.
(B) Both the Statements are true and Statement 2 is not the correct explanation of Statement 1.
(C) Statement 1 is true and Statement 2 is false.
(D) Statement 1 is false and Statement 2 is true.

Space for rough work

FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942
website: www.fiitjee.com
TG ~ @bohring_bot < 3 25
AIITS-HCT-VI(Paper-2)-PCM-JEE(Advanced)/15

Linked Comprehension Type

This section contains 2 paragraphs. Based upon each paragraph, 3 multiple choice questions have to be
answered. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

Paragraph for Question Nos. 14 to 16

Read the following write up carefully and answer the following questions:
Two consecutive numbers from 1, 2, 3, ….., n are removed. The arithmetic mean of the remaining
105
numbers is
4

14. The value of n lies in

(A) [45, 55] (B) [52, 60]
(C) [41, 49] (D) none of these

15. The removed numbers

(A) lie between 10 and 20 (B) are greater than 10
(C) are less than 15 (D) none of these

16. Sum of all numbers

(A) exceeds 1600 (B) is less than 1500
(C) lies between 1300 and 1500 (D) none of these

Paragraph for Question Nos. 17 to 19

Read the following write up carefully and answer the following questions:
P is a set containing n elements. A subset A of P is chosen and the set P is reconstructed by replacing
the elements of A. A subset B of P is chosen again

17. The number of ways of choosing A and B such that A and B have no common elements is
(A) 3n (B) 2n
n
(C) 4 (D) none of these

18. The number of ways of choosing A and B such that B contains just one element more than A is
(A) 2n (B) 2nCn – 1
2n
(C) Cn (D) (3n)2

19. The number of ways of choosing A and B such that B is a subset of A is

(A) 2nCn (B) 4n
n
(C) 3 (D) none of these

Space for rough work

SECTION – B
(Matching List Type)

This section contains 3 multiple choice questions. Each question has matching Column(s). The codes
for the Column(s) have choices (A), (B), (C) and (D) out of which ONLY ONE is correct.
2
1. If e1 and e2 are the roots of the equation x – ax + 2 = 0, then match the following Column–I with
Column–II
Column – I Column – II
(A) If e1 and e2 are the eccentricities of the ellipse, and
(p) 6
hyperbola, respectively then the values of ‘a’ are
(B) If both e1 and e2 are the eccentricities of the hyperbolas, 5
(q)
then values of a are 2
(C) If e1 and e2 are eccentricities of hyperbola and conjugate
hyperbola, then values of a are (r) 2 2
(D) If e1 is the eccentricity of the hyperbola for which there exist
infinite points from which perpendicular tangents can be
(s) 5
drawn and e2 is the eccentricity of the hyperbola in which
no such points exist then the values of a are

2. Match the following Column–I with Column–II

Column – I Column – II
2
2 2 
(A)  sin1 x    sin1 y    x3 + y3 is equal to (p) 1
2
(B)  cos1 x 2   cos1 y 2 5 5
 22  x + y is equal to (q) –2

 sin1 x 2 cos1 y 2 4
(C)   |x – y| is equal to (r) 0
4
y
(D) sin1 x  sin1 y    x is equal to (s) 2

3. Match the following Column–I with Column–II

Column – I Column – II
 
(A) z4 – 1 = 0 (p) z  cos  isin
8 8
4  
(B) z +1=0 (q) z  cos  isin
8 8
4  
(C) iz + 1 = 0 (r) z  cos  isin
4 4
4
(D) iz – 1 = 0 (s) z = cos 0 + i sin 0

Space for rough work

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ANSWERS, HINTS & SOLUTIONS
HALF COURSE TEST –VI
(Paper-2)

Q. No. PHYSICS CHEMISTRY MATHEMATICS

1. B B C
2. A D B
3. D B A
4. B C B
5. D C A
6. C B B
qualified in JEE (Advanced), 2014.

7. A A B
8. B C B
9. B C A
10. D C D
11. A B D
12. D C D
13. B C A
14. C C A
15. B A C
16. A C B
17. C B A
18. A A B
19. C C C
(A)  (q) (A) → (s) (A)  (p, s)
(B)  (p) (B) → (r) (B)  (q, r)
1.
(C)  (s) (C)→ (q) (C)  (r)
(D)  (r) (D) → (p) (D)  (p, s)
(A)  (s) (A) → (s) (A)  (q, r, s)
(B)  (p) (B) → (r) (B)  (q)
2.
(C)  (q) (C)→ (p) (C)  (r, s)
(D)  (q), (r) (D) → (q) (D)  (p)
(A)  (r) (A) → (r) (A)  (s)
(B)  (p), (q), (r) (B) → (p) (B)  (r)
3.
(C)  (s) (C) → (p, s) (C)  (p)
(D)  (r) (D) → (p, q) (D)  (q)

Physics PART – I
SECTION – A

5. Use superposition principle

v 3v
6. 1000 = 
41 4 2
  2  31
 1 = 8 cm and 2 = 24 cm
3
 minimum volume = 16  100 = 1600 cm

d
14. r2  vp
dt
 
v A  v B 10  15
15.    1 rad/sec.
r 5

P (r cos , r sin )
16. r cos  + 2 = r
y 2 m/s

2  M
 1  cos   2 sin2 p 60
r 2
1, 0 (0, 0) x
1 
 sin2
r 2 y2 = 4(x +1)

p= ar [property of parabola, here a = 1]

3
d vp v r v  1
  2  2  3/2  2   
dt r r r  2

Chemistry PART – II
SECTION – A

1. Salts of soft acid with soft base are more stable than soft acid and hard base.

2. CH2 OH CHO CH2 OH

FeSO4
CHOH 
H O
 CHOH CO
2 2

CH2 OH CH2 OH CH2 OH

Glyceraldehyde Dihydroxyacetone

3. Half-life is independent of concentration of Zn, thus order w.r.t. [Zn] = 1

At pH = 2, [H+] = 10–2, t1/2 = 10 minutes
+ –3
pH = 3, [H ] = 10 , t1/2 = 100 minutes
n 1
[t1/ 2 ]1  a2 
 
[t1/ 2 ]2  a1 
n 1
10  103 
 
100  10 2 
 n = 2, thus w.r.t. [H+] order = 2
Rate law = k [H+]2 [Zn]
Hence, (B).

4. O O

C C
O O
Br2 /Fe


Br

5. H O O

O S S O

O O H

PBo XB
6. YB 
PTotal
1
Graph YB vs is linear.
PTotal

3
7. Pc Vc  RTc
8
8 RT
i.e. times of at normal condition
3 PV

8. 

 O2  O (fast)
O3 
[O2 ][O]
K eq.  …(1)
[O3 ]
K eq. [O3 ]
From Eq. (1), [O] 
[O2 ]
slow
O  O3 
 2O2
 Rate1  k [O] [O3 ]
[O3 ]
Rate1  k . K eq.  [O3 ]
[O2 ]
k  [O3 ]2
Rate1 
[O2 ]
k [O3 ]2
Rate2   2k 
[O2 ]
2

Mathematics PART – III

SECTION – A
2 n n
1. Sum of coefficients in (1 – x sin  + x ) is (1 – sin  + 1) (putting x = 1)
This sum is greatest when sin  = –1, then maximum sum is 3n
2. Let x = p – 5, y -= q – 5 and z = r – 5, where p, q, r  0
Then the given equation reduces to p + q + r = 15
The total number of solutions to this is 17C2 = 136
3. Let A  (at2, 2at), B  (at2, –2at)
2 2
mOA  , mOB  
t t
 2  2 
Thus,      1  t2 = 4
 t  t 
Thus, tangents will intersect at (–4a, 0)
a
4. Let directrix be x  and focus be S(ae, 0). Let P(a sec , b tan ) be any point on the curve
e
x sec  y tan 
Equation of tangent at P is  1
a b
Let F be the intersection point of tangent at directrix
 a b  sec   e  
Then, F   , 
e e tan  
b  sec   e  b tan 
 mSF  , mPS 
 2
e tan  a  1  a  sec   e
 mSF  mPS  1

 1 2 A(1, 10)
5. Circumcentre O    ,  and
 3 3
 11 4 
Orthocentre H   ,   8
 3 3 G  1, 
 9
 8
Hence, co-ordinates of G are  1, 
 9
Now, AD : DG = 3 : 1 B D C
 11
Hence, the co-ordinates of D   1,  
 3
3 2
6. cos x sin 2x = cos x  cos x  sin 2x
 1  cos 2x   2sin 2x cos x 
=   
 2  2 
1
= 1  cos2x  sin3x  sin x 
4
1 1 1 
= sin3x  sin x   2sin3x  cos 2x    2cos 2x sin x  
4 2 2 
1 1 1 
= sin x  sin3x  sin5x 
4 2 2 
1 1
 a1  , a3  , n = 5
4 8

7. According to the question, 1 must lie between the roots

2
Since, 4p – p – 5 < 0, thus, f(1) > 0
 p2 – 5p + 4 < 0
1<p<4
 p  {2, 3}
8. Let z = x + iy. Then, x =  + 3 and y   5  2
 (x – 3)2 = 2 and y2 = 5 – 2
2 2 2 2
Thus, (x – 3) = 5 – y  (x – 3) + y = 5
It is a semicircle as y < 0
  1  1 1  1 1 
9. We have   1            .....
4  3   5 7   9 11 
2 2 2
=    .....
1 3 5  7 9  11
1 1 1 
    ..... 
1 3 5  7 9  11 8
10. 30 – 9  [0, ] is true but it is not principal value of cos–1(cos 30)
as cos–1(cos 30) = cos–1(cos(9 + (30 – 9)))
= cos–1(–cos(30 – 9)) =  – (30 – 9) = 10 – 30
11. Locus of point of intersection of perpendicular tangents is director circle
If there exists exactly one such point on the line 3x + 4y + 5 5  0 , then it must touch the
director circle x 2 + y2 = a2 + 1
 5 = a2 + 1
 a2 = 4
a=2
1 3
Hence, eccentricity = 1  
4 2
12. According to given data, 2a – 5 + a2 – 3 < 0
 a2 + 2a – 8 < 0
 a  (–4, 2)
13. Given equation is px2 + qx + r = 0
Let, f(x) = px 2 +qx + r
f(0) = r > 0
f(1) = p + q + r < 0
f(–1) = p – q + r < 0
Hence, one root lies in (–1, 0) and the other in (0, 1)
 [] = –1 and [] = 0
 [] + [] = –1
14.-16. Let m and (m + 1) be the removed numbers from 1, 2, ….., n
n  n  1
Then, sum of the remaining numbers is   2m  1
2
n  n  1 
 2m  1
105 2
From given conditions, 
4 n2
 2n2 – 103n – 8m + 206 = 0
Since n and m are integers, so n must be even
4k 2  103 1  k 
Let n = 2k. Then, m 
4
Since m is an integer, then 1 – k must be divisible by 4. Let k = 1 + 4t. Then we get n = 8t + 2 and
m = 16t2 – 95t + 1

Now, 1 < m < n

 1  16t2 – 95t + 1 < 8t + 2
Solving, we get t = 6. Hence, n = 50 and m = 7
Hence, the removed numbers are 7 and 8
50  50  1
Also, sum of all numbers is  1275
2

17.-19. Suppose A contains r (0  r  n) elements

Then, B is constructed by selecting some elements from the remaining n – r elements
Here, A can be chosen in nCr ways and B in n – rC0 + n – rC1 + ….. + n – rCn – r = 2n – r ways
n n–r
So, the total number of ways of choosing A and B is Cr  2 . But r can vary from 0 to n
n
n n
So, total number of ways is  Cr  2n r  1  2   3n
r 0
If A contains r elements, then B contains (r + 1) elements
Then, the number of ways of choosing A and B is nCr  nCr + 1
But r can vary from 0 to (n – 1)
n1
So, the number of ways is  n
Cr nCr 1 = nC0nC1 + nC1nC2 + ….. + nCn – 1nCn = 2nCn –1
r 0
Let A contains r(0  r  n) elements
Then, A can be chosen in nCr ways. The subset B of A can have at most r elements, and the
number of ways of choosing B is 2r. Therefore, the number of ways of choosing A and B is
n
Cr  2r. But r can vary from 0 to n
n
n n
So, the total number of ways is  Cr  2r  1  2   3n
r 0

SECTION – B
1. (A) We must have, e1 < 1 < e2
 f(1) < 0
1–a+2<0
a>3
(B) We must have both roots greater than 1
(i) D > 0 or a2 – 4 > 0 or a  (–, –2)  (2, )
(ii) 1f(1) > 0 or 1 – a + 2 > 0 or a < 3
a
(iii)  1  a > 2
2
Thus, we have a  (2, 3)
1 1
(C) We must have 2  2  1
e1 e2
 e1  e2 2  2e1e2 1

e12 e22
a2  4
 1
4
 a  2 2
(D) We must have e2  2  e1
 f  2  0
 2a 2 2 0
 a2 2

 sin1 x 2   sin1 y 2 2
2. (A) 
2
2 2 2
  sin1 x    sin1 y  
4
1  1 
 sin x   , sin y  
2 2
 x = 1 and y =  1
3 3
 x + y = –2, 0, 2
2 2
(B)  cos1 x    cos1 y   22
2 2
  cos1 x    cos1 y   
 x = y = –1
 x5 + y5 = –2
2 2 4
(C)  sin1 x   cos1 y  
4
2
2  2
  sin1 x   and  cos1 y   2
4

  sin x    and  cos1 y   
1
2
 x = 1 and y = –1
 |x – y| = 0, 2
(D) |sin–1 x – sin–1 y| = 
 
 sin1 x   and sin1 y 
2 2
 
or sin1 x  and sin1 y  
2 2
 xy = 1(–1) or (–1) = 1 or –1

3. (A) z4 – 1 = 0
 z4 = 1 = cos 0 + i sin 0
 z = (cos 0 + i sin 0)1/4 = cos 0 + i sin 0
(B) z4 + 1 = 0
 z4 = –1 = cos  + i sin 
1/ 4  
 z   cos   isin    cos  isin
4 4
(C) iz4 + 1 = 0
 
 z4 = i = cos  isin
2 2
1/ 4
    
 z   cos  i sin   cos  isin
 2 2 8 8
(D) iz4 – 1 = 0
 
z4 = –i = cos  isin
2 2
1/ 4
    
 z   cos  isin   cos  isin
 2 2 8 8

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HALF COURSE TEST – VII
ALL INDIA INTEGRATED TEST SERIES
Time Allotted: 3 Hours Maximum Marks: 360
 Pl ea s e r ea d t h e i n s t r u c t i o n s c a r ef u l l y. Yo u a r e a l l o t t ed 5 m i n ut es
s p ec i f i c a ll y f o r t h i s p u r p o s e.
 Yo u a r e n o t a l l o wed t o l ea v e t h e E xa m i n at i o n Ha l l b ef o r e t h e en d o f
t h e t es t .

B. Filling of OMR Sheet

C. Marking Scheme For All Three Parts.

(i) Section-A (01 to 30) contains 30 multiple choice questions which have only one correct answer.
Each question carries +4 marks for correct answer and – 1 mark for wrong answer.

Name of the Candidate

Enrolment No.

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Useful Data

PHYSICS
Acceleration due to gravity g = 10 m/s2

Planck constant h = 6.6 1034 J-s

Charge of electron e = 1.6  1019 C

Mass of electron me = 9.1  1031 kg

2 2
Permittivity of free space 0 = 8.85  1012 C /N-m

Density of water water = 103 kg/m3

Atmospheric pressure Pa = 105 N/m2

Gas constant R = 8.314 J K1 mol1

CHEMISTRY

Gas Constant R = 8.314 J K1 mol1

Atomic No: H=1, He = 2, Li=3, Be=4, B=5, C=6, N=7, O=8,

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Physics PART – I

SECTION – A

Straight Objective Type

This section contains 30 multiple choice questions numbered 1 to 30. Each question has 4 choices (A),
(B), (C) and (D), out of which ONLY ONE is correct

1. A block of mass m is placed on the plank of mass M as F

shown in the figure. Coefficient of friction between the m
surfaces of m and M is  and there is no friction between M
ground and M. A horizontal force F = kt is applied on the
block, where k is a constant and t is the time. Which of the
following graphs shows the acceleration of the plank?
(A) (B)
a a

O
t
O t
(C) (D)
a a

O
t
O t

2. The pulleys and strings shown in the figure are smooth and of
negligible mass. For the system to remain in equilibrium, the
T
angle  should be T
 
(A) 0 (B) 30 T T

(C) 45 (D) 60

102kg
10kg 10kg

3. A body of mass m slides down on an incline plane. It reaches at the bottom with a velocity v. If
the same mass were in the form of the ring which rolls down this inclined, the velocity of the ring
at the bottom’s
1
(A) 2v (B) v
2
(C) v (D) 2v
Space for rough work

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4. A force shown in the F-x graph is applied to a 2 kg block F(N)

horizontal as shown in the figure. The change in kinetic 5
energy is 4
(A) 15 J 3
(B) 20 J
2
(C) 25 J
(D) 30 J 1

2 4 6 8 10 x(m)

5. Two masses of 4 kg and 1 kg are moving in opposite direction with equal kinetic energies of 8 J.
If the collision between the masses are perfectly inelastic the percentage loss of kinetic energy of
the masses is
(A) 60% (B) 70%
(C) 80% (D) 90%

6. A particle of mass m and radius is moving in a horizontal circle of radius r with a centripetal force
k
F=  r̂ , where k = 1 unit. The total energy of the particle is
r2
k 1.5k
(A)  (B)
r r
k 3k
(C)  (D) 
2r 2r

7. A particle moves in the x-y plane under the action of a force F at any time t its linear momentum
is given by Px  2 cos kt  and Py  2cos  kt 
 
The angle between F and P at time t is
(A) 0 (B) 30
(C) 90 (D) 180

8. An ideal spring having force constant k is suspended from the rigid support and a block of mass
M is attached to its lower end. The mass is released from the natural length of the spring. Then
the maximum extension in the spring is
4Mg 2Mg
(A) (B)
k k
Mg Mg
(C) (D)
k 2k
Space for rough work

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9. A particle of mass m is moving with a uniform speed v in x-y plane along a straight line,
y = x+k, the angular momentum of the particle about the origin is
mvk mvy
(A) (B)
2 2
mvx
(C) (D) mvk
2

10. A disc and a ring having same mass M and radius R rotating about its axis which is normal to
their plane. If both have the same angular momentum, then the ratio of acceleration of the
particles at the circumference of the disc to the ring is
(A) 1 : 2 (B) 2 : 1
(C) 1 : 4 (D) 4 : 1

11. A spring-block system can oscillates on a smooth horizontal surface. If the spring is compressed
and is released from rest. The graph between the acceleration of the block and the distance x
travelled by it is given by
(A) a (B) a

x x
(C) a (D) a

x
x

12. A hollow sphere is placed on a rough horizontal surface. A sharp impulse is given to the sphere
horizontally towards its centre
(A) A frictional force acts on the sphere throughout the motion.
(B) kinetic energy remains constant throughout the motion.
(C) kinetic energy of translation changes into rotation but total kinetic energy decreases initially
and finally becomes constant.
(D) Work is done by the frictional force throughout the motion.
Space for rough work

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 
13.   
The initial velocity of a particle is u  2iˆ  3ˆj m/s . A constant force F  4iˆ  3ˆj N acts on the
particle. Hence
(A) Its velocity is constant. (B) Its path is parabolic.
(C) its path is straight line. (D) it moves on a circular path.

x2 t2
14. A particle moves along a curve y = and x = , where x and y are measured in metre and t
2 2
in seconds. At t = 2 sec. The velocity of the particle is
(A)  2iˆ  4ˆj  m/s (B)  2iˆ  4ˆj  m/s
(C)  4iˆ  2ˆj  m/s (D)  4iˆ  2ˆj  m/s

15. The 300 N crate slides down the curved path from A to B in the A
vertical plane as shown in the figure. If the crate has a velocity of
1.2 m/s down the inclined at point A and 8 m/s at B. Find the work
done against friction during the motion
6m
(A) 860 N m (B) 987 N m
(C) 1200 N m (D) 500 N m
B
9m

16. A block B of specific gravity 2 and another block C of specific gravity 0.5. Both are joined together
and they are floating inside water such that they are completely dipped inside water, the ratio of
the masses of the blocks B and C is
(A) 2 : 1 (B) 3 : 2
(C) 5 : 3 (D) 2 : 3

17. A source of sound emits sound waves in a uniform medium. If energy density is E and maximum
speed of particles of the medium is vmax then the graph between E and vmax is best represented
by
(A) E (B) E

vmax vmax
(C) E (D) E

vmax vmax

Space for rough work

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18. The potential energy of a harmonic oscillation of mass 2 kg in its resting position is 5 joules. Its
total energy is 9 joules and its amplitude is 1 cm. It is time period will be
(A) 3.14  102 sec (B) 3.14  102 sec
(C) 103 sec (D) data are not sufficient.

19. An object of mass 0.8 kg is attached to one end of a spring and x(m)
the system is set into simple harmonic motion. The 0.080
displacement x of the object as a function of time is shown in
the figure with the aid of the data the magnitude of the objects 2.0 3.0 4.0
acceleration at t = 1.0 is 1.0 t(s)
(A) zero. (B) 1.57 m/s2 0.080
(C) 0.197 m/s2 (D) 0.157 m/s2

20. An air bubble 0.8 mm in diameter having density 1.293 kg m3 rise in a liquid of viscosity
0.15N- s m2 and specific gravity 0.9 with terminal velocity.
(A) 0.21 cm/s (B) 0.42 cm/s
(C) 0.11 cm/s (D) 0.183 cm/s

21. Two different adiabatic paths for the same gas intersects two
a
isothermal T1 and T2 as shown in the P-V diagram. Then P

Va Vd is equal to b
T1
VC
(A) (B) VC Vb
Vb d
Vb VC Vb T2
(C) (D) c
Vc VC  Vb V

2
22. The potential energy of a particle oscillating along x-axis is given as U = 20+  x  2  where U is
in joules and x in meters. Total mechanical energy of the particle is 36 J. Maximum kinetic energy
of the particle is
(A) 24 J (B) 36 J
(C) 16 J (D) 20 J

23. Two spheres of the same material and same radius r are touching each other gravitational force
between them is directly proportional to
2 3
(A) r (B) r
(C) r 4 (D) r
Space for rough work

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8

24. Force of attraction on a particle of mass m placed at the centre of a semicircular wire of length L
and mass M is
GMm GMm
(A) (B)
r2 2r 2
2GMm
(C) zero (D)
L2

25. Starting with the same initial conditions if an ideal gas expands from volume V1 to V2 in three
different ways. The work done by the gas is W1 in the process is purely isothermal, W2 is purely
isobaric and W3 is purely adiabatic, then
(A) W2  W1  W3 (B) W2  W3  W1
(C) W1  W2  W3 (D) W1  W3  W2

26. A block of ice at 10C is slowly heated and converted to steam at 100C. Which of the following
curve represents the phenomenon qualitatively?
(A) (B)

Temperature
Temperature

Heat supplied Heat supplied

Temperature

Heat supplied
Heat supplied

Space for rough work

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27. A hollow sphere of radius R is made of metal whose specific gravity is . The sphere will float in
water if thickness of wall of the sphere is (thickness of the wall << R)
R 2R
(A) > (B) =
3 3
R 4R
(C) < (D) <
3 3

28. A tank containing a liquid, has two similar holes of area a on the opposite sides. The difference in
height of two holes is y. If the tank has an open top and liquid jets out of the two holes, the net
horizontal force experienced by the tank is given by
(A) agy (B) 2agy
1 2
(C) agy (D) agy
2 3

29. When the temperature of a body increases from T to T + T, its moment of inertia increases from
I to I + I. The coefficient of linear expansion of the body is . The ratio I/I is equal to
T 2T
(A) (B)
T T
(C) T (D) 2T

30. Water rises in a capillary tube to a height of 2.0 cm. In another capillary tube whose radius is one
third of it, water will rise to
(A) 2 cm (B) 6 cm
2
(C) cm (D) none of the above
3
Space for rough work

Chemistry PART – II

SECTION – A

Straight Objective Type

This section contains 30 multiple choice questions numbered 1 to 30. Each question has 4 choices (A),
(B), (C) and (D), out of which ONLY ONE is correct

1. Find the orbital angular momentum of an electron having quantum numbers:

1
n = 3,   2, m  0, ms  
2
3h
(A) (B) 3 
2
(C) 6 (D) 15 B.M.

2. Oxidation number of carbon in carbon suboxide is

2 4
(A)  (B) 
3 3
4
(C) +4 (D) 
3

3. Arrange the following according to their increasing order of acidic strength:

OH OH OH OH OH

(I) (II) (III) (IV) (V)

F Cl Br I
(A) I < II < III < IV < V (B) III < I < II < IV < V
(C) I < II < V < IV < III (D) V < IV < III < II < I

Space for rough work

4. 10 moles of O2 gas are contained in a cylinder of 5 litre capacity. If the cylinder leakage into
atmosphere (pressure is 1 atm), then the work done by the gas will be
(A) 1.59 kJ (B) 2.55 kJ
(C) 3.27 kJ (D) 4.89 kJ

5. Arrange the following in an increasing order of stability:

CH2

(I) (II) (III) (IV)

(A) I < II < III < IV (B) IV < III < II < I
(C) II < I < III < IV (D) III < I < IV < II

6. 90 gram of vicinal tetra bromopropane was reacted with NaNH2 to form alkyne which then passes
through ammonical AgNO3 solution. The weight of precipitate formed will be
(A) 10.25 gm (B) 36.75 gm
(C) 46.50 gm (D) 54.28 gm

7. The volume of 0.1 M HCl required to react completely with 1 gm mixture of Na2CO3 and NaHCO3
containing equimolar amount of two is
(A) 106 ml (B) 128 ml
(C) 156 ml (D) 212 ml

8. The volume of Cl2 at STP react with 1 gm Fe to produce FeCl3 will be

(A) 0.6 litre (B) 1.2 litre
(C) 1.8 litre (D) 2.2 litre

9. An electron whose total energy is 2.42  1019 J given up a photon as its energy fall to
21.76  1019 J. What is the wavelength of the emitted light?
 
(A) 806 A (B) 1027 A
 
(C) 2812 A (D) 5408 A

Space for rough work

10. What volume of N2 at 300C and 2.5 atm contains as many molecules as 2.5 L O2 at STP?
(A) 1.6 L (B) 2.1 L
(C) 2.5 litre (D) 3.2 litre

11. The solubility of Al(OH)3 at pH = 4, Ksp of Al(OH)3 = 5  1033

(A) 5  103 (B) 10  104
8
(C) 1.53  10 (D) 2.6  103

12. The dipole moment of NaCl molecule is 8.5 D and its bond length is 2.36 A . The percentage
ionic character in this molecule will be
(A) 50.25 % (B) 60.30 %
(C) 75.03% (D) 100%

13. The solubility of Cd(CN)2 in 0.050 M NaCN solution will be what? If the Ksp of Cd(CN)2 is
1.0  108 m
(A) 1  106 (B) 1  108
8
(C) 2  10 (D) 4  106

14. Calculate free energy change for the following reaction at 298 K:
2NO  g  Br2    
 2NOBr  g
Given the partial pressure of NO is 0.1 atm and the partial pressure of NOBr is 2.0 atm and
Gf0NOBr = 82.4 kJmol1, G0NO = 86.55 kJmol1
(A) 6.7 kJ (B) 8.3 kJ
(C) 1.4 kJ (D) +8.3 kJ

15. What will be the activation energy of the reaction for which the rate of reaction triples with the
temperature change by 15C, from 298 K to 313 K?
(A) 36 kJ mol1 (B) 40 kJ mole
(C) 48 kJmole (D) 57 kJ/mole1

16. A 200 ml solution of NaOH requires 100 ml of 0.44 M aqueous HCl for complete neutralization.
The temperature of the mixture increased by 1.96 K. Calculate the enthalpy change for the
neutralization of one mole of hydrogen ions. Given specific heat of water = 4.184 J/g K.
(A) 56 kJ (B) 40 kJ
(C) 90 kJ (D) 110 kJ

Space for rough work

17. For the equilibrium:

PCl5 g 

 PCl3 g  Cl2 g
If the equilibrium concentration of PCl5(g) is doubled, the concentration of Cl2(g) will be
(A) 1/2 of original value (B) 1/4 of original value
(C) twice of original value (D) none of these

18. Which of the following is correct for ClNO?

(A) shape is V – shaped and Cl – N – O bond angle is slightly less than 120.
(B) shape is T – shaped and Cl – N – O bond angle is greater than 120.
(C) shaped is linear and Cl – N – O bond angle is less than 120
(D) none of these

19. Which of the following statements is correct for compound HgCl2, SnCl2 and F2O have?
(A) all have linear shape
(B) HgCl2 and SnCl2 are linear and HgCl2 is V – shaped
(C) HgCl2 is linear while SnCl2 and F2O are V – shaped
(D) none of these

20. What is the molar mass of a gas if equal volume of oxygen gas and unknown gas take 3.25 and
8.41 minutes to effuse through a membrane respectively?
(A) 112 g/mole (B) 160 g/mole
(C) 214 g/mole (D) 306 g/mole

21. A 3.11 L container has 4.33 atm pressure at 27C. If 2.11 L of gas at 2.55 atm is added to this
container at same temperature. What will be the new pressure in the container?
(A) 4 atm (B) 4.6 atm
(C) 5.2 atm (D) 6.06 atm

22. What is the wavelength of light emitted by electron in C5+ ion during transition from n = 3 to n = 1?
(A) 2.92 nm (B) 3.6 nm
(C) 8.7 nm (D) 9.8 nm

23. Many gases when allowed to expand suddenly, there is a drop in temperature, but there is a pair
of gas which do not obey this law are
(A) H2, CO2 (B) H2, He
(C) Cl2, O2 (D) O2, He

Space for rough work

24. Calculate the weight of methane required to produce 356.4 kJ of heat on combustion. Given the
heat of combustion of methane is 890.3 kJ
(A) 16 g (B) 6.4 g
(C) 12.8 g (D) 30.6 g

25. Molecular nitrogen can be converted into NO by the following reaction sequence,
N2  3H2  2NH3
4NH3  5O2  4NO  6H2O
Calculate the mass of NO formed from 100 g of N2 with excess of H2 and O2?
(A) 100 gm (B) 150.60 gm
(C) 214.28 gm (D) 256.32 gm

26. A compound X having molecular formula C4H8 gives Y (C4H10O) when reacted with H2O/H2SO4, Y
can not be resolved in optical isomer. Compound X will be
(A) H3C CH CH CH3 (B) H3C CH2 CH CH2
(C) CH3 (D) None of these
H3C C CH2

27. Potassium salt of monocarboxylic acid containing n carbon atom undergo Kolbe’s electrolysis
then the alkane formed will contain number of carbon atom equal to
(A) 2n (B) 2n+2
(C) 2n 1 (D) 2n – 2

28. Neopentyl bromide with alcoholic KOH gives the major product
(A) But – 2 – ene (B) 2 - methyl but – 2 – ene
(C) 2 – methyl but – 1 – ene (D) 2, 2 – dimethyl but – 1 – ene

29. Which of the following equilibrium will not be affected by adding inert gas at constant volume?

(A) H2  g  I2 (g)  2HI  g
 
(B) 3H2  g   N2 (g)  2NH3  g 


(C) PCl5  g   PCl3  g  Cl2  g
 (D) all the above

30. Which of the following statement is true about boron compounds?

(A) BF3  least acidic among the boron halides
(B) BF3  does not hydrolyses
(C) B2H6  does not have banana bond
(D) all are incorrect

Space for rough work

Mathematics PART – III

SECTION – A

Straight Objective Type

This section contains 30 multiple choice questions numbered 1 to 30. Each question has 4 choices (A),
(B), (C) and (D), out of which ONLY ONE is correct

1. A relation R : A  A be defined on set A which is union of disjoint sets A1A2A3….An having equal
number of elements such that  a  Ai and b  Aj (a, b)  R if i = j and (a, b)  R, if i  j
(A) if n(A) = 15 max value of n(R) is 45 (B) R is an equivalence relation
2
(C) n(R) = [n(A)] (D) none of these

2. The value of a for which 2x 2 – 2(2a + 1)x + a(a + 1) = 0 may have one root less that a and other
root greater then ‘a’ are given by
(A) 1 > a > 0 (B) –1 < a < 0
(C) a  0 (D) a > 0 or a < –1

3. If sin6  + cos6  + k cos2 2 = 1, then k is equal to

1 1
(A) tan2 2 (B) tan2 2
2 4
3
(C) 4 cot2 2 (D) tan2 2
4

tan x tan y tan x

4. If   ( 0) and x + y + z =  then
1 2 3
(A) maximum value of tan x + tan y + tan z is –6 (B) minimum value of tan x + tan y + tan z is –6
(C) tan x tan y tan z = 1 (D) tan x + tan y + tan z = 0  x, y, z  R

3c
5. If the equation ax2 + 25x – 3c = 0 has non real roots and < (a + b), then c is always
4
(A) negative (B) 0
(C) non-negative (D) zero

Space for rough work

6. Which of the following function is surjective but not injective

(A) f : R  R f(x)  x 4  2x3  x 2  1 (B) f : R  R f(x)  x3  x  1
(C) f : R  R  f(x)  1  x 2 (D) f : R  R f(x)  x 3  2x 2  x  1

7. If 5a + 4b + 20c = t, then the value of t for which the line ax + by + c – 1 = 0 always passes
through a fixed point is
(A) 0 (B) 20
(C) 30 (D) none of these

8. The area bounded by the circles x 2 + y2 = r2, r = 1, 2 and the rays given by 2x 2 – 3xy – 2y2 = 0,
y > 0 is
 
(A) sq. units (B) sq. units
4 2
3
(C) sq. units (D)  sq. unit
4

9. The equation of the circle touching the lines |y| = x at a distance 2 units from the origin is
(A) x2 + y2 – 4x + 2 = 0 (B) x2 + y2 + 4x – 2 = 0
2 2
(C) x + y + 4x + 2 = 0 (D) none of these

10. If , ,  are the parameter of points A, B, C on the circle x 2 + (y – 2)2 = a2 and if the ABC be
equilateral, then
(A)  cos   0 (B) 
sin   0
(C)  tan   0 (D)  cot   0

11. The orthocentre of the triangle formed by the pair of lines 2x 2 – xy – y2 + x + 2y – 1 = 0 and the
line x + y + 1 = 0 is
(A) (–1, 0) (B) (0, 1)
(C) (–1, 1) (D) none of these

2 2
12. The equation  x  2   y2   x  2   y 2  4 represents
(A) pair of lines (B) a parabola
(C) a hyperbola (D) none of these

Space for rough work

13. The pole of the line  x + my + n = 0 with respect to the parabola y2 = 4ax
 h 2am  h 2am 
(A)   ,  (B)  ,  
      
 h 2am   h 2am 
(C)   ,  (D)   ,  
       

x y x2 y2
14. If   2 touches the ellipse 2  2  1 then the eccentric angle  of the point of contact is
a b a b
equal to
(A) 0 (B) 90º
(C) 45º (D) 60º
2 2
15. If e, e be the eccentricities of two conics S and S and if e + e = 3 then both S ands S can be
(A) hyperbolas (B) ellipse
(C) parabolas (D) none of these

1 1
16. Base angle of a triangle are 22 º and 112 º then find the height of triangle
2 2
(A) equal to base (B) one third of base
(C) half of base (D) one forth of base

cot A
17. If in a triangle ABC a, b, c are roots of the equation x3 – 11x2 + 38x – 40 = 0 then  is
a
equal to
3
(A) (B) 1
4
9
(C) (D) none of these
16

18. Two circles with radii a and b touch each other externally such that  is the angle between their
direct common tangents (a > b  2), then
ab ab
(A)   2cos 1   (B)   2 sin1  
ab ab
ab ab
(C)   2cos 1   (D)   2 sin1  
ab ab

Space for rough work

19. If ‘d’ is the distance between parallel tangents with positive slope to y2 = 4x and
2 2
x + y – 2x + 4y – 11 = 0
(A) 10 < d < 2 (B) 4 < d < 6
(C) d < 4 (D) none of these

20. Let P be any point on a directrix of an ellipse of eccentricity e, S be the corresponding focus and
C is the centre of the ellipse. The line PC meets the ellipse at A and the angle between PS and
tangent at A is , then  is equal to

(A) tan–1 e (B)
2
(C) tan (1 – e2) (D) none of these

21. The equation 4x2 + mxy – 3y2 = 0 represents a pair of real and distinct lines if
(A) m  R (B) m  (3, 4)
(C) m  (–3, 4) (D) m > 4

 sin2 A  sin A  1 
22. In any triangle ABC,  
 sin A
 is always greater then or equal to

(A) 9 (B) 3
(C) 27 (D) none of these

23. ABCD is a square of side length a, a  N, a > 1. Let L1, L2, L3 ..... be points on BC such that
BL1 = L1L2 = L2L3 = ..... = 1 and M1, M2, M3 ..... be points on CD such that
a 1
CM1 = M1M2 = M2M3 = ..... = 1, then   AL
n 1
2
n  LnMn2  is equal to

1 2 1
(A) a  a  1 (B) a  a  1 4a  1
2 2
1
(C)  a  1 2a  1 4a  1 (D) none of these
2

n
1
24. The value of 
r 1 a  rx  a   r  1 x
is

n a  nx  a
(A) (B)
a  a  nx x
n  a  nx  a 
(C) (D) none of these
x

Space for rough work

25. If  and  are the distinct real roots of x 2 + ax + b = 0 and 4 and 4 are the roots of
2 2 2
x – px + q = 0 then the roots of x – 4bx + 2b – p = 0 are always
(A) both non-real (B) both positive
(C) both negative (D) one positive and other negative

26. The solution set of the equation logx 2  log2x 2  log4x 2 is

(A) 2 2 , 2 2  (B)  

1
2
, 2

27. The equation of ellipse whose foci are (5, 0) and one of its directrix is 5x = 36 is
x2 y2 x2 y2
(A)  1 (B)  1
36 11 6 11
x2 y2
(C)  1 (D) none of these
6 11

28. The value of x and y satisfying the equation

sin7 y = |x3 – x2 – 9x + 9| + |x3 – 4x – x2 + 4| + sec2 2y + cos4 y are

(A) x = 6, y = n (B) x = 1, y = 2n +
2
(C) x = 1, y = 2n (D) none of these

29. The equation of the tangents to the conic 3x 2 – y2 = 3 perpendicular to the line x + 3y = 2 is
(A) y  2x  6 (B) y  6x  3
(C) y  x  6 (D) y  3x  6

30. The domain of the function f  x   log3 x  x 2  1 is

(A) (–3, –1)  (1, ) (B) [–3, –1)  [1, )
(C) (–3, –2)  (–2, –1)  (1, ) (D) [–3, –2)  (–2, –1)

Space for rough work

FIITJEE Students From All Programs have bagged 34 in Top 100, 66 in Top 200 and 174 in Top 500 All India Ranks. FIITJEE Performance in JEE

FIITJEE JEE(Main), 2015

(Advanced), 2014: 2521 FIITJEE Students from Classroom / Integrated School Programs & 3579 FIITJEE Students from All Programs have

ALL INDIA INTEGRATED TEST SERIES ANSWERS, HINTS & SOLUTIONS

HALF COURSE TEST – VII
(Main)

Q. No. PHYSICS CHEMISTRY MATHEMATICS

1. A C B
2. C B D
3. B C D
4. B B B
5. D C A
6. C B D
qualified in JEE (Advanced), 2014.

7. D C B
8. B A C
9. A B A
10. D B A
11. C A A
12. C C A
13. B D B
14. B A C
15. A D A
16. A A C
17. B D C
18. B A D
19. C C C
20. A C B
21. C D A
22. C A A
23. C B B
24. D B A
25. A C D
26. A C A
27. C D A
28. B B B
29. D D D
30. B A C

Physics PART – I

kt
1. Initial acceleration a =
Mm
After some instant upper block slide over the plank and hence acceleration of the plank remains
constant.

1
3. mgh = mv 2 . . . (1)
2
1 1
mR2  2  mv 2  mgh
2 2
v
 v 
2

1
4. Change in kinetic energy k =  8  5  20J
2

1
5. Kinetic energy, k = mv 2
2
v 1  2m/s and v 2  4 m/s
Now 4  2  1 4   4  1 v
4
 v
5
k  89 
  100     100  90%
k  16  5 

k mv 2
6. F=  
r2 r
1 k
 KE  mv 2 
2 2r
k  k
and potential energy =  Fdr  r 2
dr    
 r

7.  
P  2coskt ˆi  ˆj

 
F  2k sinkt ˆi  ˆj
 
Angle between F and P is 180

1 2
8. Potential energy = kx  Mgx
2
2Mg
x
k

9. y = mx + c and y = x+k
 = 45 and c = k
1 mvk
 Angular momentum, L = mv mv  k  
2 2

10. Angular momentum, L = I

MR2
1  mR 22  1  22
2
a1 12R
  4 :1
a 2 22R

11. F = kx = ma
 ax

x2 t4
14. y= =
2 8
v y  4 (At t = 2 sec)
t2
and x=
2
v x  2 (At t = 2 sec)
 
2iˆ  4ˆj m/s

16. From the principle of flotation

m m 
g mB  mC    B  C   1 g
 B C 
m m
 mB  mC   B  C
2 0.5
mB 2

mC 1

17. vmax = A = 2fA

1 2
Energy density E = v max
2
So the curve between E and vmax is a parabola of increasing slope and symmetric about y-axis.

1 2
18. kA  9.5  4J
2
8 8
k 2  2
 8  10 4 N/m
A 10 2
 
m 2.0
T  2  2  3.14  10 2 sec
k 8  104

19. From the figure

T = 4 sec
2 
=   1.57rad / sec
T 2
2
k  
  k   0.8     1.97N/m
m 2
a  2 x  0.197m/s2

4 3 4
20. 6rv T  r air g  r 3  g
3 3
2r 2    air  g
vT 
9
Substituting values we get,
v T  0.21 cm/s

1  1
21. T1  Va   T2  Vd 
1
T1  Vd 
 
T2  Va 
Also
1 1
T1  Vb   T2  Vc 
1
T1  Vc 
 
T2  Vb 
Va V
 b
Vd Vc

22. k max  E  Umin

=36  20 = 16 J

23. m1  m 2  volume  density

4  4 
G  r 3   r 3  2
Gm1m 2 3 3
F    
r2 r2
F  r4

M
Gm   rd.sin 
 
 L GMm
24. F =  = cos 
0 r2 Lr 0

GMm GMm 2GMm 2GMm 

=  1  = 
Lr Lr Lr L2

25. Area under PV graph = work done

2
Here area A 2  Area A1  Area A 3
1

27. Let the thickness of the wall is t

4  3 3
Weight of the sphere =  R  R  t  
3  
4
Weight of the sphere = R3
3
4 4   3 3
For flotation R3  R  R  t  
3 3  
R
t
3

28. Velocity of liquid from bottom hole v1  2g  h  y 

Velocity of liquid from top hole v 2  2gh
Force on upper hole = av 22
Force on lower hole = av12

 
Net force on the tank = a v12  v 22  2agy

Chemistry PART – II

4. V1 = 5 litre
Pext = 1 atm, T = 40 + 273 = 313 K
nRT 10  0.0821 313
V2    256.97 litre
P 1
W = Pext (V2 – V1)
= 1(256.97 – 5) = 251.97 L atm
= 251.97  101.3 J
= 25524.8 J = 2.55 kJ

6. Br Br
NaNH AgNO3
H3C C C H 
2
 H3C C C H   H3C C CAg
Br Br 1 1
mole mole
4 4
molar mass  360
90 1
mole  
360 4
1
Mole of ppt = mole
4
1
Weight =  147  36.75
4

7. Na2 CO3  2HCl  2NaCl  Na 2CO3

n 2

NaHCO3  HCl  NaCl  CO2  H2 O

n 1

Na2 CO3  NaHCO3

Wt. x gm 1 x
x 1 x
Mole =  ... 1 equimolar amount
106 86
Meq of Na2CO3 + Meq of NaHCO3 = Meq of HCl
x 1 x
2  0.1 V ...(2)
106 86
From (1)
x 1 x

106 86
86x = 106 – 106x
192x = 106  x = 0.552
1 – x = 0.448
From (2)
0.552 0.448
2  0.1 V
106 86
0.0104 + 0.00520 = 0.1  V
0.01562 = 0.1 V
V = 0.1562 litre
V = 156 ml

8. 3Cl2  2Fe  2FeCl3  s 

1
moles of Fe   0.01786
56
3
mole of Cl2 =  0.01786  0.02678
2
Requirement volume of Cl2 at STP will = 0.02678  22.4
= 0.6 litre

9. E2 2.42  1019

E1 21.76  1019
E = E2 – E1
hc
= 2.42  1019 + 21.76  1019 =

6.625  1034
or 19.34  10 19   3  108

19.875  10 26
  1.0276  10 7 m
19.34  10 19

= 1027 A

10. 2.5 litre O2 at STP

2.5
Moles of O2 =  0.1116
22.4
For N2:
PV = nRT
2.5  V = 0.1116  0.0821  573
V = 2.1 litre

12.    
 exp  e  d  4.8  10 10  2.36  10 8 e.s.u cm
18
= 11.328  10 e.s.u  cm
= 11.328 D
 obs 8.5
% ionic character =  100   100
 exp 11.328
= 75.03%

CdCN2 

 2
 Cd  2CN


13.
S 2S  0.50
NaCN  Na  CN 

0.50 M
2
Ksp = S  (0.05)
1 10 8  100  100
S
0.05  0.05
= 4  106

14. G0  2  G0f NOBr   2  G0f NO   G0f Br2 

= 2  82.4 – 2  86.55 + 1  0
3
= 8.3 kJ = 8.3  10 J
Now, 2NO  g  Br2      2NOBr
2

Qp 
pNOBr  g  
22
 4  102
2 2
pNO  g   0.1
0
G = G + RT ln Qp
= 8.3  103 + (8.314  298) ln 4  102
3 4
= 8.3  10 + 1.5  10 J
3
= +6.7  10 = 6.7 kJ

K1 E  1 1
15. ln  a   
K2 R  T 1 T2 
 1 Ea  1 1 
ln     
3 8.314  298 313 

 Ea = 5.7  104 Jmol1
= 57 kJ/mole

16. Since specific heat of solution is not given assuming dilute have the same specific heat capacity
as does water.
q = mcsT
= 300  4.18  (1.96)
= 2.46 kJ
Volume of solution = 100 + 200 ml
= 300
Weight of solution = 300 g
This heat flow is used to calculate H for 1 mole of the reaction.
HCl  aq  NaOH  aq   NaCl  aq   H2O   
Moles of HCl reacted = 0.1  0.44
= 0.044 mole
1 mole HCl yield 1 mole of H+
0.044 mole HCl yield 0.044 mole of H+
H  OH  H2 O   
+
Neutralization of 0.044 mole of H generates 2.46 kJ of heat.
The reaction is exothermic:
H = ve
+
4.4  102 mole H (aq) = 2.46 kJ
2.46
1 mole H+ (aq) = 
4.4  102
= 56 kJ
Hreaction = 56 kJ

18. N
Cl O

19.  bond pair Lone pair Hybridization Shape

HgCl2 2 0 sp Linear
SnCl2 2 1 sp2 V – shape
F 2O 2 2 sp3 V – shape

tx Mx 8.41 Mx
20.   
t O2 MO2 3.25 MO2
 Mx = 214.27

21. Initially moles of gas at 4.33 atm

PV 4.33  3.11
n1   ...(1)
RT RT
Moles of gas in 2.11 L container at 2.55 atm
2.55  2.11
n2  ...(2)
RT
On adding total moles = n1 + n2
4.33  3.11 2.55  2.11

RT RT
New pressure in the container of volume 3.11 L
nRT  4.33  3.11  2.55  2.11
P   RT = 6.06 atm
V  RT  3.11 

1 1 1
22.  Z 2RH  2  2 
  n1 n2 
RH = 1.097  107 m1
Z = 6 for carbon
n1 = 1, n2 = 3
1 2  1 1
 1.097  107   6   2  2 
 1 3 
 1
9 
 1.097  107  36  
 9 
8
= 1.097  107  36   35.1 10 7 = 2.85  109 = 2.85 nm
9

24. CH4  g  2O2  g   CO2  g  2H2 O   

H = 890.3 kJ
1 mole CH4 produce = 890.3 kJ
 890.3 kJ heat is produced by 1 mole CH4
1
1 kJ heat is produced by mole CH4
890.3
356.4
356.4 kJ heat is produced by  0.4 mole
890.3
Weight of CH4 required = 0.4  16 = 6.4 gm

N2  3H2  2NH3

25. 100 100
mole  2
28 28
4NH3  5O2  4NO  6H2O

4 mole NH3 gives 4 mole of NO

200 200
mole NH3 gives 4 moles of mole of NO
28 28
200
Moles of NO formed from 100 g N2 =
28
200
Weight of NO formed =  30gm = 214.28 gm
28

26. CH3 CH3

H2 SO4
H3C C CH2 H2 O   H3C
 C CH3
OH
Optically inactive

27. During electrolysis, two molecule of potassium salt of mono carboxylic acid combine to form
alkane along with the lost of two COO group. Hence alkane formed will contain = 2n – 2 (no of
C–atom).

