FIITJEE INTERNAL TEST

PHYSICS, CHEMISTRY & MATHEMATICS

RT–1 QP CODE: 000000.0
Time Allotted: 3 Hours Maximum Marks: 300

 Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.
 You are not allowed to leave the Examination Hall before the end of the test.
022 A LOT (NAO PAPER)

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.

A. General Instructions
1. Attempt ALL the questions. Answers have to be marked on the OMR sheets.
2. This question paper contains Three Sections.
3. Section-I is Physics, Section-II is Chemistry and Section-III is Mathematics.
4. Each Section is further divided into Two Parts: Part-A & C in the OMR. Part-B of OMR to be left unused
5. Rough spaces are provided for rough work inside the question paper. No additional sheets will be
provided for rough work.
6. No candidate is allowed to carry any textual material, printed or written, bits of papers, clip boards, log
tables, slide rule, calculator, cellular phones, pagers and electronic devices ext. except the Admit Card
inside the examination hall / room.

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 Blue/Black Ball Point Pen for each
character of your Enrolment No. and write in ink your Name, Test Centre and other details at the
designated places.
3. OMR sheet contains alphabets, numerals & special characters for marking answers.
4. Do not fold or make any stray marks on the Answer Sheet.

C. Marking Scheme for All Two Parts:

(i) Part-A (01-20) – Contains Twenty (20) multiple choice objective questions which have four (4) options
each and only one correct option. Each question carries +4 marks will be awarded for every correct
answer and -1 mark will be deducted for every incorrect answer.

(i) Part-C (01-05) contains Five (05) Numerical based questions with single digit integer as answer, ranging
from 0 to 9 (both inclusive). Each question carries +4 marks will be awarded for every correct answer and
NO MARKS will be deducted for every incorrect answer.

Name of the Candidate :______________________________________________________

Batch :______________________________ Date of Examination :___________________

Enrolment Number :_________________________________________________________

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022 A LOT (NAO Paper) (RT-1)-MAIN-PCM(000000.0)-2

SECTION – I : PHYSICS
Part A : (Only One Option Correct Type)

This section contains 20 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out
of which ONLY ONE is correct.

1. In the cube of side ‘a’ shown in the figure, the vector from the
central point of the face ABOD to the central point of the face
BEFO will be:
1 1
(A)
2
a kˆ  ˆi  (B)
2
a  ˆi  kˆ 
1 1
(C) a
2
 ˆj  ˆi  (D) a
2
 ˆj  kˆ 

2. The position vector of a particle changes with time according to the relation

r  t   15t 2 ˆi   4  20t 2  ˆj . What is the magnitude of the acceleration at t = 1?
(A) 40 (B) 100
(C) 25 (D) 50

3. Two stars of masses 3  1031 kg each, and at distance 2  1011 m rotate in a plane about their
common centre of mass O. A meteorite passes through O moving perpendicular to the star’s rotation
plane. In order to escape from the gravitational field of this double star, the minimum speed that
meteorite should have at O is : (Take Gravitational constant G = 6.67  10–11 Nm2 kg–2)
(A) 2.4  104 m/s (B) 1.4  105 m/s
4
(C) 3.8  10 m/s (D) 2.8  105 m/s

4. A particle moves on a rough horizontal ground with some initial velocity say V0. If (3/4)th of its kinetic
energy is lost in time t0, then coefficient of friction between the particle and the ground is
V V0
(A) 0 (B)
2gt 0 4gt 0
3V0 V
(C) (D) 0
4gt 0 gt 0

5. A ball is thrown upward with an initial velocity V0 from the surface of the earth. The motion of the ball
is affected by a drag force equal to m2 (where m is mass of the ball,  is its Instantaneous velocity
and  is a constant). Time taken by the ball to rise to its zenith is:
1    1   
(A) ln  1  V0  (B) ta n1  V0 
g  g 
 g  g



1    1  2 
(C) sin1  V0  (D) tan1  V0 
g  g  2 g  g 
   

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 022 A LOT (NAO Paper) (RT-1)-MAIN-PCM(000000.0)-3
th
 1
6. A uniform cable of mass ‘M’ and length ‘L’ is placed on a horizontal surface such that its   part is
n
hanging below the edge of the surface. To lift the hanging part of the cable upto the surface, the work
done should be:
MgL
(A) nMgL (B)
2n2
2MgL MgL
(C) (D)
n2 n2

7. A body of mass m1 moving with an unknown velocity of v1ˆi undergoes a collinear collision with a
body of mass m2 moving with a velocity v 2 ˆi . After collision m1 and m2 move with velocities of v 3 ˆi
and v 4 ˆi respectively. If m2 = 0.5 m1 and v3 = 0.5 v1 then v1 is:
v2 v2
(A) v 4  (B) v 4 
2 4
(C) v 4  v 2 (D) v 4  v 2
  
8. Two vectors A and B lie in a plane, another vector C do not lie in this plane, then the resultant of
  
these three vectors i.e., A  B  C
(A) can be zero (B) can not be zero
   
(C) lies in the plane containing A  B . (D) lies in the plane containing A  B .

