02-01-24

PHYSICS MAX.MARKS: 100

SECTION – I
(SINGLE CORRECT ANSWER TYPE)
This section contains 20 multiple choice questions. Each question has 4 options (1), (2), (3) and (4) for its answer,
out of which ONLY ONE option can be correct.
Marking scheme: +4 for correct answer, 0 if not attempted and -1 if not correct.
1. A very broad elevator platform is going up vertically with a constant acceleration 1 ms-2.
At the instant when the velocity of the lift is 2 m/s, a stone is projected from the platform
with a speed of 20 m/s relative to the platform at an elevation 300. The time taken by the
stone to return to the floor will be  g  10 m / s 2 

A) 30 sec B) 70 sec C) 20 sec D) 90 sec

11 11 11 11

2. The force exerted by a compression device is given by F  x   kx  x  l  for 0  x  l , where

l is the maximum possible compression, x is the compression and k is a constant. The
work required to compress the body by a distance d will be maximum when:

l
A) d  l B) d  C) d  l D) d  l
4 2 2

3. Figure shows a short magnet executing small oscillations in a uniform magnetic field
directed into page and magnitude 24  T. The period of oscillation is 0.1 s. When the
key K is closed, an upward current of 18A is established as shown. The new time period
is_____(Neglect the effect of earth’s magnetic field)

(Needle oscillates in plane normal to the page)

A) 0.1 s B) 0.2 s C) 0.05 s D) 0.4 s

4. A satellite is moved from one circular orbit around the earth to another of lesser radius.
Which of the following statement is true?

A) The kinetic energy of satellite increases and the gravitational potential energy of
satellite – earth system increases.

B) The kinetic energy of satellite increases and the gravitational potential energy of
satellite – earth system decreases.

C) The kinetic energy of satellite decreases and the gravitational potential energy of
satellite – earth system decreases.

D) The kinetic energy of satellite decreases and the gravitational potential energy of
satellite – earth system increases.

5. The relation between internal energy U, pressure P and volume V of a gas in an

adiabatic process is U = a + bPV.

Where a and b are constant. What is the effective value of adiabatic constant  ?

A) a B) b  1 C) a  1 D) b
b b a a

6. The root mean square speed of the molecules of a diatomic gas is v. when the
temperature is doubled, the molecules dissociate into two atoms. The new root mean
square speed of the individual atom is
A) 2v B) v C) 2v D) 4v
7. In young’s double slit experiment, the two slits are coherent sources of equal amplitude
and wave length  . In another experiment with the same setup, two slits are sources of
equal amplitude ‘A’and wavelength  , but are incoherent. The ratio of intensities of light
at the midpoint of the screen in the first case to that in second case, is
A) 2 : 1 B) 1 : 2 C) 3 : 4 D) 4 : 3

8. The relative error in calculating the value of g from the relation T  2 l is

g

(given the relative errors in calculating T and l are x and  y respectively)

A) x  y B) 2 x  y C) 2 x  y D) x  2 y
9. I1, I2, I3 and I4 are the moment of inertial of square plate about the axis marked 1, 2, 3
and 4 respectively and I is the moment of inertia about an axis passing through O and
perpendicular to the plate, Relation between them is given as

i) I = I1 + I2 ii) I = I1 + I3

iii) I = I2 + I4 iv) I = I1 + I2 + I3 + I4

Which of ṭhe following options is incorrect?

A) (i) and (iii) B) (ii), (iii) and (iv) C) Only (ii) D) Only (iv)

10. A cylindrical container filled with a liquid is being rotated about its central axis at a
constant angular velocity  . Four points A, B, C and D are chosen in the same plane
such that ABCD is a square of side length a and AB is horizontal while BC is vertical.
A and D lie on the axis of rotation. Let the pressure at A, B, C and D be denoted by
PA, PB, PC and PD respectively. Now, consider the following two statements.

2g
(i) PC> PA for all values of  (ii) PB> PD only if  
a

Which of these options is correct?