28. CH3 CH3 CH3

alc. KOH Ionization
H3C C CH2 Br   H3C C CH 2Br   H3C C CH 2
CH3 CH3 CH3

CH3  shift

CH3 CH3
Base
H3C C CH CH3  H3C C CH2 CH3

29. Adding inert gas at constant volume does not affect any equilibrium whether n  0 or n  0.

30. BF3 is least acidic among boron halide.

Mathematics PART – III

1. (a, b)  R if a and b  Aj
 (b, a)  R which gives symmetry
(a, a)  R gives reflexivity
Similar argument gives transitivity also max value of n(R) = 152
2
2. The given condition suggest that a lies between the roots. Let f(x) = 2x – 2(2a + 1)x + a(a + 1) for
a to lie between the roots, f(a) < 0
2a2 – 2a(2a + 1) + a(a + 1) < 0
–a2 – a < 0
a2 + a > 0
a > 0 or a < –1
6 6 2
3. sin  + cos  + k cos 2 = 1
(sin2  + cos2 )(sin4  – sin2  cos2  + cos4 ) + k cos2 2 = 1
sin4  + cos4  – sin2  cos2  + k cos2 2 = 1
(sin2  + cos2 )2 – 3 sin2  cos2  + k cos2 2 = 1
3 sin2  cos2  = k cos2 2
3 2
sin 2  k cos2 2
4
3
k  tan2 2
4

4. tan x = , tan y = 2, tan z = 3

x+y+z=
tan x + tan y + tan z = tan x tan y tan z
6 = 62  6(2 – 1) = 0
 = 1, (  0)
tan x = 1, tan y = 2, tan z = 3
tan x + tan y + tan z = 6
Maximum value 6 minimum values is –6

5. Let f(x) = ax2 + 25x – 3c

Since f(x) = 0 has non real roots, f(x) will have the same sign for all values of x
3c
Given < (a + b)
4
 4a + 4b – 3c > 0
f(2) > 0
f(x) > 0  x  R
f(0) > 0
c<0

6. (A) f(x)  x 4  2x3  x 2  1 → A polynomial of degree even will always be into say
f(x)  a0 x 2n  a1x 2n 1  a2 x 2n2  .....  a2n
 a a a    if a0  0
L imit f(x)  L imit [x 2n  a0  1  22  .....  2n 
x  x   x x x 2n    if a0  0
Hence it will never approach  / 
(B) f(x)  x3  x  1  f '(x)  3x 2  1 - injective as well as surjective

(C) f(x)  1  x 2  neither injective nor surjective (minimum value = 1)

f(x)  x 3  2x 2  x  1  f '(x)  3x 2  4x  1  D > 0
Hence f(x) is surjective but not injective

ax by
7. Given equation of line is  1 0
c 1 c 1
a b
Line contains two independent parameters and . A line will pass through a fixed point
c 1 c 1
if it contains only one independent parameter
 5a + 4b + 20c = t
5a 4b t  20c
 
c 1 c 1 c 1
b
In this equation R.H.S will the independent of c if t = 20. Then we can express in terms of
c 1
a
and vice-versa. Hence, if t = 20 then line will pass through a fixed point
c 1

8. 2x2 – 3xy – 2y2 = 0 x – 2y = 0

(x – 2y)(2x + y) = 0
Which represents a pair of perpendicular lines passing
through the origin
2 2
 2  1 3
Required area =   sq. units
4 4 4

2x + y = 0

9. As x = |y| > 0
Circle lies in I and IV quadrant
Centre of circle lies on x-axis
2 2
OC   OA    AC   2  2  2
Equation of circle is (x – 2)2 + (y – 0)2 = 2
x2 + y2 – 4x + 2 = 0

10. The point A is (a cos , 2 + a sin )

Since the triangle ABC is equilateral its circum centre (0, 2) coincides with its centroid
1 1 

3 
cos ,
3 
sin     0, 2 

 cos  = 0 and  sin  = 6

2 2
11. 2x – xy – y + x + 2y – 1 = 0
(2x + y – 1)(x – y + 1) = 0
So the sides x – y + 1 = 0 and x + y + 1 = 0 are  to each other the orthocentre is intersection of
 sides

12.  x  2 2  y 2  4   x  2 2  y 2 ..... (1)

But (x + 2)2 + y2 – [(x – 2)2 + y2] = 8 ..... (2)
Dividing (2) by (1) we get
 x  2 2  y2   x  2 2  y 2  2x ..... (3)

Adding (1) and (2) we get

2
2  x  2   y 2  4  2x
 x  2 2  y 2   x  2 2
y2 > 0
Pair of coincident lines

13. Polar of (x1, y1) w.r.t. parabola y2 = 4a is yy1 = 2a(x + x1)

2ax – yy1 + 2ax1 = 0
Compare this with  x + my + n = 0
2a y 2ax1 n
 1   x1 
 m n 
2am h 2am 
y1   pole  ,  
   

x y x2 y2
14.   2 is tangent to 2  2  1 ..... (1)
a b a b
at P(a cos , b sin )
xx1 yy1
 2 1
a2 b
xacos  by sin 
 1
a2 b2
x cos  y sin 
 1 ..... (2)
a b
x y
 1
a 2 b 2
1 
cos   sin   
2 4

15. When both are ellipse, then e, e < 1 and so e2 + e2 = 3 is not true and so it false
When both are parabolas e = 1 = e  e2 + e2 = 3 not true
When both are hyperbolas e2 + e2 = 3 possible e, e > 1

1 A
16. BD  Pcot 22 º
2
1
DC  Pcot112 º P
2
 1 1 
 cos22 2 º cos112 2 º  B D C
Base = P   
 sin 22 1 º sin112 1 º 
 2 2 
1
2P
2sin135º 2
=P   2P
1 1 cos90º  cos135º
2sin 22 º sin112 º
2 2
Base
P
2

cot A b2  c 2  a2 a2  b2  c 2  a  b  c 2  2ab 121  76 9

17.  a
=  2abc

2abc
=
2abc

80

16

18. From MLN

C
a b
sin  
ab N B
ab
  sin1 b 
ab  A
Angle between AB and AD M L
ab
2  sin1  
ab

1
19. Tangents to parabola y2 = 4x having slope m is y = mx + tangents to circle having slope m is
m
y + 2 = m(x – 1) + 4 1  m2
1
4 1  m2  m  2 
Distance between tangents = m
1  m2
2 m2  1
= 4  m>0
1  m2 m
Hence d < 4

x2 y2 (a cos , 5 sin ) a b 
P  , tan  
20. Let the ellipse be  1
a 2
b 2 e e 
A 
A = (a cos , b sin )
b
Equation of AC will be y  tan  x C S
a
a a b  (ae, 0)
Solving with x  , we get P  , tan  
e e e 
b
Slope of tangents at A is 
a tan 
b
tan 
b tan  a
Slope of PS  e   tan 
a 
 ae a 1  e
2  b
e

 
2

21. D should be b2 – 4ac > 0

22. Using A.M.  G.M.

1
sin A  1
sin A 1
3

1
sin A  1 3
sin A
1
Similarly sinB  1 3
sinB
1
sinC  1 3
sinC
 1 
  sin A  sin A  1  9
 

2 A B
23. AL21  L1M12   a2  12    a  1  12 
2
AL22  L2M22   a2  22    a  2   22  L1
2 2
AL2a 1  a   a  1  1 a  1 
 La 1Ma2 1 2 2
L2
The required sum
2 2
=  a  1 a 2  12  22  .....  a  1   2 12  22  ..... a  1 
D M M2 M1 C
   
 a  1 a2  3 a  1 a 2a  1
6
 2a  1
 a  a  1 a 
 2 

1 1
24. =  a  rx  a   r  1 x 

a  rx  a  r  1 x  x
1 n
t1  t 2  .....  tn   a  nx  a  
x a  a  nx

25.  +  = –a
 = b ..... (1)
4 + 4 = p
44 = q ..... (2)
Using (1) and (2), we get (a2 – 2b)2 = p + 2b2
a2 – 4ba2 + 2b2 – p = 0
a2 is one of roots of x 2 – 4bx + 2b2 – p = 0
2
Product of roots of equation = 2b – p
2 2 4 4 2 2
2  – ( +  ) = –( –  ) < 0
Hence one root is positive and other is negative

26. logx 2  log2x 2  log4x 2

1 1
Here x > 0 and x  1, ,
2 4
1 1 1
 
log2 x log2 2x log2 4x
log2 x 1  log2 x    2  log2 x 
Put log2 x = t
t(1 + t) = 2 + t
Hence x  2 2 , 2 2 

27. Foci (5, 0) where ae = 5

a a 36
x  
e e 5
a 36
ae   5 
e 5
a=6
5
e
6
2 2 2
b = a (1 – e ) = 11

28. For x = 1
sin7 y = sec2 2y + cos4 y
sin7 y cos2 2y = 1 + cos4 y cos2 2y
RHS  1 and LHS  1
LHS = 1
sin7 y cos2 2y = 1

sin7 y = 1 and cos2 2y = 1  y 
2

General value of y is 2n +
2

29. x + 3y – 2 = 0
Any perpendicular line 3x – y + c = 0
y = 3x + c
x2 y 2
Tangents to the hyperbola  1
1 3
c2 = a2m2 – b2  c =  6
y = 3x  6

30. f(x) is defined when x2 – 1 > 0 and 3 + x > 0 and 3 + x  1

x2 > 1 and x > –3 and x  –2
x < –1 or x > 1 and x > –3 and x  –2
Df = (–3, –2)  (–2, –1)  (1, )

Time Allotted: 3 Hours Maximum Marks: 246

INSTRUCTIONS
A. General Instructions
1. Attempt ALL the questions. Answers have to be marked on the OMR sheets.
2. This question paper contains Three Parts.
3. Part-I is Physics, Part-II is Chemistry and Part-III is Mathematics.
4. Each part is further divided into one section: Section-A.
5. Rough spaces are provided for rough work inside the question paper. No additional sheets will be
provided for rough work.
6. Blank Papers, clip boards, log tables, slide rule, calculator, cellular phones, pagers and electronic
devices, in any form, are not allowed.
B. Filling of OMR Sheet
1. Ensure matching of OMR sheet with the Question paper before you start marking your answers
on OMR sheet.
2. On the OMR sheet, darken the appropriate bubble with black pen for each character of your
Enrolment No. and write your Name, Test Centre and other details at the designated places.
3. OMR sheet contains alphabets, numerals & special characters for marking answers.
C. Marking Scheme For All Three Parts.
(i) Section-A (01 – 06) contains 6 multiple choice questions which have only one correct answer.
Each question carries +3 marks for correct answer and – 1 mark for wrong answer.
(ii) Section-A (07 – 10) contains 4 multiple choice questions having one or more than one correct
answer and each question carries +4 marks for correct answer and there is no negative
marking.
(iii) Section-A (11 – 14) contains 4 Assertion-Reasoning (multiple choice questions) which have
only one correct answer. Each question carries +3 marks each for correct answer and – 1 mark
for wrong answer.
(iv) Section-A (15 – 23) contains 3 paragraphs. Based upon paragraph, 3 multiple choice questions
have to be answered. Each question has only one correct answer and carries +4 marks for
correct answer and – 1 mark for wrong answer.

Name of the Candidate

Enrolment No.

Useful Data

PHYSICS
2
Acceleration due to gravity g = 10 m/s

Planck constant h = 6.6 1034 J-s

Charge of electron e = 1.6  1019 C

Mass of electron me = 9.1  1031 kg

Permittivity of free space 0 = 8.85  1012 C2/N-m2

Density of water water = 103 kg/m3

Atmospheric pressure Pa = 105 N/m2

Gas constant R = 8.314 J K1 mol1

CHEMISTRY

Gas Constant R = 8.314 J K1 mol1

Atomic No: H=1, He = 2, Li=3, Be=4, B=5, C=6, N=7, O=8,

Physics PART – I

SECTION – A

Straight Objective Type

This section contains 6 multiple choice questions numbered 1 to 6. Each question has 4 choices (A),
(B), (C) and (D), out of which ONLY ONE is correct.


1. Velocity of a body of mass 2kg moving in x-y plane is given by v   2iˆ  4tjˆ  m/s, where t is the
time in seconds. The power delivered to the body by the resultant force acting on it at t = 5 sec is
(A) 80 Watts (B) 160 Watts
(C) 40 Watts (D) 100 Watts

2. A solid sphere of mass m is lying at rest on a rough horizontal

surface. The coefficient of friction between ground and sphere is .
The maximum value of F, so that sphere will not slip, is equal to F

(A) 7mg/5 (B) 4mg/7

3. In the figure shown, mass of A and B is A

Fixed obstacle
equal to M each. Friction between B and B
F
lowermost surface is negligible. Initially,
both the blocks are at rest. The
/2 
dimensions of the block A are very small.
A constant horizontal force F is applied on the block B and both the blocks start moving together
without any relative motion.
Suddenly, the block B encounters a fixed obstacle and comes to rest. The block A continues to
slide on the block B. The block A just manages to reach the opposite end of the block B. What is
the coefficient of friction between the two blocks? (Required length are shown in the figure)
(A) F/Mg (B) 2F/Mg
(C) F/2Mg (D) none
Space for rough work

F = 1N
4. A block of mass 3 10 kg is placed on a rough horizontal
surface as shown in the figure. A force of 1 N is applied at one 30
end of the block and the block remains stationary. The normal
force exerted by the surface on the block acts (g = 10 m/s2)
20 cm
(A) through the centre of mass of the block.
(B) through point A.
(C) through point B. B A
(D) through the point at a distance 5 cm. from A. 20 cm

5. An insect of mass m is initially at rest at the one end of a stick of length L and mass M, which
rests on a smooth horizontal floor. The coefficient of friction between the insect and stick is k. The
insect starts moving to reach the other end in shortest possible time. Choose the incorrect
statement.
(A) The centre of mass of the stick has speed v =  kmg  t w.r.t. horizontal floor at time t.
 M 

2LM
(B) Time taken by the insect to reach the other end is t 
k M  m  g
(C) The magnitude of the linear momentum of the insect at time ‘t’ is equal to kmgt

(D) The maximum speed acquired by the stick is v max = 2kLMm2 g

M  m  m

6. A uniform wheel is rolling without slipping on a horizontal surface. At a R

certain instant, its centre of mass has velocity ‘v’ and acceleration ‘a’. P, Q
and R are three points on the rim of the disc as shown in the figure. a
Q
Choose the incorrect statement. v
(A) Acceleration of P is vertically upwards
(B) Acceleration of Q may be vertically downwards
(C) Acceleration of R can be horizontal P
(D) Acceleration of some point on the rim may be horizontal leftwards.
Space for rough work

Multiple Correct Choice Type

This section contains 4 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for
its answer, out which ONE OR MORE is/are correct.

7. A particle moves with an initial velocity v0 and retardation v, where v is its velocity at any time t.
(A) The particle will cover a total distance v 0/.
(B) The particle will come to rest after a time 1/.
(C) The particle will continue to move for a very long time.
(D) The velocity of the particle will become v0/2 after a time 1/.

8. A wave is represented by the equation y = A sin (10x + 15t + /3) where x is in meters and t is
in seconds. The expression represents
(A) a wave travelling in the positive x-direction with a velocity 1.5m/s.
(B) a wave travelling in the negative x-direction with a velocity 1.5m/s.
(C) a wave travelling in the negative x-direction having a wavelength 0.2m.
(D) a wave travelling in the positive x-direction having a wavelength 0.2m.

9. A perfectly elastic uniform string is suspended vertically with its upper end fixed to the ceiling and
the lower end is loaded with a weight. If a transverse wave is imparted to the lower end of the
string, the pulse will
(A) not travel along the length of the string (B) travel upwards with increasing speed
(C) travel upwards with decreasing speed (D) travel upwards with constant acceleration

10. A bimetallic strip is formed out of two identical strips one of copper and the other of brass. The
coefficients of linear expansion of the two metals are c and B. On heating, the temperature of
the strip goes up by T and the strip bends to form an arc of radius of curvature R. Then R is
(A) proportional to T (B) inversely proportional to T
(C) proportional to | B - C| (D) inversely proportional to | B - C |
Space for rough work

Reasoning Type

This section contains 4 questions numbered 11 to 14. Each question contains STATEMENT-1
(Assertion) and STATEMENT-2 (Reason). Each question has 4 choices (A), (B), (C) and (D) out of which
ONLY ONE is correct.

11. STATEMENT-1
Each molecule of a gas moves with rms speed if the temperature of gas is constant.
and

STATEMENT-2
3RT
The rms speed of molecules of a gas is equal to , where T and M are the temperature and
M
molecular mass of the gas. R is the ideal gas constant.
(A) Statement-1 is True, Statement-2 is True; Statement -2 is a correct explanation for
Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement -2 is NOT a correct explanation for
Statement-1.
(C) Statement -1 is True, Statement-2 is False.
(D) Statement -1 is False, Statement-2 is True.

12. STATEMENT-1
Water is filled up in a hollow cylindrical container of conducting bases and adiabatic curved
surface and kept vertical in an isolated system. If temperature of cylinder is decreased slowly
from the bottom, the ice formation will start from the bottom.
and
STATEMENT-2
The temperature of liquid which is at the top will be lowest first.
(A) Statement-1 is True, Statement-2 is True; Statement -2 is a correct explanation for
Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement -2 is NOT a correct explanation for
Statement-1.
(C) Statement -1 is True, Statement-2 is False.
(D) Statement -1 is False, Statement-2 is True.
Space for rough work

13. From the base of an incline plane inclined at an angle of 30 with the horizontal a particle is
projected with a speed u. The angle of projection of the particle is 60 with the horizontal.
STATEMENT-1: The particle will strike the inclined plane normally for a particular u.
and
STATEMENT-2: The angle at which the particle hits the incline plane is independent of u.
(A) Statement-1 is True, Statement-2 is True; Statement -2 is a correct explanation for
Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement -2 is NOT a correct explanation for
Statement-1.
(C) Statement -1 is True, Statement-2 is False.
(D) Statement -1 is False, Statement-2 is True.

14. STATEMENT-1: A ring of mass m and radius R has a mass m

m
attached to it. The ring rolls without slipping on the horizontal surface
R v0
as shown in the figure. At the instant shown the normal contact force
exerted by the floor on the ring is less than (2mg).

and
 
STATEMENT-2: Fext  macm (i.e. net external force acting on a rigid body = mass times
acceleration of centre of mass)
(A) Statement-1 is True, Statement-2 is True; Statement -2 is a correct explanation for
Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement -2 is NOT a correct explanation for
Statement-1.
(C) Statement -1 is True, Statement-2 is False.
(D) Statement -1 is False, Statement-2 is True.
Space for rough work

Comprehension Type

This section contains 3 groups of questions. Each group has 3 multiple choice question based on a
paragraph. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which ONLY ONE is
correct

Paragraph for Question Nos. 15 to 17

D
A composite spherical shell is made up of two materials having thermal
C
conductivities K and 2K respectively as shown in the figure. The B
2K A
temperature at the innermost surface is maintained at T whereas the K
temperature at the outermost surface is maintained at 10T. 3R
R
10 T
T
A, B, C and D are four points in the outer material such that AB = BC = CD.
2R
Now answer the following questions. $

15. The effective thermal resistance between the inner surface of the
shell and the outer surface of the shell for the radial heat flow is

1 1
(A) (B)
8KR 7KR

7 6
(C) (D)
48KR 49KR

16. The net rate of heat flow from the outermost surface to the innermost surface of the shell will be
(A) 63RKT (B) 72RKT
432 147
(C) RKT (D) RKT
7 2

17. Out of the segments AB, BC and CD the magnitude of the temperature difference will be
maximum across
(A) AB (B) BC
(C) CD (D) equal across all the three.
Space for rough work

Paragraph for Question Nos. 18 to 20

Two coherent point light sources are placed at x = -d/2 and x = d/2. These sources are surrounded by a
spherical screen having equation x 2 + y2 + z2 = R2 (where R >> d). The sources emit monochromatic light
of wavelength  (= d/3).
Waves from both the sources reach screen and superimpose. When waves reaching any point on the
screen have a constant path difference of n (where  is wavelength of light and n is some integer), the
superposition of these waves is constructive in nature (i.e. maximum intensity) n may be called as order
of maximum.

18. The number of maximas on the screen

(A) 7 (B) 12
(C) 6 (D) none of these

19. As screen is a bounded one, therefore the length of maxima is finite. Which of the following data
is correct about shape and length of second order maxima?
2R 5 2R 5
(A) hyperbola, (B) circle,
3 3
2R 2R
(C) hyperbola, (D) circle,
3 3

20. Which of the statements is incorrect about the interference experiment explained above
(A) if radius of spherical screen is doubled the intensity of maxima will become one fourth.
(B) All maximas are equally spaced along x-axis.
(C) The equation of first order maximum is y2 + z2 = (8/9) R2 and x = R/3.
(D) none of these
Space for rough work

Paragraph for Questions Nos. 21 to 23

A uniform sphere of mass m and moment of inertia 0

2 A
mR2 about its diameter is rotating with angular
5
velocity 0 about a horizontal diametrical axis.
It is now placed on rough horizontal ground with A 0 B
friction coefficient  . Sphere at
rest. Identical
to A

21. The time taken by sphere A to achieve pure rolling is (assuming no collision)
R0 R0
(A) (B)
g 7g
2R0 R0
(C) (D)
7g 4g

22. Eventually, when sphere A collides elastically with sphere B, then immediately after collision the
kinetic energy of sphere B is: (Assume no friction b/w spheres)
2 mR2 02
(A) (B) Zero
49
1 4
(C) mR2 02 (D) mR2 02
49 49

23. After collision, sphere B is allowed to move while sphere A is held back. The final kinetic energy
of sphere B when pure rolling starts:
1 5
(A) mR2 20 (B) mR2 20
2 343
3 10
(C) mR2 20 (D) mR2 20
343 343

Space for rough work

Chemistry PART – II

SECTION – A

Straight Objective Type

This section contains 6 multiple choice questions numbered 1 to 6. Each question has 4 choices (A),
(B), (C) and (D), out of which ONLY ONE is correct.
–1 –1
1. The catalyst decrease the Ea from 100 K mol to 80 kJ mol . At what temperature the rate of
reaction in the absence of catalyst at 500 K will be equal to rate reaction in presence of catalyst.
(A) 400 K (B) 200 K
(C) 625 K (D) None of the above

2. Which structure is most stable?

(A) O O (B) OH O

3. Total number of stereoisomer of the compound 1 – bromo – 3 – chloro cyclobutane

(A) 0 (B) 1
(C) 2 (D) 3

4. Two alkenes, X (91% yield) & Y (9% yield) are formed when the following compound is heated
CH3

CH3   X  Y
N OH  91%  9%
CH3
CH3
X and Y are
(A) (B)
CH3 & CH3 CH3 & CH3

Space for rough work

5. Potential energy of the electron at infinity is equal to

(A) 0 (B) 
Ze2 Z2e
(C)  (D) 
R R

6. The rate law has the form

3/2
rate  K  A B 
can the reaction be elementary process
(A) yes (B) no
(C) may be yes or no (D) cannot b predicted

Multiple Correct Choice Type

This section contains 4 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for
its answer, out which ONE OR MORE is/are correct.

7. Identify the correct statement (s) regarding hydrogen gas :

(A) unlike to H2, D2 do not show spin isomerism
(B) ortho hydrogen differs from para hydrogen w.r.t. their boiling point, specific heats & thermal
conductivity
(C) as the temperature is increased from absolute zero some of the para form gradually changes
to ortho form
(D) para hydrogen has lower energy than ortho hydrogen

8. Which of the following drugs are chiral?

(A)
(H3 C) 2HC CH2 CH(CH3 ) COOH (Ibuprofen)

(B)
HO CH 2 C(CH 3 )(NH 2)COOH (methyl dopa)

HO
(C)  CH3 2 C  SH  CH NH2  COOH pencillamine 
(D) CH(CH3)COOH
(Naproxen)

H3 CO

9. Which of the following are intensive properties?

(A) temperature (B) density
(C) specific heat (D) molar volume

Space for rough work

10. Which of the following mixture represent a buffer?

(A) 100 ml of 2 M HCN & 100 ml of 1M NaOH
(B) 100 ml of 2 M CH3COONa & 100 ml of 1 M HCl
(C) 100 ml of 0.2 M NH4OH & 100 ml of 0.1 M H2SO4
(D) 100 ml of 0.5 M C6H5NH2 & 100 ml of 0.2 M H2SO4

Reasoning Type

11. STATEMENT -1: CHF3 is less acidic than CHCl3

because
STATEMENT -2: Electron withdrawing groups increase the acidic nature.
(A) Statement -1 is True, Statement -2 is true; Statement-2 is a correct explanation for
Statement-1
(B) Statement -1 is True, Statement -2 is true; Statement-2 is NOT a correct explanation for
Statement-1
(C) Statement -1 is True, Statement -2 is False
(D) Statement -1 is False, Statement -2 is True

12. STATEMENT -1:

H3C CH3
C C
H H
has higher boiling point than
H3C H
C C
H CH3
but for melting point, it is the reverse.
because
STATEMENT -2:
H3C CH3
C C
H n
has greater heat of hydrogenation than
H3C H
C C
H CH3
(A) Statement -1 is True, Statement -2 is true; Statement-2 is a correct explanation for
Statement-1
(B) Statement -1 is True, Statement -2 is true; Statement-2 is NOT a correct explanation for
Statement-1
(C) Statement -1 is True, Statement -2 is False
(D) Statement -1 is False, Statement -2 is True

Space for rough work

13. STATEMENT -1: A mixture of ideal gases is cooled up to liquid He temperature (4.22K) to form
an ideal solution.
because
STATEMENT -2: An ideal gas cannot be liquefied as there exists no force of attraction among the
gas molecules.
(A) Statement -1 is True, Statement -2 is true; Statement-2 is a correct explanation for
Statement-1
(B) Statement -1 is True, Statement -2 is true; Statement-2 is NOT a correct explanation for
Statement-1
(C) Statement -1 is True, Statement -2 is False
(D) Statement -1 is False, Statement -2 is True

14. STATEMENT -1: At equilibrium at unit activity, G0 = RT ln Kc.

because
STATEMENT -2: Since G = G + RTlnQ and at equilibrium, G = 0 & Q becomes Kc at unit
activity.
(A) Statement -1 is True, Statement -2 is true; Statement-2 is a correct explanation for
Statement-1
(B) Statement -1 is True, Statement -2 is true; Statement-2 is NOT a correct explanation for
Statement-1
(C) Statement -1 is True, Statement -2 is False
(D) Statement -1 is False, Statement -2 is True

Comprehension Type

Paragraph for Question Nos. 15 to 17

An organic Lewis acid (A) which gives fumes in moist air and intensity of fumes is increased when a rod
dipped in NH4OH is brought near to it. An acidic solution of (A) on addition of NH4Cl and NH4OH gives a
precipitate (B) which dissolves in NaOH solution. An acidic solution of A does not gives precipitate with
H2S.
Read the above paragraph carefully and give the answer of following questions:

15. What will be structure of (A)?

(A) Planar (B) Trigonal planar
(C) Pyramidal (D) Octahedral

16. What will be (A) in above sequence?

(A) BF3 (B) AlCl3
(C) GaCl2 (D) ZnCl2

17. The intensity of fumes increase due to the formation of–

(A) NH4OH (B) HCl
(C) NH4Cl (D) NaAlO2

Space for rough work

Paragraph for Question Nos. 18 to 20

Lithium only forms monoxide when heated in oxygen. Sodium forms monoxide & peroxide in excess of
oxygen. Other alkali metals form superoxide with oxygen i.e. MO2. The abnormal behaviour of Li is due to
small size. The larger size of nearer alkali metals also decides the role in formation of superoxides. The
three ions are related to each other as follows:
1
O2
O2
O2  
2
 O 22  
 2O2
 Oxide ion  Peroxide ion   Superoxide ion 
All the three ions abstract proton from water.

18. Consider the following reaction:

M  O2   MO 2  M  alkali metal 
 sup eroxide 
Select the correct statement:
(A) M cannot be Li and Na (B) M cannot be Cs & Rb
(C) M cannot be Li & Rb (D) none of these

19. Lithium does not form stable peroxide because

(A) of its small size
(B) d – orbitals
(C) it is highly reactive & form superoxide in place of peroxide
(D) covalent nature of peroxide

20. Which anion is stable towards water?

(A) O2 (B) O22
(C) O2 (D) none of these

Space for rough work

Paragraph for Question Nos. 21 to 23

A reversible reaction proceeds in the direction that decrease the free energy of the system. There is an
important relationship between the equilibrium constant ‘K’ for a reaction and the standard free energy
0 G0
(G ): logK 
2.30 RT
This equation is application to all reversible processes, allows us to estimate K form standard free energy
of the reaction.
The value of equilibrium constant of an endothermic reaction rises with temperature, whereas a
temperature increase lowers the equilibrium constant of an exothermic reaction.
 k  H0  1 1 
log  2     
 k1  19.1  T1 T2 
k2
If H0 is positive (endothermic reaction) and T2 > T1 then log is positive. The opposite situation occurs
k1
0
if H is negative.

21. The equilibrium constant for the dissociation of N2O4 into NO2 is 0.155 at 25C. What is the
standard free energy change for this reaction given (log 0.155 = 0.810)
(A) 4.61 kJ/mole (B) +4.61 kJ/mole
(C) +8.12 (D) 8.12

22. The standard Gibbs free energy change for the reaction:
2NOBr g
 2NO g  Br2 g
is +11.70 kJ/mole at 25C. What is the equilibrium constant for the reaction at 25C?
(A) 8.8  103 (B) 1.6  104
3
(C) 2.6  10 (D) 4.2  105

23. The heat of formation of ammonia is 46.2 KJ/mole. What is the equilibrium constant for the
reaction:
3H2 g  N2 g 
 2NH3 g


at 500C? The equilibrium constant at 25C is 6.8  105 (given antilog (9.98) is equal to 1.1 
1010)
(A) 2.5  104 (B) 5.0  105
5
(C) 7.5  10 (D) 1.0  106

Space for rough work

Mathematics PART – III

SECTION – A

Straight Objective Type

This section contains 6 multiple choice questions numbered 1 to 6. Each question has 4 choices (A),
(B), (C) and (D), out of which ONLY ONE is correct.

|x| |x|
1. Let f(x)  e {e sgn x} and g(x)  e[e sgn x] , x  R where {x} and [.] denotes the fractional part and
integral part functions respectively. Also h(x)  ln(f(x))  ln(g(x)) then for all real x, h(x) is
(A) an odd function (B) an even function
(C) neither an odd nor an even function (D) both odd as well as even function

2. If a, b, c are three complex numbers selected randomly (without repetition) from the set
 a b
{1, 2, 3 ….. 2013) then the probability that arg    0 is
ac 
1 2
(A) (B)
3 3
3
(C) (D) none of these
2013

3. Let a, b, c  R be such that a + b + c < 0, a – b + c < 0 and c > 0, if ‘’ and ‘’ are roots of the
equation ax2 + bx + c = 0, then value of [] + [] is
(A) 2 (B) 1
(C) –1 (D) 0

4. If a – b, b – c, c – a are in A.P., then the straight line (a – b)x + (b – c)y + (c – a) = 0 will pass
through
(A) (1, 2) (B) (2, 1)
(C) (2, 3) (D) (3, 1)

Space for rough work


5. If x  loge (cos ), y  cos(loge ) where e  /2    e 2 , then
(A) x = y (B) x > y
(C) x < y (D) none of these

6. Evaluate 32log3 4
4 9
(A) (B)
9 4
(C) 36 (D) none of these

Multiple Correct Choice Type

This section contains 4 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for
its answer, out which ONE OR MORE is/are correct.

1
7. The function f(x)  x ln x
(A) is a constant function (B) has a domain (0,1)  (e, )
(C) is such that L im f(x) exist (D) is periodic
x 1

8. If log2 (32x 2  7)  2  log2 (3 x 1  1) then x equals

(A) 0 (B) 1
(C) 2 (D) none of these

2 2 2 1 2m
9. If     then
1!13! 3!11! 5!9! 7!7! n!
(A) m + n = 27 (B) m2 = 1 + 2n
(C) n2 – m2 = 27 (D) n = 1 + m

n
10. 
If 8  3 7   P  F , where P is an integer and F is a proper fraction, then
(A) P is an odd integer (B) P is an even integer
(C) F.(P + F) = 1 (D) (1  F)(P + F) = 1

Space for rough work

Reasoning Type

A B
11. STATEMENT 1: P(A  B  C)  P   .P   .P(C)
 B  C
A
STATEMENT 2: P   is probability of A when B has already happened
B
(A) Both the statements are true and Statement 2 is correct explanation of Statement 1
(B) Both the Statements are true and Statement 2 is not the correct explanation of Statement 1
(C) Statement 1 is true and Statement 2 is false
(D) Statement 1 is false and Statement 2 is true

12. STATEMENT 1: If x + y + z = xyz, then at most one of the number can be negative
STATEMENT 2: In a triangle ABC, tan A + tan B + tan C = tan A tan B tan C and there can be at
most one obtuse angle in a triangle
(A) Both the statements are true and Statement 2 is correct explanation of Statement 1
(B) Both the Statements are true and Statement 2 is not the correct explanation of Statement 1
(C) Statement 1 is true and Statement 2 is false
(D) Statement 1 is false and Statement 2 is true

13. STATEMENT 1: Diagonals of any parallelogram inscribed in an ellipse always intersect at the
centre of the ellipse
STATEMENT 2: Centre of the ellipse is the only point at which two chords can bisect each other
and every chord passing through the centre of the ellipse gets bisected at the centre
(A) Both the statements are true and Statement 2 is correct explanation of Statement 1
(B) Both the Statements are true and Statement 2 is not the correct explanation of Statement 1
(C) Statement 1 is true and Statement 2 is false
(D) Statement 1 is false and Statement 2 is true

14. STATEMENT 1: The number of non-negative integer solutions to the inequality x + y + z  2013
is 2016 C3
STATEMENT 2:   n1Cr  nCr     nCr 1 
(A) Both the statements are true and Statement 2 is correct explanation of Statement 1
(B) Both the Statements are true and Statement 2 is not the correct explanation of Statement 1
(C) Statement 1 is true and Statement 2 is false
(D) Statement 1 is false and Statement 2 is true

Space for rough work

Comprehension Type

Paragraph for Question Nos. 15 to 17

Read the following write up carefully and answer the following questions:
An equation a0  a1x  a2 x 2  .....  a99 x 99  x100  0 has roots 99 C0 , 99C1, 99C2 , ..... 99
C99

15. The value of a99 is equal to

(A) 298 (B) 299
99
(C) –2 (D) none of these

16. The value of a98 is

2198  198C99 2198  198
C99
(A) (B)
2 2
(C) 299 – 99C49 (D) none of these

17. The value of a97 is

2198  198
C99 2198  198
C99
(A) (B)
2 2
(C) 299 – 99C49 (D) none of these

Space for rough work

Paragraph for Question Nos. 18 to 20

Read the following write up carefully and answer the following questions:
There are three different lots of Mathematics, Physics and Chemistry books C5
arranged in three different lots as shown in the following diagram P4
M3 C4
P3
M2 C3
P2
M1 C2
P1
C1

18. In how many distinct ways we can pile up these books in a single lot without any restriction
12!
(A) (B) 12!
3!
12! 12!
(C) (D)
4! 5!

19. In how many distinct ways we can pile up these 12 books such that at time of piling up all these
books in a single lot we always pickup the book from the top of respective lots
12! 12!
(A) (B)
3! 5! 4! 5!
12! 12!
(C) (D)
3!  4!  5! 3! 4!

20. In how many different ways we can pile up such that no two physics book are together
(A) 9 P4  8! (B) 8 P4  8!
(C) 8 P4  9! (D) 9 P4  9!

Space for rough work

Paragraph for Question Nos. 21 to 23

Read the following write up carefully and answer the following questions:
Given the equation of a hyperbola in general 2nd degree form i.e. ax 2 + 2hxy + by2 + 2gx + 2fy + c = 0, we
can obtain the equation of its pair of asymptotes by adding a constant  to the above equation. Now since
after adding the constant , the new equation represents a pair of straight lines, therefore the determinant
a h g
h b f vanishes to zero. Hence  and the equation of pair of asymptotes can be found. If C
g f c
represents the equation of a conjugate hyperbola, then C + H = 2A, where H represents the equation of
hyperbola and A the equation of pair of asymptotes

21. The equation of a hyperbola conjugate to the hyperbola x 2 + 3xy + 2y2 + 2x + 3y = 0 is

(A) x2 + 3xy + 2y2 + 2x + 3y + 1 = 0 (B) x2 + 3xy + 2y2 + 2x + 3y + 2 = 0
2 2
(C) x + 3xy + 2y + 2x + 3y + 3 = 0 (D) x2 + 3xy + 2y2 + 2x + 3y + 4 = 0

22. A hyperbola passing through origin has 3x – 4y –1 = 0 and 4x – 3y – 6 = 0 as its asymptotes,

then the equation of its transverse and conjugate axes are
(A) x – y – 5 = 0 and x + y + 1 = 0 (B) x – y = 0 and x + y + 5 = 0
(C) x – y – 5 = 0 and x – y – 1 = 0 (D) x + y – 1 = 0 and x – y– 5 = 0

23. The tangent at any point of a hyperbola 16x 2 – 25y2 = 400 cuts off a triangle from the asymptotes
and that the portion of it intercepted between the asymptotes, then the area of this triangle is
(A) 10 sq. units (B) 20 sq. units
(C) 30 sq. units (D) 40 sq. units

Space for rough work

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ALL INDIA INTEGRATED TEST SERIES ANSWERS, HINTS & SOLUTIONS

HALF COURSE TEST –VII
(Paper-1)

Q. No. PHYSICS CHEMISTRY MATHEMATICS

1. B A A
2. D C B
3. A C C
4. B C A
5. D A C
6. C B B
qualified in JEE (Advanced), 2014.

7. A, C B, C, D A, B, C
8. B, C A, B, C, D A, C
9. B, D A, B, C, D A, C, D
10. B, D A, B, D A, D
11. D B D
12. D C D
13. D D A
14. A A B
15. C B C
16. C B A
17. A C B
18. A A B
19. B A C
20. D D A
21. C B B
22. A A C
23. D C B

Physics PART – I

d  K.E.
1. Power =
dt

2. The force of friction, when it is not slipping, is less than its limiting value : |f|  N. This leads to
the condition on F.

3. Velocity of upper block when lower block hits a obstacle,

F F
v  2a.  2. 
2M M
now after collision retardation of upper block w.r.t. earth,
Mg
a  g
M
 v 2  u2  2as
F 
 0  2g.
M 2
F
  .
Mg

1/2
4. For equilibrium
 
F = 0  = 0
3/2
3 1
N= 
2 2
….(i) F=
A
3/2 x N
Torque about A due to all forces excluding normal force is 3
found to be zero.

5. Impulse = Force  time; net momentum of the system is conserved as the net force on it is zero.

6. At A : At B :
for a rolling 2
v /R V2 B
wheel, a = R If  a then aB v2/R a
 (A) is correct. R
A
may be vertically
R a
downwards R
 (B) is correct.
At C : C R Consider this
 (C) is correct. a  (D) is correct. v2/R
2
v /R
a

R

2R 3R 2R 3R
dx dx 1  1 1  1
15. Reff =  K  4 x   2K  4x =   
R
2
 2R
2
 4K  x R 8K  x  2R
7
 Reff =
48KR

17. Consider shell with cross sectional thickness equal to AB, BC and CA respectively. Since heat
flow is same through all of them, therefore temperature difference will be maximum across the
shell with maximum resistance which will be AB.

18. As explained in figure the no of maximums are 7.

19. R2 = Rsin and (3 cos = 2)= R5/3

2R 5
Length =
3

x =0
20. OC – OB = R – R cos = R/3 x =  x = 
x = 2
OB – OA = 2R./3  Rcos’ = R/3 (3 cos’ =  x = 2
R3 R2
cos’ = 1/3) ’

x = 3 S1  x = 3
Hence R1 = Rsin’ = R8/3 O S2 A B C

21. N = mg f  mg mg
f
V  t  gt
m
2
f R  mgR  mR 2  f
5 N
5 g 5g
     0  t
2 R 2R
5
gt  R0  gt
2
2R0
 t
7g

22.
20
7

at
Before 2R0
rest
collision 7

A B
20
7
2R0
After
7
collision A B

2
1  2R0  mR2 02
K.E.  m  2
2  7  49

2 5 g mg
23. mgR  mR 2   
5 2 R
5 2R0 5 2R0
R  gt V  gt  gt
2 7 2 7
2R0 7 4 R0  mg
  gt  t . N
7 2 49 g
5 g 4 R0 10
 . .  .0
2 R 49 g 49
10 1 2 1
V R0 K.E.  I  mV 2
49 2 2

Chemistry PART – II

Ea E 100 80
1. =– a , = ,  T = 400 K
RT RT 500 T
 Option (A) is the correct choice.

2. Due to conjugation & H – bonding.

3. Br Br

Cl Cl

cis trans

4. Less substituted  - H is removed first.

5. PE of a free electron is zero at infinity i.e. out of the field of nucleus.

9. All are true except (A). D2 also show spin isomerism like H2.
Hence B, C, D is correct.

10. All compounds A, B, C, D contains chiral centres and hence optically active i.e. chiral.
Hence A, B, C, D is correct.

11. Intensive properties are independent of quantity of substance.

Hence A, B, C, D is correct.

12. A & B are acid buffers while (D) is a basic buffer. (C) is not a buffer because no base (NH4OH)
will be left after the reaction.
Hence A, B, D is correct.

Mathematics PART – III

1. h(x)  ln(f(x)  g(x))  lne[y] { y}  {y}  [y]  y  e|x| sgn x

ex if x  0
|x| 
 h(x)  e sgn x  0 if x  0
 e  x if x  0

x
e if x  0

h(  x)  0 if x  0
 e x if x  0

h(x)  h(  x)  0 for all x

ab ab
2. For arg   0 0
ac  ac
Select 3 numbers from the set and call the greatest as ‘a’ and remaining two as b and c. This can
2013
C 4 2
be done in 2013 C3  2 ways. Therefore the probability shall be 2013 3 
C3  3! 3

3. f(x) = ax2 + bx + c = 0
Now as a + b + c < 0  f(1) < 0
a – b + c < 0  f(–1) < 0
c > 0  f(0) > 0
Combining (i), (ii) and (iii) to get f(–1) f(0) < 0 and f(0) f(1) < 0
Hence one of the root lies between (–1, 0) and other root lies between (0, 1)
[] + [] = –1 + 0 = –1

4. Line is satisfied by infinite points having x coordinate 1 and y coordinate  R

(a – b)x + (b – c)y + (c – a) = 0 ….(1)
 (1) is satisfied by (1, 1)
Also as (a – b), (b – c), (c – a) are in AP
 a  b   c  a
 bc 
2
Putting in (1)
 a  b   c  a 
a  b x    y  c  a  0
 2 
 (a – b)(2x + y) + (c – a)(y + 2) = 0
Which represents family of line passing through (1, –2)
 Line (1) will pass through (1, 2) where y  R

5. x  0 and y > 0. Hence x < y

9
6. 32 log3 4  32.(3  log3 4 )  9  41 
4

7. y  xlogx e  e (constant)  A and C

as L im f(x)  e
x 1

2 2 2 1 1 14 1 14 1 213
9.     (2 C1  214 C3  214 C5 14 C7 )  2 
1!13! 3!11! 5!9! 7!7! 14! 14! 14!
 m = 13, n = 14

11. By a general Venn-diagram we can conclude that statement is not always true

(x  y)
12. Statement-1 is wrong as z can be written as
1  xy
 For any values of x and y; (xy  1) we get a value of z
And Statement-2 is correct
Hence, code (D) is the correct answer

13. Statement-2 is correct as ellipse is a central conic

And it also explains Statement-1
Hence, code (A) is the correct answer

18. We can pile up these 12 books in a single lot in 12! Ways

19. If we pickup the book always from the top of respective lot then these books will behave as alike
books in a single lot of those 12 books. It means we have to arrange 12 distinct books in which 3
are alike, 4 are alike and 5 are alike
12!
Hence required way is
3!  4!  5!

20. If we arrange Chemistry and Maths book, we can arrange in 8! Ways, and in remaining 9 slots we
can place 4 Physics books 8!  9 P4

Time Allotted: 3 Hours Maximum Marks: 243

 Pl ea s e r ea d t h e i n s t r u c t i o n s c a r ef u ll y . Yo u a r e a l l o t t ed 5 m i n u t es
s p ec i f i c a ll y f o r t h i s p u r p o s e.
 Yo u a r e n o t a ll o wed t o l ea v e t h e E xa m i n at i o n Ha ll b ef o r e t h e en d o f
t h e t es t .

INSTRUCTIONS
A. General Instructions
1. Attempt ALL the questions. Answers have to be marked on the OMR sheets.
2. This question paper contains Three Parts.
3. Part-I is Physics, Part-II is Chemistry and Part-III is Mathematics.
4. Each part is further divided into two sections: Section-A & Section-B
5. Rough spaces are provided for rough work inside the question paper. No additional sheets will be
provided for rough work.
6. Blank Papers, clip boards, log tables, slide rule, calculator, cellular phones, pagers and electronic
devices, in any form, are not allowed.
B. Filling of OMR Sheet
1. Ensure matching of OMR sheet with the Question paper before you start marking your answers
on OMR sheet.
2. On the OMR sheet, darken the appropriate bubble with black pen for each character of your
Enrolment No. and write your Name, Test Centre and other details at the designated places.
3. OMR sheet contains alphabets, numerals & special characters for marking answers.
C. Marking Scheme For All Three Parts.
(i) Section-A (01 to 09) contains 09 multiple choice questions which have only one correct answer.
Each question carries +3 marks for correct answer and – 1 mark for wrong answer.
Section-A (10 – 13) contains 4 Assertion-Reasoning (multiple choice questions) which have
only one correct answer. Each question carries +3 marks for correct answer and
– 1 mark for wrong answer.
Section-A (14 – 19) contains 2 paragraphs. Based upon paragraph, 3 multiple choice questions
have to be answered. Each question has only one correct answer and carries +4 marks for
correct answer and – 1 mark for wrong answer.
(ii) Section-B (1 – 03) contains 3 Matrix Match Type (4 × 4 Matrix) questions containing statements
given in 2 columns. Statements in the first column have to be matched with statements in the
second column. Each question carries +6 marks for all correct answer. There is no negative
marking.

Name of the Candidate

Enrolment No.

Useful Data

PHYSICS
2
Acceleration due to gravity g = 10 m/s

Planck constant h = 6.6 1034 J-s

Charge of electron e = 1.6  1019 C

Mass of electron me = 9.1  1031 kg

Permittivity of free space 0 = 8.85  1012 C2/N-m2

Density of water water = 103 kg/m3

Atmospheric pressure Pa = 105 N/m2

Gas constant R = 8.314 J K1 mol1

CHEMISTRY

Gas Constant R = 8.314 J K1 mol1

Atomic No: H=1, He = 2, Li=3, Be=4, B=5, C=6, N=7, O=8,

Physics PART – I

SECTION – A

Straight Objective Type

This section contains 9 multiple choice questions numbered 1 to 9. Each question has 4 choices (A),
(B), (C) and (D), out of which ONLY ONE is correct

1. Two uniform solid cylinders A and B each of mass 1 kg are B

A
connected by a light spring of force constant 200 N/m at their axles
and are placed on a fixed wedge as shown in the figure. The
coefficient of friction between the wedge and the cylinders is 0.2. The 60 30
angle made by the line AB with the horizontal, in equilibrium, is
(A) 0º (B) 15º
(C) 30º (D) None of these.

2. A person wants to drive on the vertical surface of a large cylindrical wooden ‘well’ commonly
known as ‘deathwell’ in a circus. The radius of the ‘well’ is 2 meter, and the coefficient of friction
between the tyres of the motorcycle and the wall of the well is 0.2. Minimum speed the
motorcyclist must have in order to prevent slipping should be (take g = 10 m/s2)
(A) 10 m/s (B) 15 m/s
(C) 20 m/s (D) 25 m/s

3. A ball of mass m falls vertically from a height h and m

collides with a block of equal mass m moving
horizontally with a velocity v on a surface. The v
coefficient of kinetic friction between the block and the m
surface is k = 0.2, while the coefficient of restitution (e)
between the ball and the block is 0.5. There is no friction k = 0.2
acting between the ball and the block. The velocity of the
block just after the collision decreases by
(A) 0.5 2gh (B) 0
(C) 0.1 2gh (D) 0.3 2gh
Space for rough work

 y
4. By applying a force F   3xy  5z  ˆj  4zkˆ a particle is moved along the (2, 4, 0)
path y = x 2 from point (0, 0, 0) to the point (2, 4, 0). The work done by the
F on the particle is y = x2
280 140
(A) (B) (0, 0, 0)
5 5 x
232 192
(C) (D)
5 5

5. A bowler throws a ball horizontally along east direction with speed of 144 km/hr. The batsman hits
the ball such that it deviates from its initial direction of motion by 74º north of east direction,
1
without changing its speed. If mass of the ball is kg and time of contact between bat and ball is
3
0.02 s. Average force applied by batsman on ball is
(A) 800 N, 53º East of North (B) 800 N, 53º North of East
(C) 800 N, 53º North of West (D) 800 N, 53º West of North.