9. A person of mass M is, sitting on a swing of length L and swinging with an angular amplitude 0. If the
person stands up when the swing passes through its lowest point, the work done by him, assuming
that his centre of mass moves by a distance  (  < < L), is close to:
(A) Mg(1  0 2 ) (B) Mg(1  0 2 )
 0 2 
(C) Mg (D) Mg  1  
 2 

10. Four identical particles of mass M are located at the corners of a

square of side ‘a’. What should be their speed if each of them revolves
under the influence of other’s gravitational field in a circular orbit
circumscribing the square?
GM GM
(A) 1.35 (B) 1.16
a a
GM GM
(C) 1.41 (D) 1.21
a a

11. A uniform cylindrical rod of length L and radius r, is made from a material whose Young's modulus of
Elasticity equals Y. When this rod is heated by temperature T and simultaneously subjected to a net
longitudinal compressional force F, its length remains unchanged. The coefficient of volume
expansion, of the material of the rod, is (nearly) equals to :
(A) 9F/(r2YT) (B) F/(3r2YT)
2
(C) 3F/(r YT) (D) 6F/(r2YT)

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022 A LOT (NAO Paper) (RT-1)-MAIN-PCM(000000.0)-4

12. A pulley one meter in diameter rotating at 600 revolutions a minute is brought to rest in 20 s by a
constant force of friction on its shaft. How many revolutions does it make before coming to rest?
(A) 200 (B) 300
(C) 400 (D) 100

13. A cubical block of side 0.5 m floats on water with 30% of its volume under water. What is the
maximum weight that can be put on the block without fully submerging it under water?
[Take density of water = 103 kg/m3]
(A) 46.3 kg (B) 65.4 kg
(C) 30.1 kg (D) 87.5 kg

14. When M1 gram of ice at –10ºC (specific heat = 0.5 cal g–1°C–1) is added to M2 gram of water
at 50ºC, finally no ice is left and the water is at 0ºC. The value of latent heat of ice, in cal g–1 is:
50M2 5M2
(A) 5 (B) 5
M1 M1
50M2 5M1
(C) (D)  50
M1 M2

15. A particle is projected vertically upward with initial velocity 25 ms1. During third second of its motion,
which of the following statement is correct? (g = 10 m/s2)
(A) Displacement of the particle is 30 m.
(B) Distance covered by the particle is 30 m.
(C) Distance covered by the particle is 2.5 m.
(D) None of these.

16. The displacement of a damped harmonic oscillator is given by

x(t) = e–0.1 t cos(10t + ). Here t is in seconds.
The time taken for its amplitude of vibration to drop to half of its initial value is close to:
(A) 13 s (B) 27 S
(C) 4 s (D) 7 s

17. A travelling harmonic wave is represented by the equation y(x, t) = 103sin(50t + 2x), where
x and y are in meter and t is in seconds. Which of the following is a correct statement about the
wave?
(A) The wave is propagating along the negative x-axis with speed 25 ms1.
(B) The wave is propagating along the positive x-axis with speed 100 ms1.
(C) The wave is propagating along the positive x-axis with speed 25 ms1.
(D) The wave is propagating along the negative x-axis with speed 100 ms1.

18. A small speaker delivers 2 W of audio output. At what distance from the speaker will one detect 120
dB intensity sound? [Given reference intensity of sound as 10–12W/m2]
(A) 30 cm (B) 10 cm
(C) 40 cm (D) 20 cm

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 022 A LOT (NAO Paper) (RT-1)-MAIN-PCM(000000.0)-5

19. A spring whose unstretched length is  has a force constant k. The spring is cut into two pieces of
unstretched lengths 1 and  2 where, 1 = n 2 and n is an integer. The ratio k1/k2 of the
corresponding force constants, k1 and k2 will be:
1
(A) n (B)
n2
1
(C) n2 (D)
n

20. A dog of mass 10 kg chases a rabbit running with a speed of 7 km/hr and mass 2 kg with a speed of
13 km/hr along a straight line. The speed of the centre of mass of the dog rabbit system will be
(A) 10 km/hr (B) 12 km/hr
(C) 16 km/hr (D) 20 km/hr

Part C : (Integer Value Correct Type)

This section contains 5 questions. The answer to each question is a single digit integer, ranging from 0 to
9(both inclusive).

1. Two thin discs each of mass 0.4 kg and radius 1 m are attached as shown in figure to
form a rigid body. The rotational inertia of this body about an axis perpendicular to the B
plane of disc B and passing through its centre is

2. A piece of copper having an internal cavity weighs 264 gm in air and 221 gm in water. The volume of
the cavity is 6.5n. Find the value of n. [Density of copper is 8.8 gm/cc.]

3. For the given cyclic process CAB as shown for a gas. The
work done is 2k joule. Find the value of k.

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022 A LOT (NAO Paper) (RT-1)-MAIN-PCM(000000.0)-6

4. A block of mass 2 kg is resting on a frictionless plane. It is struck by a jet releasing water at 1 kgs1
10
and at a speed 5 ms1. Initial acceleration of the block is m/s2. Find the value of p.
p

5. In an experiment, brass and steel wires of length 1 m each with areas of cross section
1 mm2 are used. The wires are connected in series and one end of the combined wire is connected to
a rigid support and other end is subjected to elongation. The stress requires to produced a new
elongation of 0.2 mm is k  106 N/m2. Find the value of k.
[Given, the Young’s Modulus for steel and brass are respectively 120  109 N/m2 and 60  109 N/m2]

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 022 A LOT (NAO Paper) (RT-1)-MAIN-PCM(000000.0)-7

SECTION – II : CHEMISTRY
Part A : (Only One Option Correct Type)

This section contains 20 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out
of which ONLY ONE is correct.