A) Both (i) and (ii) are correct B) (i) is correct and (ii) is incorrect

C) (ii) is correct and (i) is incorrect D) Both (i) and (ii) are incorrect
 Two point dipoles p k and p k are located at (0, 0, 0) and (1 m 0, 2m) respectively. The
^ ^
11.
2
resultant electric field due to the two dipole at the point (1m, 0, 0) is…
9p ^ 7p ^ 7p ^ 11 p
A) k B)  k C) k D)
32 0 32 0 32 0 32 0

12. A electron having kinetic energy T is moving in a circular orbit of radius R


perpendicular to a uniform magnetic induction B .If kinetic energy is doubled and
magnetic induction tripled, the radius will become
3R 3 2 4
A) B) R C) R D) R
2 2 9 3

13. A metallic rod of length ' l ' is tied an insulating string of length 2l and made to rotate
with angular speed  on a horizontal table with one end of the string fixed. If there is a
vertical magnetic field ‘B’ in the region, the e.m.f. induced across the ends of the rod is:

3Bl 2 4Bl 2 5Bl 2 2Bl 2

A) B) C) D)
2 2 2 2
14. In order to establish an instantaneous displacement current of 1 mA in the space between
the plates of 2  F parallel plate capacitor, the time varying potential difference need to
apply is
A) 100 Vs-1 B) 200 Vs-1 C) 300 Vs-1 D) 500 Vs-1
0
15. Light of wavelength 2475 A is incident on barium. Photoelectrons emitted describe a
1
circle of maximum radius 100 cm by a magnetic field of flux density 105 Tesla.
17
Work function of the barium is (nearly (Given e  1.7 1011 ), hc  12375  eV  A0 
m

A) 1.8 eV B) 2.1 eV C) 4.5 eV D) 3.3 eV

16. Two positively charged particles are projected along two parallel lines on a smooth
horizontal surface as shown. Which of the following statement is incorrect
corresponding to their subsequent motion? [Before any collision (except between the
particles) takes place]

A) The linear momentum of the system of particles is conserved in any direction

B) The angular momentum of the system of particles is conserved about any point in
space
C) The angular momentum of each particle is individually conserved about their center
of mass
D) The angular momentum of each particle is individually conserved about any point in
space
17. A particle of mass 5 x 10-5 kg is placed at lowest point of smooth parabola
x2  40 y  x and y in m  . If it is constrained to move along parabola, angular frequency of
small oscillations (in rad/s) will be approximately (g=10 m/s2)

1
A) 2 B) 10 C) D) 5
2
18. A point charge q is placed at a distance r from the center of a thin metallic neutral
spherical shell of radius R as shown in fig. electric potential at point A is

1 q 1 q 1 q q 1 1 
A) B) C) D)   
4 0 R 4 0 r 4 0 2
R r 2 4 0  r R 
19. There is a plot of binding energy per nucleon Eb, against the nuclear mass M; A, B, C, D,
E, F correspond to different nuclei.

Consider four reactions

(i) A  B  C   (ii) C  A  B   (iii) D  E  F   and (iv) F  D  E  

Where  is the energy released? In which reactions is  positive?

A) (i) and (iii) B) (ii) and (iv) C) (ii) and (iii) D) (i) and (iv)

20. An R-L-C series circuit with 100  resistance is connected to an AC source of 200 V
and  = 300 rad/s. When only capacitor is removed, the current lags behind voltage by
600. When only inductor is removed, the current leads voltage by 600. The power
dissipated in the R-L-C circuit is

A) 200 W B) 400 W C) 200 3 W D) 100 W

SECTION-II
(NUMERICAL VALUE ANSWER TYPE)
This section contains 10 questions. The answer to each question is a Numerical value. If the Answer in the
decimals , Mark nearest Integer only. Have to Answer any 5 only out of 10 questions and question will be
evaluated according to the following marking scheme:
Marking scheme: +4 for correct answer, -1 in all other cases.
21. A 15 kg block is initially moving along a smooth horizontal surface with a speed of
v  4 m / s to the left. It is acted by a force F, which varies in the manner shown.
If the velocity of the block at t = 15 seconds is ‘X’. Then the value of  X  =__