6. Lower surface of a plank is rough and lying at rest on a

rough horizontal surface. Upper surface of the plank is
smooth and has a smooth hemisphere placed over it through
a light string as shown in the figure. After the string is burnt,
trajectory of C.M. of the sphere is
(A) a circle C
(B) a ellipse
(C) a straight line Plank
(D) a parabola

7. A disc of radius R is spun to an angular speed 0 about its axis and

then imparted a horizontal velocity of magnitude 0R/4
v0
(at t = 0) with its plane remaining vertical. The coefficient of friction
between the disc and the plane is . The sense of rotation and 0
direction of its linear speed are shown in the figure. Choose the
correct statement. The disc will return to its initial position
(A) if the value of  < 0.5. (B) irrespective of the value of  ( > 0).
(C) if the value of 0.5 <  < 1. (D) if  > 1.
Space for rough work

8. A particle is projected from the horizontal surface in earth’s gravitational field at an angle  with
the horizontal then
(A) center of curvature of projectile’s trajectory at the highest point is below the horizontal surface
level if   tan1 2
(B) center of curvature of projectile’s trajectory at the highest point is above horizontal surface if
  tan1 2
(C) center of curvature of projectile’s trajectory at the highest point is below the ground level is
  tan1 2
(D) irrespective of the value of  centre of curvature always lies below the horizontal surface of
the earth.

9. A small particle is released inside the depth of a liquid in the

vessel. The volume of the particle is V. Density and viscosity
of the liquid are  and  respectively. Buoyant force on the
particle is greater than the weight of the particle. The
variation of velocity of the particle with time is

(A) (B)
VT VT

t t

Space for rough work

Reasoning Type

This section contains 4 questions numbered 10 to 13. Each question contains STATEMENT-1
(Assertion) and STATEMENT-2 (Reason). Each question has 4 choices (A), (B), (C) and (D) out of which
ONLY ONE is correct

10. STATEMENT-1: In a standing wave formed in a stretched wire, the energy of each element of the
wire remains constant.
and
STATEMENT-2: The net energy transfer in a standing wave is zero.
(A) Statement-1 is True, Statement-2 is True; Statement -2 is a correct explanation for
Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement -2 is NOT a correct explanation for
Statement-1.
(C) Statement -1 is True, Statement-2 is False.
(D) Statement -1 is False, Statement-2 is True.

11. STATEMENT-1: The molecules on the surface of liquid filled in a container always experience a
force normal to the surface if viscosity is absent.
and
STATEMENT-2: The liquid cannot sustain tangential strain.
(A) Statement-1 is True, Statement-2 is True; Statement -2 is a correct explanation for
Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement -2 is NOT a correct explanation for
Statement-1.
(C) Statement -1 is True, Statement-2 is False.
(D) Statement -1 is False, Statement-2 is True.
Space for rough work

Acoustic Acoustic
12. STATEMENT-1: If a plank is moving with speed v (v is
source Detector
greater than speed of sound). Acoustic source and detector
are fixed as shown. Acoustic detector will never detect the
sound emitted by acoustic source.
v

and

STATEMENT-2: The speed of sound is reference frame dependent.

(A) Statement-1 is True, Statement-2 is True; Statement -2 is a correct explanation for
Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement -2 is NOT a correct explanation for
Statement-1.
(C) Statement -1 is True, Statement-2 is False.
(D) Statement -1 is False, Statement-2 is True.

13. STATEMENT-1: When a displacement wave in a string is reflected back from a rigid end, its
phase is changed by .
and
STATEMENT-2: At rigid end, impulsive elastic force changes its phase.
(A) Statement-1 is True, Statement-2 is True; Statement -2 is a correct explanation for
Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement -2 is NOT a correct explanation for
Statement-1.
(C) Statement -1 is True, Statement-2 is False.
(D) Statement -1 is False, Statement-2 is True.
Space for rough work

Comprehension Type

This section contains 2 groups of questions. Each group has 3 multiple choice question based on a
paragraph. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which ONLY ONE is
correct

P1416: Paragraph for Question Nos. 14 to 16

When a particle is projected at some angle with the horizontal, the path of the particle is parabolic. In the
process the horizontal velocity remains constant but the magnitude of vertical velocity changes. At any
instant during flight the acceleration of the particle remains g in vertically downward direction. During flight
at any point the path of particle can be considered as a part of circle and radius of that circle is called the
radius of curvature of the path.

Consider that a particle is projected with velocity u = 10 m/s at an angle  = 60º with the horizontal and
2
take value of g = 10 m/s : Now answer the following questions.

14. The radius of curvature of path of particle at the instant when the velocity vector of the particle
becomes perpendicular to initial velocity vector is
20 10
(A) m (B) m
3 3 3 3
40 80
(C) m (D) m
3 3 3 3

15. The magnitude of acceleration of particle at that instant is

(A) 10 m/s2 (B) 5 3 m / s2
(C) 5 m/s2 (D) 10 3 m / s2

16. Tangential acceleration of particle at that instant is

(A) 10 m/s2 (B) 20 m/s2
2
(C) 5 m/s (D) 5 3 m / s2
Space for rough work

P1719: Paragraph for Question Nos. 17 to 19

A pendulum bob has mass m. The length of pendulum is y

 . It is initially at rest. A particle P of mass m/2 moving g
z

horizontally along –ve x-direction with velocity 2g 2 x

collides with the bob and comes to rest. When the bob 
comes to rest momentarily, another particle Q of mass m
moving horizontally along z-direction collides with the bob
and sticks to it. It is observed that the bob now moves
along a horizontal circle. The floor is a horizontal surface
at a distance 2  below the point of suspension of the
pendulum.

17. Tension in string immediately after the first collision is

(A) 2mg (B) mg
3 5
(C) mg (D) mg
2 2

18. The height of circular path above the floor is

3 4
(A) (B)
2 3
5
(C) (D) data not sufficient
4

19. Time period of the circular motion

 7
(A) 2 (B) 2
g 4g
7 3
(C) 2 (D) 2
3g 4g
Space for rough work

SECTION – B

(Matching List Type)

This section contains 3 multiple choice questions. Each question has matching Column(s). The codes
for the Column(s) have choices (A), (B), (C) and (D) out of which ONLY ONE or MORE is/are correct

1. A uniform semi–circular disc of mass M and radius R lies Y

P
in the X–Y plane with its centre at the origin as shown in
figure. Also, axis PQ is in the X–Y plane. Then, match for
the moment of inertia of semi–circular disc about different 
axis.  X
O
Q
Z
Column I Column II
(A) IX (p) MR 2

2
(B) IY (q) MR2
4
(C) IZ (r) MR2
cos2 
4
(D) IPQ (s) MR2
cos2 
2
Space for rough work

2. A block A is placed on wedge B, which

A
is placed on horizontal surface. All the
contact surfaces are rough but not B

sufficiently rough to prevent sliding at
any surface. Match Column I and II.
Column II indicates possible
direction(s) X and Y axis are along
incline and perpendicular to incline.

Column I Column II
(A) Acceleration of A (p) Vertical
(B) Net force applied by A on B (q) s y
(C) Acceleration of A relative to B (r)
(D) Net force applied by ground on B (s)

Horizontal
p

q
r 

Space for rough work

3. E and V are the gravitational field and potential respectively.

Column-I Column-II

(A) M (p) Vp
M
1
M Vq

P R 2R 3R

Concentric charged spherical shells

(B) (q) Ep
1
R/3 P
Q
Eq

solid uniformly charged sphere

(C) A charged spherical having inner (r) Vp

1
radii R/2 and outer radii R. Vq

R/2
Q P

(D) R (s) Ep
R
1
P Eq
Q
R

two identical co-axial charged rings

Space for rough work

Chemistry PART – II

SECTION – A

Straight Objective Type

This section contains 9 multiple choice questions numbered 1 to 9. Each question has 4 choices (A),
(B), (C) and (D), out of which ONLY ONE is correct
- -
1. Cl2 can displace I from KI but I displaces Cl from KClO3. The statement is
(A) True (B) Partly true
(C) False (D) Can not be

2. For a 4d electron the orbital angular momentum is

(A) 6 (B) 12 
(C) 2 (D) zero

3. K c for the reaction



A (g)  B(g) 
 C(g)  D(g) , is 20 at 25°C. If a container contains 1, 2, 4, 5 moles per litre of
A (g) , B(g), C(g) and D(g) respectively at 25°C, then the reaction shall
(A) proceed from left to right (B) proceed from right to left
(C) be at equilibrium (D) None

4. 10 ml of 0.2 M acid is added to 250 ml of a buffer solution with pH = 6.34 and the pH of the
solution becomes 6.32. The buffer capacity of the solution is:
(A) 0.1 (B) 0.2
(C) 0.3 (D) 0.4

5. Identify Z in the following reaction sequence:

CH2 HBr / Peroxide C 2H5ONa
      Z
H3C ether

CH3
(A) (B) H C O CH3
H3C 3
CH3

CH2
(C) O CH3 (D)
H3C

CH3

Space for rough work

6. Rate of electrophilic bromination in the aromatic nucleus will be maximum in

CH3 CH3

(A) (B)

CH3
H3C CH3
H3C CH3

7. The incorrect statement about the product (P) of the reaction:

H Me

Br2 / CCl4
  P

Me H
is which one of the following?
(A) P is optically inactive due to internal compensation
(B) P is optically inactive due to the presence of plane of symmetry in the molecule.
(C) The structure which P possesses can have four optical isomers possible.
(D) The structure of P can have three optical isomers possible.

8. For the gaseous reaction A (g)  4B(g)  3C(g) is found to be first order with respect to A. If at
the starting the total pressure was 100 mm Hg and after 20 minutes it is found to be 400 mm Hg.
The rate constant of the reaction is:
(A) 20 min-1 (B) 1.2  103 sec 1
(C) 5.7  104 sec 1 (D) 40 min-1

9. The root mean square speed of an ideal gas at constant pressure varies with density (d) as
(A) d2 (B) d
1
(C) (D) d
d

Space for rough work

Reasoning Type

10. STATEMENT-1: The alkali metals can form ionic hydrides which contain the H ion
and
STATEMENT-2: The alkali metals have low electronegativity.
(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for
Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for
Statement-1.
(C) Statement-1 is True, Statement -2 is False.
(D) Statement-1 is False, Statement-2 is True.

11. STATEMENT-1: Between SiCl4 and CCl4 only SiCl4 reacts with water.
and
STATEMENT-2: Carbon does not contain empty d-orbitals.
(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for
Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for
Statement-1.
(C) Statement-1 is True, Statement -2 is False.
(D) Statement-1 is False, Statement-2 is True.

12. STATEMENT-1: Spectral line would not be seen for a 2px – 2pz transition
and
STATEMENT-2: p–orbitals are degenerate orbitals
(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for
Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for
Statement-1.
(C) Statement-1 is True, Statement -2 is False.
(D) Statement-1 is False, Statement-2 is True.

13. STATEMENT-1: The heat absorbed during the isothermal expansion of an ideal gas against
vacuum is zero.
and
STATEMENT-2: The volume occupied by the molecules of a real gas is negligible.
(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for
Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for
Statement-1.
(C) Statement-1 is True, Statement -2 is False.
(D) Statement-1 is False, Statement-2 is True.

Space for rough work

Comprehension Type

Paragraph for Question Nos. 14 to 16

Read the following paragraph and answer the questions given below:

A hydrocarbon with molecular formula C10H16 undergoes reductive ozonolysis to give two moles of same
compound having molecular formula C5H8O. Even the oxidative ozonolysis of the above hydrocarbon
o
gives the same product. The hydrocarbon can react with three equivalents of H2/Pd at 120 C. Both the
hydrocarbons and the ozonolysis products are not resolvable.
Choose the correct option:

14. The hydrocarbon is

(A) (B)

(C) CH3 (D) Both (A) and (B)

H3C

15. Hydrocarbon shows

(A) geometrical isomerism (B) position isomerism
(C) functional isomerism (D) Tautomerism

16. Which of the following is correct?

(A) The ozonolysis product is optically active.
(B) The ozonolysis product is optically inactive racemic mixture.
(C) The ozonolysis product is optically inactive due to plane of symmetry.
(D) The ozonolysis product is optically inactive due to absence of chirality.

Space for rough work

Paragraph for Question Nos. 17 to 19

Read the following paragraph and answer the questions given below:
Stable equilibrium is of various types. Mechanical equilibrium is achieved when all particles are at rest
and total potential energy of the system is minimum. At any stage where particles are at rest but the
system is not at stable equilibrium as it can reduce its potential energy by reverting to another position, is
called metastable equilibrium. Thermal equilibrium results from the absence of temperature gradients in
the system. Chemical equilibrium is obtained when no further reaction occurs between reacting
substances, i.e. forward and reverse rates of reaction are equal. When steam reacts with solid iron at
high temperature, Fe3O4(s) and hydrogen gas are produced. But the reaction never goes to completion.
This is because as the products are formed, the reaction proceeds in reverse direction and when rate of
reverse reaction is equal to rate of forward reaction, the concentration of reactants and products become
constant and equilibrium is reached.
Choose the correct option:

17. The primary objective of the passage is

(A) to discuss different types of equilibrium
(B) to distinguish between different types of equilibrium
(C) to discuss concept of mechanical and chemical equilibrium
(D) to explain how changing state of equilibrium would influence internal energy

18. Which of the following is correct?

(A) Chemical equilibrium can be stable, unstable or metastable.
(B) The example given in the passage refers to thermodynamic equilibrium.
(C) Addition of catalyst to a reaction in chemical equilibrium increases the amount of heat
evolved in the reaction.
(D) Addition of catalyst to a reaction in chemical equilibrium gives a new reaction path to the
system.

19. The correct expression for equilibrium constant of the reaction, taken as an example of chemical
equilibrium in the above passage is
(pH2 )2 (pH2 )4
(A) 2
(B)
(pH2 O) (pH2 O)4
(pH2 )4 [Fe3 O 4 ] [Fe3 O 4 ]
(C) (D)
(pH2 O)4 [Fe] [Fe]

Space for rough work

SECTION – B

(Matching List Type)

This section contains 3 multiple choice questions. Each question has matching Column(s). The codes
for the Column(s) have choices (A), (B), (C) and (D) out of which ONLY ONE or MORE is/are correct

1. Match Column–I with Column–II and select the correct answer.

Column–I Column–II
Ionic species Shape
(A) SiF5 (p) Tetrahedral
(B) ICl4 (q) Square planar
(C) AsF4 (r) Linear

(D) I3 (s) Trigonal bipyramidal

2. Match Column–I (reaction) with Column–II (reaction intermediate) and select the correct answer
using the codes given below the lists:
Column–I Column–II
(A) CF  CHCl  alc. KOH
 CF  CCl (p) TS
3 2 2 2

(B) OH (q) Carbocation

H3C CH2
H3C conc. H2SO4
CH3  

CH3 CH3
(C) CH3  CH2  Br  alc. KOH / 
 CH2  CH2 (r) Carbanion
(D) Br (s) Free radical
alc. KOH
H3C CH2
H3C CH3  

CH3
CH3

3. Match the following:

Column - I Column - II
(Number of Moles) (Amount)
(A) 0.1 mole (p) 4480 mL of CO2 at STP
(B) 0.2 mole (q) 0.1 g atom of iron
(C) 0.25 mole (r) 1.5  1023 molecules of O2 gas
(D) 0.5 mole (s) 9 mL water

Space for rough work

Mathematics PART – III

SECTION – A

Straight Objective Type

This section contains 9 multiple choice questions numbered 1 to 9. Each question has 4 choices (A),
(B), (C) and (D), out of which ONLY ONE is correct

1. Vertices of a variable triangle are (3, 4), (5 cos , 5 sin ) and (5 sin , 5 cos ) then locus of its
orthocentre is
(A) (x + y – 1)2 + (x – y – 7)2 = 100 (B) (x + y – 7)2 + (x – y – 1)2 = 100
2 2 2 2
(C) (x + y – 7) + (x + y – 1) = 100 (D) (x + y – 7) + (x – y + 1) = 100

2. The probability that at least one of the events A and B occur simultaneously with probability 3/5.
If A and B occur simultaneously with probability 1/5 then P  A   P B  is
2 4
(A) (B)
5 5
6 7
(C) (D)
5 5

3. If a, b, c are three complex numbers selected randomly (without repetition) from the set
 a b
{1, 2, 3…..2013) then the probability that arg    0 is
ac 
1 2
(A) (B)
3 3
3
(C) (D) none of these
2013

4. If one of the lines given by 6x 2  xy + 4cy2 = 0 is 3x + 4y = 0 then c is equal to

(A) 3 (B) 1
(C) 1 (D) 3

5. If the quadrate equation 4x2  2(a + c  1)x + ac  b = 0 (a > b > c)

(A) both roots are greater than a
(B) both roots are less than c
c a
(C) both roots be between   and  
 2  2
c a
(D) exactly one of the roots has between   and  
 2  2

Space for rough work

6. If the focus of a parabola is at (0, –3) and its directrix is y = 3, then its equation is
2 2
(A) x = –12y (B) x = 12y
2 2
(C) y = –12x (D) y = 12x

7. Let zk; k = 1, 2, 3, 4 be four complex numbers such that |zk| = k  1 and

|30z1 + 20z2 + 15z3 + 12z4| = |z2z3z4 + z3z4z1 + z1z2z4 + z1z2z3|, then  is equal to
(A) |z1 z2 z4| (B) |z2 z3 z4|
(C) |z1 z3 z4| (D) |z1 z2 z3|

8. A circle has the same centre as an ellipse and passes through the foci F1 and F2 of the ellipse the
two curves intersect in four points. Let P be any point of intersection. If the major axis of the
ellipse is 15 and the area of PF1F2 is 26, then the distance between foci is
(A) 11 (B) 9
(C) 10 (D) 15

9. If normal chords at points P and Q on the parabola y2 = 4ax meets again on the parabola at R
then the locus of the circum-centre of PQR is
(A) y 2  4a(x  a) (B) 2y 2  a(x  a)
(C) 2y 2  a(x  2a) (D) none of these

Reasoning Type

10. STATEMENT 1: If P(2a, 0) be any point on the axis of parabola, then the chord QPR, satisfy
1 1 1
 
PQ 2 PR2 4a2
1 1
STATEMENT 2: There exists a point on the axis of the parabola y2 = 4ax, such that 2

PQ PR2
is equal to constant for all chord QPR of the parabola
(A) Both the statements are true and Statement 2 is correct explanation of Statement 1
(B) Both the Statements are true and Statement 2 is not the correct explanation of Statement 1
(C) Statement 1 is true and Statement 2 is false
(D) Statement 1 is false and Statement 2 is true

Space for rough work

11. STATEMENT 1: 3 is a multiple root of order 2 of the equation x 3 – 5x2 + 3x + 9 = 0

STATEMENT 2: If f(x) = x3 – 5x2 + 3x + 9, then f(3) = 0
(A) Both the statements are true and Statement 2 is correct explanation of Statement 1
(B) Both the Statements are true and Statement 2 is not the correct explanation of Statement 1
(C) Statement 1 is true and Statement 2 is false
(D) Statement 1 is false and Statement 2 is true

2 1 5
12. STATEMENT 1: The function f(x) = (3x – 1)|4x – 12x + 5| cos x is differentiable at x  ,
2 2

STATEMENT 2: cos(2n  1)  0, n  I
2
(A) Both the statements are true and Statement 2 is correct explanation of Statement 1
(B) Both the Statements are true and Statement 2 is not the correct explanation of Statement 1
(C) Statement 1 is true and Statement 2 is false
(D) Statement 1 is false and Statement 2 is true

13. If positive numbers x, y, z satisfy x + y + z = 1, (x, y, z < 1/2) then

 (1  2x)  (1  2y)  (1  2z)  1/3
STATEMENT 1: 
3   (1  2x)(1  2y)(1  2z)
 
STATEMENT 2: For any three positive numbers a, b, c their AM  GM
(A) Both the statements are true and Statement 2 is correct explanation of Statement 1
(B) Both the Statements are true and Statement 2 is not the correct explanation of Statement 1
(C) Statement 1 is true and Statement 2 is false
(D) Statement 1 is false and Statement 2 is true

Space for rough work

Comprehension Type

Paragraph for Question Nos. 14 to 16

Read the following write up carefully and answer the following questions:
Inequalities are very important to understand in mathematics.
a > b, a < b, a  b, a  b and a  b are called inequality terms. Many times calculus help us to prove
inequalities e.g. To check which is greater of the two from e or e we take f (x) = x 1/x and check, where
the function increases and so on. Now answer the following questions

14. If a = 3e, b = 3e, c = e3, then

(A) a < b < c (B) b < a < c
(C) c < a < b (D) a < c < b

15. The largest interval for which x12  x9 + x4  x + 3  sin x + cos x > 0 is
(A) (1, 1) (B) (, 2)
(C) (log 12, log 12) (D) (, )

16. If f(x) is a decreasing function and f(0) = 0 then

(A) f(3) > 0 (B) f(2) < 0
(C) f(–3) < 0 (D) none of these

Space for rough work

Paragraph for Question Nos. 17 to 19

Read the following write up carefully and answer the following questions:
1  x n  c0  c1x  c 2 x2 ....c n xn
n
 1 
17. Total number of term, that are dependent of the value of x, in the expansion of  x2  2  2  is
 x 
equal to
(A) 2n + 1 (B) 2n
(C) n (D) n + 1

100
 1 
18. In the expansion of  a3  3  1 number of distinct terms is
 a 
(A) 200 (B) 201
(C) 202 (D) none of these

19. An equation a0  a1x  a2 x 2  ...a99 x99  x100  0 has roots 99

C0 ,99 C1,....99 C99 . Find the value of
a98 is
2198  198 C99
(A) 2198  198 C99 (B)
2
(C) 299  99
C49 (D) none of these

SECTION – B

(Matching List Type)

This section contains 3 multiple choice questions. Each question has matching Column(s). The codes
for the Column(s) have choices (A), (B), (C) and (D) out of which ONLY ONE or MORE is/are correct

1. Match the following Column–I with Column–II

Column – I Column – II
(A) If the normal at one end of a latus rectum of an ellipse of
eccentricity e passes through one end of minor axis, then 5
(p)
e2  e4 is equal to 2
(B) The chords of contact of the pair of tangents to the circle
x2 + y2 = 1 drawn from any point on the line 2x + y = 4 pass
1 1  (q) 123
through the point  ,  , then  is equal to
  2 
(C) The eccentricity of the hyperbola x 2  4y 2  1 is (r) 1
(D) The number of lines passing through (3, 4) at a distance 5 units
(s) 2
from (6, 8) are

Space for rough work

1 1
2. Let A and B are two independent events such that P(A)  and P(B)  . Now match the
3 4
following Column–I with Column–II
Column – I Column – II
1
(A) P(AB) is equal to (p)
12
1
(B) P(AB) is equal to (q)
2
5
(C) P(BA) is equal to (r)
6
3
(D) P(AB) is equal to (s)
4

3. Match the following Column–I with Column–II

Column – I Column – II
(A) x dy = y(dx + y dy), y > 0 1
(p)
y(1) = 1 and y(x 0) = –3, then x0 = , then  is less than 4
dy
(B) If y(t) is solution of (t  1)  ty  1, y(0) = –1, and y(1) = ,
dt (q) –15
then  is greater than
(C) (x2 + y2)dx = xy dx and y(1) = 1 and y(x0) = e , then 1
(r) 
x0 = , then  is greater than 2
dy 2y
(D)   0, y(1) = 1, then y(2) is equal to (s) 3e
dx x

Space for rough work

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HALF COURSE TEST –VII
(Paper-2)

Q. No. PHYSICS CHEMISTRY MATHEMATICS

1. C A D
2. A A C
3. D A B
4. D D A
5. C B D
6. C A A
qualified in JEE (Advanced), 2014.

7. B C A
8. A C A
9. A C B
10. D A A
11. A A C
12. B A A
13. A C B
14. A C A
15. A A D
16. C D B
17. C C C
18. C D B
19. D B A
(A)  (q) (A)  (s) (A)  (r)
(B)  (q) (B)  (q) (B)  (s)
1.
(C)  (p) (C)  (p) (C)  (p)
(D)  (q) (D)  (r) (D)  (r)
(A)  (q) (A)  (r) (A)  (q)
(B)  (r) (B)  (q) (B)  (r)
2.
(C)  (p) (C)  (p) (C)  (s)
(D)  (s) (D)  (q) (D)  (p)
(A)  (q, r) (A)  (q) (A)  (p, r, s, t)
(B)  (q, r) (B)  (p) (B)  (q)
3.
(C)  (q, r) (C)  (r) (C)  (p, q, r, t)
(D)  (p, q) (D)  (s) (D)  (p)

Physics PART – I

1. In equilibrium, there will not be any friction between the cylinder and the wedge. If  be the
required angle
Cylinder A: mg sin 60º = kx cos (60º – ) ( = mg cos 30º)
Cylinder B: mg sin 30º = kx cos (30º + ) = kx sin (60º – )
 cot 30º = cot (60º – )   = 30º.

2. mv 2
N
R
f max = mg
N = mg
mv 2
 = mg
R
Rg 2  10
v=  = 10 m/s
 0.2

3. The impulse due to friction = k  impulse delivered by collision

4. The z-component of the force and the x-component of displacement are ineffective here.
dW  Fy dy  3xy.dy  z  0 
= 6x4dx (y = x2)
Integrating between x = 0 and x = 2 gives the result.

5 N
5. 144 km/hr = 144  = 40 m/s
18
|V2  V1|
  40 m/s
V2  V1 37
40
 74 53
sin 74 sin(90  37)
40 m/s E
40  2sin37 cos 37
 V2  V1  = 48 m/s
cos37

1
Change in momentum =  48 = 16
3

16
f avg  = 800 N.
0.02

7. Solve the problem using  as the coefficient of kinetic friction.

u2 cos2  u2 sin2 
8. If 
g 2g
  tan1 2
u2 cos2  u2 sin2 
if 
g 2g
  tan1 2 .

14. Time after which velocity vector becomes

53 m/s 10 m/s
perpendicular to initial velocity vector is vx

u 10 2
t   seconds 60 vy 30
v
gsin  10 sin 60 3
Let v y be the vertical component of velocity at 5 m/s
that instant then
vy = u + at
10  2
vy = 5 3 
3
5  vy = 5
vy  
3
2
 5 
2 10
 v  5    v m/s vx = 53 v = 103
 3 3
v2
gcos  
R
v2
R
gcos 
20
R m.
3 3

5
3
16. at = g sin   10  = 5 m/s2.
10
3

17.
T


g 2
19.  and T 
 cos  

Chemistry PART – II

SECTION – A

[C][D] 4  5
3. Q   10
[A][B] 1 2
K c  20
 Q  Kc
Therefore equilibrium will shift from left to right.

moles of acid or base added per litre

4. Buffer capacity =
change in pH
0.2  10 1000

Buffer capacity = 1000 250  0.4
6.34  6.32

5. CH2 HBr C2H5 ONa

  
H3C Peroxide anti M.K. addition H3C Br SN 2 H3C OC2H5

CH3

6. will be most reactive in electrophilic substitution reaction due to

hyperconjugation. It is having three  – hydrogen for hyperconjugation.

7. Me

H Br
The product is
H Br

Me
(Meso compound)

8. A (g)  4B(g)  3C(g)

P0 0 0 at initial
P0  x 4x 3x at 20 minutes
P0  x  4x  3x  400
P0  100  x  50
It shows 20 minutes is the half life time of the reaction.

3RT 3P 3P
9. uRMS =  
M M / Vm 

15. Hydrocarbon (C10H16)

CH3

H3C CH3 H3C

cis trans

SECTION – B

2. Reaction (a) is E1CB, hence RI is a carbanion.

Reaction (b) is E1, hence RI is a carbocation.
Reaction (c) is E2, hence RI is a TS.
Reaction (d) is E1, hence RI is a carbocation.

Mathematics PART – III

SECTION – A

1. All the co–ordinates of vertices are at a distance of 5 units from origin. Hence circumcentre of the
triangle is (0, 0)
 3  5cos   5 sin  4  5 sin   5cos  
Centroid =  , 
 3 3 
Centroid divides the line joining circumcentre and orthocentre in the ratio 1 : 2
Suppose co-ordinate of orthocentre is (x 1, y1)
x1 = 3 + 5 cos  + 5 sin  ..... (1)
y1 = 4 + 5 sin  – 5 cos  ..... (2)
 x1  y1  7  x1  y1  1
Solving (1) and (2) sin     cos  
 10  10
2 2
(x1 + y1 – 7) + (x1 + y1 + 1) = 100
Replace x1 by x and y1 by y
Locus of orthocentre in (x + y – 7)2 + (x – y + 1)2 = 100

3 1
2.  P(A)  P(B) 
5 5
4
  1  P(A)  1  P(B)
5
6
 P(A)  P(B) 
5

ab ab
3. For arg    0  ac  0
ac 
Select 3 numbers from the set and call the greatest as ‘a’ and remaining two as b and c. This can
2013
C 4 2
be done in 2013 C3  2 ways. Therefore the probability shall be 2013 3 
C3  3! 3

3x
4. Put y   and solve for C
4

5. f(x) = (2x  a)(2x  c) + (2x  b)

a
So f    a  b
 2
c
f   c b
 2
a c 
Now f   .f    (a  b)(c  b)  0
 2  2
c a
Hence exactly one root lies between   and  
 2  2

6. Let P(x, y) be any point on the parabola.

Then by definition (x  0)2  (y  3)2  y  3  x 2  12y

z1 z2 z3 z 4  1 1 1 1
7.     z1z2 z3 z4   
2 3 4 5 60 z1 z 2 z 3 z 4


Now, z1z1  2,z2 z2  3,z3 z3  4 and z4 z4  5 | zk | k  1 
60 60
   30  z1z 2 z 4
z1z 2 z3 z 4 2 3 4 5

8. PF1  PF2  15,PF1.PF2  52

Let PF1  x,PF2  y
x + y = 15, xy = 52
2
(FF
1 2 )  225  104
2
(FF
1 2 )  121  FF
1 2  11

9.    
Let us draw tangent at P and Q which meet in T. Let P at12 ,2at1 ,Q at 22 ,2at 2 and R at 23 ,2at 3  
So T  at1t 2 , a  t1  t 2  
Any circle which passes through points P, Q and R must also pass through the points T as the
four points are concyclic
 Circumcentre is mid-point of TR that is M(h, k)
As normal chords at P and Q meet in R
2 2
t 3  t1   t 2   t1t 2  2
t1 t1
2a  at 32 2a  a(t1  t 2 )2
h 
2bt1  0 2
2k
 t1  t 2 
a
2
 2k 
From equation (1) and (2) we get 2(h  a)  a    2y2  a(x  a)
 a 

10. Let P(h, 0) be a point on axis of parabola the straight line passing through P cuts the parabola at
a distance r
 (r sin )2 = 4a(h + r cos )
 r2 sin2 – (4a cos ) r – 4ah = 0 ….(i)
4a cos 
Where, r1  r2 
sin2 
4ah
and r1r2   2
sin 
1 1 1 r12  r22 cos2  sin2 
     
PQ2 PR2 r12  r22 r12r22 h2 2ah
2
Which is constant only, if h = 2ah i.e. h = 2a
cos2  sin2  1
 2
 2
 2
4a 4a 4a
1 1
Thus,   constant for all chords QPR, if h = 2a
PQ 2 PR2
Hence, (2a, 0) is the required point on the axis of parabola
 Statement-1 and statement-2 are true and statement-2 is correct explanation of statement-1

11. f(x) = (x – 3)2 (x + 1)

 3 is a multiple root of order 2
2
f(x) = 3x – 10x + 3
f(x) = 6x – 10
f(3) = 0, f(3)  0
Statement-1 is true, statement-2 is false
2
12. Statement-1 is correct as through |4x – 12x + 5|
1 5  1 5
is non-differentiable at x  , but cos x = 0 these points. So, f    and f    exists
2 2 2 2

13. If each of x, y, z is less than 1/2 then, statement-1 is obviously true

Also, 1 – 2x + 1 – 2y + 1 – 2z = 3 = 2(x + y + z) = 1
 The sum of the three given number is positive also at must one of x, y, z can be more than 1
1
 If one of x, y, z is more than or equal to , then their product is less than equal to zero hence
2
still remains true
Statement-2 is always true but it does not explain statement-1

SECTION – B

1. (A) Write the equation of normal and use the given condition
 b2 
Normal at  ae,  is
 a 
2 Y
b
y 2
x  ae a  a  x  ae   ea  y  b 
ae

 b2 

a 
 B 
L ae, b / a
2


a2   X
 a  O S A
b2 B L
(0, –b)
 b2 
 x  ye  e  a  
 a 
a2  b2 a2 e2
It passes through (0, -b),  b    ae2
a a
 
 b2  a2 e4  a 2 1  e 2  a 2 e 4    e 4  e2  1
(C) x 2  4y 2  1
1 5
4
 
 e2  1 e 
2
(D) 1
Find the line in terms of m and satisfy the condition to solve for m
Any line through (3, 4) is y – 4 = m(x – 3)
According to given condition
m6  3  4  8 2 3
| s |   4m  3   0  m   = 1
1 m 2 4

1
2. We have, P(A  B)  P(A)  P(B) 
12
1 1 1 1
(A) P(A  B)  P(A)  P(B)  P(A  B)    
3 4 12 2
(B) P(A  B)  1  P(B)  P(A  B) = 5/6
(C) P(BA) = 3/4
(D) P(A  B)  P(A).P(B)  1/ 12

xdy  ydx x

3. (A) x dy = y dx + y2 dy  2
 dy  d    dy
y y
x
  y  c put x = 1, y = 1  c = –2
y
x x
  y  2   5
y 3
dy t 1
(B)  y
dt t  1 t 1
t 11
 dt
I.f  e t 1  e t  n(t 1)  (t  1)e t
Solution is (t + 1)e–t y = –e–t + c put t = 0 and y = –1  c = 0
 2e–1 y = –e–1
1
y
2
(C) (x2 + y2)dy = xy dx
dy xy dy dv
 2 2
put y = vx, vx
dx x  y dx dx
1
 nv  2   nx  c
2v
1 y 1 x2 1
 c    n  2
 nx  put y = e
2 x 2y 2
 x  3e
dy y
(D) 2 0
dx x
2
x y = C put x = 1, y = 1 and we get C = 1
1
put x = 2  y 
4

Time Allotted: 3 Hours Maximum Marks: 240

INSTRUCTIONS
A. General Instructions

1. Attempt ALL the questions. Answers have to be marked on the OMR sheets.
2. This question paper contains Three Parts.
3. Part-I is Physics, Part-II is Chemistry and Part-III is Mathematics.
4. Each part is further divided into two sections: Section-A & Section-B.
5. Rough spaces are provided for rough work inside the question paper. No additional sheets will be
provided for rough work.
6. Blank Papers, clip boards, log tables, slide rule, calculator, cellular phones, pagers and electronic
devices, in any form, are not allowed.

B. Filling of OMR Sheet

C. Marking Scheme For All Three Parts.

1. Section – A (01 – 08) contains 8 multiple choice questions which have only one correct answer.
Each question carries +3 marks for correct answer and – 1 mark for wrong answer.

Section – A (09 – 12) contains 4 multiple choice questions which have more than one correct
answer. Each question carries +4 marks for correct answer and – 1 mark for wrong answer.

Section – A (13 – 18) contains 2 paragraphs. Based upon paragraph, 3 multiple choice questions
have to be answered. Each question has only one correct answer and carries +4 marks for
correct answer and – 1 marks for wrong answer.
2. Section – B (01 – 02) contains 2 Matrix Match Type questions containing statements given in 2
columns. Statements in the first column have to be matched with statements in the second
column. Each question carries +8 marks for all correct answer. For each correct row +2 mark will
be awarded. There may be one or more than one correct choice. No marks will be given for any
wrong match in any question. There is no negative marking.

Name of the Candidate

Enrolment No.

Useful Data

PHYSICS
Acceleration due to gravity g = 10 m/s2

Planck constant h = 6.6 1034 J-s

Charge of electron e = 1.6  1019 C

Mass of electron me = 9.1  1031 kg

Permittivity of free space 0 = 8.85  1012 C2/N-m2

Density of water water = 103 kg/m3

Atmospheric pressure Pa = 105 N/m2

Gas constant R = 8.314 J K1 mol1

CHEMISTRY

Gas Constant R = 8.314 J K1 mol1

Atomic No: H=1, He = 2, Li=3, Be=4, B=5, C=6, N=7, O=8,

Physics PART – I

SECTION - A
Straight Objective Type
This section contains 8 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out
of which ONLY ONE is correct.

1. The equation of a particle executing SHM is given by x = 3 cos (/2)t cm, here t is in second. The
distance travelled by the particle in first 8.5 sec is
3 3
(A) 24 + cm (B) 27  cm
2 2
3 3
(C) 24  cm (D) 27 + cm
2 2

P
2. A hollow vertical cylinder of radius r and height h has smooth internal u
surface. A small particle is placed in contact with the inner side of the
upper rim, at a point P. It is given a horizontal speed u, tangential to
h
rim. It leaves the lower rim at point Q, vertically below P. the number Q
of revolutions made by the particle will be r

h u
(A) n  (B) n 
2r 2gh

2r u  2h 
(C) n  (D) n   
h 2r  g 

3. The figure shows a semi-cylindrical gate pivoted at the point O

holding a stationary liquid of density . A horizontal force F is
h=R
applied at its lowest position to kept it stationary. The magnitude
of the force is
R O
9 3
(A) gR2 (B) gR2
2 2 F
2
(C) gR (D) zero

Space for rough work

4. An assembly of identical spring mass system is placed on k k

a smooth horizontal surface as shown. Initially the m m
springs are relaxed. The left mass is displaced to the left
while the right mass is displaced to right by two times
more than the left spring and both are released. The
resulting collision is elastic. The time period of the
oscillation of the system is

2M M
(A) 2 (B) 2
k 2k
M M
(C) 2 (D) 
k k

5. A body of mass m is free to move in a conservative field with potential energy U = U0(1  cos x)
where U0 and  are constants the time period of small oscillations is
U0 m 2
(A) 2 2
(B) 2
m U0
m U0  2
(C) 2 (D) 2
U0  2 m

6. A metallic sphere of mass 2 gm falls through a liquid with terminal velocity v. If we drop a sphere
of mass 0.25 gm of the same metal through same liquid its terminal velocity would be
(A) v (B) 2v
(C) v/2 (D) v/4

7. An artificial satellite of mass m is moving in circular orbit at a height equal to the radius R of the
earth. Suddenly due to internal explosion the satellite breaks into two parts of equal masses. One
part of the satellite stops just after the explosion and then falls to the surface of the earth. The
increase in the mechanical energy of the system (satellite + earth) due to explosion will be (g is
the acceleration due to gravity on the surface of earth)
mgR
(A) mgR (B)
2
mgR 3mgR
(C) (D)
4 4

Space for rough work

8. A particle of mass equal to half of the earth’s mass is released at a distance h(h R) from the
surface of earth. The time after which the particle will hit earth’s surface is
4h 2h
(A) (B)
3g g
h 2h
(C) (D)
g 3g

(Multiple Correct Choice Type)

This section contains 4 multiple choice questions. Each question has four choices (A), (B), (C) and (D)
out of which ONE or MORE are correct.

9. A satellite is revolving around the earth in a circular orbit. The universal gravitational constant
suddenly becomes zero at time t = 0. Then
(A) The kinetic energy for t < 0 and t > 0 is same
(B) The angular momentum for t < 0 and t > 0 is same
(C) The angular momentum keeps changing for t < 0
(D) The angular momentum keeps changing for t > 0

10. Two waves of same frequency and wavelength but amplitude A1 and A2 meet a point at the same
time. According to the superposition principle, the resultant amplitude
(A) may be A1 + A2 (B) may be A1  A2
(C) A1 + A2 or A1  A2 (D) A1  A2  A  A1 + A2

11. When tuning forks A and B are struck, a beat frequency f AB is heard. When A and C are struck a
beat frequency of f AC is heard. It was found that f AC > f AB. If f A, f B and fC are individual frequencies,
then what are the possible relations?
(A) f A > f B > fC (B) f A < f B < fC
(C) f B > f A > fC (D) fC > fA > fB

12. A body of mass M is attached to the lower end of a metal wire, whose upper end is fixed. The
mass M is brought slowly to the equilibrium position and the elongation in the wire is .
1
(A) the energy stored in the wire is Mg (B) the energy stored in the wire is Mg
2
1 1
(C) heat generated during the process is Mg (D) heat generated during the process is Mg
2 4
Space for rough work

Linked Comprehension Type

This section contains 2 paragraphs. Based upon each paragraph, 3 multiple choice questions have to be
answered. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

Paragraph for Question Nos. 13 to 15

If a body moves through a liquid or a gas then the fluid applies a force on the body which is called drag
force. Direction of the drag force is always opposite to the motion of the body relative to the fluid. At low
speeds of the body, drag force (FP) is directly proportional to the speed.
FD = kv
When k is a proportionally constant and it depends upon the dimension of the body moving in air at
2
relatively high speeds, the drag force applied by air on the body is proportional to v instead of v.
2
FD = k1v
Where this proportionally constant K can be given by
k1 = CA
Where  is the density of air
C is another constant giving the drag property of air
A is area of cross-section of the body

Consider a case an object of mass m is released from a height h and it falls under gravity. As it’s speed
increases the drag force starts increasing on the object. Due to this at some instant, the object attains
equilibrium. The speed attained by the body at this instant is called “terminal speed” of the body.
Assume that the drag force applied by air on the body follows the relation FD = kv, neglect the force of
buoyancy applied by air on the body then answer the following questions.

Read the above passage carefully and answer the following questions

13. What is the pattern of acceleration change of the body?

(A) It first increases, then decreases to zero
(B) It decreases uniformly from starting to zero
(C) It decrease non-uniformly from starting to zero
(D) It first increases then becomes constant

14. What is the terminal speed of the object?

mg
(A) (B) mgk
k
k mg
(C) (D)
mg k

15. What is the nature of change in speed of the body?

(A) It first decreases and then increases to a constant value
(B) It increases uniformly to a constant value
(C) It increases non-uniformly to a constant value
(D) It continuously decreases to zero.
Space for rough work

Paragraph for Question Nos. 16 to 18

Heat generation may occur in a variety of radial geometries. Ts

cold fluid
Consider a long, solid cylinder as shown in the figure, which could (Tf)
represent a current-carrying wire or a fuel element in a nuclear
reactor. For steady state conditions, the rate at which heat is
generated within the cylinder must equal the rate at which heat is r0
convected from the surface of the cylinder to a moving fluid. This L
condition allows the surface temperature to be maintained at a
fixed value of TS.
To determine the temperature distribution in the cylinder, we begin r
with energy conservation principle. Consider a cylindrical section of
radius r. The rate at which energy is generated within the given
volume is conducted radially outward.
 q  r2 =  K2r dT dr

where q is the energy generated per unit time per unit volume, K is the thermal conductivity and dT/dr is
the temperature gradient at radius r.
If q is constant
q
T r    r 2  C
4k
At r = r0, T(r0) = TS
q 2  r 2 
Therefore, T  r   r0 1    TS
4k  r02 
The rate of heat convected to the surrounding fluid (at temperature Tf) by the surface at temperature Ts is
proportional to the temperature difference (TS  Tf) and the surface area in contact with the fluid.
Thus, rate of heat convection = h (2 r0L) (TS  Tf)
where h is a constant called heat convection coefficient.

By overall energy balance,

q  r02L   h  2r0   TS  Tf 
 0
qr
 TS  Tf 
2h
Read the above passage carefully and answer the following questions.
16. Consider a spherical nuclear fuel material of radius r0 and rate of heat generation q (watt/m3).
Find the temperature difference between the surface and surrounding fluid in steady state (h is
the heat convection coefficient).
 0
qr  0
2 qr
(A) (B)
2h 3 h

qr 
qr
(C) 0 (D) 0
3h h
Space for rough work

17. In the above passage, the ratio of temperature gradient at r = r0/2 and r = r0 is
(A) 1 (B) 1/4
(C) 1/2 (D) 1/8

18. In the given passage, the difference in temperature at the axis and surface of the cylinder is
 02
qr  2
qr
(A) (B) 0
4k k
qr02 2qr 0
(C) (D)
2k k
Space for rough work

SECTION - B
Matrix – Match Type
This section contains 2 questions. Each question contains statements given p q r s t
in two columns, which have to be matched. The statements in Column I are
A p q r s t
labelled A, B, C and D, while the statements in Column II are labelled p, q, r,
s and t. Any given statement in Column I can have correct matching with B p q r s t
ONE OR MORE statement(s) in Column II. The appropriate bubbles
corresponding to the answers to these questions have to be darkened as C p q r s t
illustrated in the following example:
D p q r s t
If the correct matches are A – p, s and t; B – q and r; C – p and q; and D – s
and t; then the correct darkening of bubbles will look like the following:
 x 
1. Consider a situation (i) that two sound waves, y1 = (0.2m) sin 504  t   and y2 = (0.6 m) sin
 300 
 x 
496  t   are super imposed. Consider another situation (ii) that two sound waves
 300 
 x   x 
y1 = (0.4m) sin 504  t   and y2 = (0.4 m) sin 504  t  are superimposed. Match the
 300   300 
following.
Column – I Column – II
(A) In situation (i) (p) Stationary waves are formed.
(B) In situation (ii) (q) There will be phenomenon of beats

(C) When two waves of same frequency and (r) Amplitude of the resultant wave will
amplitude and travelling in opposite vary periodically with position
directions superimpose.
(D) If the intensity of sound alternately (s) Amplitude of the resultant wave will
increases and decreases periodically a vary periodically with time
result of superposition of waves of highly
different frequencies.
(t) amplitude of the resultant wave will
be constant

Space for rough work

2. Match the following:

Column I Column II
(A) A cone resting with its slant surface touching against (p) Stable Equilibrium
the smooth horizontal floor
(B) A car moving on a banked smooth circular track of (q) Unstable equilibrium
banking angle  with constant speed without its tyres
slipping
(C) A rigid body suspended in a laboratory in such a (r) Neutral equilibrium
manner that its centre of gravity lies vertically above
point of suspension and its centre of mass lies
vertically below point of suspension.
(D) A uniform cube balanced on a horizontal floor on one (s) none of the above
of its edge (only a force parallel to horizontal plane
can be applied).
(t) potential energy of the
system remains constant.

Space for rough work

Chemistry PART – II

SECTION - A
Straight Objective Type
This section contains 8 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out
of which ONLY ONE is correct.

1. At 1400 K, Kc = 2.5  103 for the following reaction:

CH4 g  2H2S g 
 CS2 g  4H2 g


If a 10 litre reaction vessel at 1400 K contains 2.0 moles of CH4, 3.0 moles of CS2, 3.0 moles of
H2 and 4.0 moles of H2S, then equilibrium concentration of CS2 will be
(A) = 0.3 M (B) < 0.3 M
(C) > 0.3 M (D) none of these

2. If the dipole moment of 1, 2, 3, 5 – tetrachloro benzene is xD, then the dipole moment of
chlorobenzene will be
(A) 0.5 xD (B) 0.25 xD
(C) xD (D) 1.25 xD

3. A sample of four hydrogen atoms, all in the nth excited state, de – excite to the ground state, and
shows a maximum of six different spectral lines. Hence the value of ‘n’ is
(A) 3 (B) 4
(C) 5 (D) 6

4. A mixture containing ferrous oxalate & ferrous sulphate required 100 ml of 0.1M acidified KMnO4
solution. This mixture is now reduced by SnCl2 so that all Fe3+ changes to Fe2+ ions & the
resultant mixture now required 50 ml of the same permanganate solution, the percentage by
mass of ferrous oxalate in the mixture is (Fe = 56, C = 12, O = 16, S = 32)
(A) 18.6% (B) 28.6%
(C) 38.6% (D) 48.6%

Space for rough work

5. Identify the incorrect statement among the following:

(A) If Auf- bau rule is not followed, then Sn(atomic number = 50) would be placed in d – block of
the periodic table
th
(B) 5 ionization energy of As (Z = 33) is less than that of Ge (Z = 32)
(C) Increasing order of first I.E. of S, Se & Cl is Se < S < Cl
2+
(D) Ionic radii of Mg (Z = 12) is smaller than F (Z = 9)

6. 150 ml of 0.1 M NaOH solution is mixed with 100 ml of 0.1 M H2CO3 solution, if K a1 and K a2 of
H2CO3 are 107 and 1013 respectively, then pOH of the resulting solution will be
(A) 13 (B) 7
(C) 1 (D) 2
A
7. Compressibility factor (Z) is plotted against pressure (p) for
B
four different gases A, B, C & D. The correct order of critical C
D
temperature of the gases shown in the above graph is (Ideal gas)
(A) A > B > C > D Z

(B) B < A < C < D

8. Which of the following compounds exhibits tautomerism?

(A) chloroethane (B) ethanol
(C) ethoxyethane (D) nitroethane

Space for rough work

(Multiple Correct Choice Type)

This section contains 4 multiple choice questions. Each question has four choices (A), (B), (C) and (D)
out of which ONE or MORE are correct.