1. Number of hybridised orbitals, which form non-bonding molecular orbitals in H 2 O, BF3 and NH3
molecules respectively, are:
(A) 2,0,1 (B) 2,1,1 (C) 1, 0,1 (D) 1, 0, 0

2. The kinetic data for the reaction 2A  B2 

 2AB are as given below.
 A  mol L1
B mol L 1
Rate mol L1 min1
0.5 1.0 2.5  103
1.0 1.0 5.0  10 3
0.5 2.0 1 10 2
Hence, the order of reaction with respect to A and B2 are, respectively,
(A) 1 and 2 (B) 2 and 1
(C) 1 and 1 (D) 2 and 2

3. A weak acid is titrated against a strong base and the volume of titrant 14
added, is plotted against pH as shown below:
Which of the following statements regarding above titration is correct?
(A) The substance in beaker is a solution of weak tri-basic acid pH 10
V1 8
(B) The pH corresponding to corresponds to pK a1 of the weak 4
2
dibasic acid being titrated 0
V1 V2
V
(C) V2  3V1
V2
(D) The pH corresponding to corresponds to pK a2 of the weak dibasic acid
2

Calculate H solution for Na 2SO 4  s  , from the following data.


4.

H f  Na  , aq.  240.1 kJ / mol

H f  SO42 , aq.  909.3 kJ / mol H f  H 2O,    285.8 kJ / mol
H f  Na2 SO4 , s   1387.1kJ / mol
(A) 2.4 kJ / mol (B) 283.4 kJ / mol
(C) 288.2 kJ / mol (D) 288.2 kJ / mol

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022 A LOT (NAO Paper) (RT-1)-MAIN-PCM(000000.0)-8

5. Which of the following will not form in the following chemical reaction?
CH3  Br2

H3C LiCl in CH 3OH

Br Cl
CH3 CH3
(A) H3C (B) H3C
OCH3 Br
Br Br
CH3 CH3
(C) H3C (D) H3C
OH Br

6. Which of the following sequence of reagents can be used to carry out the given conversion?
Br
CH3 CH3
H3C H3C
Br
CH3 CH3
1 alcoholic KOH 1 KOC  CH 3 3 or  KOtBu 
(A)  2  NBS , h (B)  2  NBS , h
 3 H 2 , Ni
1 alcoholic KOH All are correct
(C) (D)
 2  HBr, organic peroxide

7. The major product of the following reaction is:

 1 Conc . HNO , Conc . H SO
 3
 2  ( a ) Fe / HCl  b  OH 
2 4

3 excess Br2 , H 2O , NaHCO3

NH2 NO 2
Br Br

(A) (B)
Br Br
Br
NH2 NH2
Br Br

(C) (D)
Br
HCO3

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 022 A LOT (NAO Paper) (RT-1)-MAIN-PCM(000000.0)-9

8. Which of the following compounds are not matched correctly?

H
CH 3
H3C OH
OH H
enantiomers
(A)
CH2OH
HO H OH
H
CH 2 OH
H3C CH3 HO H
(B) same molecule
HO H H3C CH3
H
CH 3
H3C OH
H OH
(C) same molecule
CH2OH
HO H OH
H
CH 2 OH
Br H Br H
COOH COOH
(D) diastereomers
CH3 CH3
H Br
Br H

9. KO2 (Potassium super oxide) is used in oxygen cylinders in space and submarines because it
(A) produces ozone (B) absorbs CO2
(C) eliminates moisture (D) absorbs CO2 and increases O2 content

10. Which of the following structure is correctly named according to IUPAC nomenclature?

(A)
4-ethyl-5,6,7,9-tetramethyl dec-2,9-diene

CH3 O
H3C OEt

(B)
Ethyl-2-methyl-2-(3-nitrophenyl) propanoate

NO 2
CH3

(C)
2,6-Dimethyl octa-2,5,7-triene
CH2

CH3
H3C
(D) None of these

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022 A LOT (NAO Paper) (RT-1)-MAIN-PCM(000000.0)-10

11. Which of the following cannot be oxidized by H2 O2 ?

(A) O 3 (B) KI + HCl
(C) PbS (D) Na2 SO3

12. The major product of reaction

O
C O

HNO3  H 2 SO4
 

OMe
O O
C O C O

(A) (B)

NO 2 NO 2
OMe OMe
O O
C O C O
NO 2
(C) (D)

NO 2 OMe OMe

13. A reaction takes place in three steps. The rate constants are k1,k 2 and k 3 . The overall rate constant
k1k 3
k . If (energy of activation) E1,E2 and E3 are 60, 30 and 10 kJ. The overall energy of activation
k2
is:
(A) 40 (B) 30
(C) 400 (D) 60

14. K  sp AgCl is 2.8  10

10
at 25C . Then the solubility of AgCl in 0.1 M AgNO3 solutions:
5 9
(A) 1.67  10 M (B) 2.8  10 M
11
(C) 2.8  1010 M (D) 2.8  10 M

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 022 A LOT (NAO Paper) (RT-1)-MAIN-PCM(000000.0)-11

15. What is not correct about green house effect?

(A) it results in global warming
(B) it results in lowering of levels of ocean over the years.
(C) CO 2 is one of the main chemical species responsible for it
(D) CH4 ,O3 ,CFC also contribute to green house effect.