([ ] – greatest integer function)

 
Given that, F  40cos  t
 10 
22. A convex lens of focal length f = 20 cm is cut into two equal pieces and the pieces are
separated by 3mm as shown in the figure. A point object O is placed at a distance of 30
cm. The distance between the two image points formed will be (in mm)

23. A ball of mass m moving horizontally with a velocity  strikes the bob of a pendulum at
rest. The mass of the bob is also m. If the collision is perfectly inelastic, the height to
v2
which the system will rise is given by h  , then the value of x is
x.g

24. A copper wire is held at the two ends by rigid supports. At 300 C, the wire is just taut,
with negligible tension. Find the speed of transverse waves (in m/s) in this wire at 100 C
in decimeter/second [Given, for copper: Young’s modulus = 1.3 x 1011 N/m2, coefficient
of linear expansion = 1.7 x 10-5C-1, density = 9 x 103 kg/m3.]

25. In a meter bridge, the wire of length 1 m has a non - uniform cross section such that, the
dR
variation of its resistance R with length l is dR  1 . Two equal resistances are
dl dl l

connected as shown in the figure. The galvanometer has zero deflection when the jockey
is at point P. The length AP is X (in m), then 100X = …

26. The forward – bias voltage of a diode is changed from 0.6 V to 0.7 V, the current
changes from 5 mA to 15 mA. What is the value (in  ) of the forward bias resistance?
27. The binding energy of an electron in the ground state of the He atom is equal to
E0  24.6eV . Find the energy required (in eV) to remove both electrons from the He atom.

28. In the circuit shown in the adjoining figure, the reading of ideal ammeter A (in ampere)
is:

5
29. A solid spherical ball of radius m is connected to a point A on the wall with the help
9
of a string which makes an angle   450 with the vertical. The sphere can rotate freely
about its central axis and it is set into rotational motion against the vertical face of the
wall with an angular velocity 100 rad s-1. In how much time (in s) will it come to rest?
   0.1& g  10 m / s 2 

30. A resonance tube is old and has jagged end. It is still used in the laboratory to
determine velocity of sound in air. A tuning fork of frequency 512 Hz produces first
resonance when the tube is filled with water to a mark 11cm below a reference mark,
near the open end of the tube. The experiment is repeated with another fork of
frequency 256 Hz which produces first resonance when water reaches a mark 27 cm
below the reference mark. The velocity of sound in air (in m/s), obtained in the
experiment, is close to_____
CHEMISTRY MAX.MARKS: 100
SECTION – I
(SINGLE CORRECT ANSWER TYPE)
This section contains 20 multiple choice questions. Each question has 4 options (1), (2), (3) and (4) for its answer,
out of which ONLY ONE option can be correct.
Marking scheme: +4 for correct answer, 0 if not attempted and -1 if not correct.
31. In which of the following pairs, the hybridization of central atom is same, but shape is
not the same?

A) SO3 , CO32 B) SO32 , NH 3 C) PCl5 , POCl3 D) XeF2 , ClF3

32. Number of  –bonds in B2 , C2 , N2 respectively as per molecular orbital theory is :

A) 1,2,3 B) 0,1,2 C) 1,2,2 D) 1,1,2

33. 22.44 kJ of energy is required to convert 8 g of gaseous metal, M to M+(g). If the first
ionisation energy of the metal is 374 kJ/mol, select the incorrect statement from the
following.

C) Gram atomic mass of the metal is 133.33 g

D) 3.613  1022 atoms of M are converted to M+

34. Assertion(A):The single N–N bond is weaker than the single P–P bond

Reason(R):High inter electronic repulsion of the non–bonding electrons due to the small
N–N bond length

In the light of the above statements, choose the correct answer from the options given
below:

A) Both (A) and (R) true but (R) is not the correct explanation of (A)

B) (A) is false but (R) is true.