9. Which of the following mixture is/are buffers?

(A) 100 ml of 0.1 M Ca(CH3COO)2 & 100 ml of 0.1 M HCl
(B) 100 ml of 0.1 M (PhNH3)2SO4 & 50 ml 0.1 M Ca(OH)2
(C) 100 ml of 0.1 M (NH4)2SO4 & 100 ml of 0.1 M NH4OH
(D) 100 ml of 0.1 M borax solution (Na2B4O7.10H2O) & 25 ml of 0.05 M H2SO4

10. Which of the following equilibria shifts in the forward direction on decreasing the pressure?

(A) ice  water 
(B) water  
 water vapour
(C) S  
 S s

 

(D) diamond 
 graphite
density of diamond is more than graphite

11. The percentage by mass of chlorine in the chloride of a metal is 80%. 33.375 g of this chloride on
vaporization occupies a volume of 5.6 litre at NTP. Now select the correct options given below:
(A) valency of the metal is 2
(B) molecular weight of the metal chloride is 133.5
(C) atomic weight of the metal is 27 g
(D) 9 g of the metal can completely react with 8 g of oxygen

12. Identify the correct statement (s) among the following:

(A) The electronegativity of an element is the tendency of an isolated atom to attract an electron
(B) The second ionization potential of boron is greater than carbon
(C) Nitrogen has almost zero electron gain enthalpy
(D) The first electron affinity of fluorine is greater than that of chlorine

Space for rough work

Linked Comprehension Type

This section contains 2 paragraphs. Based upon each paragraph, 3 multiple choice questions have to be
answered. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

Paragraph for Question Nos. 13 to 15

The Lewis acid usually reacts with the acyl halide to form an acylium ion. The acylium ion is stabilized by
resonance. The acylium ion acts as an electrophile attacking the benzene ring to form an arenium ion.
The arenium ion then loses a proton to generate the final product. Powerful electron withdrawing groups
on the benzene ring such as another acyl group will retard this reaction. Naphthalene is the simplest and
most important of the fused ring hydrocarbons. Five percent of all constituents of coal tar are
naphthalene. Naphthalene can be manufactured using the Friedel-Crafts reaction via the reaction
pathway shown below:
O
O O
C
H2 C AlCl3 Zn(Hg), HCl HF
O +
Step 1 Step 2 Step 3
H2 C C C C
O OH O OH
O
Zn(Hg), HCl Pd, heat
Step 4 Step 5

Naphthalene
O
13. The Friedel-Crafts reaction occurs twice in the given sequence of reactions. Which two steps
represent Friedel-Crafts reactions?
(A) Steps 1 and 3 (B) Steps 1 and 5
(C) Steps 2 and 4 (D) Steps 3 and 5

Space for rough work

14. What are the most likely products of the following reaction?
O
H3C–C
AlCl3
O +
H3C–C
O
O O
O
CCH3
(A) C2H6 + (B) CH3COH +
C
O OH
O O
O O
COCH3 OCCH3
(C) CH3CH + (D) CH3CH +

15. Which of the following is an acylium ion?


(A) AlCl 4 (B) +

O O
||
||

(C) CH3 CO  (D) CH 3 C

Space for rough work

Paragraph for Question Nos. 16 to 18

The solubility of some sparingly soluble salts e.g. AgCl, AgBr, AgI, Ag2CrO4 etc can be appreciably
increased through complexation reaction. Let’s take the case of AgCl which is very feebly soluble in water
(Ksp of AgCl = 1  1010 at 25C) but if liquid ammonia is added into this solution, its solubility increased so

many folds due to the formation of complex  Ag NH3 2  .

  Ag NH3 2   Cl
AgCl  2NH3 aq  


A solution of Ag+ ions at a concentration of 4.0  1013 M just fails to yield a precipitate of AgCl with
concentration of 1.0  103 M of Cl ions, when the concentration of ammonia in the solution is
2.0  102 M.

Now, answer the following questions.

16. The solubility of AgCl in water (without adding NH3) at 25C is

(A) 104 M (B) 105 M
(C) 106 M (D) 108 M


17. The dissociation constant of the complex  Ag NH3 2  at 25C is
(A) 107 (B) 10+7
(C) 108 (D) 10+8


18.   Ag NH3 2 
Ag  2NH3 


On adding more NH3, the above equilibrium shifts:
(A) In the forward direction (B) In the backward direction
(C) The complex dissociation faster (D) None of these

Space for rough work

1. Match the statement in column – I with corresponding type (s) of silicates in column - II
Column  I Column  II
(A) Only one corner of SiO44 units is shared (p) Chain silicates
4
(B) At least one corner of SiO4 unit is shared (q) Sheet silicates
4
(C) Only two corners of SiO 4 units are shared (r) Pyrosilicates
4
(D) At least two corners of SiO4 units are shared (s) Cyclic silicates

2. Match the compounds in column – I with their corresponding decreasing Ka value in column – II.
Column  I Column  II
(compounds) (Acidity)
(A) o, m, p – nitrophenols (p) o > m > p
(B) o, m, p – cresols (q) p>o>m
(C) o, m, p – chlorophenols (r) p>m>o
(D) o, m, p – nitrobenzoic acid (s) m>p>o

Space for rough work

Mathematics PART – III

SECTION - A
Straight Objective Type
This section contains 8 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out
of which ONLY ONE is correct.

1. If the roots of ax2 – bx – c = 0 change by the same quantity then the expression in a, b, c that
does not change is
b2  4ac b  4c
(A) (B)
a2 a
2
b  4ac a  b2
2
(C) 2
(D)
a c2

3 
2. The point of intersection of the curves arg  z  3i   and arg  2z  1  2i   is
4 4
1 1
(A) 3  9i  (B)  3  9i 
4 4
1
(C)  3  2i  (D) none of these
2

3
3. Suppose a, b, c are in A.P. and a2, b2, c2 are in G.P. If a < b < c and a + b + c = , then the value
2
of a is
1 1
(A) (B)
2 2 2 3
1 1 1 1
(C)  (D) 
2 3 2 2

Space for rough work

200
 5 
4. If {x} represents fractional part of x, then   is
 8 
1 1
(A) (B)
4 8
3 5
(C) (D)
8 8

5. The maximum number of points of intersection of 8 circles is

(A) 16 (B) 24
(C) 28 (D) 56

6. If a sin x + b cos(x + ) + b cos(x – ) = d for some real x, then the minimum value of |cos | is
1 1
(A) d2  a2 (B) d2  a2
2b 2a
1 1
(C) d2  a2 (D) d2  a2
2d a

7. The number of real solutions of the equation tan1 x 2  3x  2  sin1 4x  x 2  3   is

(A) one (B) two
(C) zero (D) infinite

8. If in a triangle a2 + b2 + c2 = ac + 3ab , then the triangle is

(A) equilateral (B) right angled isosceles
(C) right angled scalene (D) none of these

Space for rough work

(Multiple Correct Choice Type)

This section contains 4 multiple choice questions. Each question has four choices (A), (B), (C) and (D)
out of which ONE or MORE are correct.

9. The roots of ax2 + bx + c = 0, are non–real complex and a + c < b, then

(A) 4a + c > 2b (B) 4a + c < 2b
(C) a + 4c > 2b (D) a + 4c < 2b

10. A circle of radius r ( 0) touches the parabola y2 + 12x = 0 at the vertex of the parabola. The
centre of the circle lies to the left of the vertex and the circle lies entirely within the parabola, then
the interval(s) in which r lies can be
(A) (1, 7) (B) (0, 6)
 13 
(C)  1,  (D) (2, 8)
 2 
2iz1  z1  z2 cos   isin 
11. If  and z1, z2 are non-zero complex numbers then
2iz1  z1  z2 cos   isin 
z1  z 2 z1
(A) is purely real (B) is purely real
z1 z2
z1  z 2
(C) is purely real (D) none of these
z1  z 2

2 2
12. The equation x 2   y  1  x2   y  1  k will represent a hyperbola for

(A) k  (0, 2) (B) k  (0, 1)

Space for rough work

Linked Comprehension Type

This section contains 2 paragraphs. Based upon each paragraph, 3 multiple choice questions have to be
answered. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

Paragraph for Question Nos. 13 to 15

Read the following write up carefully and answer the following questions:
Let r be the position vector of a variable point such that r . (10j – 8i – r ) = 40 and P1 = max{| r + 2i –
8
3j|}, P2 = min{| r + 2i – 3j|2}. A tangent is drawn to the curve y  2 at the point A with abscissa 2. The
x
drawn line cuts the x-axis at B

13. The value of P2 is equal to

(A) 9 (B) 2 2  1
(C) 6 2  3 (D) 9  4 2

14. The value of P1 + P2 is equal to

(A) 2 (B) 10
(C) 18 (D) 5
 
15. AB · OB is equal to
(A) 1 (B) 2
(C) 3 (D) 4

Space for rough work

Paragraph for Question Nos. 16 to 18

Read the following write up carefully and answer the following questions:
2 2 2
Consider the complex numbers z1 and z2 satisfying the relation |z1 + z2| = |z1| + |z2|

16. Complex number z1z2 is

(A) purely real (B) purely imaginary
(C) zero (D) none of these

z1
17. is
z2
(A) purely real (B) purely imaginary
(C) zero (D) none of these

z 
18. One of the possible argument of i  1  is
 z2 
 
(A) (B) 
2 2
(C) 0 (D) none of these

Space for rough work

SECTION - B
Matrix – Match Type
This section contains 2 questions. Each question contains statements given p q r s t
in two columns, which have to be matched. The statements in Column I are
A p q r s t
labelled A, B, C and D, while the statements in Column II are labelled p, q, r,
s and t. Any given statement in Column I can have correct matching with B p q r s t
ONE OR MORE statement(s) in Column II. The appropriate bubbles
corresponding to the answers to these questions have to be darkened as C p q r s t
illustrated in the following example:
D p q r s t
If the correct matches are A – p, s and t; B – q and r; C – p and q; and D – s
and t; then the correct darkening of bubbles will look like the following:
1. Match the following Column–I with Column–II
Column – I Column – II
(A) The locus of mid-points of chords of an ellipse which are
(p) pair of straight line
drawn through an end of minor axis, is
(B) The locus of an end of latus rectum of all ellipse having a
(q) circle
given major axis is
(C) The locus of the foot of perpendicular from a focus of the
(r) parabola
ellipse on any tangent is
(D) A variable line drawn through a fixed point cuts axes at A
(s) ellipse
and B. The locus of mid-point of AB is
(t) hyperbola

2. Match the following Column–I with Column–II

Column – I Column – II
(A) If the equation x2 + 4 + 3 sin(ax + b) – 2x = 0 has atleast
one real solution, where a, b  [0, 2], then sin(a + b) can (p) –1
be equal to
(B) sin–1 x  cos–1 x, then x can be equal to (q) 0
(C) The number of the ordered pairs (x, y) satisfying |y| = cos x
(r) 1
and y = sin–1(sin x), where –2  x  3, is equal to
n 2
(D) If n  N and the set of equations cos–1 x + (sin–1 y)2 =
4
2  2 (s) 4
 
and sin1 y  cos1 x 
16
is consistent, then n can be

equal to
(t) 5

Space for rough work

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HALF COURSE TEST –VIII
(Paper-1)

Q. No. PHYSICS CHEMISTRY MATHEMATICCS

1. B B C
2. D C D
3. D A D
4. C D B
5. C A D
qualified in JEE (Advanced), 2014.

6. D C A
7. C C C
8. A D C
9. A, B A, B, C, D B, D
10. A, B, D B, D A, B, C, D
11. A, B, C, D B, C, D A, B
12. B, C B, C, D A, B
13. C A D
14. A B C
15. C D C
16. C B B
17. C C B
18. A A C
(A)  (q, s) (A)  (r) (A)  (s)
(B)  (p, r) (B)  (p, q, r, s,) (B)  (r)
1. (C)  (p, r) (C)  (p, s) (C)  (q)
(D)  (q, s) (D)  (p, q, s) (D)  (t)

(A)  (r, t) (A)  (q) (A)  (p)

(B)  (t) (B)  (s) (B)  (p, q)
2.
(C)  (q) (C)  (p) (C)  (t)
(D)  (q) (D)  (p) (D)  (r)

Physics PART – I

SECTION – A

1. In one oscillation distance covered = 4  3 = 12

 2
Here  =   T = 4 sec
2 T
3
 8.5 sec = 4  2 + 0.5 and distance covered = 2  12 +3 3 cos[(/2)  0.5] = 27 
2

3. The line of force due to pressure every where passes through O.

 Net torque = 0
 F=0

4. Each block completes half the oscillation and time period is independent of amplitude

dU
5. F  U0  sin x  U0  sin x
dx
for small angle
F  U0  2 x
m
T  2
U0  2

4 3
r1 
m1 2 3
6. Since  8 
m2 0.25 4 3
r2 
3
r
 1 2
r2
vT  r 2
v
hence v T 
4

7. Conservation of linear momentum during the explosion

m m
mv   0   v '  v = 2v
2 2
Increase in the mechanical energy = K + U
1m 1 1
U  0 K = (2v)2  mv 2  mv 2
22 2 2
GM
v
2R
mgR
K 
4

8. Acceleration for particle = g

Acceleration for earth = g/2
g 3g
grel  g  
2 2
1 4h
h  grel t 2  t 
2 3g

9. The satellite will fly out tangentially with same speed.

10. Use vector sum for superposition principle

11. f AC > fAB

 fA  fC  fA  fB
 All of them are possible.

1 1 F 
12. Energy stored per unit volume =  stress  strain =  
2 2 A 
1 1
Total Energy =  F    Mg
2 2

Chemistry PART – II

SECTION – A
1. The above reaction is not at equilibrium as,
Qc > Kc, So the reaction will move in the backward direction.
Hence (B) is correct.

2. Let the dipole moment of chloro benzene = Cl

1D1 then x = 2  1  cos120/2 Cl

 1 = x D
Hence (C) is correct. Cl Cl

3 nth excited state means that electron is in the (n + 1)th level six, these 4 – H – atoms show a total
of six different spectral lines, while returning to the ground level, that is possible only when
(n + 1) = 4.
Hence (A) is correct.

4. Let moles of FeC2O4 be x & moles of FeSO4 be y.

Then, 3x + y = 0.05 ...(i)
also, x + y = 0.025 …(ii)
Hence (D) is correct.

5. Sn  50   1s2 2s2 .2p6 3s2 3p6 .3d10 .4s2 .4p 6 .4d10 .5s2 .5p 2  by obeying Auf  bau rule 
Sn  50   1s2 2s2 .2p6 3s2 3p 6 .3d10 .4s 2 .4p6 .4d10 .4f 4  if not following Auf  bau rule 
So, if Auf – bau rule is not followed, then Sn would be placed of in f – block.
Hence (A) is correct.

6. 

H2CO3  NaOH 
 NaHCO3  H2O
m. mol 10 15 0 0
after
the reaction 0 5 10
Also,


NaHCO 3  NaOH 
 Na2CO3  H2O
10 5 0
5 0 5
So, it is a buffer,

pH  p
k a2
 log
Naa CO3 
NaHCO3 
k a2
p  13
 pH =
 pOH  1
Hence (C) is correct.

7. The more is the derivation of a real gas from ideal behaviour the more is the critical temperature.
Hence (C) is correct.

O O
8. CH 2  N  O CH 2  N  OH

10. In case of physical equilibria, equilibria always shifts from low density to high density on
increasing the pressure.

11. Since 5.6 littre are occupied by ¼ mole of NTP, so the molecular weight of metal chloride
= 133.5 g.
Let the formula of the chloride be MClx
x  mole of metal = 1  mole of Cl
Let the total mass of metal chloride = 100 g
Then mass of Cl = 80 g
& mass of metal = 20 g
& let mol weight of metal = MA
Now,
20 80
x  1 ...  i 
MA 35.5
Also, MA + 35.5x = 133.5 ….(ii)
On solving, x = 3 & MA = 27 g
Hence (B, C, D)

16. AgCl s 

 Ag aq  Cl aq

  

Ksp = x2
 x  K sp  10 5 M
Hence (B) is correct.

17. Ksp = [Ag+] [Cl]

10 10   Ag  103    Ag   10 7 M
Now,

 Ag NH   


 Ag

 2NH3
 3 2 

Initial conc. 4 103 107 2 102

at equilibrium 4 10 3
 x 10 7
 x 2 10 2
 2x 
 4 103  107  2 102
2 2

So, K d  
 Ag  NH3 


10 7  2  10 2   108
 Ag NH3 2 

4  10 3 
Hence (C) is correct.

Mathematics PART – III

SECTION – A
2 2
1. Let ,  be the roots of ax – bx – c = 0 and let ,  be the roots of ax – bx – c = 0 such that
     '  ' , i.e., ( + )2 – 4 = ( + )2 – 4
b2  4ac b'2  4a'c '
 
a2 a'2
b2  4ac
Hence, the expression does not vary in value
a2

3 y
2. Let z = x + iy, then arg  z  3i   arg  x  iy  3i  y = –x + 3
4
(0, 3)
 3 
 x < 0, y – 3 > 0  is in II quadrant  3
 4  y  x
2
y3 3 1
and  tan = –1
x 4
x
 y = –x + 3  x < 0 and y > 3 ….. (1) –1/2

and arg  2z  1  2i   arg  2x  1  i  2y  2   
4
  
 2x + 1 > 0, 2y – 2 > 0  is in I quadrant 
 4 
2y  2 
and  tan  1  2y – 2 = 2x + 1
2x  1 4
3 1
 y  x  x   , y 1 ….. (2)
2 2
From equations (1) and (2), we get graph. It is clear from the graph that two lines do not intersect
 No point of intersection

3 1
3. Let a = b – d ad c = b + d, then a + b + c =  b
2 2
1 1 1
 The numbers are  d , ,  d [d > 0 as a < b < c]
2 2 2
2 2 2
2  1 1  1 
2 2 2
Now a , b , c are in G.P.  b2    a2 c 2       d    d 
2
   2   2 
2
1 1  1 1 1 1
    d2    d2    d    d ( d > 0)
16  4  4 4 2 2

1  24 100 1
4.   Integer
8 8

5. Maximum number of point of intersection = 2  8C2 = 56

6. a sin x + b{cos(x + ) + cos(x – )} = d

 a sin x + 2b cos x cos  = d

d2  a2 d2  a2
cos2    cos  
4b2 2b

 
7. tan1 x 2  3x  2 and cos 1 4x  x 2  3 
2 2
 0 < LHS <   no solution

2
 a 2
  
8.   c    b  3a    0
 2   2  
 
c b
 a 
1
   3
   
 2   2 

2
9. b – 4ac < 0
 f(x) = ax2 + bx + c must have same sign for all real x
f(–1) = a – b + c < 0
 f(–2) < 0  4a – 2b + c  4a + c < 2b
 1 a b
f     0    c  0  a – 2b + 4c < 0  a + 4c < 2b
 2 4 2

10. Circle ( + r)2 + y2 = r2

Solving with the parabola, x = 0 and 12 – 2r  0
 rmax = 6

z1  z2
z1  i
11. 2 1
 z1  z2 
z1  i  
 2 
2z1
i
z1  z2
 1
2z1
i
z1  z2
2z1
 is purely real
z1  z 2
z1  z 2
 is purely real
z1
z2
 is purely real
z1

12. Equation is equivalent to |S1P – S2P| = constant

 2a = k and 2ae = S1S2 = 2
2
 e
k
Also, k > 0
 k  (0, 2)


13.-15. r  xi  yj
x2 + y2 + 8x – 10y + 40 = 0
c = (–4, 5), r = 1
2 2
P1 = max{(x + 2) + (y – 3) }
2 2
P2 = min{(x + 2) + (y – 3) }
 P = (–2, 3)
CP = 2 2
2 2
 
P2 = 2 2  1 , P 1 = 2 2  1

 
OA  2i  2j

OB  3i
 
 OA  OB  3

2 2
16.-18.  z1  z2   z1  z2   z1  z 2
 z1z2  z1z2  0
z1  z1 
   0
z 2  z2 

SECTION – B

hx ky h2 k 2
1. (A)    which passes through (0, b)
a2 b2 a2 b2
k h2 k 2
  2  2
b a b
 Ellipse
(B) h = ae, k = a(1 – e2)
 h2 = -a(k – a)
 Parabola
(C) y  mx  a 2m2  b 2
Perpendicular from (ae, 0) is my + x = ae
 x2 + y2 = a2
 Circle
x y
(D)   1 passes through (, )
a b
 
  1
a b
a b
h , k
2 2
 
  2
h k
 
 2
x y
     
  x   y   
 2  2 4
 Rectangular hyperbola

2. (A) (x – 1)2 + 3{1 + sin(ax + b)} = 0

  = 1 and sin(a + b) = –1

(B)  sin1 x  sin1 x
2
1
 –1  x 
2
–1
(C) The graph of |y| = cos x and y = sin (sin x)
intersect at 5 points in [–2, 3]

–2 3   O  3 5 2
 
2 2 2 2 2

2 4n  1 2

(D) Adding we get, sin1 y  
32


4n  1 2 2
 0  
32 4
–1 4n  1 2
Also, cos x = 
32
n=1

Time Allotted: 3 Hours Maximum Marks: 240

INSTRUCTIONS
A. General Instructions

B. Filling of OMR Sheet

C. Marking Scheme For All Three Parts.

1. Section – A (01 – 04) contains 4 multiple choice questions which have only one correct answer.
Each question carries +3 marks for correct answer and – 1 mark for wrong answer.
Section – A (05 – 09) contains 5 multiple choice questions which have more than one correct
answer. Each question carries +4 marks for correct answer and – 1 mark for wrong answer.

2. Section – B (01 – 02) contains 2 Matrix Match Type questions containing statements given in 2
columns. Statements in the first column have to be matched with statements in the second
column. Each question carries +8 marks for all correct answer. For each correct row +2 mark will
be awarded. There may be one or more than one correct choice. No marks will be given for any
wrong match in any question. There is no negative marking.
3. Section – C (01 – 08) contains 8 Numerical based questions with answers as numerical value
and each question carries +4 marks for correct answer and – 1 mark for wrong answer.

Name of the Candidate

Enrolment No.

Useful Data

PHYSICS
2
Acceleration due to gravity g = 10 m/s

Planck constant h = 6.6 1034 J-s

Charge of electron e = 1.6  1019 C

Mass of electron me = 9.1  1031 kg

Permittivity of free space 0 = 8.85  1012 C2/N-m2

Density of water water = 103 kg/m3

Atmospheric pressure Pa = 105 N/m2

Gas constant R = 8.314 J K1 mol1

CHEMISTRY

Gas Constant R = 8.314 J K1 mol1

Atomic No: H=1, He = 2, Li=3, Be=4, B=5, C=6, N=7, O=8,

Physics PART – I

SECTION - A
Straight Objective Type
This section contains 4 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out
of which ONLY ONE is correct.

1. A block of mass m attached to a spring executes SHM with angular frequency 0. The spring is
m
cut into 4 equal parts and another block with mass is attached to the two pieces in parallel
2
combination. The angular frequency of oscillations of the mass will be
(A) 20 (B) 40
(C) 2 20 (D) 0 / 4

2. An artificial satellite is first taken to a height equal to half the radius of Earth. Let E1 be the energy
required. It is then given the appropriate orbital speed such that it goes in a circular orbit at that
E1
height. Let E2 be the energy required. The ratio is
E2
(A) 4 : 1 (B) 3 : 1
(C) 1 : 1 (D) 1 : 2.

3. A thin homogeneous rod of length 2L floats partly immersed in water,

being supported by a string fastened to one of its end, as shown. The
specific gravity of the rod is 0.75. the length of the rod that extends out of
water in equilibrium is
(A) L (B) L/2
(C) L/4 (D) 3L

4. A particle of mass m is executing SHM as shown in

diagram (1) and (2). In diagram (2) a force F is
applied on the particle in the upward direction k
F = mg. If their velocities are same at mean position m
and let A1 and A2 be their amplitudes and T1 and T2 m F
be their time periods respectively then, (i) (ii)
(A) A1 = A2, T1 = T2
(B) A1 = A2, T1 > T2
(C) A1 > A2, T1 = T2
(D) A1 > A2, T1 > T2

Space for rough work

(Multiple Correct Choice Type)

This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D)
out of which ONE or MORE are correct.

5. A hollow sphere and a solid sphere of same radius and same material fall through a liquid from
same height (neglect viscous force). Then, the correct option(s) from the following is/are:
(A) The solid sphere reaches the ground earlier than the hollow sphere.
(B) The hollow sphere reaches the ground earlier than the solid sphere.
(C) Buoyant forces on both solid as well as hollow sphere are same.
(D) Buoyant forces on the solid and hollow spheres are different.

6. A spherical black body has radius r. If P is the power radiated by the black body and R is the rate
of cooling of the black body, then
1
(A) P  (B) P  r2
r
1
(C) R  (D) R  r2
r

7. Two blocks connected with the spring of force constant 4 m/s 3 m/s
ˆ ˆ
100 N/m are given velocities 4 i m/s and 3i m/s when
the spring is in natural length as shown in figure 5kg 2kg
(A) The velocity of 2kg block is maximum when 5 kg
block is at instantaneous rest.
(B) The maximum and minimum velocities of 2kg block is 7 ˆi m/s and –3iˆ m/s respectively.
(C) The maximum and minimum velocities of 5kg block is 4 m/s and zero respectively.
(D) None of these

8. A particle is rotating in a conical pendulum with help of a string of length  as

 
shown in the figure. The speed of the particle is constant and angle  is also
constant with time. It can be said that
(A) Angular momentum of the particle about the point of suspension is not
constant.
(B) Only the direction of angular momentum of the particle about the point of suspension is
constant.
(C) Only the magnitude of angular momentum about the point of suspension is constant.
(D) Net torque on the particle about the point of suspension is zero.

/4
9. A light rod of length  is pivoted at distance /4 from the
left end has two masses m 0 and m attached to its ends
P
such that rod is in equilibrium. Find m in terms of m0. m0 m

m0 m0
(A) (B)
2 3
(C) Torque about P is zero (D) Torque about P is not zero

Space for rough work

SECTION - B

Matrix – Match Type

This section contains 2 questions. Each question contains statements given p q r s t
in two columns, which have to be matched. The statements in Column I are
A p q r s t
labelled A, B, C and D, while the statements in Column II are labelled p, q, r,
s and t. Any given statement in Column I can have correct matching with B p q r s t
ONE OR MORE statement(s) in Column II. The appropriate bubbles
corresponding to the answers to these questions have to be darkened as C p q r s t
illustrated in the following example:
D p q r s t
If the correct matches are A – p, s and t; B – q and r; C – p and q; and D – s
and t; then the correct darkening of bubbles will look like the following:

1. A sphere of mass m, radius R is left on a rough wedge A m

(coefficient of friction ) fixed with the ground. Now, match the

following B

Column A Column B
(A) If  = 0 (p) P.E. > K.E.
(B) If   0; and before rolling starts (q) Friction is acting upward
(C) If   0; and after rolling starts (r) Friction is acting downward
(D) If   0; when the sphere is at (s) P.E  K.E.
bottom most point.
(t) Translational (K.E.) > (Rotational) (K.E.)

Space for rough work

D
2. A sphere of mass M, radius R kept on a horizontal frictionless surface. A
small ball of mass m hits the sphere at a height h = R and rebounds A
C
E
elastically as shown in the figure. Consider the situation mentioned and
match the following. B

Column A Column B

Linear momentum for the sphere is conserved

(A) (p) A
along tangent at

Angular momentum for the system is

(B) (q) B
conserved about point.

(C) Net torque on the sphere is zero about (r) C

Linear Impulse is acting about line joining

(D) (s) D
points A and

(t) E

Space for rough work

SECTION – C
(Integer value correct Type)

This section contains 8 questions. The answer to each question is a single digit integer, ranging from
0 to 9 (both inclusive).
1. The variation of density () with temperature (T) of an ideal gas of 
molecular mass (M) at constant pressure is plotted as shown in figure. P
RT 2
The pressure of the gas at state P is Find the value of K. 30
KM
T

2. The two ends A and B of a uniform of length  = 1m and mass m vA = 20 m/s

are moving with velocities VA and VB as shown. The length AP, A = 53
where P is point on the rod with velocity 12 m/s, is 8K cm. Find the A B B = 37
value of K. vB
2 2 2
3. A particle moving on the sphere X  Y  Z  1 on the convex side. XY plane is vertical plane
and ‘g’ acts along –ve Y axis. The speed of the particle is 2ms–1 always. The mass of the particle
is 1kg. The normal reaction acting on the particle at point (0, 1m, 0) is _________.

4. M1 M2
M1 M2

M1 fixed
Frictionless surface
If M1 & M2 are free to oscillate together If M1 is fixed, and only M2 oscillates horizontally
about their centre of Mass, then time period then time period is T2
is T1.
T M
If 2  3 Then 2 is ________
T1 M1

Space for rough work

 2X 
5. The displacement of a string is given by y (x1,t)  .06 sin   cos 120t 
 3 
–2
Where X and Y are in m and t in sec. The length of the string is 1.5m and its mass is 3.0×10 Kg.
This is a result of super position of two waves travelling in opposite directions.
–1
The speed of any one of the component waves is 36X ms . Then X = ?

6. A hollow sphere of volume V is floating on water surface half immersed. The minimum volume of
water poured on surface so that the sphere now sinks into the water is V/K. Then K is:

7. In determination of Young’s Modulus

by Searle’s Method, the extension Vs 15.92
load curve is shown below. The radius
of the wire is 1mm and length of the Extension 7.96
–6
(10 m)
wire used is 2m. The Young’s modulus
is X  1011N / m2 . Then X = _________. 1kg 2kg
Load

8. A particle starts falling from the top of a tower at t = 0 where g = 10ms–2. Simultaneously, a car
situated on ground some distance away starts moving towards the tower at a constant velocity of
5ms1. The radius of curvature of the trajectory of particle as seen by an observer in the Jeep at
5
some time ‘t’ is (17)3/2 metre. Find the time ‘t’ where ‘t’ is in seconds.
2

Space for rough work

Chemistry PART – II

SECTION - A
Straight Objective Type
This section contains 4 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out
of which ONLY ONE is correct.

1. Which of the following is the correct nature of a solution obtained by mixing 100 ml of 0.10 M HA
and 100 ml of 0.1 M NaOH?
(I) Neutral, if HA is a strong acid
(II) Basic if HA is a weak acid
(III) Neutral if HA is a weak acid.
(A) I only (B) I and II only
(C) I and III only (D) II and III only

2. The structures shown below,

COOH COOH

H C OH H C CH3

CH3 OH
are:
(A) Identical
(B) different conformations of the same compound
(C) Enatiomers
(D) Diastreomers

3. Consider the following statements:

(i) 4 g if O2 gas occupies 11.2 litres at 00C and 0.25 atm pressure assuming ideal behaviour.
(ii) Vander Waal’s constant ‘a’ for NH3(g) is higher than CO2(g).
(iii) Average distance travelled by a gas molecule between two successive collisions is directly
proportional to the square of its collision diameter.
3
(iv) Average kinetic energy of two molecules of an ideal gas is kT , where k is Boltzmann
2
constant.
Identify the correct statements of the above
(A) I, ii and iii (B) I, ii, iii and iv
(C) ii, iii and iv (D) I, ii and iv

4. If we wish to prepare 250 ml of 0.1 M NaOH solution, it is advisable to use more than 1.0 g (as
required) NaOH because,
(A) NaOH is hygroscopic in nature (B) NaOH is easy to purify
(C) NaOH is less soluble in water (D) NaOH is a strong base

Space for rough work

(Multiple Correct Choice Type)

This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D)
out of which ONE or MORE are correct.

5. Which of the following reaction product is not correct?

(A) O
COCl2
H2N 
anhyd.AlCl3
 H2N C NH2

(B)
OH

i conc. H SO / 
 2
ii  Ni / H2
4



(D)
OH
conc. H SO / 

2 4

OH
HO

6. Which of the following compounds exhibit tautomerism?

(A) CH=CHOH (B) O O

O O

Space for rough work

7. In which of the following reaction(s), H2O2 can convert the underlined atom into the product with
oxidation state = +6.
-
(A) CrCl3 + H2O2 + OH  (B) K2Cr2O7 + H2O2 + H2SO4 
(C) SO2 + H2O2 + OH-  (D) H2O2 + Na3AsO3 + OH- 

8. From the following data for the reaction between A and B:

[A] [B] Initial rate at (mol-1sec-1)
-1 -1
[mol ] [mol ] 300 K 320 K
2.5  10-4 3.0  10-5 5.0  10-4 2.0  10-3
-4 -5 -3
5.0  10 6.0  10 4.0  10 -
1.0  10-3 6.0  10-5 1.6  10-2 -
Now select the correct statement(s) of the following:
(A) Overall order of this reaction is 3.
(B) Units of rate constant of the above reaction is mol-2 l2 sec-1
(C) Rate constant of this reaction at 300 K is 2.66  108 mol-2 l2 sec-1
(D) Rate constant of this reaction at 320 K is 12.66  107 mol-2 l2 sec-1

9. The correct statement regarding H3BO3 is/are

(A) Monobasic and weak Lewis acid
(B) when mannitol is added to aqueous solution of H3BO3 pH of the solution decreases
(C) titration of H3BO3 against NaOH is done in presence of mannitol
(D) Tribasic and weak Bronsted acid.

Space for rough work

SECTION - B

Matrix – Match Type

1. Match the species given in Column – I with their corresponding characteristics in Column – II:
Column – I Column – II
(Species) (Characteristics)
(A) ICl2 (p) possesses permanent dipole moment
(B) XeOF4 (q) at least two lone pairs on the central atom
(C) SeF4 (r) distorted structure
(D) Ba(BrO3)2 (s) sea-Saw shaped
(t) trigonal bipyramidal geometry

Space for rough work

2. Match the reaction in Column – I with the characteristics of corresponding organic product in
Column – II:
Column – I Column – II
(Reaction) (Characteristics of organic product)
(A) (p) Aromatic
OH
(i) Conc.H2 SO4 / 
 
(ii)NBS / h  1 equivalent

(iii) alc.KOH(excess) / 

(B) O (q) All carbon atoms in the ring are sp2
hybridised
(i) anhyd. AlCl3
(ii) Zn(hg) / Conc HCl
Ph Cl 
(iii) NBS / h  1equivalent 

(iv )alc.KOH(excess) / 

(C) (i) aq. K2 CO3 solution (r) Non-aromatic

(ii) Cl

(iii) NBS / h 1equivalent 

(iv) alc.KOH / 
(D) CH3 (s) Bicyclic

(i) COCl2 / anhyd.AlCl3


(ii)Zn(Hg) / conc.HCl



(t) At least one carbon atom is sp3

hybridised

Space for rough work

SECTION – C
(Integer value correct Type)

This section contains 8 questions. The answer to each question is a single digit integer, ranging from
0 to 9 (both inclusive).
1. Two moles of helium gas ( = 5/3) are initially at a temperature of 270C and occupy a volume of
20 litre. The gas is first expanded at constant pressure until the volume is doubled. It then
undergoes adiabatic change until the temperature returns to its initial value. If total work done by
the gas is – xkcal, then what is x?

2. Magnesium carbide on hydrolysis gives a product which on heating with Na followed by treatment
with ethyl chloride gives another product (A). (A) on catalytic reduction gives (B). Calculate the
number of monochlorinated product (s) of compound (B).

3. 100 ml of tap water containing Mg(HCO3)2 was titrated with M/50 HCl with methyl orange as
indicator. If 30 ml of HCl were required, calculate ‘X’, if temporary hardness as parts of CaCO3
per 106 parts of water is X  102.

4. Sodium borohydride on reaction with iodine in the presence of diglyme solution gives a gas (X)
which is used as an important rocket fuel. The gas (X) on heating with ammonia forms an adduct
which on further heating to high temperature forms another compound (Y).
Calculate the number of dichloro derivative of compound (Y).

5. How many isomeric dinitrobenzoic acids can exist? C6H3(NO2)2CO2H

6. What is the number of monochloro substituted derivatives formed from diphenyl methane?

7. In an experiment, 50 ml of 0.1 M solution of a metallic salt reacted exactly with 25 ml of 0.1 M

solution of Na2SO3. In the reaction SO32– is oxidized to SO42–. If the original oxidation number of
the metal in the salt was 3, what would be the new oxidation number of metal?

8. For the reaction at 25oC

A  g  2B  g  2C  g   D  g
The initial concentration of B is 1.5 times the initial concentration of A. If at equilibrium, the
equilibrium concentration of A and D are equal. The equilibrium constant at 25oC is

Space for rough work

Mathematics PART – III

SECTION - A
Straight Objective Type
This section contains 4 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out
of which ONLY ONE is correct.

1. A straight line L with negative slope passes through (8, 2) and cuts the positive axes at P and Q.
As L varies, the absolute minimum value of OP + OQ is (O is origin)
(A) 10 (B) 18
(C) 16 (D) 12

2. If (, ) is a point on the circle whose centre lies on the x-axis and which touches the line x + y =
0 at (2, –2), then the greatest value of  is
(A) 4  2 (B) 4  2
(C) 4  2 2 (D) 6

3. Maximum number of common normal of y2 = 4ax and x2 = 4by may be equal to

(A) 2 (B) 3
(C) 5 (D) infinite

x2 y2
4. The locus of foot of perpendicular from the centre on any tangent to the ellipse   1 is
a2 b2
(A) a circle (B) ellipse
(C) hyperbola (D) none of these

Space for rough work

(Multiple Correct Choice Type)

This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D)
out of which ONE or MORE are correct.

5. The range of values of ‘a’ such that the angle  between the pair of tangents drawn from (a, 0) to

the circle x2 + y2 = 1 satisfies     , lies in
2
(A) (1, 2) (B) 1, 2  

(C)  2,  1  (D)    
2,  1  1, 2 
6. In a ABC, if (a + b + c)(b + c – a) = xbc then x can be equal to
(A) 1 (B) 2
(C) 3 (D) 4


7. If the equation sin–1(x2 + x + 1) + cot–1(ax + 1) = has exactly 2 solutions, then a can not have
2
the integral values
(A) –1 (B) 0
(C) 1 (D) 2

8. If cos4  + , sin4  +  are the roots of x 2 + b(2x + 1) = 0 and cos2  + , sin2  +  are the roots
of the equation x2 + 4x + 2 = 0, then b is equal to
(A) 1 (B) –1
(C) 2 (D) –2

9. The sum of the coefficients in the expansion of (2 + 3cx + c2x2)12 vanishes then c equals
(A) –2 (B) 2
(C) 1 (D) –1

Space for rough work

SECTION - B

Matrix – Match Type

1. Match the following Column–I with Column–II

Column – I Column – II
sin x cos3x
(A) The function can take the values (p) 4
sin3x cos x
1
(B) (sin 12º)(sin 48º)(sin 54º) is equal to (q)
8
(C) In an acute angled ABC, the least values of  sec A (r) 6
(D)  
3 cosec 20º  sec 20º is equal to (s) 0
(t) 2

2. ABC has side lengths 18, 24, 30 respectively. Match the following Column–I with Column–II
Column – I Column – II
(A) The ratio of circum radius to in-radius is (p) 2
(B) The distance between circum-centre and orthocentre is 4 2
(q)
3
(C) The distance between circum-centre and in-centre (r) 2 2
(D) The area of triangle whose vertices are the in-centre, circum- (s) 4 2
centre and centroid respectively, is
3
(t)
2

Space for rough work

SECTION – C
(Integer value correct Type)

This section contains 8 questions. The answer to each question is a single digit integer, ranging from
0 to 9 (both inclusive).

1. If z0 lies on the curve zz   6  i  z   6  i z  28  0 then minimum of |z0 + 5 – i| is equal to _____

M
2. |z| = 2 and the maximum value of |z – 1| + |z – 2| + |z – 3| is M, then is equal to _____
4

3. Let k be the number of 8-digits number such that the product of any two consecutive digits is
n  n  1 a
prime and na   k , where a, n  N. If n is maximum, then is equal to _____
2 64

4. 6 boys, 5 girls and 3 teachers are arranged in a line such that boys are in ascending order, girls
are in descending order, no two teachers are together. The number of such arrangement is
 220k   11C5 , then k is equal to _____
5. If 3 positive real numbers a, b, c are in A.P. and abc = 4 then the value of [b], (where [.] denotes
greatest integer function) is _____

 2 sin2 
6. If cot( – ), 3 cot , cot( + ) are in A.P. and  is not an integral multiple of , then is
2 sin2 
equal to _____

7. If x, y, z  (0, 1] such that (log x)(log y) = log(xy) and 2(log x)(log z) = log(zx), then the value of
2x + 3y + 4z is equal to _____

8. |z2 – 9| + |z2| = 41, then the maximum value of |z| is _____

Space for rough work

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HALF COURSE TEST –VIII
(Paper-2)

Q. No. PHYSICS CHEMISTRY MATHEMATICCS

1. B B B
2. C C C
3. A D C
4. A A D
5. A, C A, B, C D
6. B, C A, C, D A, B, C
qualified in JEE (Advanced), 2014.

7. A, B, C A, B, C A, C, D
8. A, C A, B, C B, C
9. B, C A, B, C A, D
(A)  (t) (A)  (q, r, t) (A)  (p, q, r, s)
(B)  (q) (B)  (p, r) (B)  (q)
1. (C)  (q) (C)  (p, r, s, t) (C)  (r)
(D)  (p) (D)  (p, r) (D)  (p)

(A)  (p, q, r, s, t) (A)  (p, q) (A)  (r)

(B)  (p, q, r, s, t) (B)  (p, q, s) (B)  (s)
2 (C)  (p, q, r, s, t) (C)  (p, q, s) (C)  (q)
(D)  (r) (D)  (q, r, s, t) (D)  (p)

1. 3 3 8
2. 8 4 3
3. 6 3 8
4. 8 4 6
5. 5 6 1
6. 2 4 3
7. 8 2 9
8. 2 4 5

Physics PART – I

SECTION – A

K 8K
1. 0  , '  = 4 0
m m/2

GM e m  GM e m  GM e m
2. E1    
R Re  3R e
Re  e 
2
1 1 GM e GM e m
E 2  mv 02  m.  .
2 2 R 3R e
R e
2
 (C)

3. For vertical equilibrium T

uptrust
T = weight  upthrust

3  L
T  2LA   g  (2L  x)Ag  Ag  x   (2Lx)/2
O
4  2
where  is the density of water weight
for rotational equilibrium about O
  2L  x   L
(2L  x)Ag L    cos   Ag   x   L cos 
  2   2
 x=L

4. If v be the velocity at mean position in two cases then

1 1 m
mv 2  kA 12 and T1  T2  2
2 2 k
A1 = A 2
5. Buoyant forces on both the spheres will be same because volume of both the spheres is same
and weight of the hollow sphere is less. If ‘A’ and ‘a’ are the accelerations of solid and hollow
spheres respectively, then A > a.

6. P = (4r2)T4
dT  3T 4 1
R  
dt  s r

SECTION – C

1. At point P slope is tan 150

d 1
  tan 150  
dT 3
PM d PM
Since    
RT dT RT 2

V2
3.  4ms 2 Mg
R
V2 N
Mg – N = 10 – N  1. 4
R
 N = 6 Newtons

K M
4.  T  2
M K
M1M2 M2
T1  2 T2  2
K(M1  M2 ) K
M2
T2  2
K
2
M1  M2  T2  M  M2 1  M2
  9 1 
M1  T1  M1 M1
M2
 8
M1

5  = 120 
K = 2 / 3

V   180  36X
K
X=5

6. While floating net force = zero

Increase in buoyancy force = increase in weight
V
g  xg
2
V
 x
2

 15.92  7.96  6 7 m
7. tan     10  7.96  10 Newton
Mg  1 10 
2m N N
Y 3 2 7
 8  1011 2
3.14(10 m) 7.96  10 m m
8. 5
–1
5ms
10t Jeep  Vrel
10t

Vrel  5 1  4t 2
5 1 10
sin    gsin   aN 
2 2
5 1  4t 1  4t 1  4t 2
Vrel225 5
R (1  4t 2 )3 /2  (17)3/ 2  1  4t 2  17
aN 10 2
t = 2 seconds.

Chemistry PART – II

SECTION – A

3. (i) use PV = nRT

w
PV = RT
m
(ii) due to hydrogen bonding aNH3  aCO2
1
(iii) mean free path () 
2
3
(iv) Average K.E per molecule = kT.
2

5. (A) Aniline does not undergo Fridel Craft reaction due to lone pair on N-atom which directly
attacks on AlCl3.
(B) formation of carbocation at bridge head position is not favoured.
(C) CHCl3  alc. KOH  :CCl2 (dichlorocarbene)
Cl Cl
: CCl2 C Cl

Cl

(D) Three times protonation followed by ring expansion gives benzene.

6. In the option (b), the H atom is linked to an sp2 hybridised C atom.

8. r = K [A]m [B]n
on calculation, m = 2, n = 1 & K = 2.66  108 mol-2 l2 sec-1 at 300 K
so, K will be > 2.66  108 at 320 K.
Hence, (A, B, C)

SECTION – C

1. The PV diagram for the given changes is as:

(P1, 300 K) (P1, T2)

A B
P (in atm)

(P2, 300 K)
C

20 40 V1
V (in litres)
for isobaric process AB,

V1 V2
 (from Charle’s law)
T1 T2
20 40
 ,
300 T2
 T2 = 600 K
wt = wAB + wBC
Now, wAB = nRdT (Isobaric process)
= 2  2  (600 – 300) = 1200 cal.
wBC = for adiabatic process
w = n CV T
3
= 2   R  (600 – 300)
2
= 1800 Cal
 wT = 1200 – 1800 = 3000 Cal = 3 kcal.

2. H2 O
Mg2 C3   Mg(OH)2  H3C C C H  (i) Na
 H3C C C CH2 CH3
(ii) CH3 CH2  Cl

(A)

Ni/H2
H
Cl / h
C C C C C+ C C C C C 
monochlorination
 CH3 CH2 CH2 CH2 CH3
(Enantiomeric)
Cl Cl (B)
+
C C C C C Cl

3. 30 ml of N/50 HCl = 30 ml of N/50 Mg(HCO3)2

= 30 ml of N/50 CaCO3 = 100 ml of tap water
E N V
 mass of CaCO3 in 100 ml tap water =
1000
50  30
=  0.03 g
50  1000
 Hardness = 300 ppm
X=3
diglyme
4. NaBH4  I2 
solution
 B 2H6  H2  2NaI
 X
Cl2
B 2H6  2NH3  B2H6 .2NH3  

 B3N3H6   B3N3H4 Cl2
 adduct  Y  dichloro derivatives 
So, dichloro derivatives of (Y) = 4

Mathematics PART – III

SECTION – A

1. Let the equation of the line L be y – 2 = m(x – 8), m < 0

 2 
 P   8  , 0  , Q = (0, 2 – 8m)
 m 
2
 OP + OQ = 8   2  8m  18
m

2. Let centre C = (a, 0)

COP = 45º
OP = 2 2  CP A C B
 OC = 4 O
The point on the circle with greatest abscissa is B P
 = OB = 4  2 2 (2, –2)

b
3. y = mx – 2am – am3, y = mx + 2b +
m2
b
For common normal, 2b   2am  am3  0
m2
 am5 + 2am3 + 2bm2 + b = 0
 Maximum 5 normal

4. Equation of tangent y  mx  a 2m2  b 2

1
Equation of a perpendicular from centre (0, 0) on the tangent is y   x
m
x x2
 Required locus y   x  a2 2  b2
y y

5. Pair of tangents SS = T2

tan  
2 h2  ab
=
2 a2  1  
ab a2  2
tan  < 0
2 a2  1
 0
a2  2

 a   2,  1  1, 2  
6. 2s(2s – 2a) = xbc
4s  s  a  A
 x = 4cos2
bc 2
 x  (0, 4)
2 2
7. x + x + 1 = ax + 1 and –1  x + x + 1  1
a<1

8. cos4  – sin4  = cos2  – sin2  = cos 2

2 2
 (–2b) – 4b = (–4) – 4  2
 b = –1, 2

9. Put x = 1
2 12
 (2 + 3c + c ) = 0 (as the sum of coefficients vanishes)
2
 c + 3c + 2 = 0
 c = –1, –2

SECTION – B

1 
1. (A) Can take any value except  , 3 
3 
 A  B  C sec A  sec B  sec C  C
(C) Centroid of ABC is  ,  y = sec x
 3 3 
sec A  secB  sec C A BC
  sec   B
3  3 
A
sec A  sec B  sec C
 2  Least value of  sec A = 6
3
2sin 20º cos 20º
(D) 4
1
 sin20º cos 20º
2

2. R = 15
1
 18  24

r = 2 =6
s 18  24  30
2
Circum-centre = (9, 12), orthocentre = (0, 0), in-centre = (6, 6), centroid = (6, 8)

SECTION – C
1. x2 + y2 – 12x – 2y + 28 = 0
C = (6, 1), r = 3
 |z0 + 5 – i|min = 8

2. |z – 1| + |z – 2| + |z – 3|  |3z – 6| = 3|z – 2| = 3|2ei – 2| = 12

9
3. k=2
n  n  1
 na   29
2
10
 n(2a + n – 1) = 2
Which is only possible for n = 1
 a = 512

11!
4. 6 boys and 5 girls in
6! 5!
Teachers are at remaining places in 12P3 ways
11!
 12 P3 
5! 6!

5. b  ac
 b3  abc
 b3  4
 b  41/3
 [b] = 1

sin2
6. 6 cot  = cot( – ) + cot( + ) =
sin      sin     
3cos  2sin  cos 
 
sin  cot 2  cos 2
 4 sin2  = 6 sin2 
2 sin2 
 3
sin2 

7. x=y=z=1
 2x + 3y + 4z = 9

8. 2zz  2 | z  3 || z  3 | 82
 |z + 3| + |z – 3| = 10
 |z|max = 5

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PART TEST –III
ALL INDIA INTEGRATED TEST SERIES
Time Allotted: 3 Hours Maximum Marks: 432
 Pl ea s e r ea d t h e i n s t r u c t i o n s c a r ef u l l y. Yo u a r e a l l o t t ed 5 m i n ut es
s p ec i f i c a ll y f o r t h i s p u r p o s e.
 Yo u a r e n o t a l l o wed t o l ea v e t h e E xa m i n at i o n Ha l l b ef o r e t h e en d o f
t h e t es t .