16. Select the correct statement(s) among the following:

(A) For single electronic atom or ion, the most probable distance of electron in an orbital having no
n2 a0
radial node is from the nucleus, where a0 is the first Bohr radius.
Z
(B) Outside an orbital contour, the probability of finding electron is zero
(C) The angular wave function of any s-orbital depends on  and  .
(D) None of these

17. Which of the following is the correct order of bond length of O-O in KO 2 ,O 2 ,O 2 AsF6 molecules?  

(A) KO 2 <O2 AsF6  O2  (B) KO 2 <O 2 <O 2 AsF6  
(C) KO 2 >O 2 >O 2 AsF6   
(D) KO 2 >O 2 AsF6 >O 2 
18. One mole of helium and one mole of neon are taken in a vessel. Which of the following statements
are correct?
(A) Helium exerts lesser pressure than neon
(B) Molecules of helium strike the wall of vessel more frequently
(C) Molecules of helium have lesser average molecular speed than neon
(D) Helium exerts higher pressure than neon

fused
19. Ca2 B6 O11  Na2 CO3   A  B  CaCO3
A  CO2 
 B  Na2CO3
H 2O strongly
B  Conc. HCl 
 NaCl  Acid   Acid  C  
heated
D
Which of the following is incorrect about the compounds A, B, C or D?
(A) ‘D’ is boric anhydride (B) ‘A’ is Na2BO 2
(C) ‘B’ is used in borax bead test (D) ‘C’ is tetra boric acid

20. Consider this equilibrium, for which ΔH <0 ,

HgO( s)  4I  (aq.)  H 2O(l ) HgI 4 2 (aq.)  2OH  (aq.)
2-
Which of the following will increase the equilibrium concentration of HgI 4 ?
(B) Increasing the mass of HgO  s  present
–
(A) Increasing [I ]
(C) Adding 1 M NaOH (D) By Adding catalyst

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022 A LOT (NAO Paper) (RT-1)-MAIN-PCM(000000.0)-12

Part C : (Integer Value Correct Type)

This section contains 5 questions. The answer to each question is a single digit integer, ranging from 0 to
9(both inclusive).

1. How many different alkenes can be used to prepare 3-methylpentane by catalytically hydrogenation?

2. 5.7 g of bleaching powder was suspended in 500 ml of water 25 ml of this solution on treatment with
KI in the presence of HCl liberated iodine which is reacted with 24.35 ml of N/10 Na2 S2O3 . The % of
x
‘available’ chlorine in the bleaching powder is x % . Then the value of is?
10.1
3. Solid phosphorus melts and vapourizes at high temperature. Gaseous phosphorus effuses at a rate
that is 0.567 times that of Ne in the same apparatus under the same conditions. How many atoms
2
are in a molecule of gaseous phosphorus? (Given that  0.567   0.32 , atomic mass of P = 31 g/mol,
atomic mass of Ne = 20 g/mol)

4. 0.8g of an organic acid of molecular weight 192 requires 0.5 g of NaOH for complete
neutralization. The number of replaceable hydrogen atoms in the acid is:

5. In a reaction, C(s)  CO2(g) 2CO(g) , the equilibrium pressure is 12 atm. If 50% of CO2 reacts at
equilibrium then the partial pressure (in atm) of gas CO is?

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 022 A LOT (NAO Paper) (RT-1)-MAIN-PCM(000000.0)-13

SECTION – II : MATHEMATICS
Part A : (Only One Option Correct Type)

This section contains 20 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out
of which ONLY ONE is correct.

2
 x  1 x  1 x  2   e3x  1
1. If 3 2
 0, the complete solution set of values of x is
x  3  x   x  1  x  3 
(A)  , 3    3,   (B)  , 3    1,1   3,    2
(C)  , 3    1,1   3,   (D)  ,  3    1, 0    0, 1   3,    2

  y
2. If sin x  cos ec x  tan y  cot y  4 where x and y  0,  , then tan is a root of the equation
 2 2
(A)  2  2  1  0 (B)  2  2  1  0
(C) 2 2  2  1  0 (D)  2    1  0

3. The equation of images of pair of lines y = |2x – 1| in y-axis is

(A) 4x2 – y2 – 4x + 1 = 0 (B) 4x2 – y2 + 4x + 1 = 0
2 2
(C) 4x + y + 4x + 1 = 0 (D) 4x2 + y2 – 4x + 1 = 0

4. The area of the triangle formed by joining the origin to the points of intersection of the line
5x  2y  3 5 and circle x2 + y2 = 10 is
(A) 6 (B) 5
(C) 4 (D) 3

5. The mirror image of the parabola y 2  4x in the tangent to the parabola to the point (1, 2) is
2 2
(A)  x  1  4  y  1 (B)  x  1  4  y  1
2 2
(C)  x  1  4  y  1 (D)  x  1  4  y  1

6. The number of non-negative integral solutions of x1 + x2 + x3 + 4x4 = 20 is

(A) 535 (B) 536
(C) 537 (D) 538


7. Let P  a sec ,b tan   and Q  a sec ,b tan   , where     , be two points on the hyperbola
2
x2 y2
2
  1 . If (h, k) is the point of the intersection of the normals at P and Q, then k is equal to
a b2
a2  b2  a2  b2 
(A) (B)   
a  a
 
a2  b2  a2  b2 
(C) (D)   
b  b
 

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022 A LOT (NAO Paper) (RT-1)-MAIN-PCM(000000.0)-14