B. Filling of OMR Sheet

C. Marking Scheme For All Three Parts.

(i) Section-A (01 to 03 and 10 to 12) contains 6 multiple choice questions which have only one
correct answer. Each question carries +8 marks for correct answer and – 2 mark for wrong
answer.

Section-A (04 to 09 and 13 to 30) contains 24 multiple choice questions which have only one
correct answer. Each question carries +4 marks for correct answer and – 1 mark for wrong
answer.

Name of the Candidate

Enrolment No.

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Useful Data

PHYSICS
Acceleration due to gravity g = 10 m/s2

Planck constant h = 6.6 1034 J-s

Charge of electron e = 1.6  1019 C

Mass of electron me = 9.1  1031 kg

Permittivity of free space 0 = 8.85  1012 C2/N-m2

Density of water water = 103 kg/m3

Atmospheric pressure Pa = 105 N/m2

Gas constant R = 8.314 J K1 mol1

CHEMISTRY

Gas Constant R = 8.314 J K1 mol1

Atomic No: H=1, He = 2, Li=3, Be=4, B=5, C=6, N=7, O=8,

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Physics PART – I

SECTION – A
Single Correct Choice Type

This section contains 30 multiple choice questions. Each question has four choices (A), (B), (C) and (D)
out of which ONLY ONE is correct.

1. ABCD is a square of side 4R and EFGH is a square loop of D

C
side 2R. The loop carries a current I. The magnetic field at the G F
centre O of the loop O
 0I μoI
(A)  (B)  E
2R 2πR H B
A
μ0I  0I
(C)  (D) 
2πR 2R

2. In the given circuit, if the voltmeter is ideal its reading is V1 volts. 4 4

If the voltmeter has a resistance of 100, the reading is V2 .
Then V1  V2 is equal to :-
10 volts V
(A) Zero (B) 0.3volts.
(C) 0.2volts (D) 0.1volts
O

3. The charge on the capacitors A and B in steady state is :- P C

R R
CE CE CE A C
(A) ,CE (B) , B
2 2 2 R
R
L
CE
(C) CE, (D) CE,CE R
2 E
S


4. In a region, uniform electric field exists as E  10iˆ  10 ˆj  N/ C. If the potential of origin is 0 volts,
 
the potential of point (10m, 10m, 10m) is
(A) 20 volts (B) – 200 volts
(C) 10 volts (D) – 10 volts

5. Find the current in resistor R at steady state R

2E E
(A) (B) zero C1
R
3E EC1 E C2
(C) (D)
R RC2 E

Space for rough work

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6. The charges on the plates of the capacitor in the steady state in the 10V 2
given circuit is
1F
(A) zero. (B) 5 C
(C) 10 C (D) none of these
5

10 

5V 1

7. An infinite number of charges of magnitudes q, q/2, q/4, q/8 and so on are placed along the
circumference of a circle of radius 1 unit with centre as origin. The potential at the origin will be
q q
(A) (B)
20 40
q
(C) (D) infinite
30

8. From a uniform solid sphere an eccentric cavity has been cut out as shown in P
the figure. C1 and C 2 represent their centres respectively. If the size of the C2

C
cavity is increased keeping the centre at C 2 , then the field strength at any 1

internal point P will

(A) be zero
(B) remain constant in magnitude but direction of field will change with the change in position
of P.
(C) remain constant in magnitude and direction for all positions of point P.
(D) gradually decrease in magnitude as the size of cavity increases.

9. A regular pentagon having four identical charges of equal magnitude placed B C



at the four corners A, B, C and D produce a field E  3iˆ  4 ˆj  2kˆ units at  O
the centre O of the pentagon. If all the four charges are removed from the A D
corners and reassembled at the fifth point P, find the field at O due to this
rearrangement. P

(A) 10iˆ  5 ˆj  9kˆ (B) 3iˆ  4ˆj  2kˆ

1m
10. Figure shows a meter bridge. If there is no current through galvanometer
1
then  1 is equal to
Y
(A) 200/3 cm (B) 100/3 cm G

(C) 50 cm (D) None of the above X X/2

Y/2

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11. A bulb is connected across a voltage source. A resistor R is connected R S

in parallel to the bulb through a switch S. If switch S is closed then
(A) Brightness of bulb decreases
(B) Brightness of bulb increases
(C) Brightness remains unchanged.
(D) None of the above

O O
V

12. Three similar plates are arranged ‘d’ meters apart as shown in the d d
figure. If area of each plate is A and a charge +Q is given to
middle plate then potential of middle plate is
Qd Qd
(A) (B)
A 0 2A 0
Qd 2Qd
(C) (D)
4A 0 A 0
13. A conducting shell having inner radius R1 and outer radius
R 2 contains a charge + q which is placed at a distance x from its +q

P
centre. Field at an exterior point P which is situated at a distance r R2
R1 r
from centre of shell  r  R2 
(A) depends on x
(B) depends on R1 and R 2
(C) depends only or r
(D) None.

14. The potential deference across the resistor as measured by a voltmetre, V

is 60  1.5V and the current through the resistor as measured by an

ammeter is  2  0.01 A. what is the maximum% error in the measured R
value of R?
(A) 3% (B) 1.5%
(C) 2.5% (D) None of these
A
V

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15. Find the value of r so that maximum power is generated in r

external circuit.
r r r
(A) r = 10  (B) r = 20 
r
(C) r = 30  (D) r = 40 
r r
r
r r
r

r
10V

40

16. Consider a parallel plate capacitor having an electric field E inside it as shown +
+
in the adjacent figure. A particle of mass m and charge q is hanging inside the +
+

capacitor through a light inextensible string of length . The time period of this 
+
+
pendulum is   +

  

(A) T = 2 (B) T = 2 2 2
g   qE m  qE 2qgE
g2  2 
m m

(C) T = 2 (D) none of the above
2 q2E2 2qgE
g  2 
m m

17. A cube made from wires of equal length is connected to a battery as

shown in the figure. The magnetic field at the centre of the cube is
L
12 0I 6 0 I
(A) (B)
2 L 2 L + 
6 I
(C) 0 (D) zero
L

18. Find the equivalent resistance across AB: A 2

(A) 1  (B) 2  2
2
(C) 3  (D) 4  2
B 2

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19. The resistance per unit length of the wires in segments AB, BC, B a
C
CD, and AD are the same. The magnetic field due to the current
at point O is a a
30I 30I a O
(A) (B) I A D
4 5a 5 a
0I
(C) (D) none of the above
2 5a

20. A charge Q1 is placed at O, inside a hollow conducting sphere having O1

inner and outer radii as 10 m and 11 m as shown. The force experienced
5m Q1 Q2
by Q2 at P is F1 and force experienced by Q2 when Q1 is placed at O1 is
O 12m P
F2. Then F1/F2 is equal to 10m
2 11m
 12 
(A) 1 (B)  
 13 
2
 13 
(C)   (D) none of the above
 12 

21. In the circuit shown in the figure k1 is open. The charge on R1=2 K1
capacitor C in steady state is q1. Now key is closed and at
steady state charge on C is q2. The ratio of charges q1/q2 is R2=3
(A) 5/3 (B) 3/5 E C
(C) 1 (D) 2/3

22. In the shown in the figure AC and BD are straight lines E

r
and CED and AFB are semicircular with radii r and 4r A C D B i
O
respectively. The entire setup is lying in the same plane. i
4r
If i is current entering at A what fraction of i will flow in the
ACEDB such that resultant magnetic field at O is zero.
F
(A) i/5 (B) 4i/5
(C) 3i/5 (D) no current through bigger semicircular part.
Space for rough work

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23. Consider the network shown in the figure. All resistances are g
equal to 2. What is the potential difference between points
e and h?
(A) zero (B) 5V h
(C) 2.5 V (D) 10 V

c f d
e
a b
10 V

24. What is the energy stored in the capacitor between terminals a and 10 V

b of the network shown in the figure? (Capacitance of each

capacitor C = 5F).
C C
(A) 1 J (B) 0.25 J
a
(C) zero. (D) 15.6 J C
b
C

25. Two capacitors having equal capacitance C, are having same charges q +q R
q each. They are connected across a resistor R as shown in the figure.
What is the total heat generated in the resistor upto the instant when C
the steady state has reached? +q q
(A) q2/2C (B) q2/C
C
(C) zero (D) none of the above.

26. In the circuit shown, switch is placed in position1, till the capacitor is E R1
charged to half of the maximum possible charge in this situation.
Now the switch S is placed in position 2. The maximum energy lost S 1
by the circuit after switch S is placed in position 2 is C 2
1 1
(A) CE2 (B) CE2 E R2
2 8
7 9
(C) CE2 (D) CE2
8 8
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27. The current in branch AB is 2 2

B
(A) infinite (B) 1.33A
(C) 2 A (D) 1.5 A 2
3

28. Calculate the magnetic field at point P due to three arcs as

r I
shown in the figure
45 45
0I  0I
(A) (B) r
2r 4r r
I
(C) 0 (D) zero
2r

29. A point charge +q is located at the origin and a point charge  2q II I x=a
III
is located at x = a (a > 0). In which of the following region or
regions might there exists a point where the electric potential (V) +q
2q
is zero?
(A)  < x < a (B) 0 < x < a
(C) a < x <  (D) V does not vanish in the region   < x < 

30. What is the potential difference between points a and b, in the a

steady state in the circuit given below? (E1 = 2.1V, E2 = 6.3 V, R1 R1
R1 = 1.7 , R2 = 3.5 , C = 2F) R2 E2
(A) 4.9 V (B) 2.8 V E1 R1
(C) 5.9 V (D) 4.2 V E2
R1 C
b

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Chemistry PART – II

SECTION – A

Straight Objective Type

This section contains 30 multiple choice questions numbered 1 to 30. Each question has 4 choices (A),
(B), (C) and (D), out of which ONLY ONE is correct.
1. Consider the reaction,
O17
16 H+
CH3C18OH + CH3CH2OH (A) and (B)
(A) and (B) respectively are
16
O 17 O
17
18 16 18
(A) CH3C OC2H5 + H2O (B) CH3COC2H5 + H 2 O
O17
16
18
(C) CH3COC2H5 + H2O (D) Both (A) and (B)

I /Δ
2. The product of the given reaction CH3COOAg 
2
 [X]
[X] is
(A) CH3I (B) CH3COOCH3
(C) CH3CH3 (D) CH3COOC2H5

3. Which halo derivative(s) on treatment with OH will give carboxylic acid?

O O
(A) CH3–CH–C–CH2–CH3 (B) Cl
Br
O
(C) CH3–C–CCl3 (D) All of these

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+
NNBF4

4. A
In the above process product A is
(A) Fluorobenzene (B) Benzene
(C) 1,4-difluorobenzene (D) 1,3-difluorobenzene

5. Treatment of NH3 with excess of ethyl chloride gives:

(A) diethylamine (B) ethane
(C) tetraethylammonium chloride (D) methylamine

6. The product of the given reaction,

CH3CH3
HNO2
CH3 – C – C – CH3 Product
OH NH2
is
CH3CH3 O CH3
(A) CH3 – C – C – CH3 (B) CH3– C – C – CH3
OH OH CH3
CH3
(C) CH3 – C – CH – CH3 (D) CH3 – C = C – CH3
OH CH3 CH3CH3

7. Which among the following amines will give carbylamine reaction?

(A) CH3–CH2–NH2 (B) CH3–NH–CH3
(C) (C6H5)3N (D) CH3–CH2–NH–OH

8. Silver benzoate reacts with bromine in acetone to form

Br COBr
(A) (B)

COOAg
(C) (D)
Br

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9. Which of the following compound is most rapidly hydrolyzed by SN 1 mechanism?

(A) C 6 H 5 Cl (B) Cl CH 2  CH  CH 2
(C) C 6 H 5 3 CCl (D) C 6 H 5 CH 2 Cl

Me Me Me
10. OH– H OH–
OH H Cl I
H OH ( Racemic mixture)
II
Et Et Et
B A
Steps I and II are:
(A) both SN1 (B) both SN2
(C) I SN1, II SN2 (D) I SN2, II SN1

 
O
11. An alkyl halide (X) having molecular formula C6H13Cl reacts with (CH3)3 CO K to form two
isomeric alkenes (Y) and (Z) of molecular formula (C6H12). Both the alkenes give
2,3Dimethyl butane on hydrogenation. The halide (X) is
CH3 CH3
(A) CH3 CH CH CH2 Cl (B) CH3 CH2 CH2 CH CH2 CH3
Cl
CH3 CH3 CH3
(C) CH C CH CH3 (D) CH3 C CH2 CH2 CH3
3

Cl Cl
12. The formation of cyanohydrin from a ketone is an example of:
(A) nucleophilic substitution. (B) electrophilic substitution.
(C) electrophilic addition. (D) nucleophilic addition.

13. Acetone is mixed with bleaching powder to give:

(A) chloroform (B) acetaldehyde
(C) ethanol (D) phosgene

14. Which of the following compounds is stable?

(A) CH3CH(OH)2 (B) (CH3)2CH(OH)2
(C) CCl3CH(OH)2 (D) none of these

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15. Which of the following does not respond to haloform reaction?

(A) CH 3  CH 2  OH (B) CCl 3 CHO

O
(C) CH 3  C  CH 2  COC 2 H 5 (D) CH 3  CH  CH 3

16. In the given reaction

O
D2O (P) will be
NaOD

O O
D D D
(A) (B)

O O
D D D D
(C) D (D) D D

17. A compound (X) C4H8O which gives a 2, 4-DNP derivative and a positive iodoform test is:
O
(A) CH3–CH2–C–CH3 (B) CH3–CH2–CH2–CHO
(C) CH3–CH–CHO (D) All of these
CH3
18. The repeat unit of polystyrene is
(A) —CH2—CH— (B) —CH2—CH—

C6H5 CN
(C) —CH2—CH— (D) —CCl2—CCl2—

Cl
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19. Which one of the following polymers may be classified as step growth polymer?
(A) Teflon (B) Polythene
(C) PVC (D) Nylon 66

20. Propyne and propene can be distinguished by

(A) conc. H2SO4 (B) Br2 inCCl4
(C) AgNO3 [Ammonical] (D) dil. KMnO4

21. Identify an aromatic compound A(C7H8O) which gives following tests with the given reagents.
Na metal
Positive
FeCl3 (neutral)
A(C7H8O) Negative
Lucas reagent
Positive
CH2–OH OCH3

(A) (B)

OH OH
CH3
(C) (D)

CH3
22. Predict the final product (S) formed in the following reaction sequence.
LiAlH4 PCl
KCN H O
   (P) 5  (Q)  (R) 3  (S)
N COOH
H
(A) (B)

CH2CN H2N(CH2)3CH2CH2COOH
N
H
(C) CN (D)

N CH2COOH
N COOH
H
H
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23. Carboxylic acids react with diazomethane to yield:

(A) Amines (B) Alcohols
(C) Esters (D) Amides

24. Which carbonyl group of the given compound is most reactive for nucleophilic addition reaction?
O
1

3 2
O
O
(A) 1 (B) 2
(C) 3 (D) All have equal reactivity

25. Identify the products (B) and (C) formed in the following set of reactions.
O

O
(A)
O O
(A) (B): OH O (B) (B): O OH
(C): (C):
OH OH
OH OH OH OH
O
OH O
(C) Both (B) and (C): (D) Both (B) and (C):
OH
OH OH
26. Identify the unknown products (A) and (B) in the given reaction:
H / 
Glucose + 3 C6H5NH–NH2  Osazone + (A) + (B)
(A) C6H5NH2 and NH3 (B) C6H5NH2 and NH2OH
(C) C6H5NH—NHOH and NH3 (D) NH2OH and H2O

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27. Buna–S rubber is a polymer of

(A) 1, 3-butadiene and styrene (B) vinyl acetate
(C) Acrylonitrile (D) none of these

28. Identify the final product (Z) in the following sequence of reactions:
CH3I AgOH 
CH3NH2 excess
 X   Y   Z  CH3 OH
(A) CH3  CH2  CH2  NH2 (B) (CH3)3N
(C) (CH3)2NH (D) (CH3)4N+OH–

29. The order of reactivity of the following alcohols towards conc. HCl is
CH3 H3C CH3
F CH3 Ph OH
F OH HO OH
I II III IV
(A) I > II > III > IV (B) I > III > II > IV
(C) IV > III > II > I (D) IV > III > I > II

30. Which of the following will be oxidized by HIO4?

(1) R R (2) R R

O O O OH
(3) R R (4) R R

OH OH HO OH
Select the correct answer using the codes below:
(A) 1, 2 and 3 (B) 1, 3 and 4
(C) 1, 2 and 4 (D) 2, 3 and 4

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Mathematics PART – III

SECTION – A

Straight Objective Type

This section contains 30 multiple choice questions numbered 1 to 30. Each question has 4 choices (A),
(B), (C) and (D), out of which ONLY ONE is correct.

1  7 cos2 x f x
1.  sin7 x cos2 x dx   sin x 7  c , then f(x) is equal to
(A) sin x (B) cos x
(C) tan x (D) cot x

1
2.  tan  tan x d is equal to
–1
(A) tan  sec x + c (B) tan–1  n (sec x) + c
 1   1 
 2 

(C) tan x  tan1   n 1   2   c   2
 
(D) tan   x tan1 x  n 1  x 2   c


1 1
et t
3. If k  
1 t
dt , then  e n 1  t  dt is equal to
0 0
(A) k (B) 2k
(C) en2  k (D) none of these

n I3
4. If In   e x  sin x  dx , then is equal to
I1
3 1
(A) (B)
5 5
2
(C) 1 (D)
5

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2k 100
5. If k  N and Ik   sin x  sin x  dx , (where [.] denotes the greatest integer function) then  Ik
2k k 1
equal to
(A) –10100 (B) –40400
(C) –20200 (D) none of these

6. Consider a function ‘f’ defined on the set of all non-negative integers such that f(0) = 1, f(1) = 0
and f(n) + f(n – 1) = nf(n – 1) + (n – 1)f(n – 2) for n  2, then f(5) is equal to
(A) 40 (B) 44
(C) 45 (D) 60
3 2
7. Let R be the set of all real numbers the function f: R  R defined by f(x) = x – 3x + 6x – 5 is
(A) one-to-one but not onto (B) onto but not one-to-one
(C) neither one-to-one nor onto (D) one-to-one and onto

 1
8. A function f: R  R is differentiable and satisfies the f    0 for all integers n  1, then
n
(A) f(x) = 0 for all n  (0, 1] (B) f(0) = 0 = f (0)
(C) f(0) = 0 but f (0) need not be equal to (D) |f(x)|  1 for all x  [0, 1]

9. The maximum value of the sum of the intercepts made by any tangent to the curve (a sin2 , 2a
sin ) with the axes is
a
(A) 2a (B)
4
a
(C) (D) a
2

x x
10. f  x  and g  x   where 0 < x  1, then in this interval
sin x tan x
(A) f(x) and g(x) both are increasing (B) f(x) is decreasing and g(x) is increasing
(C) f(x) is increasing and g(x) is decreasing (D) none of these

11.
1 
log x  1  x 2  f  x  dx 
1 
log x  1  x 2  f   x  dx is equal to
 
1 x 
 log x  1  x 2  1 x 
 log x  1  x 2 
(A) 0 (B) 2
1 
log x  1  x 2  f  x   f  x  dx
0 
x  log x  1  x 2 
(C) 2f(x) (D) none of these

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 
x 1 2x 
12. e
  tan x  2 
dx is equal to

1  x2 
  
 1   1 
(A) e x  tan1 x   c (B) e x  tan1 x   c
 1  x2   1 x2 
 1   2 
(C) e x  cot 1 x   c (D) e x  tan1 x   c
 1  x2   1 x2 

1/10
 x  
2
13.  x x 8  2x 9 dx is
5 2 11/10 5
(A) x  2x  c  (B)  x  111/10  c
11 11
6 11/10 11 2 10/10
(C)  x  1
7
c (D)
5

x  2x  c

1
 x2
14. If  e sin  x  k  dx  0 (for some ,   R),   0, then the value of k can belong to
0

  5   
(A)  ,  (B)  , 
 3 12  3 2
 3 5    
(C)  ,  (D)   ,  
4 6   2 3

 x /3
8x
15.  dx (where [.] denotes the greatest integer function) is equal to
2 
3x
0

(A)
x (B)
 x
n 2 n 4
2x x
(C) (D)
n 2 n 8

x 1
16. Domain of 
x 2  4x  3  1 log5   
5 x
  
8x  2x 2  6  1  0 are

(A) (–, 1]  [3, ) (B) [1, 3]

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17. Let f: R  R and g: R  R be two one-one and onto function such that they are the mirror
images of each other about the line y = a. If h(x) = f(x) + g(x), then h(x) is
(A) bijective function (B) injective but n of surjective
(C) surjective but not injective (D) neither injective nor surjective

x 1  1
18. The set of value of  so that f  x   2
does not take any value in the interval  1,   is
  x 1  3 
 1
(A)  ,   (B) [2, )
 4
 1   1
(C)   , 2  (D)  ,    [2,  )
 4   4

19. Period of the function f(x) = cos 2{2x} + sin {2x} is ((where {.} denotes the fractional part of
function)

(A) 1 (B)
2
(C) 1/2 (D) 

1  9 
20. If f  n  2    f  n  1   , n  N and f(x) > 0 for all n  N then nlim f  n  is equal to
2  f  n   

(A) 3 (B) –3
1
(C) (D) none of these
2

 x2 
21. The value of lim   (where [.] denotes the greatest integer function) is
x 0 sin x tan x
 
(A) 0 (B) 1
(C) limit does not exist (D) none of these

22. The left hand derivative of f(x) = [x] sin (x), at x = k, where k is an integer is (where [.] denotes
the greatest integer function)
(A) (–1)k (k – 1) (B) (–1)k – 1 (k – 1)
k
(C) (–1) k (D) (–1)k – 1 k

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 2 
 x sin   , 1  x  1  x  0 
23. Let f  x    x then
x x , x  1 or x  1

(A) f(x) is an odd function (B) f(x) is an even function
(C) f(x) is neither odd nor even (D) none of these

 1    1   x 
24. Given the function f(x) such that 2f(x) + xf   – 2f  2 sin    x      4 cos2  x cos
x    
4   2 x

 1  1
(A) f  2   f    0 (B) f(1) = –1, but f(2), f   can’t be determined
2 2
 1
(C) f  2   f 1  f   (D) f(2) + f(1) = 1
2

1 1
25. Let f  x   and g  x   then
x x
(A) f{g(x)} and g{f(x)} have different domain (B) f{g(x)} and g{f(x)} have same domain
(C) g{f(x)} is a bijective mapping (D) f{g(x)} is neither odd nor even

26. The period of the function f(x) = sin 3x cos [3x] – cos 3x sin [3x] (where [.] denotes the greatest
integer function) is
1 1
(A) (B)
2 3
(C) 1 (D) 3

27. The integer n for which lim 


  cos x  1 cos x  e x  
 is a finite non-zero number is
x 0  xn 
 
(A) 1 (B) 2
(C) 3 (D) 4

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28. If f(x) = x3 – x2 + 100x + 2015, then

 1   1 
(A) f    f  (B) f(2015) > f(2016)
 2015   2016 
 1   1 
(C) f(2016) = f(2015) (D) f    f 
 2015   2016 

29. The lines tangent to the curve y3 – x2y + 5y – 2x = 0 and x4 – x3y2 + 5x + 2y = 0 at the origin
intersect at an angle . Then  is equal to
 
(A) (B)
6 4
 
(C) (D)
3 2
3 2
30. If the function f(x) = ax + bx + 11x – 6 satisfies conditions of Rolle’s theorem in [1, 3] for
1
x  2 , then value of a and b are respectively
3
(A) –3, 2 (B) 2, –4
(C) 1, –6 (D) none of these

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ALL INDIA INTEGRATED TEST SERIES

ANSWERS, HINTS & SOLUTIONS
PART TEST–III
(Main)

Q. No. PHYSICS CHEMISTRY MATHEMATICS

1. D C C
2. D B C
3. C D C
4. B A A
5. B C C
qualified in JEE (Advanced), 2014.

6. A B B
7. A A D
8. C A B
9. C C A
10. A C C
11. C C A
12. B D A
13. C A A
14. A C C
15. C C D
16. D D D
17. D A D
18. A A A
19. A D C
20. A C A
21. A A A
22. A D A
23. B C A
24. C B C
25. B A B
26. D A B
27. D A C
28. B B B
29. A C D
30. D C C

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Physics PART – I

SECTION – A

μ0I 1 1  μ0I    1  μ0I

1. 4 . sin 450  sin 450      2  
4π  R 2R  π    2R  2πR

10
2. V1   4 volts = 5 volt
44
100  4
10 
V2  100  4
100  4
4
100  4
= 4.9 volts  V1  V2  0.1 volt

3. In steady state :-
E E
VP  VQ  R 
2R 2
E E
VQ  VS  R 
2R 2
VS  VR
VP  VR  E
 Q A  CE
CE
QB 
2

4. VP  V0   10  10  10  10 
= - 200 volts.

5.
R C1+ C2
R

E i
i

10 10V 2
6. I1 +2I  10 = 0  5 + I  I1 + 2I  10 = 0
3 I 1F
After solving we get, I1 = 0, Q = 0
I1
10/3 
I  I1

5V 1

1  q q q  q  1  q
7. V=  q     ... =  
40  1  2 4 8  40 1  1 2   20

8. Field at any internal point of the cavity.

 4 
E  G  C2 C1
3
 
Since C1 and C 2 are fixed (cavity being concentric with C 2 all the time), field remains constant
both in magnitude and direction.

9. If the fifth point P has identical charges the field at the centre of symmetry O would be zero.
  
Hence the field due to one single such charge at P would produce a field E , such that ( E + E ) =
0
 
 E =  E = ( 3iˆ  4 ˆj  2kˆ )

this field will be four times E  due to this rearrangement, hence the required field at O is

4 E = ( 12iˆ  16ˆj  8kˆ ) units

10. If IG  0 then it becomes a balanced whetstone bridge

2 2 200
1   1m   100  cm
3 3 3

 Q d d
12. E1   + +
0 2A 0 + +
Q/2
Qd + +
V  E1d 
2A 0 + +
+ +

13. Apply gaus theorem

qin q 1

E 4r 2   0
E 
4  0 r 2
 E  2  field depends only on r.
r

R V I
14.  
R V I
R V I
  100   100   100
R V I
 3%
17. Magnetic field is zero at the centre due to symmetrical current distribution.

18. A
2 2
2

2 2
B

2  0 I/ 4   a
2   0 I / 4   2 2 
19. B at O  0     
4  a / 2   5 4  a  5 5 a a
20I 0I 30I 
  
4 5a 4a 5 4 5a

20. Due to electrostatic shielding, the ratio will be 1.

21. When key is open

q1 = CE
when key is closed
 E 
q2  C  R2 
 R1  R2 
q  CE  R2  R1  5
 1   R1  R2    
q2  CER2  R2 3

22. If I1 is current in ACDB then

 0 i1  0 i2
.  .
4 r 4 4r
i1 1

i2 4
1 i
i1  i 
5 5

23. From symmetry the currents in segment gh is zero. Thus, the equivalent resistance R is given by
1 1 1 1 1 3 2
     R  
R 4 4 2 2 2 3
10
Current in eh =  2.5A
4
 Potential difference across eh = 5V

24. It is a balanced Wheatstone bridge for capacitor. Consequently the potentials of points a and b
are equal. Thus energy stored in the capacitor is zero.

25. Let after steady state the charges on the capacitor be q1 and q2
q1 = CV, q2 = CV where v is the potential difference between the plates.
Also from charge conservation
q1 + q2 = 0
 V=0
1 1
Thus, final energy stored in the capacitors = CV2 + CV2 = 0
2 2
q2 q2 q2
Initial energy stored =  
2C 2C C
Thus, heat developed in the resistor = q2/C

26. Initial capacitor is charged to Q/2

2
1Q 1 1 Q2 1
Energy stored in capacitor =    =  CE2 = Ui
2 2  C 8 C 8
1
Finally energy stored in the capacitor, Uf = CE2
2

3  3
Work done by the battery =  CE   E  CE 2
 2  2
3 3 1 1 
 Heat energy lost = CE2  (Uf  Ui) = CE2   CE2  CE2 
2 2 2 8 
 3 1 1 9
= CE 2      CE 2
2 2 8 8

27. The equivalent circuit is P

2 2
Q 0.5 A 2
P I2 2 Q
Current flowing through 3 resistance 2
= 3/3 = 1 Amp 3 1A 0.5 A 2
B
Current flowing through P to Q is also A I1 3
1 amp. 1A
From the figure current flowing through
3V 3V
AB is 1 + 0.5 = 1.5 Amp

28. By using the relation

 0 i
B where  is in radian
4 r
Here  = 180 i.e. 
0I
So, Bp =
4r

29. Let x = a potential is zero (region II)

q 2q
k k
a  a  a
 a = a
So, in the region II,
Potential is zero and can be proved.
A solution can also found in region I (x = a/3).

30. In the steady state i i

a
E2  iR1 iR1 E1 iR1iR1 = 0
R1 R1
E2 E1 = 4iR1 E2
R2
E 2  E2 2.1 i
I= = (6.3  2.1)/(41.7) = E1 R1 i
4R1 3.4 E2
R1 C
Solving, we get, Va  Vb =  iR1 + E2  iR1 b
2.1 i
= 6.3  21.7 = 6.3  2.1 = 4.2 V
3.4

Chemistry PART – II

SECTION – A

+ 17 ..
17 16.. H 17
OH :O OH :OH
1. 18 CH2CH3 18
+
18
CH3C OH CH3C OH
.. CH3C OH
..
+
H OC2H5 OC2H5 H
16
+
(i) H2O18 (ii) H

17
O
16
CH3C OC2H5

O
O O C
3. OH– OH
CH3–CH–C–CH2–CH3 –H2O CH3–CH–C–CH–CH3 CH3–CH–CH–CH3
Br Br
O
C–OH COOH COOH
H2O
CH3–CH–CH–CH3 CH3–CH–CH–CH3 CH3–CH–CH2–CH3

4. This is Baltz Schiemann reaction.

C2H5
+ –
5 4C2H5–Cl + NH3  C2H5–N–C2H5 Cl
C2H5

CH3CH3 CH3CH3 CH3

HNO2 
6. CH3 – C – C – CH3 CH3 – C – C – CH3 CH3 – C – C – CH3

OH NH2 OH OH CH3
–H+
O CH3
CH3 – C – C – CH3
CH3

7. Carbyl amine test is responded by only 1° aliphatic/aromatic amines.

8. This is Hundsdicker reaction.

O O
Ph C O Ag Ph C O + AgBr Ph + CO2
Br Ph–Br

9. (C6H5)3CCl is hydrolysed most rapidly by SN1 because (C6H5)3C+ is most stable.

10. SN1 reaction is accompanied by racemisation while SN2 reaction is accompanied by inversion of
configuration.

CH3 CH3
CH3 CH3
CH2 C CH CH3
11. CH3 C CH CH3 (major) CH3 CH3
H2/Ni
CH3 CH3 CH3CHCHCH3
Cl
2, 3 Dimethyl Butane
CH3 C CH CH3
(minor)

12. Nucleophilic addition.

13. Chloroform is formed when acetone is mixed with bleaching powder (CaOCl2).

14. Due to intramolecular H–bonding.

15. Because –CH2– hydrogen is more acidic than –CH3 hydrogen.

O D
O O O O
DO– D2O
D
16. –HOD –OD–

These steps are continued unless all the four -H’s are replaced by D.

O O
17. CH3–CH2–C–CH3 contain –C– and keto methyl group both.

18. In polystyrene the monomeric unit is styrene which is having the formula of H2C=CH—C6H5.

19. Step growth polymers (or condensation polymers) are prepared by reaction between two
functional groups and thereby eliminating small molecules during polymerization. Nylon-66 is an
example of step growth polymer.

20. Test of terminal alkynes.

21. Compound 'A' give positive test with Na metal. That means it contains polar Hydrogen i.e.,
alcoholic group. It gives negative test with neutral FeCl3 implies it does not contain of a phenolic
group. It gives positive Lucas test means it is an alcohol.
22
LiAlH4 PCl5 KCN

N COOH N CH2OH N CH2Cl N CH2 – CN

| H H H
H H3O+

N CH2COOH
H

23. RCOOH + CH2N2  RCOOCH3 + N2 ; methyl esters are formed.

24. ‘2’ position is attached with two electron withdrawing groups.

25. LiAlH4 can reduce both ketone and ester.

27. n CH2=CH–CH=CH2 + n C6H5–CH=CH2 (styrene)

(1,3-butadiene)
 [CH2–CH=CH–CH2–CH(C6H5)–CH2]n
(Buna–S)

 
CH3I (excess) AgOH
28. CH3NH2  (CH3 )3 NI  (CH3 )4 N OH 

(CH3 )3 N  CH3 OH

30. R R
R OH HO R
HIO4
(1)  
O O O O
R R
R OH
HIO4
(2)  
O R
O OH O
HIO4
(4) R  CHOH  CHOH  R   R  CH  O  R  CH  O

Mathematics PART – III

SECTION – A

1  7 cos2 x sec 2 x 7 tan x

1.  sin7 x cos2 x dx   sin7 x dx   sin7 x  sin7 x  c S
    1 
2. tan x  tan1  d  tan x  tan1   
 1   2
d  = tan x  tan1   n 1   2   c
  2 
 
1 1 1
et 1 et
3. k dt =  e t n 1  t  dt =  n 1  t  e t    dt = e n2  k
0
1 t 0
0
0
1 t

  
n n n 1
4. In   e x  sin x  dx  e x  sin x   n  sin x  cos x ex dx
0
0 0

n  n  1 In  2 In n  n  1
In   
1  n2 In  2 1  n2
2k 2k
5.  sin x  sin x  dx   sin x  sin x     sin x   dx
2k 0


= 2k  sin x dx   2k   cos x  0  4k
0
Ik = –4k
100
 Ik  20200
k 1

6. f(n) – nf(n – 1) = (–1)[f(n – 1) – (n – 1) f(n – 2)] = (–1)n – 1 [f(1) – f(0)] = (–1)n

n
f  n f  n  1  1
   and this gives
n!  n  1! n!
n
 1 1  1 
f  n   n! 1    ..... 
 1! 2 ! n! 

1
8. xn  n  N  0 as n   by the definition of continuity f(xn)  f(0) as n  . But f(xn) = 0 for
n
all x  N, so f(0) = 0
f  x   f 0 f  xn   f  0 
Also lim  lim  f is derivable at x = 0
x 0 x0 x 0 xn  0

y  2a sin  1
9. Equation of tangent 2

x  a sin  sin 
x y
 2
 1
a sin  a sin 

Sum of intercepts = a sin2   sin  = 2a when sin  = 1 
sin x  x cos x
10. f 'x 
sin2 x

Let h(x) = sin x – x cos x

h(x) = x sin x > 0  0 < x  1
h(x) is increasing
h(0) < h(x)  sin x – x cos x > 0 for 0 < x  1
f (x) > 0
Similarly g(x) is decreasing
 
x 1 1 1 2x   1 
12.  e  tan x  1  x2  1  x 2  2 
dx  e x  tan1 x  c
1  x2 
 
1  x2 
  

1/10 1/10 5 2 11/10

13.  2
8
 x  x x  2x
9
 dx =   x  1 x 2  2x  
dx =
11

x  2x  c
14. Value of definite integral can be zero when either the lines coincide or f becomes identically zero
for given interval or f(x) = 0 has some root in the given interval.
Here sin(x + k) = 0 in n  [0, 1]
17. Let for particular x0
g(x0) = a + k then g(x) = a – k
 h(x0) = 2a
 h(x) = 2a  x  R
 h(x) is many one and into

x 1 1 
18. y 2
; put y = –t  t   , 1
x    1 3 
 1 
Hence, we get    ,  
 4
20. As n  
lim f  n   lim f  n  2   lim f  n  1  k (say)
n  n  n 

1  9 
lim f  n  2   lim  f  n  1  
n  n  2
 f  n  
1 9
k  k    k = 3
2 k
21. f(x) = x2 – sin x tan x
f (x) = 2x – sin x (sec2 x + 1)
 f(x) < 0  f(x) is decreasing
f(x) < f(0) as x > 0
 x2 – sin x cos x < 0
1
24. Put separately x  , x = 1 and x = 2 we get
2
 1  1
(A) f  2   f    1 (C) f(2) + f(1) = f   (D) f(2) + f(1) = 0
2  2

28. f(x) = 3x2 – 2x + 100 > 0

30. f(1) = f(3) ….. (1)

f = 3ax2 + 3bx + 11 = 0 ….. (2)
Solving we get a = 1 and b = –6

Time Allotted: 3 Hours Maximum Marks: 180

 Pl ea s e r ea d t h e i n s t r u c t i o n s c a r ef u ll y . Yo u a r e a l l o t t ed 5 m i n u t es
s p ec i f i c a ll y f o r t h i s p u r p o s e.
 Yo u a r e n o t a ll o wed t o l ea v e t h e E xa m i n at i o n Ha ll b ef o r e t h e en d o f
t h e t es t .

B. Filling of OMR Sheet

C. Marking Scheme For All Three Parts.

(i) Section-A (01 to 10) contains 10 multiple choice questions which have only one correct answer.
Each question carries +2 marks for correct answer. There is no negative marking.

Section-A (11 to 15) contains 5 multiple choice questions which have more than one correct
answer. Each question carries +4 marks for correct answer and – 1 mark for wrong answer.

(ii) Section-C (01 to 05) contains 5 Numerical based questions with answers as numerical value
and each question carries +4 marks for correct answer and – 1 mark for wrong answer.

Name of the Candidate

Enrolment No.

Useful Data

PHYSICS
Acceleration due to gravity g = 10 m/s2

Planck constant h = 6.6 1034 J-s

Charge of electron e = 1.6  1019 C

Mass of electron me = 9.1  1031 kg

2 2
Permittivity of free space 0 = 8.85  1012 C /N-m

Density of water water = 103 kg/m3

Atmospheric pressure Pa = 105 N/m2

Gas constant R = 8.314 J K1 mol1

CHEMISTRY

Gas Constant R = 8.314 J K1 mol1

Atomic No: H=1, He = 2, Li=3, Be=4, B=5, C=6, N=7, O=8,

Physics PART – I

SECTION – A
Single Correct Choice Type

This section contains 10 multiple choice questions. Each question has four choices (A), (B), (C) and (D)
out of which ONLY ONE is correct.

1. Three concentric charged metallic spherical shells A, B and C have radii a, b

and c; charge densities ,  and ; and potentials VA, VB and VC
respectively. Then which of the following relations is correct? b
c
(a  b  c) [(a 2 / b)  b  c] A
a
(A) VA  (B) VB  C B
0 0
[(a2  b2 ) / c  c]
(C) VC  (D) VA  VB  VC  (a  b  c)
0 0

2. Four identical charges are placed at the four vertices of a square lying in YZ plane. A fifth charge
is moved along X axis. The variation of potential energy (U) along X axis is correctly represented
by
(A) (B)
U U

x O +x x O +x
(C) (D)
U U

x O +x x O +x

3. Two resistances 300 ohm and 400 ohm are connected in series with a 60 volt supply. A voltmeter
connected across 400 ohm reads 30V. The same voltmeter when connected across the other
resistance will read.
(A) 22.5 V (B) 50 V
(C) 30 V (D) 10 V
Space for rough work

4. The equivalent capacitance between P and Q in the figure is (area of A

each plate is A and separation between two consecutive plates is d) d P
A 3A0
(A) 0 (B)
2d 2d
2A0 5 A 0 d Q
(C) (D)
d 3 d

5. Two long parallel straight conductors carry currents i1 and i2 (i1 > i2). When the current are in the
same direction, the magnetic field at a point midway between the wires is 20 T. If the direction of
i2 is reversed, the field becomes 50 T. the ratio of the currents i1/i2 is
(A) 5/2 (B) 7/3
(C) 4/3 (D) 5/3

6. A uniform constant magnetic field B is directed at an angle of 45 y
to the axis in the xy –plane. PQRS is a rigid, square wire frame S R
carrying a steady current I0, with its centre at the origin O. At time
t = 0, the frame is at rest in the position as shown in figure, with its
sides parallel to the x and y axes. Each side of the frame is of
mass M and length L. The torque  about O acting on the frame O x
due to the magnetic field will be?
BI0L2 ˆ ˆ BI0 L2 ˆ ˆ
(A)  j  i  (B)  j  i  P Q
2   2  

(C) BI0L2   ˆj  ˆi  (D) BI0L2  ˆj  ˆi 

7. A coil is in the form of equilateral triangle of side ‘a’. The current in the coil is I. The magnetic field
at a distance ‘a’ equidistant’ from all the vertices will be
0I 30I
(A) (B)
2 3a 2a
0I 0I
(C) (D)
3a 4 3a

Space for rough work

8. Two capacitor each of capacitance C = 1 F each charged to potential 100 V

100 V and 50 V respectively and connected across a uncharged +q1
C
capacitor ‘C’ = 1F as shown in the figure if switch ‘S’ is closed then
charge in the uncharged capacitor at equilibrium will be
50 50
(A) C (B) 50 C C C V
3 +q2
100
(C) 25 C (D) C
3

 a a   a a 
9. Three point charges +q, 2q and +q are placed at points  0, ,  ; (0,0,0) and  0, , 
 2 2  2 2
respectively. Find the net electric dipole moment vector of this charge assembly.
(A) 2qa(ˆj) (B) 2qa( k)ˆ
(C) 2qa( ˆj) (D) 2qa(  ˆj)

10. if resistance of each resistor is R, then effective resistance

between points A and B of the shown network is
8R 6R
(A) (B) A B
11 11
6R 2R
(C) (D)
5 3

Multiple Correct Answer(s) Type

This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D)
out of which ONE or MORE are correct.

11 The diagram show is a modified meter bridge, which is R R1 R2

used for measuring two unknown resistances R1 and R 2

at the same time. When only the first galvanometer is
used, balance point is found at point C. Now the first G1 G2
galvanometer is removed and the second galvanometer is
used, which gives balance point D. Using the details given A 25cm
C D 100
3
cm
B

in the diagram, find out the value of R1 and R2

(A) R1  5 R / 3 (B) R 2  4 R / 3
(C) R1  4 R / 3 (D) R 2  5 R / 3

Space for rough work

12. 10C of charge is given to a conducting spherical shell and another– 3C point charge is placed
inside the shell. For this arrangement mark out the correct statement(s).
(A) The charge on the inner surface of the shell will be +3C and it can be distributed uniformly or
non uniformly
(B) The charge on the inner surface of the shell will be +3C and its distribution would be uniform
(C) The net charge on outer surface of the shell will be +7 C and its distribution can be uniform or
non uniform
(D) The net charge on outer surface of the shell will be +7C and its distribution would be uniform

13. A parallel plate capacitor of plate area A and plate separation d is charged to a potential
difference V and then the battery is disconnected. A slab of dielectric constant K is then inserted
between the plates of the capacitor so as to fill the space between the plates. If Q, E and W
denote the magnitude of charge, the electric field between the plates (after the slab is inserted)
and magnitude of the work done on the system in question during the process of inserting the
slab, then :
 AV  KAV
(A) Q  0 (B) Q  0
d d
V 0 AV 2  1
(C) E  (D) W  1  
Kd 2d  K 

 
14. A proton is fired form origin with velocity v  v 0 ˆj  v 0 kˆ in a uniform magnetic field B  B 0 ˆj . In the
subsequent motion of the proton
(A) its z coordinate can never be negative
(B) its x coordinate can never be positive
(C) its x and z coordinates cannot be zero at the same time
(D) its y coordinate will be proportional to its time of flight.

15. Two bulbs 25W, 100V (upper bulb in figure) and 100W, 200V A 100 B
(lower bulb in figure) are connected in the circuit as shown in
figure. Choose the correct answer (s)
(A) Heat lost per second in the circuit will be 80J 600
(B) Ratio of heat produced per second in bulbs will be 1: 1 C D
(C) Ratio of heat produced in branch AB to branch CD will be 1:2
(D) Current drawn from the cell is 0.4 A (500/3) 200V

Space for rough work

SECTION – C
(Integer Answer Type)
This section contains 5 questions. The answer to each of the questions is a single digit integer, ranging
from 0 to 9. The bubble corresponding to the correct is to be darkened in the ORS.

1. In the network shown in the figure, the potential difference 12V

F A
(in Volt) across points A and B is

2.0 6.0 4.0

C G

4.0 6.0 2.0

B
12V

2. Two circular coils of radii 5cm and 10 cm carry equal current of 2A in opposite sense . The coils
have 50 and 100 turns respectively and are placed in such a way that they lie in same plane and
their centres coincide. Magnitude of magnetic field(in Tesla) at the common centre of coils is

3. A circuit is connected as shown in the figure 3F 6F

X
with the switch S open. When the switch is
closed, if total amount of charge that flows
Q S
from Y to X is Q(in C), then 
3
Y
3 6

4. Plate A of a parallel plate air filled capacitor is connected to a A B

spring having force constant k and plate B is fixed. They rest + -
on a frictionless table top as shown in the figure. if a charge +q + -
is placed on plate A and a charge –q on plate B, by how much
does the spring expand in equilibrium (q2 = 6kA0)

5. A uniform magnetic field with a slit system as shown in the figure

is to be used as a momentum filter for high energy charged
particles (enter and exit perpendicular to PQ) . With a field of B
tesla it is found that the filter transmits  particle each of energy
2.2 MeV. The magnetic field is increased to 2.13B tesla and
P Q
deuteron ions are passed into the filter. What is the approximate
Source Detector
energy (In MeV) of each deuteron ionstransmitted by the filter?
Space for rough work

Chemistry PART - II

SECTION – A
Single Correct Choice Type

This section contains 10 multiple choice questions. Each question has four choices (A), (B), (C) and (D)
out of which ONLY ONE is correct.

1. In alkaline medium alanine exits predominantly as

(A) H3C CH2 CH NH2 (B) H3C CH CH2 NH2

COO COO
(C) H3C CH NH2 (D)
H3C CH2 CH NH3
COO
COO

2. The Cannizzaro reaction is not given by:

(A) trimethyl acetaldehyde (B) benzaldehyde
(C) acetaldehyde (D) formaldehyde

3. NH2


NaNO2 HCl
 
05 C
 A 
CuCN
 B  
H2 O/H
C

(C) CN (D) CH3

Space for rough work

4. O
H3C C NH2  
NaOH/Br2

(A) (B)
H3C CH2 NH2 H3C CH3

NH2
(C) (D)
H3C H3C NH2

NH2

5.
Cl
 
NaCN
DMF

(A) I (B) I

CN
CN
(C) I (D) I
CN

Space for rough work

6. COOH

  An alcohol
NaBH4


O
An alcohol is
(A) (B)

CH2OH CH2OH
O OH
(C) (D)
CH2OH
COOH
OH O

7.
 CO 
AlCl3
HCl


(A) CH2OH (B) CO

(C) (D) COOH

CHO

Space for rough work

8. H2C CH CH CH2 CH2 OH  

2
 A,
MnO

OH
A is
(A) H2C CH C CH2 CH2 OH (B) H2C CH CH CH2 CHO
O OH
(C) H2C CH C CH2 CHO (D) O
O H2C CH C CH2 C OH
O

9. What is the final product of the following reaction?

OCH3
i H3O
ii


OCH3

COOCH3
(A) (B) OCH3
O
OCH3

COOH COOH
(C) (D)
O O

COOCH3

Space for rough work

10. What is the final product of the following reaction?

COOH

1. NaOBr
 

?
2. H

CONH2
(A) COOH (B)

NH2
(C) O (D)

NH
N
H

Multiple Correct Answer(s) Type

This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D)
out of which ONE or MORE are correct.

11. H3C CH CHO 

HCN
 A  
95% H2 SO 4

 B pH  7

CH3
Which of these substances appear in reaction sequence?
(A) CH3 (B) CH3 OH
H3C C CHO H2C C C H
CH3 OH CH3 CN
(C) H OH (D) H3C C CH COOH
H3C C CH COOH CH3
CH3

Space for rough work

12. Which of the following statement(s) are correct?

(A) CH3
H3C C CH2 CH3  
alc. KOH
 alkene with 6  H
Cl
(B) CH3
H3C C CH CH2 CH3  
alc. KOH
 alkene with 8  H

H Cl
(C)
CH2 CH CH3    alkene with 3  H
alc. KOH

Br
(D) CH3
H3C C CH CH3    
CH3 OH/conc. H2 SO4
cold
 carbonium ion with 7  H leading to alkene
CH3 OH

13. O OH
H3C C CH2 CH3 KCN
 A 
H2 SO4
LiAlH4
 H3C C CH2CH3
CH2NH2
Which of these statements are correct?
(A) One of these chemical substances gives iodoform test
OH
(B)
H3C C CH3 is formed
CN
(C) The final mixture is optically active
(D) The final mixture is racemic

Space for rough work

14. O
Ph C Cl
H3C NH2    organic compound with 2 benzene rings

Which of the following statements is/are correct?