1 1 2
8. If ,  be the roots x2 – a(x – 1) – b = 0, then the value of 2
 2
 is
  a   a ab
4 1
(A) (B)
ab ab
4b
(C) 0 (D) 2
b  a2

9. Let f(x) = x2 + bx + c, b is negative odd integer, f(x) = 0 has two distinct prime number as roots, and
b + c = 15, then least value of f(x) is
233 233
(A) (B)
4 4
225
(C)  (D) none of these
4

n 
1 1 1
10. Given Sn = 2
r 0
r
,S= 2
r 0
r
. If S – Sn <
1000
, then least value of n is

(A) 8 (B) 9
(C) 10 (D) 11

1 3 7 15
11. The sum of the series     ...... upto n terms
2 4 8 16
1 1
(A) upto n terms is : n  1  n (B) upto n term is : n  n
2 2
(C) upto n terms is 2 (D) upto infinity is 3

1 1 1 1
12. The Value of    is
1.15 3. 13 5.11 7. 9
214 215
(A) (B)
15 16
210 213
(C) (D)
15 15

13. If a, b, c be three natural numbers in A.P. and a + b + c = 30, then the possible number of values of
(a, b, c) is
(A) 19 (B) 24
(C) 30 (D) 36

6
14. The greatest integer less than or equal to  
2 1 is
(A) 194 (B) 195
(C) 196 (D) 197

15. If 5 40 is divided by 11 then the remainder is

(A) 2 (B) 3
(C) 5 (D) 1

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 022 A LOT (NAO Paper) (RT-1)-MAIN-PCM(000000.0)-15

16. The number of real values of m from for which the equation z3   3  i  z 2  3z  m  i  0 has at
least one real root is (where z = x + iy)
(A) 1 (B) 3
(C) infinite (D) 2

17. Number of different nine digits numbers can be formed from the number 223355888 so that the odd
digits occupy even position is
(A) 50 (B) 55
(C) 60 (D) 65

18. The number of solution of the equation tan x  cot x  cos ecx in the interval 0, 2 are
(A) 1 (B) 2
(C) 3 (D) 4

19. If cos B cos C  sin B sin Csin2 A  1; then

(A) A  450 (B) A  600
(C) A  B (D) A  900 & B  C

20. If 0 < x < 1, the number of solutions of the equation tan1  x  1  tan1 x  tan1  x  1  tan1 3x is
(A) 0 (B) 1
(C) 2 (D) 3

Part C : (Integer Value Correct Type)

This section contains 5 questions. The answer to each question is a single digit integer, ranging from 0 to
9(both inclusive).

1. The number of values of x satisfying (log2 x)(log3 x) = log3(x3) + log2(x2) – 6 is


2. Let A  , so that tan A.tan2A  tan 2A.tan 4A  tan 4A.tan A  K  0 . Then K is equal to
7

3. The number of integer values of m, for which the x-co-ordinate of the point of intersection of the lines
3x + 4y = 9 and y = mx + 1 is also an integer, is

4. The intercept made by the circle x2 + y2 – 14x – 10y + 23 = 0 on the line 7x + y – 4 = 0

5. The equation of the line touching both the parabolas y 2  4x and x 2  32y is x  2y    0, then 
is

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022 A LOT (NAO Paper) (RT-1)-MAIN-PCM(000000.0)-16

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 022 A LOT (NAO Paper) (RT-1)-MAIN-PCM(000000.0)-17

FIITJEE INTERNAL TEST

PHYSICS, CHEMISTRY & MATHEMATICS
RT–1 CODE: 000000.0

ANSWERS
PHYSICS (SECTION –I)
1. C 2. D 3. D 4. A
5. B 6. B 7. C 8. B
9. A 10. B 11. C 12. D
13. D 14. A 15. C 16. D
17. A 18. C 19. D 20. B
1. 2 2. 2 3. 5 4. 4
5. 8

CHEMISTRY (SECTION –II)

1. A 2. A 3. B 4. A
5. C 6. B 7. A 8. C
9. D 10. B 11. A 12. B
13. A 14. B 15. B 16. A
17. C 18. B 19. B 20. A
1. 4 2. 3 3. 2 4. 3
5. 8

MATHEMATICS (SECTION –III)

1. D 2. B 3. B 4. B
5. C 6. B 7. D 8. D
9. C 10. C 11. A 12. C
13. A 14. D 15. D 16. D
17. C 18. B 19. D 20. B
1. 2 2. 7 3. 2 4. 2
5. 4

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022 A LOT (NAO Paper) (RT-1)-MAIN-PCM(000000.0)-18

HINTS & SOLUTIONS

PHYSICS (SECTION –I)
PART – A

a a
1. 1 ,0, 
2 2
 a a
2  0, , 
 2 2
  a a
r2  r1   ˆi  ˆj
2 2

2. r  (15t 2 )iˆ  (4  20t 2 )ˆj

 dr
v  (30t)iˆ  (40t)jˆ
dt

 dv
a  (30)iˆ  (40)jˆ
dt

a  50

1 2( GMm)
3. mv 2  0
2 r
4GM 4  6.67  10 11  3  1031
V2  
r 2  1011
V  20 2  10 4 m / s
= 2.828 × 105 m/s