(A) One of the benzene rings is ortho & para directing
(B) One of the benzene rings is meta directing
(C) Organic compound is

H3C NH

C O
Ph
(D) Organic compound is
O
H3C NH C Ph

15. Select the correct statement(s), among the following given statements
(A) CH3CH2CH2I will react more readily than (CH3)2CHI for SN2 reactions
(B) CH3CH2CH2Cl will react more readily than CH3CH2CH2Br for SN2 reactions
(C) CH3CH2CH2CH2Br will react more readily than (CH3)3C – CH2Br for SN2 reactions
(D) CH3 – O – C6H5 – Br will react more readily than NO2 – C6H5 – CH2Br for SN2 reaction

Space for rough work

1. What is the total number of basic groups in the following?

H3N CH2CH2
CH NH2
O C
O

2. O

 product s


OH



O
How many product(s) are possible?

3. How many structures of A are possible?

CH3
H
H3C   A 
H2 O
Br2 / CCl4
 C 4H8Br2
5 such products possible 
OH

4. O
H3C C CH3   Iodoform reaction
NaOH

How many molecules of I2 are required for 3 molecules of CH3COCH3?

5. O

CH2CH2CH2Cl
 
CH3MgI
 product containing 1 cyclic ether

How many C atoms are part of cyclic ether?

Space for rough work

Mathematics PART – III

SECTION – A
Single Correct Choice Type

This section contains 10 multiple choice questions. Each question has four choices (A), (B), (C) and (D)
out of which ONLY ONE is correct.

1. If (a, 0); a > 0 is the point where the curve y  sin2x  3 sin x cuts the x-axis first, A is the area
bounded by this part of the curve the origin and the positive x-axis, then
(A) 4A + 8 cos a = 7 (B) 4A + 8 sin a = 7
(C) 4A – 8 sin a = 7 (D) 4A – 8 cos a = 7

ln x h  ln x

2. Let f  x   lim
 sin  x  h    sin x   
then f   is
h 0 h 2
(A) equal to 0 (B) equal to 1

(C) ln (D) none existent
2

1/m
 x  
7m
3. For any natural number m,  x 2m  xm 2x 6m  7xm  14 dx (where x > 0) equals
m 1 m 1

(A)
 7x 7m  2x 2m  14xm  m
c (B)
 2x 7m  14x 2m  7xm  m
c
14  m  1 14  m  1
m 1 m 1

'
u x
 ux  u'  x 
4. Let u(x) and v(x) are differentiable function such that  7 . If  p and  q
v x v 'x  v  x  
 
pq
then has the value equal to
pq
(A) 1 (B) 0
(C) 7 (D) –7

Space for rough work

3
1 t 3 2 dy  dy 
5. A function is represented parametrically by the equation x  3 ; y  2  then  x 
t 2t t dx  dx 
has the value equal to
(A) 2 (B) 0
(C) –1 (D) –2

x10
6. Let g is the inverse function of f and f '  x   . If g(2) = a then g(2) is equal to
1  x  2

5 1  a2
(A) (B)
210 a10
a10 1  a10
(C) (D)
1  a2 a2

  1  
7. Let f : 0,   R be continuous and satisfy f '  x   for all x   0,  . If f(0) = 3 then
 2 1  cos x  2

 
f   has the value
2
13
(A) (B) 2
4
(C) 4 (D) none of these

 dy 
8. For the curve represented implicitly as 3x – 2y = 1, the value of lim   is
x   dx 

(A) equal to 1 (B) equal to 0

  sin 101x  sin 

99
9. x dx equals
100 100
sin 100x  sin x  cos 100x  sin x 
(A) c (B) c
100 100
100 101
cos 100x  cos x  sin 100x  sin x 
(C) c (D) c
100 101

Space for rough work

1  x x
10. lim  a arc tan  b arc tan  the value equal to
x 0 
x x a b 
ab
(A) (B) 0
3
a2  b2 a2  b2
(C) (D)
6a2b2 3a2b2

Multiple Correct Answer(s) Type

This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D)
out of which ONE or MORE are correct.

dy
11. If 2x + 2y = 2x + y then has the value equal to
dx
2y 1
(A)  (B)
2x 1  2x

12. Two functions f and g have first and second derivative at x = 0 and satisfy the relations
2
f 0   ; f(0) = 2g(0) = 4g(0); g(0) = 5f(0) = 6f(0) = 3 then
g 0 
f x 15
(A) if h  x   then h'  0   (B) if k(x) = f(x) g(x) sin x then k(0) = 2
g x  4
g'  x  1
(C) lim  (D) none of these
x 0 f ' x 2

1 1  1
13. Let f '  x   3x2 sin x cos ; if x  0 f(0) = 0 and f    0 then
x x 
(A) f(x) is continuous at x = 0 (B) f(x) is non derivable at x = 0
(C) f(x) is continuous at x = 0 (D) f(x) is non-derivable at x = 0

Space for rough work

14. Consider the function f(x) and g(x), both defined from R  R and are defined as f(x) = 2x – x2 and
1
g(x) = xn where n  N. If the area between f(x) and g(x) is then n is a divisor of
2
(A) 12 (B) 15
(C) 20 (D) 30

 
cos x 0x
2

15. Consider f  x    2 such that f is periodic with period x, then
   x  
x
 2 
 2
 2 
(A) the range of f is 0, 
 4
(B) f is continuous for all real x, but not differentiable for some real x
(C) f is continuous for all real x
 3 
(D) the area bounded by y = f(x) and the x-axis from x = –n to x = n is 2n  1   for n  N
 24 

/ 4 n
1. Let In   tann x dx (n = 0, 1, 2, 3, …..) and Sn   In In1  In In 3  In 1 In 2  then the value of
0 n0

lim  Sn  is _____
x 

Space for rough work

1  sin x 
2. Let L denotes the value of lim cos1   and U denotes the value of f(0) where
x 0 x  x 
x
2U
 
f  x    sin t 2  t  x dt , compute the value of
L2
_____
0

 /2
1 32V
3. Suppose V   x sin2 x  dx , then the value of _____
0
2 


cos 4x  cos 4
4. If the value of the integral  dx is k( cos 2 cos ) then the value of k _____
0
cos x  cos 

 m 1
x sin x  0, m  N
5. A function f(x) is defined as f  x    x . The least value of m for which f(x) is

0 if x  0
continuous at x = 0 _____

Space for rough work

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ALL INDIA INTEGRATED TEST SERIES

ANSWERS, HINTS & SOLUTIONS
PART TEST –III
(Paper-1)

Q. No. PHYSICS CHEMISTRY MATHEMATICS

B C A
1.
B C A
2.
A B C
3.
D D A
4.
qualified in JEE (Advanced), 2014.

B B C
5.
A C B
6.
A C C
7.
A A C
8.
C C A
9.
D C D
10.
A, B C, D A, B, C, D
11.
A, D B, C A, B, C
12.
A, C, D A, D A, C, D
13.
B, D A, B, D B, C, D
14.
A, D A, C A, D
15.
0 2 1
1.
0 1 6
2.
9 3 4
3.
3 9 4
4.
5 4 3
5.

Physics PART – I

2. Potential energy will be finite and maximum at x = 0.

5. B1  B2 = 20 T, B1 + B2 = 50 T and B  i
 
6. Use   M  B

7. Use symmetry arguments. Net B will be perpendicular to the plane of the loop
 
9. Use P = qi ri

Chemistry PART – II

SECTION – A

1. In alkaline medium, alanine exists as anion:

OH
H3C CH NH2   H3C CH NH2
COOH COO

2. CH3 O
H3C C C H O H

CH3
O
C H O H

O
H3C C H 3  H(not appropriate for Cannizzaro reaciton)
O
H C H O H

3.
NH2 N NCl NH2 COOH

NaNO2 HCl 
 
05 C
  
CuCN
 
H2 O/H


4. Hoffmann bronamide degradation.

5. CN I

Cl
 
NaCN
DMF
SN 2 reaction
polar approtic solvent 

I CN

6. NaBH4 selectively reduceds

O
C

7. Gattermann Koch reaction.

8. MnO2 oxidized allylic alcohol.

9. OCH 3 OH O
H O H O
3   2
 
O 
CH OH 3
OCH 3 OH

COOCH 3 COOCH 3 COOH

10. O
COOH
C O
OH

BrO 
 
NH
CONH 2 NH 2

11. OH OH
H /H2 O H2 O
H3C CH CHO 
 HCN
H3C CH CH  
 H3C CH CH COOH 
 H3C C CH COOH

CH3 CH3 CN CH3 CH3

12. (A)
CH3
H3C C CH CH3 E2
9  H
(B)
CH3
H3C C CH CH2 CH3 E2

8  H
(C)
CH CH CH3 E2
3  H
(D) C C C
CH3 shift
C C C C C C C C C C C C
C OH C C
H
H3C OH

C
C C C C
O C
CH3

ether and NOT an alkalene

13.
O H
OH OH
LiAH4
H3C C CH2 CH3   C C C C  
C C C C
CN CH2 NH2
sp2 CN A B
hybridised  above & below  R as well as S racemic mixture racemic mixture
   
planar  the plane 
LiAH4

14. Organic compound is

O
H3C NH C

o and p directing
meta directing

SECTION – C

1. NH2 si basic.
COO is (conjugate) base.
+
NH3 is acidic.

2. 2'
OH O 1' 3' O
H
O 1 O 2'
2 4'
3'

aldol
 1' 
 


 5'
3 4' O
4 5'
O O
H OH

3. The compound A is represented by the 3 structures as follows:

OH H CH3 H3C CH3

H3C CH2 CH CH3 H
dehydration
 C C C C H3C CH2 CH CH2
H3C H H H

trans cis
Br2 / CCl4
Br2 / CCl4 Br2 / CCl4

meso d and  forms d and  forms

racemic  racemic

5 products

4. O O O O
CH2 
 H3C CH2 
 H3C CH2  
 H3C
NaOH/I2 2 more times
H3C C C C C CI3
I OH
I I

CH3  COO  CHI3   CI3 H3C C OH

1 molecules of H3C C CH3 requires 3 molecules of I .

5. O O

CH2CH2 CH2Cl

CH3 MgI
 OH

Mathematics PART – III

SECTION – A

1. (a, 0) lies on the given curve

 o  sin 2a  3 sina
3
 sin a = 0 or cosa 
2

 a (as a > 0 and 1st point of intersection with positive x-axis) and
6
 /6
7 7
A   sin 2x  
3 sin x dx =
4
 3   2cosa  4A + 8 cos a = 7
4
0

2. Let g(x) = (sin x)ln x = e ln x ln(sin x)

ln x  ln  sin x  
f  x   g'  x    sin x  cot x ln x   
 x 
 
Hence, f    g'    0
 2 2

3. Put 2x7m + 7x2m + 14xm = tm

14 m(x7m – 1 + x2m – 1 + xm – 1) dx = mtm – 1 dt

4. u(x) = 7v(x)  p = 7
'
 u x 
  0 q=0
 v  x  
 
pq
 1
pq

dx  3  2t 
5.   4 
dt  t 
dy  3  2t 
  3 
dt  t 

6. f[g(x)] = x  f[g(2)] g(2) = 12  f(a) g(2) [putting x = 2]

a10 1  a2
Given f '  a   2
 g'  2   10
1 a a

1
7. f 'x  integrating
1  cos x
x
f  x   tan  c
2
f(0) = 3  c = 3
x
f  x   tan  3
2

 f   4
 2

8. 2y = 3x –1  y ln 2 = ln (3x – 1)
dy 3x ln3 ln3
ln2  
dx 3x  1 1  3 x
dy ln3 1
lim  lim  log2 3
x  dx ln2 x  1  3  x

99 99
9. I    sin100x  x  sin x  dx =   sin 100x  cos x  cos100x sin x    sin x  dx
99 100
=  sin 100x  cos x  sin x  dx   cos 100x  sin x  dx
 
I II
100
sin 100x  sin x  100 100 100
=  cos 100x  sin x  dx   cos 100x  sin x  dx
100 100 
100
sin 100x  sin x 
= c
100

10. Using Lopital Rule

 a tan1 x / a  b tan1 x / b 
lim  
x 0 x x
 
a 1 1 b 1 1
    
= lim
1 x / a 2 a 2 x  1 x / b 2 b 2 x  
x 0 3
x
2
  2 2
a2 b2  1 = a b
= lim  2  2

x 0  a  x x
 
b  x x 3
  3a2b2 
 3 1
 x sin if x  0
13. f x   x
0 if x  0

14. 2x – x2 = xn
 x = 0 and 1
1 1
x 3 x n 1  2 1
A  2
2x  x  x dx  x  n
 
3 n  1
 = 
3 n1
2

0 0

2 1 1
Hence,  
3 n 1 2
n=5
Hence, n is devisor of (20, 15, 30)

 
cos x 0x
2

15. Given f  x    2 and f is periodic with period 
   x  
x
 2 
 2
 We draw the graph of y = f(x)

x x
–    3 2 5 3
–
2 2 2 2
y
 2 
From the graph the range of the function is 0, 
 4
n
It is discontinuous at x = n, n  I at x  ,nI
2
Area bounded by y = f(x) and the x-axis from –n to n for n  N
2
   /2 
    3 
= 2n  f  x  dx  2n   cos x dx     x  dx  = 2n  1  
0  0  /2 
2    24 

SECTION – C

 /4  /4
1. We have In  In 2   tann xdx   tan
n 2
xdx
0 0
 /4 1
1
=  
tann x 1  tan2 x dx   un du   n1
(where u = tan x)
0 0
1 1
So, In  In 2  which implies that In1  In3 
1 n n2

Hence,  In In 1  In In 3  In1 In 2  In 2 In 3 
n0
  
 1  1   1 1 
=  In  In 2 In1  In 3  =   n  1   n  2     n  1  n  2 
n0 n0 n0

1  1 
 Sn  1   lim  1  1
n2 x   n  2 

 x 2  sin2 x 
 sin x  sin1  
cos1    x2 
 x   
2. L : lim = lim
x 0 x x  0 x
sin1  M M
= lim  lim (Using cos1 t  sin1 1  t 2 )
x 0 M x 0 x

x 2  sin2 x  x  sin x   x  sin x 

= 1 lim 2 2
 1 lim
x0 x x x0 x3 x
1 1 1
= lim 2  L
x 0 6 3 3
x

  sin  t    
2
U : f  x   t cos x  cos t 2  t sin x dt
0
x x

 
or f  x   cos x  sin t 2  t dt  sin x  cos t 2  t dt  
0 0

x  x 
     
f '  x   cos x  sin x 2  x    sin t 2  t dt  sin x    cos t 2  t dt  cos x  cos x 2  x sin x
     
0  0 
 
f "  x     sin x 2  x sin x  cos x  cos x 2  x  2x  1 
   
 x  
 
   sin t 2  t dt  cos x  sin x sin x x 2  x 
   
 0  

  
 cos x 2  x cos x  sin x sin x2  x  2x  1 
  
 x  
  
 cos x cos x 2  x    cos t 2  t dt  sin x 
  
 0  
 f(0) = 1  U = 1

 /2  /2
1 1
3. V  x 2 sin2 x  1dx =  x cos2x dx
2 0
2 0
dt
Put 2x = t, dx 
2

4. Consider the numerator cos 4x – cos 4

= (2 cos2 2x – 1) – (2 cos2 2 – 1)
= 2(cos2 2x – cos2 2)
= 2(cos 2x + cos 2)(cos 2x – cos 2)
= 4(cos 2x + cos 2)(cos x – cos )(cos x + cos )
 
 I  4  cos2x  cos 2  cos x  cos   dx = 4   cos 2  cos   dx (all other integral vanish)
0 0
 4 cos ·cos 2

Time Allotted: 3 Hours Maximum Marks: 180

 Pl ea s e r ea d t h e i n s t r u c t i o n s c a r ef u l l y. Yo u a r e a l l o t t ed 5 m i n ut es
s p ec i f i c a ll y f o r t h i s p u r p o s e.
 Yo u a r e n o t a ll o wed t o l ea v e t h e E xa m i n at i o n Ha ll b ef o r e t h e en d o f
t h e t es t .

INSTRUCTIONS
A. General Instructions
1. Attempt ALL the questions. Answers have to be marked on the OMR sheets.
2. This question paper contains Three Parts.
3. Part-I is Physics, Part-II is Chemistry and Part-III is Mathematics.
4. Each part has only one section: Section-A.
5. Rough spaces are provided for rough work inside the question paper. No additional sheets will be
provided for rough work.
6. Blank Papers, clip boards, log tables, slide rule, calculator, cellular phones, pagers and electronic
devices, in any form, are not allowed.
B. Filling of OMR Sheet
1. Ensure matching of OMR sheet with the Question paper before you start marking your answers
on OMR sheet.
2. On the OMR sheet, darken the appropriate bubble with black pen for each character of your
Enrolment No. and write your Name, Test Centre and other details at the designated places.
3. OMR sheet contains alphabets, numerals & special characters for marking answers.
C. Marking Scheme For All Three Parts.
(i) Section-A (01 to 08) contains 8 multiple choice questions which have one or more correct
answer. Each question carries +3 marks for correct answer and – 1 mark for wrong answer.
Section-A (09 to 16) contains 4 paragraphs with each having 2 questions. Each question carries
+3 marks for correct answer and – 1 mark for wrong answer.
Section-A (17 – 20) contains 4 Matching Lists Type questions: Each question has four
statements in LIST I & 4 statements in LIST II. The codes for lists have choices (A), (B), (C), (D)
out of which only one is correct. Each question carries +3 marks for correct answer and
– 1 mark for wrong answer.

Name of the Candidate

Enrolment No.

Useful Data

PHYSICS
Acceleration due to gravity g = 10 m/s2

Planck constant h = 6.6 1034 J-s

Charge of electron e = 1.6  1019 C

Mass of electron me = 9.1  1031 kg

2 2
Permittivity of free space 0 = 8.85  1012 C /N-m

Density of water water = 103 kg/m3

Atmospheric pressure Pa = 105 N/m2

Gas constant R = 8.314 J K1 mol1

CHEMISTRY

Gas Constant R = 8.314 J K1 mol1

Atomic No: H=1, He = 2, Li=3, Be=4, B=5, C=6, N=7, O=8,

Physics PART – I

SECTION – A
Single Correct Choice Type
This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D)
out of which ONLY ONE is correct.

1. A conducting ring of mass 2 kg and radius 0.5 m is

placed on a smooth horizontal plane. The ring carries a
current of I = 4A. A horizontal magnetic field B = 10T, B

coplanar with ring exist in the region. if the ring is

released from rest then
(A) angular acceleration of the ring is 40 rad s2 (B) torque on the ring is 20 Nm
(C) angular acceleration of the ring is 20 rad s2 (D) torque on the ring is 10 Nm

2. A sphere has a positive charge. Figure shows variation of electric field E

(E) with distance x from its centre. From this figure, we conclude that
(A) Sphere is made of a non conducting materials
(B) Diameter of sphere is equal to R/2
(C) Electric potential, due to sphere, is maximum at its centre
(D) Density of charge is uniform throughout the volume of sphere R
O x

3. Electric charge +q is uniformly distributed over the entire length of a ring of radius r rotating with
constant angular velocity  about its own axis. Assuming mass of ring to be m, its magnetic
moment to be M and angular momentum to be L.
qr 2 2qr 2
(A) M  (B) M 
2 5
M q M 2q
(C)  (D) 
L 2m L m

4. A battery of emf E is connected across a conductor as shown as one B

observe from A to B. then A
r1 r2
(A) Current remains same
(B) Drift velocity of electron decreases
i0
(C) Electric field decreases
(D) Potential drop across the length decreases E

Space for rough work

5. To convert galvanometer into a voltmeter, a large resistance is connected in series with the
galvanometer. Due to the use of this large resistance is
(A) Accuracy increases (B) Least count increases
(C) Least count decreases (D) Range increase

6. A point charge q is placed at corner D of a cube of side d. Flux of electric E C

field passing through face q
F D
(A) ABCD is zero (B) CDFE is zero
q H
(C) DFGA is zero (D) ABHG is B
240
G
d A

7. Select the correct statement(s) from the following.

The path of a charged particle moving with an initial velocity u
(A) parallel to the uniform electrostatic field is a straight line.
(B) perpendicular to the uniform electrostatic field is a parabola.
(C) parallel to the uniform magnetic field is a helix with its axis parallel to the direction of B.
(D) perpendicular to the uniform magnetic field is a circle in a plane normal to the field.

8. In the figure shown, area of each plate is A. Match the following 1 2 3 4 5 6

 A 2d d
(A) Charge on plate 5 is 0 V
d
(B) Potential difference between plates 2 and 3 is V
V
(C) Potential difference between plates 2 and 5 is zero
(D) Charge on the plate 3 is zero.

Linked Comprehension Type

This section contains 4 paragraphs. Based upon each paragraph, 2 multiple choice questions have to be
answered. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

Paragraph for Question Nos. 9 to 10

A coil (ABCDEFGH) is in the form of prism having each side equal to F

length  carries a current i0 (Assume gap between parallel wires AH and i0
DE to be negligibly small)
C E z
G H
y
D
x
B A

9. The direction of magnetic field at the geometric centre of three dimensional prism shaped coil will
be
(A) along y-axis (B) along x-axis
(C) along z-axis (D) none of these

Space for rough work

10. If an external magnetic field B0 is applied along +ve z-axis then the force on the coil ABCDEFGH
due to B0 only will be equal to
(A) zero (B) L0B
(C) 6L0B (D) 2L0B

Paragraph for Question Nos. 11 to 12

Two long straight parallel wires are 2 metres part, perpendicular to the A
plane of the paper. The wire A carries a current of 9.6 amps, directed
into the plane of the paper. The wire B carries a current such that the
1.6 m
10
magnetic field of induction at the point P, at a distance of metre from 2m
11
the wire B, is zero. Answer the following questions. S
1.2 m

B
10/ 11m

11. The magnitude of direction of the current in B.

(A) 4.0 A (B) 3.0 A
(C) 2.0 A (D) 1.0 A

12. The magnitude of the magnetic field of induction at the point S.

(A) 1.1 T (B) 1.3 T
(C) 1.4 T (D) 2.8 T

Paragraph for Question Nos. 13 to 14

In the circuit shown in fig. E, F, G and H are cells of emf 2, 1, 3 and 1 Volts E
A + B
respectively and their internal resistance are 2, 1, 3 and 1 respectively.
Calculate
+ 2
H
F +

D + C
G

13. The potential difference between B and D and

2 4
(A) V (B) V
13 13
6 8
(C) V (D) V
13 13

Space for rough work

14. The potential difference across the terminals of the cell G.

19 17
(A) V (B) V
13 13
21 23
(C) V (D) V
13 13

Paragraph for Question Nos. 15 to 16

Measurements and interpretations of voltage and electric current signals are common in modern
medicine. Occasionally, a situation arises in which a voltmeter or an ammeter is needed but it is not
available. A galvanometer is an instrument that can be used to construct an ammeter. It can also be used
to construct a voltmeter. In both cases, a resistor R must be connected to galvanometer to effect the
change.
r
P1 G P2
r R
IG G
IG
IR R
(ii)
(i)
To turn the galvanometer into an ammeter, the resister R is connected in parallel, as shown in the figure
(i), The resistor R is connected in series with the galvanometer in order to turn it into a voltmeter, as
shown in the figure (ii). The current required to produce a full scale deflection in a galvanometer is 10 mA.
The internal resistance of galvanometer is 100 . Let Vr be the voltage across r and Vr the voltage
across R.
15. Which of the following formulas correctly applies to the ammeter concept?
(A) Vr > VR (B) Vr < VR
(C) Vr = VR (D) more information needed

16. What resistance must be connected in parallel to the galvanometer to turn it into an ammeter
capable of reading electric current upto 10.01 A?
(A) 0.1  (B) 1 
(C) 10  (D) none of these

Space for rough work

SECTION – A
Matching List Type

This section contains 4 multiple choice questions. Each question has matching lists. The codes for the
lists have choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

17. List–I specifies a point P at distance r from the center/axis of a symmetrical distribution of charge.
List–II gives the variation of electric intensity at P as a function of r(>0). Match the entries of List–I
with possible entries of List–II.
List – I List - II
(P) P lies outside a long cylinder having uniform volume charge (1) 1
density E 2
r
(Q) P lies inside a spherical charged conductor (2) 1
E
r
(R) P lies inside a spherical body having uniform volume charge (3) Er
density
(S) P lies outside of a nonconducting solid sphere of radius R and its (4) E  r0
2
0 x
volume charge density varies as   , where x is distance
R2
from centre of the sphere
Codes:
P Q R S
(A) 2 4 3 1
(B) 4 1 3 2
(C) 2 3 2 4
(D) 4 1 2 3

18. In the circuit shown the capacitor has an initial charge q = 100 C and a 10 F

capacitor C = 10 F. The emfs of the batteries are shown in the figure and the
5 S1
batteries have negligible internal resistance. The switches are closed at t = 0.
Match the following. The quantities on the right columns are given in SI units. 5 S2

10 V
List – I List - II
(P) Time constant of the circuit (1) 0
(Q) Current through S1 at t = 0 (2) 25
(R) Current through S2 when t  0 (3) 2
(S) Final charge on the capacitor (4) 50
Codes:
P Q R S
(A) 3 1 4 2
(B) 2 1 3 4
(C) 4 3 2 1
(D) 2 1 4 3

Space for rough work

19. Side of the cube is ‘a’ and wire AB, BC, CH and CE Z given that OF = a
are insulated with each other, and each having
E H
current I, then match the following.
External magnetic field applied along positive x-axis
 B
C i
as B = B0x î i

i
i
O
I F E X

A D
Y
List – I List – II
Magnitude of force due to external magnetic field only
(P) On AB (1) zero
2
(Q) On BC (2) 22iB0a
2
(R) On CH (3) iB0a
(S) On CE (4) 2iB0a2
Codes:
P Q R S
(A) 3 1 4 2
(B) 2 1 3 4
(C) 4 3 2 1
(D) 1 4 2 3

20. Match the following:

List – I List – II
(P) Self–energy of a uniformly charged spherical shell (1) 1 Q2
80 R
(Q) Energy stored per unit volume of a charged air filled (2) 1 2Q2
capacitor.
40 R2
(R) Force between two circular plates of a charged parallel plate (3) 1 Q2
capacitor, radius of the plates being R and separation R/32.
40 R
(S) Work done by electric field in moving a point charge Q from (4) 1
the surface of a uniformly charged sphere of radius R, and 0E2
2
moving it to infinity, total charge on the sphere  Q .
Codes:
P Q R S
(A) 3 1 4 2
(B) 2 1 3 4
(C) 1 4 2 3
(D) 4 1 2 3

Space for rough work

Chemistry PART - II

SECTION – A
Single Correct Choice Type
This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D)
out of which ONLY ONE is correct.

1. Regarding the given compound:

CH2 CH2 OH

HO
Choose the correct option (s)
(A) The above compound exhibit geometrical as well as optical isomerism
(B) Total number of stereoisomers possible for above compound is 4
(C) One of isomers of the above compounds shows plane of symmetry as well as axis of
symmetry
(D) The above compound does not exhibit mesomerism

2. In the given compound choose the correct statement (s) regarding percentage enol content in
H2O
O O
O O O

(I) (II) (III) (IV)

(A) I > III (B) I > III > IV
(C) II > III (D) IV > I > II

3. Which of the following option(s) is/are correct regarding given chemical reaction?
CH3

Br2 / h
  Monobromo derivative

(A) Number of possible mono bromo product (including stereo) is 7

(B) Number of possible mono bromo product (excluding stereo) is 5
(C) Total number of optical active isomers per mono bromo products are 4
(D) Total number of optical inactive isomers for mono product are 2

Space for rough work

4. OH

OsO 4

NaHSO
 X Y
3 95% 5%

Which of the following statements are correct regarding product of above reaction?
(A) OH OH
OH OH

Compound Y is and compound X is

OH OH
(B) Compound X and Y are enantiomers
(C) Compound X and Y both are optically active
(D) OH OH
OH OH

Compound X is and compound Y is

OH OH

5. Consider the following reaction sequence

CH3
2CH MgBr /H O H O
|
3 3 3
(B)   (A)   CH3  CHCOOH
Which of the following statements are true?
(A) Compound (A) can be CH3CHCOOC2H5
CH3
(B) Compound (A) can be CH3CHCOOC6H5
CH3
OH
(C) Compound (B) can be CH3CH C–CH3
CH3 CH3
CH3
(D) Compound (B) can be CH3C–COOC6H5
MgBr
Space for rough work

6. In the reaction sequence,

2 NaOEt H /H O
CH2(COOEt)2 + BrCH2CH2CH2Br  (A)  ? The possible final product(s) is/are
excess 
COOH COOH
(A) (B) COOH
COOEt COOH
(C) (D)

7. In a reaction sequence
cold dil. KMnO HIO
   4 (A) 4 (B) OH
(C).

O OH
(A) (A) is O
(B) (B) is

OH
(C) (A) is (D) (C) is
OH
O

8. Which of the following isomer is an example of synthetic rubber?

(A) Buna–N (B) ABS
(C) Neoprene (D) cis–polyisoprene

Space for rough work

Linked Comprehension Type

This section contains 4 paragraphs. Based upon each paragraph, 2 multiple choice questions have to be
answered. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

Paragraph for Question Nos. 09 to 10

Nucleophilic aliphatic substitution reaction is given by those compounds which have nucleophilic groups
as leaving group. Less is the basicity of the leaving group, more is its leaving power. Leaving group
leaves in the form of nucleophile (charged or neutral).

9. Leaving power of which group is maximum?

O
(A) O S CF 3 (B) N N
O
O
(C) O S CH3 (D) Cl
O

10. Leaving power of which group is maximum?

O O

(A) O S CF 3 (B) O S C 4H 9

O O
O
(C) O S CH3 (D) Br
O

Space for rough work

Paragraph for Question Nos. 11 to 12

Reimer-Tiemann reaction involves an aryl carbanion/enolate anion and also –CCl3 derived from the action
of strong bases on CHCl3, though the latter has only a transient existence decomposing to : CCl2 , a
highly electron deficient electrophile that attacks the aromatic nucleus; the product from phenoxide ion is
after acidification, very largely the o-aldehyde plus just a small amount of p-isomer.
11. If the above reaction is carried out on the anion of p-hydroxytoluene, what will be
theproduct/products?
(A) OH (B) OH

CHO

Me Me
(C) Both (A) and (B) (D) None of the above

12. Name the electrophile attacking carbanion in the above reaction.

(A) Chloroform (B) Trichloromethyl anion
(C) dichlorocarbene (D) hydrogen ion

Paragraph for Question Nos. 13 to 14

Aldehyde having no -hydrogen when treated with strong base, undergoes redox reaction. This reaction
is known as Cannizzaro reaction. Tischenko reaction is also a redox reaction which takes place in the
presence of aluminium ethoxide.

13. Which of these compounds will give Cannizzaro reaction?

(A) C6H5CH2CHO (B) CHO–CHO
(C) C6H5CH2CHO (D) All of these

14. Internal crossed Cannizzaro reaction is given by:

(A) C6H5COCHO (B) C6H5CHO
(C) CH3–CHO (D) C6H5–CH2CO–CHO

Paragraph for Question Nos. 15 to 16

Aryl and vinyl halides are less reactive towards nucleophilic substitution reactions. Chlorobenzene itself
and other aryl halides having electron–donating substituents undergo nucleophilic substitution reactions
by benzyne mechanism while aryl halides having electron – withdrawing substituents at o- and p-
positions undergo bimolecular nucleophilic substitution reactions.
Based on the given passage, reply the following questions:

15. In the reaction of p-chlorotoluene with KNH2 in liq NH3, the major product is
(A) o–toluidine (B) m–toluidine
(C) p–toluidine (D) p–chloroaniline

Space for rough work

OCH3
NaNH2/liq NH3
16. In the reaction A. The major product is

Cl
OCH3 OCH3

(A) (B)
NH2
NH2
OCH3 OCH3

SECTION – A
Matching List Type

This section contains 4 multiple choice questions. Each question has matching lists. The codes for the
lists have choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

17. Match the List – I with List – II:

List – I List – II
(P) H C C CH 
i  BH
 3 (1) CH3 – CH2 – CH = CH2
3 2 ii H2O2 / OH

CH3
(Q) H3C C CH2 
 i Hg OAc 
2


/HOH
(2) CH3 – CH = CH – CH3
 ii NaBH4

CH3
(R) Cl (3) CH3
CH3 ONa
H3C CH2 CH CH3 
 H3C CH CH2 OH
(S) Cl (4) OH
 CH3 3 CONa H3C C CH3
H3C CH2 CH CH3 


CH3
P Q R S
(A) 3 4 2 1
(B) 2 3 4 1
(C) 4 1 2 3
(D) 1 2 3 4

Space for rough work

18. Match the following:

List – I List – II
(P) RCONH2  Br2 /KOH
 RNH2 (1) Curtius rearrangement
(Q) N3H
RCOOH   R  NH2
 (2) Lossen rearrangement
H2SO4

(R) NaN3
RCOCl   R  NH2 (3) Schmidt rearrangement


(S) O (4) Hoffmann bromamide degradation

||
 i NH2 OH
R  C Cl 
OH / 

P Q R S
(A) 3 2 1 4
(B) 2 3 4 1
(C) 4 3 1 2
(D) 1 2 3 4

19. Match the following List – I with List – II:

List – I List – II
(Reaction) (Reagent)
(P) Br (1) HBr

CH3–CH=CH2  CH3–CH–CH3
(Q) CH3–CH=CH2  CH3–CH2–CH2Br (2) NBS
(R) CH3–CH=CH2  BrCH2–CH=CH2 (3) Br2/CCl4
(S) CH3–CH=CH2  CH3–CHBr–CH2Br (4) HBr (Peroxide)

P Q R S
(A) 3 2 1 4
(B) 2 3 4 1
(C) 4 1 2 3
(D) 1 4 2 3

Space for rough work

20. Match the following List – I with List – II:

List – I List – II
(Substrate + reagent) (Product)
(P) CH –CH –CH –Cl 
C2H5 ONa
 (1) CH3–CH2–CH=CH2
3 2 2
C2H5 OH

(Q) CH3 (2) CH3–CH=CH–CH3

C H ONa 2 5
CH3CCH3 
C H OH

2 5

Cl
(R) (3) CH3–CH=CH2
Cl
CH 3  CH 2  CH  CH 3 tBuOK
 
BuOH

(S) (4) CH3C=CH2

Cl
CH 3  CH 2  CH  CH 3 C
2 H 5 ONa
  CH3
C 2H 5OH

P Q R S
(A) 3 4 1 2
(B) 2 3 4 1
(C) 4 1 2 3
(D) 1 2 3 4

Space for rough work

Mathematics PART – III

SECTION – A
Single Correct Choice Type
This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D)
out of which ONLY ONE is correct.

1. The curve y = ax 2 + bx + c passes through the point (1, 2) and its tangent at origin is the line y =
x. The area bounded by the curve, the ordinate of the curve at minima and the tangent line is
1 1
(A) (B)
24 12
1 1
(C) (D)
8 6

2. Primitive of I   1  2 cosec x cot x  2cot 2 x dx w.r.t. x is

x x
(A) 2lncos c (B) 2lnsin c
2 2
1 x
(C) lncos  c (D) ln sin x – ln (cosec x) + c
2 2

3. A non-zero polynomial with real coefficients has the property that f(x) = f(x)·f(x), then the leading
coefficient of f(x) is
1 1
(A) (B)
6 9
1 1
(C) (D)
12 18

etan x  e x  ln  sec x  tan x   x

4. Let f  x   be a continuous function at x = 0. The value of f(0)
tan x  x
equals
1 2
(A) (B)
2 3
3
(C) (D) 2
2

Space for rough work

5. For the function y = f(x) = (x2 + bx + c)ex which of the following holds?
(A) if f(x) > 0 for all real x  f(x) > 0 (B) if f(x) > 0 for all real x  f(x) > 0
(C) if f(x) > 0 for all real x  f(x) > 0 (D) if f(x) > 0 for all real x  f(x) > 0

x2  1
6. Suppose f is defined from R  [–1, 1] as f  x   where R is the set of real number. Then
x2  1
the statement which does not hold is
(A) f is many one onto
(B) f increases for x > 0 and decreases for x < 0
(C) minimum value is not attained even though f is bounded
(D) the area included by the curve y = f(x) and the line y = 1 is  sq. unit

7. People living at mars, instead of the usual definition of derivative D f(x) define a new kind of
f 2  x  h  f 2  x 
derivative, D* f(x) by the formula D * f  x   lim where f 2(x) means [f(x)]2. If f(x) =
h 0 h
x ln x then D * f  x  x  e has the value
(A) e (B) 2e
(C) 4e (D) 8e

 /2  /2  /2
8. Suppose I1    
cos  sin2 x dx , I2    
cos 2 sin2 x dx , I3   cos   sin x  dx then
0 0 0
(A) I1 = 0 (B) I2 + I3 = 0
(C) I1 + I2 + I3 = 0 (D) I2 = I3

Linked Comprehension Type

This section contains 4 paragraphs. Based upon each paragraph, 2 multiple choice questions have to be
answered. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

Paragraph for Question Nos. 9 to 10

Read the following write up carefully and answer the following questions:
Let the function f satisfies f(x)·f(–x) = f(–x)·f(x) for all x and f(0) = 3

9. The value of f(x)·f(–x) for all x is

(A) 4 (B) 9
(C) 12 (D) 16

Space for rough work

51
dx
10.  has the value equal to
51
3  f x
(A) 17 (B) 34
(C) 102 (D) 0

Paragraph for Question Nos. 11 to 12

Read the following write up carefully and answer the following questions:
A curve is represented parametrically by the equations x = et cos t and y = et sin t where t is a parameter
then

11. The relation between the parameter t and the angle  between the tangent to the given curve and
the x-axis is given by ‘t’ equals
 
(A)   (B)  
2 4
 
(C)   (D)  
4 4


12. If f(t) =   x  y  dt then the value of f    f  0  is
2
(A) 1 (B) –1
(C) e/2 (D) 0

Paragraph for Question Nos. 13 to 14

Read the following write up carefully and answer the following questions:
A curve is represented parametrically by the equations x = f(t) = a
  and y = g(t) = bna  a, b > 0 and
n b t t

a  1, b  1 where t  R

dy
13. Which of the following is not a correct expression for ?
dx
1
(A)  2
(B) –(g(t))2
f t
gt  f t 
(C)  (D) 
f t  gt 

Space for rough work

d2 y
14. The value of at the point where f(t) = g(t) is
dx 2
1
(A) 0 (B)
2
(C) 1 (D) 2

Paragraph for Question Nos. 15 to 16

Read the following write up carefully and answer the following questions:
Consider a quadratic function f(x) = ax2 + bx + c (a, b, c  R, a  0) and satisfying the following conditions
(i) f(x – 4) = f(2 – x)  x  R and f(x)  x  x  R
2
 x  1
(ii) f x     x  (0, 2)
 2 
(iii) The minimum value of f(x) is zero

15. The value of the leading coefficient to the quadratic polynomial is

1 1
(A) (B)
4 3
1
(C) (D) 1
2

16. f(1) has the value equal to

1 1
(A) (B)
4 3
1
(C) (D) 1
2

Space for rough work

SECTION – A

Matching List Type

This section contains 4 multiple choice questions. Each question has matching lists. The codes for the
lists have choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

17. Match the following List–I with List–II

List – I List – II

(P)
2 6 5
x x  x 1   1.
x
c
 6 5 2 dx x 2

1/2

2x  3x  2   4x  1

(Q)
x 5
 x4 x  2

dx 2.
1

1  x 2  x 5 
2
c
 7
4x  5x  10x 6 4 2

 2x  5x 
12 9
1 1
(R)  3
dx 3.  
2x3  3x 2  2x 3  c
 x  x  1
5 3 6

 2x  1 dx  x 3 x2 1 
1/2
(S)  4. x   c
3/2 
x 2
 4x  1   25 20 10 
Codes:
P Q R S
(A) 4 1 2 3
(B) 3 4 2 1
(C) 1 3 4 2
(D) 3 2 1 4

Space for rough work

ex e x
18. Let I   dx and J   e4x  1 dx . Then for any arbitrary constant C match the following
e4x  1
List–I with List–II
List – I List – II
2x
1  e 1 
(P) I 1. tan1  x 
c
2  2 e 
1  e2x  2e x  1 
(Q) J+I 2. ln  2x c
2 2  e  2ex  1 
1   e2x  1  1  e2x  2e x  1  
(R) J–I 3.  tan1  x 
 ln  2x
   c
2 2   2e  2  e  2ex  1  
1   e2x  1  1  e2x  2e x  1  
(S) J 4.  tan1  x 
 ln  2x
   c
2 2   2e  2  e  2e x  1  
Codes:
P Q R S
(A) 4 1 2 3
(B) 3 4 1 2
(C) 1 3 4 2
(D) 3 1 2 4

19. Match the following List–I with List–II

List – I List – II
g x  cos x
dt
 1  sin t  dt
2
(P) If f  x    where g  x   then the
3
0 1 t 0 1. 3

value of f '  
 2
(Q) If f(x) is non-zero differentiable function such that
x
2 2. 2
 f  t  dt   f  x   for all x then 2 + f(2) equals
0
b

  2  x  x  dx
2
(R) If is maximum then (a + b) has the value equal
3. 1
a
to
 sin 2x b 
(S) If lim  3  a  2   0 then (3a + b) has the value equal to 4. –1
x 0  x x 
Codes:
P Q R S
(A) 4 1 2 3
(B) 3 4 1 2
(C) 4 1 3 2
(D) 2 1 4 3

Space for rough work

20. Match the following List–I with List–II.

List – I List – II
3
(P)  3 3

If I    x  1   4  x   x cos xdx then |50 2 I| is equal to
2
1. 0

10
(Q) If J   sgn  sin x  dx then 10J is equal to where sgn x denotes
2. 100
0
signam function of x
102
(R) If K   cot 1 x  dx then 70 + [K] is equal to (where [y]
  3. 50
0
denotes greatest integer equal to or less than y
51

  x  25 dx L
0
(S) If L  then is equal to (where [y] and {y}
51
2 4. 70
 x  25 dx
0
denotes greatest integer function and fractional part of function
Codes:
P Q R S
(A) 4 1 2 3
(B) 3 4 1 2
(C) 2 1 4 3
(D) 2 1 3 4

Space for rough work

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ALL INDIA INTEGRATED TEST SERIES

ANSWERS, HINTS & SOLUTIONS
PART TEST –III
(Paper-2)

Q. No. PHYSICS CHEMISTRY MATHEMATICS

A, D A, B, D A
1.
A, C, D A, B, C B
2.
A, C A, B, C D
3.
A, B, C, D A, B, C C
4.
qualified in JEE (Advanced), 2014.

A, B, D A, B, C A, C
5.
A, B, C, D D A, C, D
6.
A, B, D C, D C
7.
A, B, C A, B, C A, B, C
8.
D B B
9.
A A A
10.
B C C
11.
B C C
12.
A B D
13.
C A D
14.
C B A
15.
A B D
16.
A A B
17.
B C D
18.
A D C
19.
C A C
20.

Physics PART – I

9. Use symmetry arguments.

10. zero, due to symmetry

11. B due to infinite wire  I/

  
12. B  B1  B2

13-14. Use Kirchoff’s Law

Chemistry PART – II

SECTION – A

OH
(i) 2CH3MgBr
|
5. (CH3 )2 CHCOOR  (CH3 )2 CH  C  CH3
(ii) H3 O |
CH3
R = C2H5, C6H5

COOEt COOH COOH

COOEt (i) NaOEt H3O+
6. CH2 COOEt COOH 
(ii) BrCH2CH2CH2Br
COOEt

7.
OH
dil
. KMnO
4  HIO 4
OH 
O
OH 
O aldol condensati on

O
(A) (B) (C)

8. ABS is a polymer of acrylonitrile, butadiene and styrene.

cis – polyisoprene is a natural rubber.

13. Aldehyde, without  hydrogen, can respond to Cannizzaro reaction.

14. Ph CO CHO contains one-CO-and another –CHO group adjacently without having
–hydrogen.

Cl
– – – NH2
NH2 NH2 NH2 (1)NH
15. 3
(2) NH3

CH3 CH3 CH3 CH3

Cl NH2 NH2
–
NH2 – NH3
16. + NH2

OCH3 OCH3 OCH3 OCH3

Mathematics PART – III

1. x = 1; y = 2
2=a+b+c ….. (1)
x = 0, y = 0  c = 0  a + b = 2
dy
Now,  2ax  b  1
dx  0, 0
 b = 1, a = 1
Hence, the curve is y = x2 + x
0 0
1
 x    x  dx  24 sq. units
2 2
A  x  x dx 
1/2 1/2

2. I   1  2 cosec x cot x  2cot 2 x dx

2 2
=  cosec x  2 cosec x cot x  cot x dx
=  cosec x  cot x dx
3. Degree of f(x) = n
Degree of f(x) = n – 1
Degree of f(x) = n – 2
 n = (n – 1) + (n – 2) = 2n – 3
n=3
Hence, f(x) = ax3 + bx2 + cx + d, (a  0)
f(x) = 3ax2 + 2bx + c; f(x) = 6ax + 2b
1
a
18
4. For continuity of f at x = 0, we have
sec x  1
k  f  0   lim f  x   1  3 lim using L.H. Rule
x0 x 0 3x 2
1 3
f 0  1 
2 2
5. f(x) = (x2 + bx + c)ex
 f(x) = (x2 + (b + 2)x + (b + c))ex
f(x) > 0 iff D = b2 – 4ac < 0
2
Now, f(x) > 0 iff D = (b + 2) – 4(b + c) = D + 4 < 0
Thus for f(x) > 0 D + 4 < 0 holds
 D < 0  f(x) > 0
x2  1 2
6. y  f  x  2  1 2
x 1 x 1
4x
f 'x  2

x2  1 
x > 0, f is increasing
x < 0, f is decreasing
range is [–1, 1]  into (A – false) minimum value occurs at x = 0
f(0) = –1 (C – false)

 x2  1
A  2 1  2  dx  2 (D – false)
0 x  1

7. D* f(x) = 2f(x) f(x)

D* (x ln x) = 2x ln x (1 + ln x)

9. Given f(x)·f(–x) = f(–x)·f(x)

f '  x  f '  x 

f x f  x 
Integrating ln f(x) = –ln f(–x) + c
ln (f(x)·f(–x)) = c
f(x)·f(–x) = c
2
f(0) = 3  f (0) = c  c = 9
 f(x)·f(–x) = 9
51
dx
10. Let I  
51
3  f  x
51
dx
I  (Using king)
51
3  f  x 
 I = 17
dy
11. y = et sin t   et  cos t  sin t 
dt
dx
x = et cos t   et cos t  sin t 
dt
dy cos t  sin t
   tan 
dx cos t  sin t
 
 tan   t   tan 
4 

 t 
4

t
4

12. f(t) =  e t  cos t  sin t  dt = et sin t + c


 
 f    f  0   e /2  c  c  e /2
2

13. x = f(t) = a
  = a tnb
n b t
….. (1)
n a  t t t
y = g(t) = b =  bna    anb   a tnb

 y = g(t) = a
n b  t
= f(–t)
  ….. (2)
From equation (1) and (2) we get, xy = 1

14. f(t) = g(t)  f(t) = f(–t)  t = 0

{ f(t) is one-one function}
At t = 0; x = y = 1
 xy = 1
dy 1 d2 y 2
   2 and 2
 3
dx x dx x
d2 y
At x = 1, 2
dx 2

15.-16. Since, minimum value is zero

Hence, y = f(x) touches the x-axis and mouth opening upward i.e. a > 0
Given f(x – 4) = f(2 – x) (x  x + 3)
f(x – 1) = f(–1 – x)
f(–1 + x) = f(–1 – x)
Hence, f is symmetric about the line x = –1
 f(x) = a(x + 1)2
Now, also given f(x)  x  x  R
 f(1)  1 ….. (1)
2
 x  1
and f  x     in (0, 2)
 2 
f(1)  1 ….. (2)
From (1) and (2), we get f(1) = 1
Now f(x) = a(x + 1)2
1
f(1) = 4a = 1  a 
4
2

 f x 
 x  1 (Now proceed)
4

17. (P) Let I1  


x 2 x6  x 5  1  dx =
 
x 2 x 6  x5  1
dx 
x 2
 x  x 4  dx
2  2  2
 2x 6
 3x 5
 2  2 
x  2x 3  3x 2  3 
6  2x 3
 3x 2  2x 3 
 x 
Putting 2x3 + 3x2 + 2x–x = t
x6  x5  1 1
 dx  dt
x4 6
1 dt 1
Hence, I1   2  c
6 t 6t
(Q) Putting 4x5 + 5x4 + 10x2 = t
20(x4 + x3 + x)dx = dt
(R) Putting 1 + x–2 + x–5 = t
(–2x–3 – 5x–6)dx = dt
2x  1 2x 2  x 3
(S)  3/2
dx =   4 1 3/2 dx
3 4 1
x 1   2  1   2 
 x x   x x 
1 4
Now, put 2   1  t 2
x x

ex e x
18. Given I   dx and J   e4x  1 dx
e4 x  1
e4 x  e x e3x
J dx =  1  e4x dx
1  e4x

 IJ  

e x 1  e2x
dx

1  e4x
Put ex = t  ex dx = dt
1
1 2
1  t2 t
 I J   dx =  dt
1 t4  1
2

t    2
 t

1  1
Put t   y  dy   t  2  dt
t  t 
dy 1 y 1 t2  1
 IJ   2  tan1 = tan1
y 2 2 2 2 2t
1  e2x  1 
I+J= tan1  x 
c ….. (1)
2  2e 
1 e2x  2e x  1
Similarly, J – I = ….. (2)
2 2 e2x  2e x  1
1  1 e2x  1 1 e2x  2e x  1
Equation (1) – (2) gives I   tan  
2 2  2ex 2 e2x  2e x  1
1  1 e2x  1 1 e2x  2ex  1
Equation (1) + (2) gives J   tan  
2 2  2ex 2 e2x  2e x  1

g'  x 
19. (P) f '  x   and g(x) = [1 + sin(cos2 x)](–sin x)
3
1 g x

1  sin cos2 x  sin x  
Hence, f '  x    
3
1 g  x 
 1 0 1
f '     1
2  1 0
1  g3  
 2
(R) Maximum when a = –1, b = 2
a+b=1
sin2x b
(S) If lim 3
a 2  0
x 0 x x
sin2x  ax3  bx
 lim  0 for limit to be exist 2 + b = 0
x 0 x3
 b = –2
sin2x  ax3  2x
 lim 0
x 0 x3
Applying L.H. rule
2cos 2x  3ax 2  2
lim 0
x 0 3x 2
2 1  cos 2x 
 a  lim
x0 3x 2
2
4 sin x 4
 a 
3x 2 3

3
20.  3 3
(P) I1    x  1   4  x   x cos xdx
2
 ….. (1)

 3 3

Applying king property we get I1    4  x    x  1  5  x cos    5  x   dx
2

2
 3 3

I1     4  x    x  1  5  x cos xdx ….. (2)

3
Now, (1) + (2), gives 2I1    2x  5  cos xdx
2
3 3
 2I1   2x cos xdx   5 cos xdx
2 2
3 3
2I1  2 x cos x dx x  5  cos x dx
2 2
3 3
 x sin x sin x   sin x  1 2
2I1  2   dx   5   = 0  2  1  1   2
   2   2  
 502 I1 = –100
2
Hence, |50  I1| = 100

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PART TEST –IV
ALL INDIA INTEGRATED TEST SERIES
Time Allotted: 3 Hours Maximum Marks: 432
 Pl ea s e r ea d t h e i n s t r u c t i o n s c a r ef u l l y. Yo u a r e a l l o t t ed 5 m i n ut es
s p ec i f i c a ll y f o r t h i s p u r p o s e.
 Yo u a r e n o t a l l o wed t o l ea v e t h e E xa m i n at i o n Ha l l b ef o r e t h e en d o f
t h e t es t .