4. (3/4)th K.E. is lost, therefore (1/4)th KE is left. Hence, its velocity becomes (V0/2) under a retardation
g in time t0.
V0 V
 V0  gt 0    0
2 2gt 0

dv
5. (g  v 2 ) 
dt
 
g  dv 
gdt   
 g 2 
  v 
 

Integrating 0  t and V0  0:

 
 
g 1  V0 
gt   tan
  g
 
  

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 022 A LOT (NAO Paper) (RT-1)-MAIN-PCM(000000.0)-19

1   
t tan1  V
g  g 0 
 

M
6. Mass of the hanging part =
n
L
hCOM 
2n
 M   L  MgL
Work done W = mghCOM    g   2
 n   2n  2n

7. Applying linear momentum conservation

m1v1ˆi  m 2 v 2 ˆi  m1v 3 ˆi  m 2 v 4 ˆi
m1v 1  0.5 m1v 2  m1(0.5 v1 )  0.5 m1v 4
0.5 m1v1 = 0.5 m1(v4 – v2)
v1 = v4 – v2
  
8. If A  B  C  0 , it will form a triangle and triangle is a planar figure.

9. Angular momentum conservation

MV0L = MV1(L –  )
 L 
V1  V0  
L   
wg + wp = KE
1
mg  w p 
2

m V12  V02 
1   L 2 
w p  mg  mV02    1
2  L    
 
2
1  L  
= mg  mV02  1   1
2   L    
 
Now,  << L

10. Net force on particle towards centre of circle is

GM2 GM2
FC   2 2
2a 2 a
2
GM  1 
= 2 
 2
a 2 
This force will act as centripetal force. Distance of
a
particle from centre of circle is .
2
a mv 2
r , FC 
2 r
2 2
mv GM  1 
 2   2
a a 2 
2
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022 A LOT (NAO Paper) (RT-1)-MAIN-PCM(000000.0)-20

GM  1 
v2    1
a 2 2 
GM GM
v2  1.35  ; v  1.16
a a

11.  Length of cylinder remains unchanged

F F
so    
 A Compressive  A  Thermal
F
 YT ( is linear coefficient of expansion)
r 2
F
 
YTr 2
 The coefficient of volume expansion  = 3
F
  3
YTr 2

600  2
12. Given D = 1 m, 0 = 600 rev/min =  20 rad / sec
60
f = 0 in 20 sec
therefore, f = 0 t
 20
 = 0    rad / sec 2
t 20
so the no. of revolution before coming to rest is
2f  20  2
02 4002
=  = 200  rad = 100 revolutions
2 2

13. 0.3 3  = 3  ××

kg
 = 300
m3
m + 3 = 3 0.3 3

M = 3(w – ) = (5)3 {1000 – 300} = 700 × (5)3

= 87.5 kg

14. Heat lost = Heat gain

 M2 × 1 × 50 = M1 × 0.5 × 10 + M1Lf
50M2  5M1 50M2
 Lf  = 5
M1 M1

15. Displacement of the particle during third second of the motion (i.e. between t = 2 s and t = 3 s) is
zero. Hence, t = 2.5 sec is the turning point of the motion.
1
For distance; St = 2 = 25  2   10  22 = 30 m
2
1
and St=2.5 = 25  2.5   10  2.52 = 31.25
2

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 022 A LOT (NAO Paper) (RT-1)-MAIN-PCM(000000.0)-21

Hence, distance covered by the particle during third second of motion

= 2 (31.25  30) = 2.5 m.

A0
16. A = A0e–0.1 t =
2
ln2 = 0.1 t
t = 10 ln2 = 6.93  7 sec.

17. y = 10–3 sin (50t + 2x)

Wave is travelling along negative x-axis
 50
Wave speed =  = 25 m/s.
k 2

18. Loudness of sound is given by

I
dB  10log (I is intensity of sound, I0 is reference intensity of sound)
I0
I
 120  10 log  
 I0 
 I = 1 W/m2
P 2
Also, I  2

4r 4 r 2
2 I
 r  m = 0.399 m  40 cm
4 2
C
19. k1 
1
C
k2 
2
k1 C 2  1
  2 
k 2  1C n 2 n

20. Velocity of centre of mass,

m v  m2 v 2 10  13  2  7
vcm = 1 1 = = 12 km/hr
m1  m2 10 2
PART– C

1. IB = I of sphere B about cm + I of sphere A about cm + mass of sphere A  (distance

between A and B)2 B

mR2 mR2
=   m(2R)2 = 5mR2
2 2 A

2. Weight of copper in air = Vcug = 264 g

Weight of copper in water = Vcug  Vg = 221

Hence, outer volume of the copper piece V =

 264  221 g
= 43 cc
1 x g
The volume of the material of the copper piece
massof copper piece 264
Vcu =   30 cc
density of material 8.8
Hence, volume of the cavity = V – Vcu = 43 – 30 = 13 cc

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022 A LOT (NAO Paper) (RT-1)-MAIN-PCM(000000.0)-22

3. Work done = Area of loop

1
= (4) (5) = 10 J
2

v dm
4. F  5 1  5 N
dt
F 5
a    2.5 ms2
M 2

5. =1M
A = 10–6 M2
F
Stress = Ys = 120 × 109 Ys = 120 × 109
A
F
Stress
Stress = YB = 60 × 109
Y
F
 
AY
F  F
 1   2  1  2  0.2  10 3
AY1 AY2
F 0.2  10 3

A  

Y1 Y2
0.2  10 3 0.2  10 3  109  120
= 
1 1 1 2
9

120  10 60  109
0.2  10 6  120
=  8  106
3

CHEMISTRY (SECTION –II)

PART – A

1. A
Two lone pairs of oxygen are present in sp3 hybridized orbitals will form non-bonding M.O.
One lone pair of nitrogen is present in sp3 hybridized orbital, will form non-bonding M.O.