B. Filling of OMR Sheet

C. Marking Scheme For All Three Parts.

(i) Section-A (01 to 03 and 10 to 12) contains 6 multiple choice questions which have only one
correct answer. Each question carries +8 marks for correct answer and – 2 mark for wrong
answer.

Section-A (04 to 09 and 13 to 30) contains 24 multiple choice questions which have only one
correct answer. Each question carries +4 marks for correct answer and – 1 mark for wrong
answer.

Name of the Candidate

Enrolment No.

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Useful Data

PHYSICS
2
Acceleration due to gravity g = 10 m/s

Planck constant h = 6.6 1034 J-s

Charge of electron e = 1.6  1019 C

Mass of electron me = 9.1  1031 kg

Permittivity of free space 0 = 8.85  1012 C2/N-m2

Density of water water = 103 kg/m3

Atmospheric pressure Pa = 105 N/m2

Gas constant R = 8.314 J K1 mol1

CHEMISTRY

Gas Constant R = 8.314 J K1 mol1

Atomic No: H=1, He = 2, Li=3, Be=4, B=5, C=6, N=7, O=8,

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Physics PART – I

SECTION – A

Straight Objective Type

This section contains 30 multiple choice questions numbered 1 to 30. Each question has 4 choices (A),
(B), (C) and (D), out of which ONLY ONE is correct

1. In a standard meter bridge experiment conducted to measure the specific resistance of a wire the
following data was recorded:-
Length (L)  (1  01)m measured by a meter scale
Radius of wire = 1mm .01mm. (measured by screw gauge).
Resistance of wire (R) = (5.01) 
The maximum possible error in the measurement of specific resistance is
πr 2
(You may use the formula:- ρ  R)
L
(A) .1m (B) .2m
(C) .5m (D) .6m

a 3b 2
2. A physical quantity P is related to four measurable quantities as P  . If the errors in
c 3 d4.
measurement of a, b, c, d are 0.1%, 0.2%, 0.3%, 0.1% respectively, what is the maximum
percentage error in measurement of P.
(A) 1% (B) 2%
(C) 3% (D) 4%

3. When N0 molecules of a radioactive nuclide are taken at t = 0, the activity reduces to ‘A’ in time
t 0 . When 3No molecules of the same nuclide are taken, activity will become 3Ae in time t equal
to
1
(A) t 0 (B) t0 
λ
1
(C) t0  (D) None of these
λ
Where;  is decay constant of the Nuclide
Space for rough work

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4. The plot of characteristic frequency ‘f’ in case of Ky –X-rays, with the atomic number ‘Z’ of an
element may be best represented by
(A) (B)
f f

Z
Z→
(C) (D)
f f

Z Z

5. Two waves of Intensity I1 and I2 interfere. The maximum intensity produced to the minimum
intensity is in the ratio α. Then, I1 / I2 is equal to
α 1 1 α
(A) , if I1 > I2 (B) , if I1 > I2
α 1 α 1
2 2
 1 α   α  1
(C)  , if I1 < I2 (D)  if I1 > I2
 α  1   1  α 
   

6. The current i in the circuit at any time t is 5 L = 0.1 H

 
(A) 20 A (B) 20 sin 50t  A
4 
5

  C = 0.004F

(C) 20 sin 50t A (D) 20 sin  50t  A
 4  I
100 sin (50t) volts.

7. A conducting square loop of length ‘L’ is placed in a magnetic

X X X X X X X AX X X X X X X X
field of Induction B0  End ‘A’ and End ‘C’ are moved away
L B0
from each other. So that finally the square loop becomes a X X X X X X X X X X X X X
straight line. The total charge flowing through the loop, if its B D
resistance is ‘R’ then X X X X X X X X X X X X X
B L2 B L2
(A) 0 (B)  0
πR R X X X X X X X CX X X X X X X XX
2
BL B L2
(C) 0 (D)  0
R R
Space for rough work

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8. A thin biconvex lens of focal length 30cm is Kept 90 cm away

from a plane mirror kept parallel to the plane of the lens. A point O F 90cm
object is placed on the optical axis of the lens 60cm away from 30cm 30cm

the lens as shown in the figure. Now, if the point object is slowly Y
moved towards the lens with a small velocity, the image of the
object in the mirror. X
(A) Moves in the + ve x direction (B) Moves in – ve x direction
(C) Does not move (D) Data insufficient

9. The ratio of de-Broglie wavelength associated with an electron moving in nth orbit of a hydrogen
atom and its first Bohr’s radius is
(A) 2n : 1 (B) n : 2
(C) n3 : 1 (D) n : 1

10. A hemispherical bowl of radius 10 cm is filled with liquid of refractive index

 = 4/3. A glass plate of refractive index 1.5 is placed on the top of bowl. If
for the observer above the plate the shift in position of a point P on the bottom
is 3 cm find the thickness of glass plate. P
(A) 1.5 cm (B) 1 cm
(C) 7 cm (D) 10 cm

11. An electron is knocked out from K shell of tungsten atom. P1, P2, P3 are probabilities that an
electron jumps from L, M and N shell to K shell, then which is correct
(A) P1 =P2 = P3 (B) P3 > P2 > P1
(C) P1 = P2 > P3 (D) P1 > P2 > P3

12. A potential difference of 2 kV is applied across a X-ray tube and X-rays thus produced fall on a
metallic plate at a potential of 100 V. What is the maximum K.E. of photoelectrons emitted from
the plate if its work function is 5.8 eV ?
(A) 2094.2 eV (B) 2105.8 eV
(C) 1906.8 eV (D) 1894.2 eV

13. The velocity of a real object, moving along the principal axis of a lens, at an instant is 4 times the
velocity of the virtual image for a lens. Then the ratio of focal length to image distance is
(A) 2 : 1 (B) 1 : 2
(C) 4 : 1 (D) 1 : 4
Space for rough work

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14. There are 10 atoms of C14 in a sample of carbon atoms. After one half life period the number of
14
radioactive atoms of C in sample will be
(A) exactly 5 (B) certainly more than 5
(C) certainly less than 5 (D) can not find exactly

15. An arrangement for YDSE is shown in f =20 cm

figure. S is a light source then find the
position of central maxima from point O on S O1
screen B.
1mm
(A) 5 mm below O 3 mm
O
(B) 5 mm above O
O2
(C) 30 mm below O
(D) 12 mm above O 80 cm 1m
20 cm
Screen A Screen B

16. An object starts moving at an angle of 450 with the f = 10 cm

principal axis as shown in the figure in front of a
biconvex lens of focal length + 10 cm. If  denotes the 0
angle at which image starts moving with principal axis 45 
then O I

20 cm

3 
(A)  = (B)  =
4 2
 
(C)  = (D)  = –
4 4

17. A thin equiconvex lens, made of material of refractive index 1.5, has a focal length of 10 cm in air.
One side of the lens is replaced by a medium (B) of refractive index 1.3. When a point object is
placed in air at a distance of 20 cm from the lens on its principal axis, its image is formed at
distance of x cm in the medium (B), from the lens. When the same object is placed in the medium
B, at a distance of 20 cm from the lens on its principal axis, its image is formed at a distance of y
cm in air, from the lens. Then
fair =10 cm fair =10 cm
 =1  =1
 =1.3  =1.3

O I I O

20 cm
20cm x y

 =1.5  =1.5
(A) x > 1.3 y (B) x < 1.3 y
(C) x = 1.3 y (D) none of the above is true.
Space for rough work

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18. Planar light waves and sound waves incident on the air – water A
interface along the direction of the line are AO as shown in the figure. 450 air
Assume that AO represents the direction of the incident ray (for both
light and sound). Given that the speed of sound wave in air and water O
is 330 m/s and 1500 m/s respectively and of the light wave it is
8 8
3  10 m/s and 2.26  10 m/s. respectively then, (assume Snell’s
law is valid for both waves):
(A) light wave and sound wave will refract through same angle.
0 B
(B) angle of refraction for light wave will be 30 . Water
(C) water is acoustically denser medium than air for sound wave.
(D) the angle between the two rays when they get separated by the air water surface is
 3 1  0.75  
  sin    rad ..
4  2 

19. Find the value of reactance XC as shown in the figure. i1= 1A 15 XL = 20 
(A) 3 R
0.5
(B) 6 
i2=10A XC
(C) 2
(D) none of these 
V, f, Hz

20. A conducting cross bar of mass m and length L can slide, without friction, L
vertically up and down on two conducting rails. The rails are connected on v 

the bottom via a variable resistance R and a battery E, so that rails and B
the bar form a rectangular conducting loop. A uniform magnetic field B
points into the plane of the loop. Whole system is placed in constant  
h
gravitational field g. Find the value of E for which rod will move up with a
constant velocity v.  R E
mgR mg
(A) (B)  BV 
B RB
mg
(C)  B V (D) zero
RB

21. The value of the current in the resistor R in the following circuit will be C L C L

(Given C = 1 F, L = 1 H, R = 1 )
100 R
(A) zero (B) sin(50t )
R C C
(C) 100/R (D) can’t be calculated

V = 100 sin50t

Space for rough work

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22. A concave mirror is placed in a medium of refractive index 1  1.2. An

O 1 = 1.2
object placed at 30 cm from pole of the mirror and its image coincides.
30cm
Now a liquid of refractive index μ is placed in the concave mirror in the
shaded region while the medium of refractive index 1.2 is retained in
the remaining region. It is found that object needs to be shifted by 7.5
cm below O for the object and image to coincide. The refractive index
of the liquid μ is
4
(A) (B) 1.2
3
(C) 1.6 (D) none of the above

23. An object is placed at 30 cm from a convex lens of focal length 15cm. On the other side of the
lens a convex mirror of focal length 12cm is placed so that the principal axis of both coincide. It is
observed that the object and image coincide. What is the separation between the lens and
mirror?
(A) 6 cm (B) 30 cm
(C) either (A) or (B) (D) none of the above

24. Initially two capacitors each of capacitors C have charges as shown and
the inductor L has zero current. After the switch S is closed, the L Q
C
Q
maximum current in the inductor is
3Q 3Q 2Q
C
(A) (B) 2Q
2LC LC
4Q 5Q S
(C) (D)
LC LC

25. The given figure shows a convergent lens placed inside a cell filled  = 1.6
with liquid. The lens has focal length +20 cm when in air, and its
material has refractive index 1.50. If the liquid has refractive index
1.60, then focal length of the lens in the cell is  = 1.5
(A) 80 cm (B) 80 cm
(C) 24 cm (D) 160 cm
Space for rough work

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26. An equiconvex glass lens (a) has focal length f and power P. It is cut into two
symmetrical halves (b) by a plane containing the principal axis. The two
pieces are recombined as shown in figure (c). The power of the new
combination is (c)
P (a) (b)
(A) zero (B)
2
(C) 2P (D) none of the above

27. Choose incorrect statements:

An inductor of inductance L, capacitance C and resistance R are connected in series to a variable
sinusoidal voltage supply, whose frequency is variable, but rms voltage is constant. At resonance
(I) Average power will be maximum (II) Average power will be minimum
(III) Impedence will be maximum (IV) current and voltage will be in same phase
(A) I and IV (B) II and III
(C) I and III (D) II, III, and IV

28. Which of the following is accompanied by the characteristic X-ray emission?

(A) -particle emission (B) Electron emission
(C) Positron emission (D) K-electron capture

29. The switch Sw of the circuit shown in the figure is closed at t = 0. i1 and i2 are L1
the electric current in the branch containing two ideal inductors L1 and L2
respectively, at a certain instant. Choose the most appropriate statement.
di  di  L2
(A) At t = 0,(just after the switch is closed) 1  and 2 
dt L1 dt L 2
(B) At t = 0, ,(just after the switch is closed) i1 = 0, i2 = 0
L 2 L1
(C) After long time i1  and i2  r Sw
(L1  L 2 ) (L1  L2 )r 
(D) All the above are true

30. Monochromatic light is used in Young’s double slit experiment. When one of the slits is covered
by a transparent sheet of thickness 1.8  105 m, made of material of refractive index 1 number
of fringes which shift is 18. when another sheet of thickness 3.6  105 m, made of material of
refractive index  2 is used, number of fringes which shift is 9. Relation between 1 and  2 is
given by
(A) 4  2  1 = 3 (B) 4 1   2 = 3
(C) 3  2  1 = 4 (D) 2 1   2 = 4

Space for rough work

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Chemistry PART – II

SECTION – A

Straight Objective Type

This section contains 30 multiple choice questions numbered 1 to 30. Each question has 4 choices (A),
(B), (C) and (D), out of which ONLY ONE is correct

1. If a metal crystal exists in fcc and bcc having the radius r. What is the edge length ratio (a1/a2) in
two crystal structure, a2, a1 are the edge lengths of the cube in fcc and bcc respectively
(A) 3 : 2 (B) 2 : 3
(C) 6 : 1 (D) 1: 6

2. What is the oxidation number and coordination number of metal in K 3 Fe  C 2 O 4 3 

(A) O.N = +2, C.N = 3 (B) O.N = +3, C.N = 3
(C) O.N = +3, C.N = 6 (D) none of these

3. Orange red vapours of X is formed by heating the chloride salt with K2Cr2O7 and concentrated
H2SO4. When X on passing in NaOH gives yellow solution of Na2CrO4. Then X =
(A) CrO3 (B) CrCl3
(C) CrO2Cl2 (D) H2CrO4

4. The emf of the following three galvanic cells:

I. Zn / Zn2 1 m || Cu2 1 m | Cu
II. Zn / Zn2  0.1 m  || Cu2 1 m  | Cu
III. Zn / Zn2 1 m || Cu2 0.1 m  | Cu
Are represented by E1, E2 and E3. Which of the following statements is true?
(A) E1 > E2 > E3 (B) E3 > E2 > E1
(C) E3 > E1 > E2 (D) E2 > E1 > E3

5. In which of the following reactions, there is no change in valency?

(A) 4KClO3   3KClO4  KCl (B) SO2  2H2 S 
 2H2 O  3S
(C) BaO2  H2SO4 
 BaSO4  H2O2 (D) 2BaO  O2 
 2BaO2

6. The case of liquefaction of noble gases decreases in the order

(A) He  Ne  Ar  Kr  Xe (B) Xe  Kr  Ar  Ne  He
(C) Kr  Xe  He  Ar  Ne (D) Ar  Kr  Xe  He  Ne

Space for rough work

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7. Which of the following is used as a foaming agent in froth floatation process is

(A) pine oil (B) amyl xanthenes
(C) CuSO4 (D) KCN

8. The plot of 1/xA versus 1/yA (where xA and yA are the mole fractions of A in liquid and vapour
phases respectively) is linear whose slope and intercept respectively are given as
(A) PA0 / PB0 ,  PA0  PB0  / PB0 (B) PA0 / PB0 ,  PB0  PA0  / PB0
(C) PB0 / PA0 ,  PA0  PB0  / PB0 (D) PB0 / PA0 ,  PB0  PA0  / PB0

9. K sp of AgBr  8  10 13
Ag,AgNO3 1.0M // KBr 1.0M AgBr, Ag
The emf of the cell is
(A) 0.71 (B) +0.42
(C) 0.42 (D) +0.71

10. The voltage of a certain cell at 25C and 20C are 0.3525 and 0.3533 V respectively. If the
number of electrons involved in the overall reaction are two, what is the value of S0 at 25C
(A) 30.88 JK1 (B) 60.11 JK1
1
(C) 30.88 JK (D) 60.11 JK1

11. Which of the following is/are paramagnetic?

NO2 ,NO,N2O 4 ,N2 O2 ,N2 O5
(A) only NO2 (B) NO2, NO
(C) NO, NO2, N2O5 (D) all are paramagnetic

12. KClO4 is made in the following sequence of reactions:

Cl2  KOH  KCl  KClO  H2O ...(i)
KClO  KCl  KClO 3 ...(ii)
KClO3  KClO4  KCl ...(iii)
What mass of Cl2 is needed to produce 1.0 kg of KClO3?
(A) 2.05 kg (B) 1 kg
(C) 4.02 kg (D) 5 kg

13. A solution containing 6 g urea per litre isotonic with a solution containing
(A) 5.85 g NaCl per litre (B) 34.2 g sucrose per lire
(C) 36.0 g glucose per litre (D) 6.0 g of sucrose per litre

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14. In the sequence of reaction

k1 k2 k3
A   B  C  D
k3 > k2 > k1, then the rate determining step of the reaction is
(A) A  B (B) B  C
(C) C  D (D) A  D

15. Chemically rust of iron is

(A) hydrated ferrous oxide
(B) hydrated ferric oxide
(C) only ferric oxide
(D) a mixture of ferric oxide with a little of Fe(OH)3

16. Which of the following will give four isomers?

(A)  Co  en NH3  2 Cl2  Cl (B)  Co  PPh3 NH3  2 Cl2  Cl
(C)  Co  en 3  Cl3 (D)  Co  en 2 Cl2  Br

17. The colourless species is

(A) VCl3 (B) VOSO4
(C) Na3VO4 (D)  V H2 O 6 SO 4  H2 O

18. At 25°C, the standard emf of a cell having reaction involving two electron exchange is found to be
0.295 V. The equilibrium constant of the reaction is approximately
(A) 9.51  108 (B) 10
(C) 1  1010 (D) 9.51  109

19. A silver cup is plated with silver by passing 965 coulombs of electricity, the amount of silver
deposited is: (Mass no. of Ag = 108)
(A) 1.08 g (B) 1.002 g
(C) 108 g (D) 9.89 g

20. E° values of Mg2+/Mg is –2.37 volts of Zn+2/Zn is –0.76 volt and Fe+2/Fe is –0.44 volt. Which of the
following statements is correct?
(A) Zn oxidize Fe (B) Zn will reduce Fe2+
2+
(C) Zn will reduce Mg (D) Mg oxidizes Fe

21. Lowering of vapour pressure P1 of aqueous solution of acid HA1 and P2 of aqueous solution of
acid HA2 are related as P1 > P2. Correct relation is (Assume same concentration for both acids)
(A) pK a > pK a (B) pK a < pK a
1 2 1 2

(C) pK a = pK a (D) none of these

1 2

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22. The correct relationship between the boiling points of very dilute solutions of AlCl 3(T1), and
CaCl2(T2) having the same molar concentrations is
(A) T1 = T2 (B) T1 > T2
(C) T2 > T1 (D) T2  T1

23. The vapour pressure of pure benzene at 25°C is 639.7 mm Hg and the vapour pressure of a
solution of solute in benzene at 25°C is 631.9 mm Hg. The molality of the solution is
(A) 0.156 (B) 0.108
(C) 0.518 (D) 0.815

24. The stabilization of the dispersed phase in a lyophobic sol is due to

(A) the viscosity of the medium.
(B) the surface tension of the medium.
(C) affinity for the medium.
(D) the formation of an electrical layer between the two phase.

25. The process which is catalysed by one of the products is called

(A) acid-based catalyst. (B) auto catalyst.
(C) negative catalyst. (D) positive catalyst.

26. Which of the following statements is correct regarding the slag obtained during the extraction of a
metal like copper or iron?
(A) The slag is lighter and has higher melting point than the metal.
(B) The slag is lighter and has lower melting point than the metal.
(C) The slag is heavier and has higher melting point than the metal.
(D) The slag is heavier and has lower melting point than the metal.

27. In oxoacids of chlorine, ClO bond contains

(A) pp bond (B) dd bond
(C) pd bond (D) df bond

28. The compound that is not a Lewis acid is

(A) BaCl2 (B) AlCl3
(C) BCl3 (D) SnCl4

29. The formula of sodium nitroprusside is

(A) Na4[Fe(CN)5NO2] (B) Na2[Fe(CN)5NOS]
(C) NaFe[Fe(CN)6] (D) Na2[Fe(CN)6NO2]

2 3 2
30. Amongst  TiF6  ,  CoF6  ,Cu 2 Cl2 & NiCl4  . The colourless species are
2 3 2
(A)  TiF6  & Cu2 Cl2 (B)  CoF6  & NiCl4 
2 3 2
(C)  TiF6  & CoF6  (D) Cu 2 Cl2 & NiCl4 

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Mathematics PART – III

SECTION – A

Straight Objective Type

This section contains 30 multiple choice questions numbered 1 to 30. Each question has 4 choices (A),
(B), (C) and (D), out of which ONLY ONE is correct

1. The number of values of a for which ˆ  a ˆi  2ajˆ  3ak, ˆ   (2a  1)iˆ  (2a  3)jˆ  (a  1)k,
ˆ and
ˆ  (3a  5)iˆ  (a  5)jˆ  (a  2)kˆ are coplanar
(A) 0 (B) 1
(C) 3 (D) 6

2.  
The differential equation xy3 1  cos x   y dx  xdy  0 represents the curve
x2 x3
2
  x 2 sin x  cx cos x  dsin x  k . Then a + b + c + d is equal to
ay b
(A) 8 (B) 9
(C) 10 (D) 11

3. Let p and q be unit vectors such that p  q  p  r and r  p  q . Then the value of p q r
is
(A) 0 (B) 1
(C) 2 (D) 3

a b c 
4. The matrix A  c a b  is an orthogonal matrix. Then a  b  c  k , where k is
b c a 
(A) 0 (B) 1
(C) 2 (D) 3

5. Seven digit numbers are formed using digits 1, 2, 3, 4, 5, 6, 7, 8, 9 without repetition. The
probability of selecting a number such that product of any 5 consecutive digits is divisible by
k
either 5 or 7 is , then k is
12
(A) 2 (B) 4
(C) 6 (D) 7

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 1 1 a11
6. If a singular matrix A  aij  always commute with B    such that  k , then k is
2 2
2 1 a12
(A) 0 (B) 1
(C) 2 (D) 3

1 2 a b 
7. Let A    and B    be two matrices such that they are commutative and c  3b then
 3 4  c d
ad
the value of is
3b  c
(A) 0 (B) 1
(C) 2 (D) 3

8. Two natural numbers x and y are chosen at random. If P is the probability that x 2 + y2 is divisible
by 5 and 7 than the value of 25 P is
(A) 3 (B) 6
(C) 9 (D) none of these

2 2 2 2
9. p  ˆi  p  ˆj  p  kˆ is K p . Then K is
(A) 0 (B) 1
(C) 2 (D) 3

10. ˆ ˆ ˆ are non – zero and coplanar then

Given a,b,c
ˆ ˆ  c,c
(A) aˆ  b,b ˆ ˆ  aˆ are always coplanar ˆ ˆ  c,c
(B) aˆ  b,b ˆ ˆ  aˆ are always coplanar
ˆ ˆ
(C) 2aˆ  3b,2b  3c,2c
ˆ ˆ  3aˆ are always coplanar (D) none of these

x 1 y z
11. The lines   and x – y + z – 2 = 0 =  x  3z  5 are coplanar for  is equal to
2 1 2
97 43
(A)  (B) 
11 5
73
(C)  (D) –5
9

12. If A is a square matrix of order n with |A| = 9 and |adj adj adj A| = 316 then number of elements in
A is
(A) 0 (B) 1
(C) 2 (D) 3

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1 1
 z1 z2   z1 z2  1/ 4 0 
13. For two unimodular complex numbers z1 and z2 z k  then k
 2 z1    z2 z1   0 1/ 4 
is
(A) 4 (B) 2
(C) 0 (D) 1

a b c
14. If a, b, c are the roots of x3 + 2x2 + p = 0, then the value of b c a is
c a b
(A) 2 (B) 4
(C) 6 (D) 8

y z
15. A line L is passing through (1, 1, 1) intersects line x   at A and line
2 3
L2  2 (x + 1) = 6 (y + 2) = 3 ( z + 3) at B. Then distance of B from (0, 0, 0) is
(A) 12 (B) 14
(C) 18 (D) none of these

16. The statements are not true

(A) det(A  B)  det A  det B for all square matrices A, B of order n
(B) det(AB)  det A.detB for all A, B of order n
(C) det(A.B)  det A  det B
(D) none of these

17. There are n urns u1, u2, ….., un. Each urn contains 2n  1 balls. The ith urn contains 2i number of
balls P  ui  , i.e., probability of selecting ith urn is proportional to i2  2 . If we randomly select one
of the urns and draw one ball and probability of ball being white be P  A  , then lim P  A  is
n
3 1
(A) (B)
4 4
2 3
(C) (D)
3 16

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18. Let p, q, r be three vectors such that p q r   2 . If a  x  q  r   y  r  p   z  p  q  is

perpendicular to p  q  r . Then the value of x  y  z is
(A) 2 (B) 1
(C) 0 (D) none of these

19. PQRSP’Q’R’S’ (where P, Q, R, S are vertices corresponding to P, Q, R, S respectively) is a
cube of edges 1 unit. A and B are midpoints of the edges Q’P’ and Q’R’ respectively. The
distance of vertex S from the plane AQB is
3 3
(A) units (B) units
4 2
4
(C) units (D) 1 unit
3
    
20. If a,b,c are unit vectors perpendicular to each other such that r.a  1, r.b  2 and r.c  3 then
       
(A) r  3a  2b  c (B) r  a  2b  3c
      
(C) r  3a  b  2b  c  c  a (D) none of these

21. Four of the eight vertices of a regular octagon are chosen at random. The probability that the
figure with these four vertices is square, equals
1 1
(A) (B)
14 7
1 1
(C) (D)
28 35

dx 1 x2
22. The differential equation  determines a family of circles with
dy x
(A) variable radii and a fixed centre (1, 0)
(B) variable radii and a fixed centre (–1, 0)
(C) fixed radius 1 and variable centres along x-axis
(D) fixed radius 1 and variable centres along y-axis

x 1 y  2 z  3 x4 y6 zk

23. If the lines   and   intersect then the value of k is
2 3 4 1 2 2
(A) – 7 (B) – 8
(C) – 9 (D) – 10

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4 sinB
1 cos A
b
24. The value of determinant  = 2a 8 sin A 1 is (where a, b and c are the sides of a triangle
3a 12sin A cosB
and A, B and C are angles opposite to a, b and c respectively)
(A) 0 (B) 1
(C) 2 (D) 3

25. The equations ax + a2y + z = 0, bx + b2y + z = 0, cx + c2y + z = 0 have only zero solution if
common ratio of G.P. is not equal to (here a, b, c are in G.P.)
(A) 0,  1 (B) 2,  3
(C) 4,  5 (D) none of these

1/ x
 
lim tan   x  x3 e2
x 0 4 
x
26. The value of  = lim 1 x4  1 1 equals
x 
1
x 2 sin
lim x x5  9 0
x 0 sin x
(A) 1 (B) 2
(C) 3 (D) none of these

27. If a,ˆ b,ˆcˆ are three non-coplanar unit vectors, then

     
aˆ p q aˆ  bˆ p q bˆ  cˆ p q cˆ is equal to
   
 
(A) aˆ  bˆ  cˆ   p  q (B) aˆ  bˆ  cˆ  p  q
   
(C) p  q (D) p  q

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 
28. If a and b are two unit vectors and  is the angle between them, then the unit vector along the
 
angular bisector of a and b will be given by
   
ab ab
(A) (B)
 
2cos 2cos
2 2
 
ab
(C) (D) none of these

2cos
2

29. The equation of the line passing through the point (1, –2, 3) and parallel to the line
x – y + 2z = 5 and 3x + y + z = 6 is
x 1 y  2 z  3 x 1 y  2 z  3
(A)   (B)  
3 5 4 1 1 2
x 1 y  2 z  3 x 1 y  2 z  3
(C)   (D)  
3 1 6 3 1 2

1
30. The area of the region bounded by x = , x = 2, y = ln x and y = 2x is
2
3 5 4 2 3 5 4 2
(A)  ln2 + (B)  ln2 +
2 2 ln2 2 2 ln2
3 5 4 2 3 5 4 2
(C)  ln2  (D)  ln2 –
2 2 ln2 2 2 ln2

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ALL INDIA INTEGRATED TEST SERIES

ANSWERS, HINTS & SOLUTIONS
PART TEST – IV
(Main)

Q. No. PHYSICS CHEMISTRY MATHEMATICS

1. A B B
2. B C B
3. C C B
4. D D B
5. C C D
qualified in JEE (Advanced), 2014.

6. D B C
7. C A B
8. B B C
9. A A C
10. A C B
11. D B D
12. D A D
13. A B B
14. D A D
15. A D B
16. D D A
17. B C A
18. D C C
19. B A C
20. C B B
21. A B D
22. C B D
23. C A C
24. A D A
25. D B A
26. C B D
27. A C D
28. D A B
29. D B A
30. A A A

Physics PART – I

3  14  10 6  5
1.   15.7 ohm metre.
1
 2r L R 
      ohm metre.
 r L R 
 01 01 01
 15  7 2    ohm metre
 1 1 5 
 15.7μ 02  01  002 ohm – metre
 15.7  032  m
= 0.50 m

P
2. 100   3  0 1 2  0  2  3  0  3  4  0  1
P
 3  4  9  4  2%

3. A  λN0 e  λt0
3Ae  3λN0 e  λt
1
 e  0   e 1
λ t t

e
1
t  t0  .
λ
2
4. f  a2  Z  b  .
 (D)

5.
Imax

 I1  I2  α
2
Imin
 I1  I2 
2
I1  α  1 
  if I1 > I2
I2  α  1 

5 2 10 2
100  π
6. Ic  sin  50t   /4
5 2  4
/4
100  π I=Ic+IL
IL  sin 50t  /4
/4

5 2  4 
I = 20sin(50t) 5 2

8. When u decreases v increases. For plane mirror u decreases.

the image in plane mirror moves towards the left.

9. n  2rn and rn = n2r1

  :r1  2n :1

t1 t 2
10. Apparent depth = 
1  2
t t 
Shift =  t1  t 2    1  2 
 1 2 
 t 10  3 
3.0   t1  10    1  
 1.5 4 
t1
3.0 =  2.5
3
t1 = 1.5 cm
11. In case of X-rays characteristic intensity of K, K, K . . . . goes on decreasing while photon
energy increases. This indicates that probabilities of P1, P2 and P3 should be in increasing order.

12. Maximum energy of X – ray photon = 2000 eV

KEmax = 2000 – (100 + 5.8) = 1894.2 eV
v0 x2
13.  2
vi y
x=2y
1 1 1
  
v u f
1 1 1
 
y 2y f
f
2
y
y S
15. x  dsin   d 0
D
 y = D sin 
1 
 1000   5 mm. below.
200

16. m = 1 and Vxo = Voy

VxI = m2 Vxo
VyI = -m Vyo


4
 2 1  2  1
17.  
v' 4 R
3  2  3   2 3 1  3  1  2 2
    
v v' R v u R
1.3 1   1  22 1 1.3 3  1  2 2
  3   
x 20 R y 20 R
1.3 1 1 1.3
Hence,   
x 20 y 20
 1.3 y > x

18. The sound wave will be totally internally reflected on the surface. Sound
For light wave 0 0
45 45
3  108
1 sin 450 = sin r
2.2  108 O

 0.75 
 r = sin-1   rad . r
 2 
  0.75   3  0.75  
 =-  sin1   =   sin1    rad. Water
4  2   4  2  
19. 152 + 202 = 52 + X2C  100  X2C 100 = 600  XC = 6

20. FT  Fm  mg  ILB  mg
E  BV  E  BV 
so, I  =   B  mg  0
R  R 
mg
E=  B V
RB
21. Using symmetry, current will not flow in the resistor.

22.
1 1 2
  
1 2

 m

  1 1

m
   1 m 

15
 feq  m
feq f1 f2 15 R1 15 15 15 
O μ  1.2
 30  7.5 
22.5
and feq  
2 2
22.5 15 4
This  m m 1  .
2 1 3
m  4  4  1.2
Also       1.6
m 3 1.2 3
23. For image to be coincident, either the rays should retrace or the image due to the lens should
formed just at the pole of the mirror in thin case. The image formed due to lens is at 30 cm (2f) be
from the lens. Thus either this image should be at centre of curvature of the convex mirror or at
the pole of the mirror. Hence 6cm or 30 cm should be the separation between the lens and the
mirror.
24. The charges on each capacitor will be equal to Q/2 (how?) when the current in the circuit is
maximum.

29. Initially the current in the circuit is zero, and there is no potential drop across the internal
resistance.
di di
L1 1  L 2 2  , at t = 0.
dt dt
At any time t,
di di
L1 1  L 2 2    ir
dt dt

30. (  1)t = n
(1  1)  1.8  10 5 18

( 2  1)  3.6  10 5 9
(1  1)  4(  2  1)
4 2   1  3

Chemistry PART – II

SECTION – A

1. In fcc a2 = 2 2r
4r
bcc a1 =
3
a1 4r 1 2
   
a2 3 2 2r 3

2. In K 3 Fe  C 2 O 4 3  ; O.N. of Fe = 3 + x + 3(2) = 0

x = +3
C2O42 is a bidentate ligand.
C.N of Fe = 3(2) = 6

3. 4NaCl  K 2Cr2O7  3H2SO4  2CrO2Cl2  2Na2SO4  K 2SO4  3H2O

Cr2 O2 Cl2  4NaOH  Na2 CrO4  2NaCl  2H2 O

4. Cell reaction is Zn  Cu2 

 Zn2  Cu
2
0 RT  Zn 
Ecell  Ecell  ln
nF Cu2 

Greater the factor  Zn2  / Cu 2  , less is the EMF.

6. He has the weakest vander waal’s forces of attraction due to its smallest size and hence lowest
boiling point. In other words, its liquefaction is difficult. As the size of noble gas increases from He
to Xe, vander waal’s forces increase and consequently boiling point increases and liquefaction
becomes easy.

8. PA0  x A PA0 PB0  XBPB0

PA x APA0 x APA0
yA   
PA  PB x APA  xBPB x APA  1  x A  PB0
0 0 0

1


x A PA0  PB0  PB0 
yA x APA0
1 PA0 1 PB0  PA0
 
x A Pb0 y A PB0
PA0 PB0  PA0
 slope  Iint ercept 
PB0 PB0

9.  Ag   8  10 13  Ag   1.0M

R L

 Ag 
E  0  0.059log    L  0.71 V
 Ag 
R

10. G0 = nFE0

H0 - TS0 = nFE0
S 0 H0
E0  T
nF nF
Solving for E0 at 25C and 20C gives
0
S = 30.88 JK1

12. Cl2  2KOH  KCl  KClO  H2O ...(i)

3KClO  2KCl  KClO 3 ...(ii)
4KClO3   3KClO4  KCl ...(iii)
Multiply equation (i) by 12 and equation (ii) by 4 and add equation (i), (ii) and (iii) to obtain
12Cl2  24KOH  21KCl  3KClO 4  12H2O
852 415.5
852  1000
1000 g KClO4 will be produced from = g
415.5
= 2.05 kg

13. Isotonic solutions have same molar concentration.

6g urea/litre = 0.1 M
34.2 g sucrose/litre = 0.1 M

k1
14. As k1 is lowest, hence A   B is slowest.

16.  Co  en 2 Cl2  Br shows ionization isomerism, cis and trans isomerism and optical isomerism.

17. Na3VO4 contains vanadium in +5 oxidation state which has vacant d – orbitals and is thus
colourless.

18. G° = –2FE° = –RT ln K

K = 1010

108
19. m = ZQ ; m =  965 = 1.08
1  96500

20. For Zn + 2Fe

+2 +2
 Zn + 2Fe E ocell  0.44  0.76   ve and hence, the reaction is
spontaneous.

21. Lowering, P  ‘a’ number of solute particles.

Hence, HA1 is stronger acid than HA2 (Given P1 > P2) and pK a < pK a .
1 2

(639.7  631.9) 1000

23. m=  = 0.156.
639.7 78
24. The stabilization of the dispersed phase is due to adsorption of the ions.

25. Auto catalyst

30. In case of Ti, it is in +4 oxidation state and has no d-electrons in case of Cu d-orbitals are filled so
in both cases d-d-transition will not take place and compound will be colourless.

Mathematics PART – III

2
 31  1259 
1. Because |  | 15a   a    0
 30  900 

Only if a = 0

dy y
2.  y 3 1  cos x    0
dx x
1 dy 1
   1  cos x 
y 3 dx y2 x
1
Let u
y2
1 du u
    1  cos x 
2 dx x
2
 dx
IF  e x  x2
 ux 2  2  x 2 1  cos x  dx
x2 x3
 2
  x 2 sin x  2x cos x  2sin x  k
2y 3

3. p  q  p  p  q
 p  q  p  q
  p.q   0
  p q r   p.  q  r   p.  q   p  q  p     q2  .  p 2   1

4. AA T  I
a b c  a c b   1 0 0 
 c a b  b a c   0 1 0 
b c a  c b a 0 0 1
 a2  b2  c 2 ac  ab  bc ab  bc  ca   1 0 0 
 
 ac  ab  bc a2  b2  c 2 ab  bc  ca   0 1 0 
 2 2 2  
ab  bc  ac bc  ac  ab a  b  c  0 0 1
a2  b2  c 2  1  ab  bc  ca  0
 a  b  c 2  1  2  0   1  a  b  c  1

 a11 a12   1 1  1 1 a11 a12 

6. a    
 21 a22  2 1 2 1 a21 a22 
 a11  2a12  a11  a21 ..... (i)
a11  a12  a12  a22 ..... (ii)
a21  2a22  2a11  a21
a21  a22  2a12  a22

 From (ii), a11  a22

a21  2a12
 Given matrix is singular
 a11a22  a21a21
2 2
  a11   2  a12 
 a11  2a12

7. AB = BA

9
8. P=
25

9. Let p  p1ˆi  p2 ˆj  p3kˆ

p  ˆi  p kˆ  p ˆj
2 3
2
p  ˆi  p22  p32
2
p  ˆj  p12  p32
2
p  kˆ  p12  p22
2

Required solution = 2 p12  p22  p23  2 p 
11. The lines should intersect or be parallel
Let  2t  1,  t,2t  be any point on the first line
Putting it in both the planes we get   5

3
|adj adj adj A| = | A |
n 1
12.  316  n  3

2 2 4  3  2 1 1
21. P= 8
  .
C4 8765 35

2xdx
22.  2dy
1  x2
1  x2    y  c 
2 2
x + (y + c) = 1
 centre (0, – c).

x 1 y  2 z  3
23.   =a
2 3 4
x4 y6 zk
  =b
1 2 2
x = 2a + 1, y = 3a + 2, z = 4a + 3
x = b + 4, y = 2b + 6, z = 2b – k
2a + 1 = b + 4
 2a – b = 3 …(1)
3a – 2b = 4 …(2)
a = 2, b = 1

4a + 3 = 2b – 12
 k = 2b – 4a – 3 = 2 – 8 – 3 = – 9.

sin A sinB sinC

24.   k
a b c
1 4k cos A
 = 2a 8ak 1 = 0.
3a 12ak cosB

2
25. a, b, c are in G.P. with common ratio r, such that b = ar, c = ar
Now system of equation have only zero (trivial) solution.
a a2 1
If  = b b2 1  0
c c2 1
(a – b)(b – c)(c – a)  0
(a – ar)(ar – ar2)(ar2 – a)  0
a3r(1 – r)3(a + r)  0
r  0,  1.

1/x
 
26. lim tan   x   e2
x 0 4 
lim 1x  1
x 
1
x 2 sin
lim x 0
x 0 sin x
  = 0.
     
27. â p q = projection of p  q in the direction of â . Hence the given vector is p  q


  a  b
28. Vector in the direction of angular bisector of a and b =
2 
  a
ab
have magnitude cos(/2)
2 /2
 
ab
So, the unit vector in this direction will have magnitude
2cos
 
2 b

29. If l, m, n be the d.c.’s of the line, them as it lies in both the given planes it is perpendicular to their
l m n
normals i.e. l – m + 2n = 0 and 3l + m + n = 0 or = 
3 5 4
x 1 y  2 z  3
 Equation of the line is  
3 5 4

2
x 3 5 4 2
30. Required area is  (2  ln x)dx =  ln 2 
2 2 ln 2
1/2

Time Allotted: 3 Hours Maximum Marks: 210

 Pl ea s e r ea d t h e i n s t r u c t i o n s c a r ef u ll y . Yo u a r e a l l o t t ed 5 m i n u t es
s p ec i f i c a ll y f o r t h i s p u r p o s e.
 Yo u a r e n o t a ll o wed t o l ea v e t h e E xa m i n at i o n Ha ll b ef o r e t h e en d o f
t h e t es t .

B. Filling of OMR Sheet

C. Marking Scheme For All Three Parts.

(i) Section-A (01 to 10) contains 10 multiple choice questions which have only one correct answer.
Each question carries +3 marks for correct answer and – 1 mark for wrong answer.

Section-A (11 to 15) contains 5 multiple choice questions which have more than one correct
answer. Each question carries +4 marks for correct answer. There is no negative marking.

(ii) Section-C (01 to 05) contains 5 Numerical based questions with answers as numerical value
and each question carries +4 marks for correct answer. There is no negative marking.

Name of the Candidate

Enrolment No.

Useful Data

PHYSICS
Acceleration due to gravity g = 10 m/s2

Planck constant h = 6.6 1034 J-s

Charge of electron e = 1.6  1019 C

Mass of electron me = 9.1  1031 kg

2 2
Permittivity of free space 0 = 8.85  1012 C /N-m

Density of water water = 103 kg/m3

Atmospheric pressure Pa = 105 N/m2

Gas constant R = 8.314 J K1 mol1

CHEMISTRY

Gas Constant R = 8.314 J K1 mol1

Atomic No: H=1, He = 2, Li=3, Be=4, B=5, C=6, N=7, O=8,

Physics PART – I

SECTION – A

Straight Objective Type

This section contains 10 multiple choice questions numbered 1 to 10. Each question has 4 choices (A),
(B), (C) and (D), out of which ONLY ONE is correct.

1. Binding energy per nucleon vs mass number curve for nuclei is

nucleon in MeV
shown in the figure W, X, Y and Z are four nuclei indicated on

Binding energy
Y
8.5
the curve. The process that would release energy is 8.0
X
W
(A) Y  2Z (B) W  X + Z 7.5
5.0 Z
(C) W  2Y (D) X  Y + Z
0 30 60 90 120
Mass number of nuclei

2. A wire is bent to form the double loop shown in the figure. There is a x x x x
A B
uniform magnetic field directed into the plane of the loop. If the x x
C Dx x
magnitude of this field is decreasing, current will flow from x x x x
(A) A to B and C to D (B) B to A and D to C
x x x x
(C) A to B and D to C (D) B to A and C to D

3. A cubical room is formed with 6 plane mirrors. An insect moves along the diagonal of the floor
with uniform speed. The speed of its image in two adjacent walls is 202 cm/s. Then the speed of
the image formed on the roof is
(A) 20 cm/s (B) 40 cm/s
(C) 202 cm/s (D) 102 cm/s

4. A prism of refractive index 2 refracting angle A produces minimum deviation Dm of a ray on one
face at an angle of incidence 45 . The values of A and Dm are respectively.
(A) 45, 45 (B) 45, 60
(C) 60, 30 (D) 60, 45

5. Which of the following statement is correct?

(A) The rest mass of a stable nucleus is less than the sum of the rest masses of its separated
nucleons.
(B) The rest mass of a stable nucleus is greater than the sum of the rest masses of separated
nucleons.
(C) In nuclear fission, energy is released by fusing two nuclei of medium mass (approximately
100 amu)
(D) In nuclear fission, energy is released by fragmentation of a very heavy nucleus.
Space for rough work

6. Shown in the figure is a circular loop of radius r and resistance R. A B

variable magnetic field of induction B = et is established inside the X X

coil. If the key (K) is closed at t = 0, the electrical power developed in X X X
R
the external resistor at this instant is equal to X rX K
r 2 10r 3
(A) (B)
R R
 2r 4R  2r 4
(C) (D)
5 4R

7. In a YDSE apparatus, two films of thickness tA and tB, having refractive indices A and B are
placed in front of slits A and B respectively. If tAA = tBB the central maximum will
(A) not shift (B) shift towards A if tA < tB
(C) shift towards B if tA < tB (D) shift towards A if tA > tB

8. Velocity of image of fish in the mirror with respect to ground is Thin

(A) zero glass
Fish 4m/s
(B) 2 m/s towards mirror.
(C) 2 m/s away from mirror =2
Observer
(D) 4 m/s towards mirror. 1 m/s

I B
9. The network shown in figure is a part of a complete circuit. If
A 1 5mH
at a certain instant the current (I) is 5A, and is decreasing at a 15 V
rate of 103 A/s. Then VB – VA is equal to
(A) 15V (B) 25V
(C) 20V (D) 10 V

10. A planoconvex lens made of a material of refractive index 1.5 is silvered on the convex surface.
Calculate the radius of curvature of the curved surface (Given : silvered lens which behaves as a
mirror of focal length of -20 cm):
(A) 10 cm (B) 20 cm
(C) 30 cm (D) 60 cm
Space for rough work

Multiple Correct Choice Type

This section contains 5 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for
its answer, out which ONE OR MORE is/are correct.

11. Parallel rays of light are falling on convex spherical surface of air   1.5
radius of curvature R = 20 cm. Refractive index of the medium is
  1.5. After refraction from the spherical surface parallel rays :
(A) actually meet at some point
(B) appears to meet after extending the refracted rays backwards
(C) meet (or appears to meet) at a distance of 30 cm from the spherical surface
(D) meet (or appears to meet) at a distance of 60 cm from the spherical surface

12. A variable voltage V  2t is applied across an inductor of inductance L=2H

L  2H as shown in figure. Then,

(A) current versus time graph is a parabola V = 2t
(B) energy stored in magnetic field at t = 2s is 4J
(C) potential energy at time t = 1 s in magnetic field is increasing at a rate of 1 J/s
(D) energy stored in magnetic field is zero all the time.

13. In the given circuit, initially the charge on capacitor is Qo . At t = 0, C

the switch S is closed. Which of the following statement is/are

correct?
Qo S
(A) Maximum current through the inductor is L
LC

(B) Charge on the capacitor is zero at t  LC
2

(C) Current through the inductor is aero at t  LC
2
(D) Energy stored in magnetic field is zero all the time.

14. The potential difference applied to an X-ray tube is increased. Then which of the following is
incorrect.
(A) The intensity increases. (B) The minimum wavelength increases.
(C) The intensity decreases. (D) The minimum wavelength remains constant.

15. Which of the following is correct?

(A) Young’s double slit experiment uses in coherent source
(B) Incoherent sources may having variable phase difference.
(C) Young’s double slit experiment uses in incoherent source
(D) Incoherent sources must having same phase difference.
Space for rough work

SECTION – C

Integer Answer Type

This section contains 5 questions. The answer to each of the questions is a single-digit X Y Z W

integer, ranging from 0 to 9. The appropriate bubbles below the respective question 0 0 0 0

numbers in the ORS have to be darkened. For example, if the correct answers to question 1 1 1 1
numbers X, Y, Z and W (say) are 6, 0, 9 and 2, respectively, then the correct darkening of 2 2 2 2
bubbles will look like the as shown. 3 3 3 3

4 4 4 4

5 5 5 5

6 6 6 6

7 7 7 7

8 8 8 8

9 9 9 9

1. If a thin convex lens of focal length 20 cm is B

cut into two pieces from a plane passing
through the pole, then both the pieces A
(touching each other) kept as shown in 30 cm
figure. If a object AB of length 2 cm is kept
at distance of 30 cm. Find the height (in
cm) of the image.

2. A solenoid of inductance 100 mH and resistance 20  is connected to a cell of emf 10 V at t = 0.

1
If the energy stored in the inductor when the time t = 5 ln2 milliseconds is J, then find the
64 x
value of “x”?
3. A wire of length  = 6  0.06 cm and radius r = 0.5  0.005 cm and mass m = 0.3  0.003 gm.
Then find the maximum percentage error in density.
r 
4. If the attractive potential between electron and proton were given by V(r) = V0 ln   (where V0,
 r0 
r0 are constants), the radius rn of the nth Bohr orbit in Hydrogen atom would be proportional to nK.
Then find the value of K.
o
5. An isosceles prism of angle 120 C has a refractive index 1.44. Two parallel
monochromatic rays enter the prism parallel to each other in air as shown. The
rays emerge from the opposite faces make an angle K [sin1(0.72)  30] with 120

each other. Then find the value of K.

Space for rough work

Chemistry PART – II

SECTION – A

Straight Objective Type

This section contains 10 multiple choice questions numbered 1 to 10. Each question has 4 choices (A),
(B), (C) and (D), out of which ONLY ONE is correct.

1. The packing efficiency of the two dimensional square unit cell shown below is
(A) 39.27 % (B) 68.02%
(C) 74.05 % (D) 78.54 %

2. The freezing point of aqueous solution contains 5% by mass urea, 1% by mass KCl and 10% by
mass of glucose is (Kf (H2O) = 1.86 K molality)
(A) 290.2 K (B) 295.5 K
(C) 269.9 K (D) 250 K

3. Zn gives H2 gas with H2SO4 and conc. HCl but not with conc. HNO3 because
(A) NO3 ion is reduced in preference to hydronium ion
(B) conc. HNO3 is a weaker acid than conc H2SO4 and conc. HCl
(C) conc. HNO3 acts as a reducing agent
(D) Zn is more reactive than H2

2 4
4. Fe H O  and Fe CN  differ in
 2 6   6 

(A) magnetic moment and colour

(B) geometry and magnetic moment
(C) oxidation state of central metal ion and number of unpaired electron
(D) geometry and hybridization

5. HClO4  P2O5   A B
(A) and (B) are
(A) HClO3, H3PO4 (B) Cl2O6, HPO3
(C) Cl2O7, HPO3 (D) ClO2, H3PO4

Space for rough work

6. Which of the following minerals does not contain aluminium?

(A) Cryolite (B) Mica
(C) Feldspar (D) Fluorspar

7. Copper metal on treatment with dil HNO3 produces a gas X. X when passed through acidic
solution of stannous chloride a nitrogen containing compound Y is obtained. Y on reaction with
nitrous acid produces a gas Z. Gas Z is
(A) NO (B) N2
(C) NO2 (D) N2O

8. Alum helps in purifying water by

(A) Coagulating the mud particles
(B) Making mud water soluble.
(C) Forming Si complex with clay particles.
(D) Sulphate part which coming with dirt and remove it.

9. Which of the following order is correct for bond angles in given compound?
OF2 OCl2 OH2
(I) (II) (III)
(A) I > II > III (B) III > I >II
(C) II > III > I (D) II > I > III

10. Which of the following complex is sp3 hybridized and paramagnetic?

(A) Ni(CO)4 (B) [Ni(CN)4)]2
2-
(C) [NiCl4] (D) [Cu(NH3)4]2+

Space for rough work

Multiple Correct Choice Type

This section contains 5 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for
its answer, out which ONE OR MORE is/are correct.

11. Which of the following compounds are coloured due to charge transfer spectrum?
(A) K2CrO7 (B) KMnO4
(C) [Co(NH3)6]Cl3 (D) [Cu(NH3)4]SO4

12. Nitrogen dioxide can be obtained by heating

(A) KNO3 (B) Pb(NO3)2
(C) LiNO3 (D) AgNO3

13. Which of the following statement is correct?

(A) Fe2+ gives brown colour with ammonium thiocyanate
2+
(B) Fe gives blue precipitate with potassium ferricyanide
3+
(C) Fe gives brown colour with potassium ferricyanide
(D) Fe3+ gives red colour with potassium ferrocyanide

14. Which of the following are ambidentate ligands?

(A) CN (B) SCN

(C) NO2 (D) CH3COO

15. In which of the following pairs of solutions will be the values of the Van’t Hoff factor be the same
(A) 0.05 M K4[Fe(CN)6] and 0.10 M FeSO4
(B) 0.10 M K4[Fe(CN)6] and 0.05 M FeSO4. (NH4)2.SO4.6H2O
(C) 0.20 M NaCl and 0.l M BaCl2
(D) 0.05 M FeSO4(NH4)2SO46H2O and 0.02 M KCl.MgCl2.6H2O

Space for rough work

SECTION – C

Integer Answer Type

This section contains 5 questions. The answer to each of the questions is a single-digit X Y Z W

integer, ranging from 0 to 9. The appropriate bubbles below the respective question 0 0 0 0

4 4 4 4

5 5 5 5

6 6 6 6

7 7 7 7

8 8 8 8

9 9 9 9

1. The total number of P – O – P and P – O – H bonds present in pyro phosphoric acid.

2. How many lone pairs are present in H3O+?

3. How many bond angles are 90C in XeF4?

4. Number of salts among the following that will give metal on heating is Zn(NO3)2, AgNO3,
Cu(NO3)2, Hg(NO3)2.

5. What is the value of spin only magnetic moment value of Fe(CO)5?

Space for rough work

Mathematics PART – III

SECTION – A

Straight Objective Type

This section contains 10 multiple choice questions numbered 1 to 10. Each question has 4 choices (A),
(B), (C) and (D), out of which ONLY ONE is correct.

3yx 2
1. Solution of the differential y '  is
x 3  2y 4
2 3 2 3
(A) x 3 y 1  y c (B) x 2 y 1  y c
3 3
2
(C) xy 1  y3  c (D) none of these
3

[x] {x} 1
2. Let f(x)  {x} 1 [x] , where [x] = greatest integer function and {x} = fractional part of x.
1 [x] {x}
Then which of the following is incorrect
(A) f(x)  0 for x  (, 1]
(B) f(x) has no real root for x  (0,1)
(C) for x  [0,1), range of f(x) is (2, 1]
(D) for x  [1, 2), f(x) has exactly one local minima

xdy  y 
3. The solution of  2  1 dx is
x  y2
2
x y
2

y
(A) y  x cot (c  x) (B) cos1     x  C
x
2
y
(C) y  x tan(C  x) (D) 2  x tan(C  x)
x

Space for rough work

cos 2 63 cos 3 90 cos 4  77

4. Value of cos   72  cos  40 cos 3  88 is equal to
1 cos  
90  cos  2
99 
(A) 0 (B) –1
(C) –2 (D) none of these

sin A sinB sinC

5. If   cos A cosB cos C  0 and A, B, C are angles of a triangle
cos3 A cos3 B cos3 C
Then
(A) ABC is always isosceles (B) ABC is always equilateral
(C) ABC is always scalable (D) none of these

6. Solution of ydx  xdy  (1  x 2 )dx  x2 sin ydy  0

y 1 y 1
(A)   x  cos y  c  0 (B)   x  cos y  c  0
x x x x2
2y 1
(C)   x  cos y  c  0 (D) none of these
x x2
15
1  3
7. The simplification of   is equal to
 3 1
(A) 215 I (B) 215 I
(C) ( 3)15 I (D) none of these

8. If A and B are two 3 × 3 symmetric matrix such that X = AB – BA then det(X) is equal to
(A) 0 (B) 1
(C) 2 (D) 3

dy
9. The curve whose equation satisfies x  4y  x 2 y passes through (1,(n2)2 ), then the value
dx
of y(2) is
(A) 6 n2 (B) (2 n2 )2
2
(C) (6 n2 ) (D) none of these

Space for rough work

1 1
10. The probability of scoring a century in a cricket match by Sachin and Saurabh are and
4 5
respectively and probability of getting out of a player by bowled, LBW, Catch, Run and stumping
1 2 2 1 1
are , , , and respectively, if Sachin score a century while Saurabh does not, then the
7 7 7 7 7
probability Sachin is catch out and Saurabh is bowled, is (given both players got out)
4 1
(A) (B)
49 19
2 2
(C) (D)
49 19

Multiple Correct Choice Type

This section contains 5 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for
its answer, out which ONE OR MORE is/are correct.

n! n  1!  n  2!   
11. For fixed integer n,    n  1 ! n  2 !  n  3 ! , then   4  is divisible by
  n!3 
n  2 ! n  3 ! n  4 !
(A) n (B) n2  4n  5
(C) n2  4n  5 (D) 2n2

0 1 2
12. A  0 3 0  and 6A 1  aA 2  bA  cI , where a, b, c  integers and I is identity matrix. Then
 1 1 1
 3 1 6 
1
(A) a + b + c = – 2 (B) A   0 2 0 
1
6
 3 1 0 
3 3 2 
1 
(C) a  b  c  18 (D) A  1
 2 4 1
12
5 1 4 

Space for rough work

k k
13. If E and F are two independent events such that P(E  F)  and P(E  F)  (k  I) and
3 6
(P(E)  P(F))(1  P(F)) > 0. Then
2 1
(A) P(E)  (B) P(F) 
3 2
1 7
(C) P(E)  (D) P(E)  P(F) 
2 6

         
14. If a, b, c are non-coplanar unit vectors also b, c are non-collinear and 2a   b  c   b  c , then
   
(A) angle between a and c is 60 (B) angle between b and c is 30
   
(C) angle between a and b is 120 (D) b is perpendicular to c
  
15. If a, b and c are non-coplanar vectors such that a.a  b.b  c.c  3 and a.b  b.c  c.a  1, then
5
(A)  a  b b  c c  a   4 5 (B) volume of tetrahedron is
3
5
(C)  a  b b  c c  a   20 (D)  a b c  
3

SECTION – C

Integer Answer Type

This section contains 5 questions. The answer to each of the questions is a single-digit X Y Z W

integer, ranging from 0 to 9. The appropriate bubbles below the respective question 0 0 0 0

4 4 4 4

5 5 5 5

6 6 6 6

7 7 7 7

8 8 8 8

9 9 9 9

1. Three boxes are labelled as X, Y, Z and each box contains 5 balls numbered 1, 2, 3, 4, 5. The
balls in each box are well mixed. One ball is chosen at random from each of the three boxes X, Y,
Z respectively. If , ,  are the numbers on the balls chosen, then the probability that  =  +  is
K
equal to then K is equal to _____.
25

Space for rough work

2. There are 3 different pairs (i.e. 6 units say a, a, b, b, c, c) of shoes in a lot. Now three persons
come and pick the shoes randomly (each gets 2 units). Let p be the probability that no one is able
13p
to wear shoes (i.e. no one gets a correct pair), then the value of , is _____.
4p

  
3.
        
Let a  b  c, b  d  0, c.d  0, | d | 0 and | c | 2 then the value of
d a d   is equal to _____.
2 
d c

    
4. Let a  cos  î  ˆj  k,b ˆ   î  ˆj  kˆ , a.(b
ˆ  î  cos ˆj  k,c  c) is maximum, then square of the
     
shortest distance between r  a  1b and r  b  2 c is _____.

5.    
The differential equation of y x 2  y 2  x dx  x x 2  y 2  y dy  0 represents a curve given

by f  x, y   0 . The point (0, 2) satisfies the equation of the curve and the value of x when y = 1 is
h
. Then h is equal to _____.
4

Space for rough work

(Advanced), 2014: 2521 FIITJEE Students from Classroom / Integrated School Programs & 3579 FIITJEE Students from All Programs have
FIITJEE JEE(Advanced)-2015
FIITJEE Students From All Programs have bagged 34 in Top 100, 66 in Top 200 and 174 in Top 500 All India Ranks. FIITJEE Performance in JEE

ALL INDIA INTEGRATED TEST SERIES

ANSWERS, HINTS & SOLUTIONS
PART TEST –IV
(Paper-1)

Q. No. PHYSICS CHEMISTRY MATHEMATICS

1. C D A
2. C C D
3. C A C
4. C A A
5. A C B
qualified in JEE (Advanced), 2014.

6. D D B
7. B D A
8. A A A
9. A C C
10. D C C
11. A, D A, B A, B
12. A, B, C B, C, D A, B
13. A, B B, C A, B, D
14. A, B, C, D A, B, C A, C
15. A, B B, D A, B, C
1. 1 5 2
2. 5 1 2
3. 4 4 1
4. 1 2 2
5. 2 0 3

Physics PART – I

2. Use Lenz’s Rule

3. Speed of image = speed of object in all the mirrors

 A  m 
sin  
4. Use  =  2 
A
sin  
2
d 2
6. =  and P =
dt R

dy
7. For CBF P = 0 = tA(A  1)  tB(B  1) +
D

9. VA  iR + 15 + (5  103)  103 = VB

Chemistry PART – II

SECTION – A

2r 2 2r 2 
1. PF   
2 2r 
2 2
8r 4


So packing efficiency = 100  78.54%
4

2. Tf  Tf for glucos e  Tf for KCl  Tf for urea
1.86 1000 10 1.86 1000 1 1.86 1000  5
Tf   2 
100  80 74.5 100 100  60
Tf = 3.069
 F. Pt = 273 – 3.069 = 269.93 K

3.  ZnNO3 2  2NO2  2H2 O

Zn  4HNO 3 
NO3 ion is reduced to NO2 in preference to H3O+ ion.

4. H2O is a weak field ligand and CN is strong, field ligand.

5. 2HClO4  P2 O5 
 Cl2O7  2HPO3

7. Cu  dil HNO3 
 NO 
SnCl2 /HCl
 NH2 OH.HCl 
HNO2
 N2O
X Y  Z

9. O O O
 
F 103 F Cl 110 Cl H 105 H
Because of larger size of Cl atm.
OCl2 have largest bond angle.

2
10. NiCl4  is sp3 hybridized, because Cl is weak field ligand. Hence there is two unpaired
electrons.

15. Number of particles from K4[Fe(CN)6] = 5

Number of particles from FeSO4.(NH4)2SO4.6H2O = 5
Number of particles from KCl.MgCl2.6H2O = 5

SECTION – C

1. O O
H O P O P OH
OH OH

2.
O H
H H

F F
Xe

F F

1
4. AgNO3   Ag  NO2  O2
2
HgNO3 2  Hg  2NO2  O 2

5. In Fe CO 5  , there is no unpaired electron.

Mathematics PART – III

2 3 4 3 –1 2 3
1. (3 yx ) dx + (–x – 2y ) dy = 0, x y = y +c
3

 1 x 
2. f(x)   
 2 
 2 2
 [x]  1  ({x}  1)  ([x]  {x})
2

x dy  ydx
3.   dx  Solution is y = x tan (C – x)
x2  y2

cos   7   9  cos   7   10  cos   7   11

4. Given determinant = cos   8   9  cos   8  10  cos   8   11
cos   9   9  cos   9   10  cos   9  11
cos  7 sin  8 0 cos  9 sin  9 0
 cos  8 sin  8 0 cos  10 sin  10 0 0
cos  9 sin  9 0 cos  11 sin  11 0

sin A sinB sinC tan A tanB tanC

5. Let   cos A cosB cos C  cos A cosB cos C 1 1 1
cos3 A cos3 B cos3 C cos  A cos  B cos  C

  sin(B  C)sin(C  A).sin(A  B)  0

ydx  xdy  1  x 2 
6.   2  dx  sin ydy  0
x2  x 
 1 
d  y x   d   x   d(  cos y)  0
 x 
1
 y x   x  cos y  c  0
x

n
cos   sin   cosn  sinn 
7. We know    
 sin  cos   sinn cosn 
15
1  3   1 0
Let   3   2


2
 I
3
 2
1 
2  0 1
15
1  3
   215 I
 3 1 

8. X is a skew symmetric matrix of odd order.

 det(X) = 0

dy 1 dy 2 y x 1 dy dt
9. x  4y  x 2 y    now let y t 
dx 2 y dx x 2 2 y dx dx
dt 2 x
  t  this is linear differential equation so
dx x 2
2
 dx 1 11 x y 1 x2
I.F.  e x
2
 e2nx 
2
  2 . dx  2  nx  c 
solution t y nx  x 2c
x x x 2 x 2 2
Given curve passes through (1,(n2)2 )
x2
( n2)2  c  c  n2  nx  x 2 .n2 nowy y(2)  6n2  y(2)  (6n2)2
2
 1 4 1 1
     (1 2) 2
10. Required probability 4 5 7 7 
 1 4 1 1 49
      2(19)  11
4 5 7 7

1 n  1  n  1n  2 
3
11.    n! n1  1n  2 
n   1n  2 n  3 
n 
n  1n  2  n  1 n  2 n  3  n  4 n  3 n  2 n  1
1 n  1 n  2 n  1 1 n  1 n  2 n  1
 2 2
3
  n  1  n  2  1 n  2  n  2  n  3    n  1  n  2  0 1 2 n  2
n! 1 n  3  n  3 n  4  0 1 2 n  3
2
  n  1  n  2  .2  n  3  n  2 


 2n n2  4n  5  4 
0 1 2  0 1 2  2 1 2
12. A  0 3 0  0 3 0   0 9 0 
2

 1 1 1  1 1 1  1 1 3 
 3 1 6 
adjA 1 
A 1
  0 2 0 
A 6
 3 1 0 
 Comparing L.H.S and R.H.S., we get a = 1, b = 4, c = –1, d = 12
Alternate method
By Cayley Hamilton theorem, A must satisfy A  I  0
 3  2 2  5  6  0
A must satisfy the above equation
A 3  2A 2  5A  6I  0
Multiplying by A–1
A 2  2A  5I  6A 1

k
13. P(E  F)  P(E).P(F)  xy  ….. (1)
3
Let P(E) = x and P(F) = y
k
And P(E  F)  P(E).P(F)  (1  P(E))(1  P(F)) 
6
k
 (1  x)(1  y)  ….. (2)
6
Solving equation 1 and 2, we get 6x 2  x (k + 6) + 2k = 0
As x  Q; then D should be perfect square, which is possible for k = 1
2 1 1 2
Therefore we get x  or and y = or
3 2 2 3
But given (P(E)  P(F)) (1P(F)) > 0  P(E) > P(F)
2 1 2 1
Hence x  and y   P(E) = and P(F) =
3 2 3 2
 
      bc
14. a  c b  a  b c 
2
   1    1
ba  c    c a b    0
 2  2
  1   1
 a  c  and a  b  
2 2

 1 is and 2 = 120
3

a.a a.b a.c 3 1 1

2
15.  a b c   b.a b.b b.c  1 3 1  3  9  1  1 3  1  11  3   24  2  2  20
c.a c.b c.c 1 1 3
1 5
Volume of tetrahedron = a b c  
6   3

      
 a  b b  c c  a    a  b  b  c  . c  a   a  b  a  c  b  b  b  c   c  a 
  
=  a  b  .c   a  c  .c   b  c .c   a  b  .a   a  c .a   b  c  .a  2  a b c 
 a  b b  c c  a 
Let d  a  b
   
  d.c .b  d.b .c  .  c  a     a b c  b  0   c  a 
 
2
 a b c  b c a   a b c 

SECTION – C
1. n(S) = 5  5  5 = 125
n(A) = 514, 541, 523, 532, 413, 431, 422, 312, 321, 211
10 2
Required probability = 
125 25

6c 2 4c 2 2c 2  3c13c1 4c 2 2c 2  3c 2 3c 2 2  3 90  54  18  6 8
2. p   .
6c 2 4c 2 2c 2 90 15

   2
3. 
d a d  d c 
  
4. a.(b  c) is maximum then cos  = 1.
 (S.D.)2 = 2

5. y x 2  y 2 dx  x x 2  y 2 dy  xdx  ydy
xdx  ydy
ydx  xdy 
x2  y2
Integrating both sides, xy  x 2  y 2  c , c  2
3
 x 2  y 2  xy  2  y  1  x 
4

Time Allotted: 3 Hours Maximum Marks: 198

 Pl ea s e r ea d t h e i n s t r u c t i o n s c a r ef u l l y. Yo u a r e a l l o t t ed 5 m i n ut es
s p ec i f i c a ll y f o r t h i s p u r p o s e.
 Yo u a r e n o t a ll o wed t o l ea v e t h e E xa m i n at i o n Ha ll b ef o r e t h e en d o f
t h e t es t .

B. Filling of OMR Sheet

C. Marking Scheme For All Three Parts.

(i) Section-A (01 to 08) contains 8 multiple choice questions which have only one correct answer.
Each question carries +3 marks for correct answer and – 1 mark for wrong answer.

Section-A (09 to 14) contains 3 paragraphs with each having 2 questions. Each question carries
+3 marks for correct answer and – 1 mark for wrong answer.

Section-A (15 to 20) contains 6 multiple choice questions which have more than one correct
answer. Each question carries +4 marks for correct answer. There is no negative marking.

Name of the Candidate

Enrolment No.

Useful Data

PHYSICS
Acceleration due to gravity g = 10 m/s2

Planck constant h = 6.6 1034 J-s

Charge of electron e = 1.6  1019 C

Mass of electron me = 9.1  1031 kg

2 2
Permittivity of free space 0 = 8.85  1012 C /N-m

Density of water water = 103 kg/m3

Atmospheric pressure Pa = 105 N/m2

Gas constant R = 8.314 J K1 mol1

CHEMISTRY

Gas Constant R = 8.314 J K1 mol1

Atomic No: H=1, He = 2, Li=3, Be=4, B=5, C=6, N=7, O=8,

Physics PART – I

SECTION – A

Straight Objective Type

This section contains 8 multiple choice questions numbered 1 to 8. Each question has 4 choices (A),
(B), (C) and (D), out of which ONLY ONE is correct

1. Two mirrors, one concave and the other convex, are placed 60 cm apart with their reflecting
surfaces facing each other. An object is placed 30 cm from the pole of either of them on their
axis. If the focal lengths of both the mirrors are 15cm, the position of the image formed by
reflection, first at the convex and then at the concave mirror, is
(A) 19.09 cm from the pole of the convex mirror
(B) 19.09 cm from the pole of the concave mirror
(C) 11.09 cm from the pole of the concave mirror
(D) 11.09 cm from the pole of the convex mirror

2. An alternating voltage E = E0 sin t, is applied across a coil of inductor L. The current flowing
through the circuit at any instant t is
E E
(A) 0 sin (t + /2) (B) 0 sin (t - /2)
L L
(C) E0 L sin (t - /2) (D) E0 L sin (t + /2)

3. Consider a usual set up of Young’s double slit experiment with slits y

of equal intensity as shown in the figure. Take ‘O’ as origin and the
D S1
Y axis as indicated. If average intensity between y1 =  and
4d
d O
D
y2 = + equals 'n' times the intensity of maxima, then 'n' equals
4d S2
D
1 2  2
(A) 1   (B) 2 1  
2   
 2 1 2
(C)  1   (D)  1  
  2 

Space for rough work

4. One of the combination from the fundamental physical constants is hc/G. The unit of this
expression is
2 3
(A) kg (B) m
(C) s1 (D) none of the above

5. Let [0] denote the dimensional formula of the permittivity of vacuum. If M = mass, L = length,
T = time and I = electric current. Then
(A) [0 ]  [M1L3 T 2I] (B) [0 ]  [M1L3 T 4I2 ]
(C) [0 ]  [M1L2 T 1I2 ] (D) [0 ]  [M1L2 T 1I]

6. A coil of resistance R and inductance L is connected across an a.c. power supply of r.m.s.
voltage V. The average power dissipated in the coil is,
2 V 2R
(A) V /R (B)
(R2  2L2 )
V2
(C) (D) zero
(R2  2L2 )

7. The figure shows a plane wave front at a time ‘t’ and at a at “t” at “t1”
time ‘t1’
In the time interval (t1  t) the wave front must have
passed through

(A) a prism (B) a prism and a convex lens

8. A vertical ring of radius r and resistance R falls vertically. It is in contact with P Q

two vertical rails which are joined at the top. The rails are without friction x x
and resistance. There is a horizontal uniform magnetic field of magnitude B
perpendicular to the plane of the ring and the rails. When the speed of the
x x
ring is v, the current in the section PQ is
2Brv
(A) zero (B) x x
R
4Brv 8Brv
(C) (D)
R R
Space for rough work

Comprehension Type

Paragraph for Question Nos. 9 to 10

A metal wire PQ slides on parallel metallic rails having separation 0.25 m, each P
having negligible resistance. There is a 2  resistor and 10 V battery as shown in 10 V
figure. 0.25 m
2
There is a uniform magnetic field directed into the plane of the paper of magnitude
0.5T. A force of 0.5 N to the left is required to move wire PQ with constant speed to
the right. Q
Then find the answer of following questions based on above paragraph.

9. With what speed is the wire PQ moving?

(A) 8 m/s (B) 4 m/s
(C) 16 m/s (D) 32 m/s

10. Taking this as a simple motor, what is its efficiency?

(A) 0.3 (B) 0.2
(C) 0.1 (D) 0.4

Paragraph for Question Nos. 11 to 12

In the figure shown C1 = 1F, C2 = 2F and L = 5H. Initially C1 is charged to 50 V and +  + 

C2 to 10 V. Switch S is closed at time t = 0. Suppose at some instant charge on C1 is C1 C2
20 C with the same polarities as shown in the figure.
L S
Then find the answer of following questions based on above paragraph.

11. Energy stored in capacitor C2 at this instant will be:

(A) 10 J (B) 15 J
(C) 25 J (D) 40 J

12. Current in the circuit at this instant will be

(A) 102 A (B) 152 A
(C) 10 A (D) 20 A
Space for rough work

Paragraph for Question Nos. 13 to 14

Spherical aberration in spherical mirrors is a defect


which is due to dependence of focal length ‘f’ on Principle axis  Pole (P)
angle of incidence ‘’ as shown in figure is given by C F
R f
f  R  sec 
2
Where R is radius of curvature of mirror and  is the angle of incidence. The rays which are closed to
principle axis are called paraxial rays and the rays far away from principal axis called marginal rays. As a
result of above dependence different rays are brought to focus at different points and the image of a point
object is not a point.
Then find the answer of following questions based on above paragraph.

13. If f P and f m represent the focal length of paraxial and marginal rays respectively, then correct
relationship is:
(A) f p = f m (B) f p > f m
(C) f p < f m (D) none

14. If angle of incidence is 60, then focal length of this ray is:
(A) R (B) R/2
(C) 2R (D) 0

Multiple Correct Choice Type

This section contains 6 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for
its answer, out which ONE OR MORE is/are correct

Ke 2
15. Suppose the potential energy between electron and proton at a distance r is given by  .
2r 3
Using Bohrs theory choose the correct statements.
(A) Energy in the nth orbit is proportional to n3
(B) Energy in the nth orbit is proportional to n6
(C) Energy is proportional to m 2 (m : mass of electron)
(D) Energy is proportional to m 3 (m : mass of electron)
Space for rough work

16. Let [o] denote the dimensional formula of the permittivity of the vacuum, and [o] that of the
permeability of the vacuum. If M = mass, L = length, T = time and I = electric current,
(A) [o] = M1L3T2I (B) [o] = M1L3T4I2
(C) [o] = MLT2I2 (D) [o] = ML2T1I

17. Which of the following (referred to a spherical mirror) do (does) not depend on whether the rays
are paraxial or not?
(A) Pole (B) Focus
(C) Radius of curvature (D) Principal axis

18. A resistance-less rod of length '  ' and mass ‘m’ can slide without
friction on long rails of zero resistance which are connected to
circuit element named ‘X’ as shown. This arrangement is in the B
horizontal plane in a uniform vertical magnetic field ‘B’ as shown
below. Then match what happens to the rod X 2m
(A) X is resistor and rod is thrown with some velocity towards right
and it comes to rest
(B) X is a resistor and the rod is pulled to right with constant force
and it attains terminal velocity
(C) X is resistor and rod is thrown with some velocity towards right and it never comes in rest
(D) X is a resistor and the rod is pulled to right with constant force and it never attains terminal
velocity

19. A long wire carrying a steady current I lies in the plane of a circular conducting loop placed at a
certain distance from the wire. There will be an induced current in the loop if it is
(A) moved parallel to the wire
(B) rotated about an axis perpendicular to the plane of the loop, passing through the centre of the
loop
(C) rotated about an axis parallel to the wire passing through the centre of the loop
(D) moved away from the wire

20. A point object ‘O’ is kept is front of a concave mirror of focal

length 20 cm at a distance of 30 cm from the pole. The object is
C O F P
moving on the principal axis away from the pole with a velocity
of 1 cm/second. The instantaneous velocity of the image is
(A) 2 cm/sec towards the pole
(B) 4 cm such that the image distance tends to increase
20 cm
(C) 4 cm away from the pole parallel to the principal axis
(D) 4 cm/sec towards the pole and parallel to the principal axis
Space for rough work

Chemistry PART – II

SECTION – A

Straight Objective Type

This section contains 8 multiple choice questions numbered 1 to 8. Each question has 4 choices (A),
(B), (C) and (D), out of which ONLY ONE is correct

1. Aqueous solutions of Na2S2O3 on reaction with Cl2 gives

(A) Na2S4O6 (B) NaHSO4
(C) NaCl (D) NaOH

2. Both [Ni(CO)4] and [Ni(CN)4]2 are diamagnetic. The hybridization of nickel in these complexes,
respectively, are
(A) sp3, sp3 (B) sp3, dsp2
2 3
(C) dsp , sp (D) dsp2, dsp2

3. At a certain temperature the dissociation constants of formic acid and acetic acid are 1.8  104
and 1.8  105 respectively. The concentration of acetic acid solution in which the hydrogen ion
has the same concentration as in 0.001 M formic acid solution is equal to
(A) 0.001 M (B) 0.01 M
(C) 0.1 M (D) none of these

4. The increasing order of thermal stability for following salts will be

(A) FeC2O4 < CaC2O4 < CaCO3 < K2C2O4 (B) K2C2O4 < CaCO3 < FeC2O4 < CaC2O4
(C) FeC2O4 < K2C2O4 < CaC2O4 < CaCO3 (D) CaCO3 < K2C2O4 < CaC2O4 < FeC2O4

5. Which one of the following pairs of ions have the same electronic configuration?
(A) Cr3+, Fe3+ (B) Fe3+, Mn2+
2+ 2+
(C) Mn , Fe (D) Sc3+, Cr3+

6. A mono atomic gas is suddenly compressed to 1/8 of its volume adiabatically. The pressure of
gas will change to
(A) 24/5 times (B) 8 times
(C) 40/3 times (F) 32 times
3+ 3+
7. How do we differentiate between Fe and Cr in group III?
(A) By adding excess of NH4OH solution (B) By increasing NH4 ion concentration
(C) By decreasing OH ion concentration (D) Both (B) and (C)

Space for rough work

8. The compound which does not respond to ring test is

(A) Pb(NO3)2 (B) LiNO3
(C) NaNO3 (D) KNO3

Comprehension Type

Paragraph for question Nos. 9 to 10

 A   NaCl  B  white ppt.

B  KI  C green ppt.
C  KIexcess D  Ecolourless solution
E  NH3  KOH  F

9. Compound (A) and (B) are respectively

(A) AgNO3 and AgCl (B) Pb(NO3)2 ad PbCl2
(C) Hg2(NO3)2 and Hg2Cl2 (D) Cu2(NO3)2 and Cu2Cl2

10. When compound (B) and reacts with NH3 solution, the colour of the compound formed is
(A) black (B) red
(C) yellow (D) white

Paragraph for question Nos. 11 to 12

The binary compounds of oxygen with other elements are called oxides. They are classified either
depending upon their acid – base characteristics or on the basis of oxygen content.

11. Which of the following pairs contains mixed oxides?

(A) Pb3O4, Fe3O4 (B) MnO2, BaO2
(C) KO2, Na2O2 (D) Mn3O4, N2O5

12. Which of the following pairs contain amphoteric oxides?

(A) BeO, BaO (B) BeO, Al2O3
(C) Al2O3, P2O5 (D) FeO, CuO
Space for rough work

Paragraph for Question Nos. 13 to 14

A metal is regarded as an assembly of metal ions and free electrons. When the metal is in contact with
water, some metal ions enter into the liquid due to a tendency in the metal, called by Nernst as
“Electrolytic solution tension”. An alloy of 1.45 gm Ag and Si was dissolved in desired amount of HNO3
and volume made upto 500 ml. An Ag electrode was dipped in solution and Ecell of the cell is
Pt H2 | H || Ag / Ag was 0.503 V at 298 K. EoAg / Ag  0.80 V 
 
(1 atm) (1 M)

Read the above paragraph carefully and answer the questions given below:

13. Observed concentration of Ag+ in the cell

(A) 2.25 × 10–9 M (B) 9.25 × 10–6 M
–7 –5
(C) 9.25 × 10 M (D) 9.25 × 10 M

 
14. The emf of the cell involving the reaction, 2Ag (aq)  H2 (g)  2Ag(s)  2H (aq), is 0.80 V.
The standard oxidation potential of Ag electrode is
(A) 0.80 V (B) –0.80 V
(C) 0.40 V (D) 0.20 V

Multiple Correct Choice Type

This section contains 6 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for
its answer, out which ONE OR MORE is/are correct

15. Which of the following exhibit Frenkel defects?

(A) AgI (B) AgCl
(C) KBr (D) ZnS

16. A d-block element forms octahedral complex but its magnetic moment remains same either in
strong field or in a weak field ligand. Which of the following is/are correct?
(A) Element always form colourless compound
(B) Number of electrons in t2g orbitals are higher than in eg orbitals
(C) It can have either d3 configuration
(D) It can have either d7 configuration

17. The compound (s) which have peroxo linkage is (are)

(A) H2S2O3 (B) H2SO5
(C) H2S2O7 (D) H2S2O8
Space for rough work

18. Identify the incorrect statement(s) among the following?

(A) CN is a pseudo halide
(B) Cl2O(gas) on dissolving in KOH solution produces KClO3
(C) Bleaching action of chlorine occurs in presence of moisture
(D) H3P3O9 contains three PP bonds

19. For a cell reaction represented by the equation

Cu(s)  2Ag (aq)  Cu2  (aq)  2Ag(s)
The value of Go could be obtained by
2+ +
(A) setting up Cu(s)/Cu (aq) and Ag(s)/Ag (aq) half-cells (using molar solution and measuring
the emf of the resulting cell)
(B) adding powdered copper to a solution of Ag+ ions and determining the enthalpy change,
repeating this experiment at different temperatures
(C) setting up Cu(s)/Cu2+(aq) and Ag(s)/Ag+(aq) half-cells and determining the molarities of the
ions which gives an emf of zero for the cell
(D) setting up Cu(s)/Cu2+(aq) and Ag(s)/Ag+(aq) half-cells, using the resultant emf to drive a
motor and measuring the work which it can do

20. In the depression of freezing point experiment, it is found that the

(A) vapour pressure of the solution is less than that of pure solvent
(B) vapour pressure of the solution is more than that of pure solvent
(C) only solute molecules solidify at the freezing point
(D) only solvent molecules solidify at the freezing point
Space for rough work

Mathematics PART – III

SECTION – A

Straight Objective Type

This section contains 8 multiple choice questions numbered 1 to 8. Each question has 4 choices (A),
(B), (C) and (D), out of which ONLY ONE is correct

1. Find the area enclosed between x2 + y2  92 and sin(x – y)  0

9 3 9 3
(A) (B)
4 2
(C) 33 (D) 3
–1
2. Consider an invertible function f(x) with f(x) > 0 and f(x) < 0  x  R. Also f (x) is the inverse
function of f(x). Let f(x) and f –1(x) intersect each other at x = 0 and x = ( > 0) only and if
 
1 2
f
 x   dx  A ;  f  x  dx  B and 2  C then
0 0
(A) A > B > C (B) A > C > B
(C) A < B < C (D) A < C < B

3. The area bounded by the curve y2  4x and the circle x 2 + y2 – 2x – 3  0 is equal to

8 8
(A) 2 + (B) 4 +
3 3
8 8
(C)  + (D)  –
3 3

4. If Pth, qth, rth term of a GP are the positive numbers a, b, c then angle between the vectors
loga3 ˆi  logb3 ˆj  logc 3kˆ and  q  r  ˆi  r  p  ˆj  (p  q) kˆ is
 
(A) (B)
6 2
  1 
(C) (D) sin1  
3  2 2 2
 a b c 

9
5. If A is a square matrix of order n such that adj  adjA   A , then the value of n can be
(A) 1 (B) 2
(C) 3 (D) 4
Space for rough work

 1 0 0
6. Given that A   1 0 1 and A n  An 2  A 3  A n  4 and A 3  A 2  A  I  0 .
0 1 0 
 1 0 0
A  k  1 1 0  . Then k is equal to
20

k  1 0 1
(A) 3 (B) 6
(C) 9 (D) none of these

7. Let A be the event that P (A) = 0, then

(A) A is impossible event (always) (B) A could be possible
(C) A can be said (D) none of these

8. Forces 2iˆ  7ˆj,2iˆ  5ˆj  6k,

ˆ ˆi  2ˆj  kˆ act at a point P whose position vector is 4iˆ  3 ˆj  2k.
ˆ Then
the vector moment of the resultant of three forces acting at P about the point Q, where position
vector is 6iˆ  ˆj  3k.
ˆ
(A) 24iˆ  13ˆj  4kˆ (B) 24iˆ  13ˆj  2kˆ
(C) 24iˆ  13ˆj  4kˆ (D) none of these

Comprehension Type

Paragraph for Question Nos. 9 to 10

Read the following write up carefully and answer the following questions:
 3 1 
3 4   x 4  
2 2  any matrix.
Let M1    and M2    (x  0, 1, x  R) are two matrices and let P  
 1 1 0 1  1 3
 
 2 2 
Also matrix A  PM1PT , B  PT A 2012P , C  PM2PT and D  PT C2012P .
Now answer the following questions

9. Trace (B + D) = 0 for
(A) x = 2 (B) x = 1
(C) more than 100 values of x (D) no value of x
Space for rough work

10. Let matrix E  (M1B1M2D1 )1 , then sum of all elements of second row of E is equal to
(A) 0 (B) 2012
(C) 2010 (D) none of these

Paragraph for Question Nos. 11 to 12

Read the following write up carefully and answer the following questions:
If P(A1  A 2 )  P(A1 ).P(A 2 ) then A1 , A 2 are called independent events for each pair is A1 , A 2 , A 3 if
(A i  A j )  P(A i ).P(A j ) then A1 , A 2 , A 3 are called mutually pair-wise independent events

11. An contains four tickets, each is marked as aab, aba, baa, bbb. And one ticket is drawn. Let
A i  event that ith digit of the mark on ticket is a. Then
(A) A1 , A 2 , A 3 are pair-wise independent
(B) A1 , A 2 , A 3 are mutually independent
(C) A1 , A 2 , A 3 are need not be pair-wise independent
(D) none of these

1
12. A and B are two independent events. The probability that both occur simultaneously is and the
6
1
probability the neither occurs is , then
3
1 1 2 1
(A) P(A)  , P(B)  (B) P(A)  , P(B) 
2 3 3 3
1 1
(C) P(A)  , P(B)  (D) none of these
3 3

Paragraph for Question Nos. 13 to 14

Read the following write up carefully and answer the following questions:
If P(x, yz) is varying such that ax  by  cz  d  0 where a2  b2  c 2  0 then P traces a plane with
normal a ˆi  bjˆ  ckˆ and a(x  x )  b(y  y )  c(z  z )  0 gives plane equation passing through
1 1 1

A(x1 ,y1 ,z1 ) with normal vector a ˆi  bjˆ  ckˆ (a  b  c 2  0)

2 2

13. A variable plane with a constant distance ‘d’ from origin cuts x, y, z axes at P, Q, R. Then locus of
centroid of triangle PQR is
1 1 1 1 1 1 1 9
(A) 2  2  2  2 (B) 3  3  3  3
x y z d x y z d
1 1 1 1 1 1 1 9
(C)    (D) 2  2  2  2
|x| |y| |z| | d| x y z d

Space for rough work

14. A variable plane at a constant d from origin, intersects the axes at A, B, C. Planes are drawn to
the coordinate planes and passing through A, B, C. Then locus of point of intersection of all
planes is
1 1 1 1 1 1 1 2
(A) 2  2  2  2 (B) 2  2  2  2
x y z p x y z p
1 1 1 3
(C) 2  2  2  2 (D) none of these
x y z p

Multiple Correct Choice Type

This section contains 6 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for
its answer, out which ONE OR MORE is/are correct

p1 : ax  by  cz  1
15. Given three distinct planes p2 : bx  cy  az  1 , then
p3 : cx  ay  bz  1
1
(A) if the planes intersect at the single point (3,3,3) then a + b + c =
3
(B) if a + b + c= 0, then the planes will not intersect at a common point
x 1 y z 1 x 1 y z 1
(C) if a + b + c  0 then the lines   and   will never intersect
a b c c b a
x 1 y 1 z 1 x 1 y 1 z 1
(D) the lines   and   will never intersect
a b c b c a

16. Which of the following is/are correct?

(A) P is orthogonal iff PT is orthogonal
(B) if P is orthogonal, then P–1 is orthogonal
(C) if P and Q are orthogonal, then PQ is orthogonal
(D) none of these

0 i  2
17. Let A denotes the matrix   , where i = –1
 i 0 
(A) I + A + A2 + ........ + A2012 = I (B) A4n + 1= A, n  N
4n + 2
(C) A = I, n  N (D) I + A + A2 + .... + A2012 = A

sin2 A sin A cos A cos 2 A

18. If A, B, C are the angles of a triangle and   sin2 B sinB cosB cos2 B  0 , then
sin2 C sinC cos C cos 2 C
(A) triangle must be equilateral (B) triangle must be isosceles
(C) triangle can be equilateral (D) triangle must be right angled
Space for rough work

19. Let the vectors  ˆi  3ˆj  k,

ˆ ˆi  (  1)jˆ  2k,
ˆ 3iˆ  5ˆj  2k.
ˆ Then
(A) the number of planes determined above 3 vectors = 2
 17 
(B) angle between planes  cos1  
 3 23 
(C) angle between planes  tan1  
3
5

(D) none of these

x
20. Let two curves y = f(x) with f(0) = 1 and g(x)   f(t)dt with g(0) = 4 are such that, their tangents

at the points with equal abscissa intersects on x-axis, then
1
x2  4 13
(A)  dx 
1
4  g(x) 3
(B) area bounded by y = f(x), y = g(x), x = 1 and y axis is 12(e1/ 4  1) sq. unit
(C) f(x)  g(x) is decreasing in [1, 1]
  x 
(D) period of f     is 8 ({.} denotes fractional part of x)
 2 

Space for rough work

(Advanced), 2014: 2521 FIITJEE Students from Classroom / Integrated School Programs & 3579 FIITJEE Students from All Programs have
FIITJEE
FIITJEE Students From All Programs have bagged 34 in Top 100, 66 in Top 200 and 174 in Top 500 All India Ranks. FIITJEE Performance in JEE

JEE(Advanced)-2015
ALL INDIA INTEGRATED TEST SERIES
ANSWERS, HINTS & SOLUTIONS
PART TEST –IV
(Paper-2)

Q. No. PHYSICS CHEMISTRY MATHEMATICS

1. B B B
2. B B D
3. A B A
4. A A B
5. B B D
qualified in JEE (Advanced), 2014.

6. B D C
7. B C B
8. D A C
9. C C D
10. B A C
11. C A A
12. B B A
13. B B D
14. D A A
15. B, D A, B, D A, B, D
16. B, C B, C A, B, C
17. A, C, D B, D A, B, C
18. A, B B, D B, C
19. C, D A, C A, B
20. D A, D B, C

Physics PART – I

1 1 2 st nd
1. Use   . Image after 1 reflection will act as object for 2 reflection.
v u R

2. In inductor current lags voltage by /2

D
y
4d
dy
 4I0 cos2 dy
D
D
y 
4d
3. n(4I0 )  d
4d

D
dy

4d

hc GM2
4. E 
 L
hc LM2
 dimensions of   M2
G L

5. <P> = VRMSIRMScos

14-16. iB = 0.5

0.5 10  Bv
 i= 
B 2
(Bv)i
= = efficiency
(0.5)v

17. Q2  20 = 20  50
 Q2 = 10
Q22
U2 =
2C2

18. Energy conservation

Q12 Q22 1 2
   Li = constant
2C 2C 2

19. Use energy conservation

Chemistry PART – II

SECTION – A

2. In Ni(CO)4 = Ni  3d104s04p0 (sp3 hybridization)

[Ni(CN)4]2 = Ni2+  3d84s04p0 (dsp2 hybridization)
CN is strong field ligand.

 
3. 

 H  HCOO
HCOOH 
2
H 
K HCOOH   
HCOOH
 


 H  CH3 OO
CH3 COOH 
2
H 
K CH3 COOH 
CH3 COOH
As [H+] in both cases are equal:
CH3COOH  KHCOOH  CH COOH  1.8  10 4  103  10 2 M
 3 
HCOOH K CH3COOH 1.8  10 5

5. Fe3+  3d53s0
Mn2+  3d54s0

6.  PV   cons tan t
5/3
 V 
P0  V0   P  0 
 8 
 P = 32 P0

16. Irrespective of nature of ligand field, magnetic moment of octahedral complex having metal ions
in d1, d2, d8, d9 configuration, remains same.
Elements may or may not form colourless complex.

17. (B) O
H O O S OH
O
(D) O O
HO S O O S OH
O O

Mathematics PART – III

SECTION – A

1. It is clear by symmetry 2A = (3)2

9 3
A=
2

2. Clearly A < C < B

y = f(x)

–1
y = f (x)

y=x

3. Circle intersect the y2 = 4x at (1, 2) end of latus rectum

1
8
Required curve = 2 2 x  2   2
0
3

 
4. a.b  0

n 1
5. adjA  A
n1 n12
adj  adjA   adjA  A

6.  
A 20  A18  A A 2  I

A  A  A  A  I ………..
18 16 2

A  A  A  A  I
4 2 2

 A  A  9A  A  I
20 2 2

 9A 3  9A  9A 2  9I   A A 2  I  A 2  I    
20 2
 A = 10A – 9I
1 0 0 
 A = 10 1 1 0   9I
20

1 0 1
 k  1  10
k 9

7. Consider S = {1, 2, 3, …..}

Consider event A = selecting the no. 1 from S
1
Probability of A = Let  0.
n n
But still. Event A can occur

i j k
8. Required vector  r 3  f 3  2 4 1
3 4 5

 k 4(xk  1) 
1  2k 4k  x
10. k
M1    , M2  
k
x 1 

 k 1  2k  
 0 1 
Also, PPT  PTP  I and B  M12012 , D  M22012 , E  M22011 M12011

2 1 2 1 2 1
11. Clearly P(A1 )   , P(A 2 )   , P(A 3 )   ,
4 2 4 2 4 2
1 1 1
P(A1  A 2 )  , P(A 2  A 3 )  , P(A1  A 3 )   P(A 1  A 2  A 3 )  0
4 4 4

12. P(A), P(B) are roots of x 2  5 6 x  1

6 0
 x  12 , 13

P(a,0,0) Q(0,b,0) R(0,0,c)

13.    a 3 ,  b 3 ,   c 3
0 6 b
  1
a b c
d
1 1 1
 
a2 b2 c 2
1 1 1 1
Locus is 2  2
 2  2
d 9 9 9
1 1 1 9
 Locus of , ,  is 2  2  2  2
x y z d

14. Equation of plane is lx  my  nz  p (say)

A  p l ,0,0  ,B  0, p m ,0  ,C  0,0, p n 
Let (, ,  ) is a point in locus
   p  ,  p
m,   p
n

  2  m 2  n2  1

15. D = 3abc  (a3 + b3 + c3)

D1 = D2 = D3 = (a2 + b2 + c2  ab  bc  ac)
As plane are distinct then a  b  c
 1 1 1 
Point of intersection  , , 
 a  b  c a  b  c a  b  c 

x 1 y z 1 x 1 y z 1
The lines   and   have shortest distance zero.
a b c c b a
x 1 y 1 z 1 x 1 y 1 z 1
The lines   and   will intersect if a = b = c which is not
a b c b c c
possible

T
16. (A) P   P , P is orthogonal
T T

T
 P  P  I
T T

 PP T  I
 P is orthogonal
(B) P is orthogonal
 PPT  I
 P 1PP T  P 1I
 PT  P 1
  P T is orthogonal  P 1 is orthogonal
(C) PT  P1 and QT  Q 1
T
  PQ  PQ   PQQTPT  PQQ 1P 1  PIP 1  PP 1  I
T 1
 PQ    PQ 
So, PQ is orthogonal

17. A2 = –I, A3= –A, A4 = I also I + A + A2 + A3 = 0

18.   sin  A  B  sin  B  C  sin C  A   0

 Either A = B or B = C or C = A or A = B = C
Hence, the result follows

20. y  f(x) = f(x) (y  x) and y  g(x) = g(x) (y  x)

f (x) g(x) 1
   f(x)  g(x)  f(x)  e x/4 ,g(x)  4e x/4
f(x) g(x) 4
1
x2  4 13
I  4  4ex/4 dx  12
1
1
Area  (4e x/4  e x/4 )dx  12(e1/4  1)
0

FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942
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