3. B
There are two sharp changes in the pH value as seen in the graph. So it is a dibasic acid. After
adding V1 volume of base, H2A convert to HA- and at V1/2 we have a buffer with equal concentration
of H2A and HA- hence, pH = pKa1.

4. A

H solution  2   240.1   909.3   1387.1
 2.4 kJ / mol

5. C

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 022 A LOT (NAO Paper) (RT-1)-MAIN-PCM(000000.0)-23
+
Br
CH3
H3C
Nucleophilic species like Br-, Cl-, CH3OH present in the reaction mixture can attack on the halonium
ion shown above.

6. B

7. A
Sol.
NO 2 NH2
1 Conc . HNO3 , Conc. H 2 SO4  2  ( a ) Fe / HCl b  OH 
  


 3 excess Br2 , H 2O , NaHCO3

NH 2
Br Br

Br

8. C

9. D

10. B

11. A
Sol. O3 is more powerful oxidizing agent than H2 O2 . So H2 O2reduces O3 to O2 .
O3  H2O 2 H2 O  2O2

12. B
Sol. OMe is stronger activating group. Hence nitration will take place at ortho position to the OMe
group.

14. B
Sol. K  sp AgCl = 2.8  10
10
= s×(s+0.1)
s+0.1 0.1
Hence, s  2.8  109 M

15. B
Sol. It results in rising of levels of ocean over the years

16. A
Sol. The probability of finding electron is 90% inside an orbital contour. Wave function of s-orbital does
not contains angular part having  and  .

17. C
Sol. KO2 has O2
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022 A LOT (NAO Paper) (RT-1)-MAIN-PCM(000000.0)-24

O2 bond order = 1.5

O2  AsF6  has O 2 bond order = 2.5
O 2 : bond order = 2
 
Bond length in increasing order: O2  O2  O2

18. B
Sol. Both He and Ne will exert equal pressure because number of moles is same for both the gases.

19. B
Sol.  A   NaBO2 , B   Na2B4O7 ,  C   H2B4 O7 , D  B2O3
20. A
Sol. Le chatelier principle

PART – C
1. [4]
Sol.

 1
 2  2

2. [3]
Sol. Milli eq. of Na 2S2 O3  24.35  0.1  2.435
Milli eq. of Cl2in 500 ml,  2.435  500 / 25  48.7
Meq. Of Cl2 = meq of bleaching powder = meq of available Cl2 in the bleaching powder
% of Cl2 = (48.7/1000)×(35.5/5.7)×100=30.33%

3. [2]
rP 20
Sol.  0.567  x  number of P atoms
rNe x  31

4. [3]
Sol. Moles of organic acid = 0.8/192 = 0.00417
Moles of NaOH = 0.5/40 = 0.0125

5. [8]
Sol.
C(s)  CO2(g) 2CO(g)
Initial a 0
a
At eq. a a
2
 a
Total moles and pressure at equilibrium  a   and 12 respectively
 2
2
p a
K p  CO  pCO   12  8
pCO2 a
a
2
a
a
88 2  12  4
Kp   16 pCO2 
4 a
a
2

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 022 A LOT (NAO Paper) (RT-1)-MAIN-PCM(000000.0)-25

MATHEMATICS (SECTION –III)

PART – A

 x  1 x  1 x  2 2  e3x  1

1. 2 2
0
x  3  x   x  1  x  3 

 x  1 x  2 2  e3x  1 + - + + - - +
 0 -3 -1 0 1 2 3
3
x  x  3   x  1 x  3 
 x   , 3    1,0    0,1   3,    2

2. sin x  cos ec x  2 (Using A.M  G.M )

Also tan y  cot y  2
 
So, sin x  cosec x  tan y  cot y  4 is possible when x  and y 
2 4
y
2 tan
 2 y y
tan y  tan  1   1  tan2  2 tan  1  0
4 y 2 2
1  tan2
2
y
So, tan is a root of the equation  2  2  1  0
2

3. Equation of images of pair of lines is (2x – y + 1)(2x + y + 1) = 0

4. The equation of lines joining origin and point of intersection of 5x  2y  3 5

2
 5x  2y 
and x2 + y2 = 10, is x 2  y2  10  or x 2  y2  8 5xy  0
 3 5 
 
 These lines are perpendicular
1 1 1
 Area of triangle  × base × height  × radius × radius   10  10  5
2 2 2

5.  
Any point on the given parabola is t 2 ,2t . The equation of the tangent at (1, 2) is x – y + 1 = 0. The
2
image (h, k) of the point (t , 2t) in x – y + 1 = 0 is

Given by
2
h  t 2 k  2t 2 t  2t  1
 
 
1 1 1 1
 h  t  t  2t  1  2t  1 and k  2t  t 2  2t  1  t 2  1
2 2

Eliminating t from h = 2t – 1 and k = t2 + 1 we get, (h + 1)2 = 4(k – 1)

2
The required equation of reflection is  x  1  4  y  1

6. Number of non-negative integral solutions of the given equation

= Coefficient of x20 in (1 – x)–1(1 – x)–1(1 – x)–1(1 – x4)–1
= Coefficient of x20 in (1 – x)–3(1 – x4)–1
= Coefficient of x20 in (1 + 3C1x + 4C2x2 + 5C3x3 + 6C4x4 + .....)(1 + x4 + x8 + .....)
= 1 + 6C4 + 10C8 + 14C12 + 18C16 + 22C20 = 536

7. Equation of normal at P   is


ax  by cosec  a2  b2 sec   ..... (1)

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022 A LOT (NAO Paper) (RT-1)-MAIN-PCM(000000.0)-26

And at Q   is

ax  bycosec  a2  b2 sec  
 ax  by sec   a2  b2 cosec   ..... (2)

a2  b2
From (1) and (2) y   k
b

1 1 2
8. Satisfying  and  with the given equation we get, 2
 2

  a   a ba
4b
Hence, required result is
b  a2
2

9. Let root is , 
 +  = –b,  = C
  even,  odd   = 2
Now b + c = 15
 –2 –  + 2 = 15   = 17
225
So f(x) = x2 – 19x + 34 and least value of f(x) is 
4

1
1
n 1
2 1 1
10. Sn =  2 n , S 2
1 2 1
1 1
2 2
1 1
 S – Sn = n   2n > 1000
2 1000
Now, 512 < 1000 < 1024  29 < 1000 < 210
If 2n > 1000, then least value of n is 10

 1  1   1   1 
11. Sn   1     1  2    1  3   .....  1  n 
 2   2   2   2 
1 1 1  1
 n    2 .....  n   n  1  n
2 2 2  2

12. Let S be the required Sum. Then we have 2S  16  16C1  16

C3  16
C5  16
C7  ...  16
C15

13. a + c = 2b  3b = 30  b = 10  a + c = 20
Number of positive integral solution is 19C1 = 19

6
14. Let  2 1  = 1 + F, where I is an integer and 0 < F <1
6 1
Let f =  
2  1 . We have 2 1
2 1
 0  2 1 1  0 < f < 1
6 6
Also 1 + F + f =   
2 1  2 1
= 2[6Co.23 + 6C2.22 + 6C4.2 + 6C6] = 2(8 + 60 + 30 + 1) = 198
Hence, F + f = 198 – I is an integer. But 0 < F + f < 2
Therefore F + f = 1, and thus, I = 197

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 022 A LOT (NAO Paper) (RT-1)-MAIN-PCM(000000.0)-27
10
15. 5 40  62510   56  11  9  = multiple of 11  910
10
 910  11  2  = multiple of 11  210
210  1024  93  11  1

16.   
z3   3  i  z2  3z   m  i  0 ; z3  3z2  3z  m  i z2  1  0 
If z is a real root, then z3  3z2  3z  m  0 and z2  1  0
 z  1 ; z  1  m  1 ; z  1  m  5

17. 1 (2) 3 (4) 5 (6) 7 (8) 9

4!
Four odd digits 3, 3, 5, 5 occupy four even positions is  6 ways
2!2!
5!
Rest five digits 2, 2, 8, 8, 8 can occupy rest five positions is  10 ways
2!3!
 Required number of numbers = 6  10 = 60

18. sin2 x  cos 2 x  cos x  cos 2x   cos x  cos    x   2x  2n     x 

 5  5
in 0,2  x  , , but x  , hence x  , are the solutions.
3 3 3 3

19. cosBcosC  sinB sinCsin2 A  cosBcosC  sinBsinC

 cos B  C   1  cos B  C   1
 B  C  cos  B  C  cannot be greater than1 & A  900
 
20. The given equation can be written as tan1  x  1  tan1  x  1  tan1 3x  tan1 x
2x 2x
 2
  x  3x3  2x  x3
2x 1  3x 2
 4x 3  x  0  x 4x2  1  0 
1 1
 x = 0, x   . Thus x =
2 2
PART– C

1. For x > 0, put log2 x = y, log3 x = z to obtain yz = 3z + 2y – 6

 (y – 3)(z – 2) = 0  x = 8, 9

2. cos  A  B  C   cos A cosBcosC 1 

  tan A tanB 
3. Given lines are 3x + 4y= 9 ….. (1)
y = mx + 1 ….. (2)
From (1) and (2) we get, 3x  4  mx  1  9
  3  4m  x  9  4  5
5
x
2  4m
Since x is an integer
 3  4m  1, 1,5, 5
4m  2, 4,2, 8  4m  2, 4,2, 8
m  1 or –2 [ m is an integer]
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022 A LOT (NAO Paper) (RT-1)-MAIN-PCM(000000.0)-28

4. Radius = 51 ,  dist. of the chord 7x + y – 4 = 0

| 78  5  4 |
from centre = = 50
50
So, length of chord = 2 × 51  50 = 2

5. Equation of tangent of parabola y 2  4x is

1
y  mx  ..... (1)
m
Equation (1) is also a tangent of x 2  32y
 1 32
Then x 2  32  mx    x 2  32mx  0
 m m
Condition of tangency (D = 0)
1 1
m3   m 
8 2
x
From Equation (1), y   2  x  2y  4  0
2

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