C) Both (A) and (R) true and (R) is correct explanation of (A).

D) (A) is true but (R) is false.

35. A coordinate complex of formula CrCl3 .6H 2O has green colour. 1L of 0.1 M solution of
the complex when treated with excess of AgNO3 gave 28.7g of white precipitate. The
formula of the complex would be: [At.wt. Ag : 108 amu; At.wt. of Cl = 35.5 amu]

A) Cr  H 2O 6 Cl3  B) CrCl  H 2O 5  Cl2 .H 2O

C) CrCl2  H 2O 4  Cl.2 H 2O D) Cr  H 2O 3 Cl3 

36. Which type of Isomerism cannot be shown by Co  en 2  NO2 2  Cl ?

A) Geometrical B) Ionisation C) linkage D) hydrate

37. Which of the following does not involve d  d transition for causing colour ?
4
A)  Fe  CN 6  B) CuSO4 .5H2O C) KMnO4 D) CrCl3 .4 NH 3

38. The NH3 evolved from 0.5 gm of the organic compound in KJeldhal’s estimation of
Nitrogen neutralizes 10 ml of 1M H2 SO 4 . Identify the incorrect statement out of the
following.
A) Percentage of nitrogen in the organic compound is 56%
B) 20 milli moles of NH3 is produced
C) 10 milli moles of NH3 is produced
D) if the evolved NH3 were neutralized by 10 ml of 1M HCl, the % of nitrogen, would
have been 28%.
39.
CH3

KMnO4 H HBr (C)

CH3 CH (A) (B)
  H 2O  ROOR (Major)

CH3

The product “C” in the above reaction formed is

A) B) C) D)
 Br Br Br Br Br

40. O NH
O
O
(I) (II) (III) (IV) (V)

Ease of SN1 reactions among these compounds upon treatment with aqueous NaOH will
be in the order as:

A) (I) > (II) > (III) > (IV) > (V) B) (IV) > (I) > (III) > (II) > (V)

C) (I) > (IV) > (III) > (II) > (V) D) (V) > (IV) > (III) > (II) > (I)

41.

 A 
 i  BH /THF
3
 ii  H O / OH
2 2
  B

 B  
HF
C 

The compound(C) is

A) B)

C) D)

42. When neopentyl alcohol is heated with an acid, it slowly converted into an 85 : 15
mixture of alkenes A and B, respectively. Then, the ratio between number of hyper
conjugated structures for A and B is :

A) 5 : 9 B) 5 : 1 C) 1 : 5 D) 9 : 5
43. End product in the following sequence of reactions is :

A) B) C) D)
44.

A) B)

C) D)
45. Given below are two statements, one is labelled as Assertion (A) and other is labelled
as Reason(R):
Assertion (A): Gabriel phthalimide synthesis cannot be used to prepare aromatic
primary amines.
Reason (R): Aryl halides do not undergo nucleophilic substitution reaction at room
temperature.
In the light of the above statements, choose the correct answer from the options given
below:
 A) Both (A) and (R) true but (R) is not the correct explanation of (A)

B) (A) is false but (R) is true.

C) Both (A) and (R) true and (R) is correct explanation of (A).
D) (A) is true but (R) is false.
46. Which of the following statement is false?
A) In fibrous proteins, poly peptide chains are held by hydrogen & disulphide bonds.
B) In Globular proteins, chains of polypeptides coil around to give a spherical shape
C) Keratin & myosin are fibrous proteins and soluble in water
D) Insulin & albumin are globular proteins and soluble in water
47. The conductivity of a saturated aq.solution of AgCl at 298 K is found to be
1.382  10 2  1 m 1 . The ionic conductance of Ag  and Cl  at infinite dilution are
61.9  1cm 2 mol 1 and 76.3 1cm 2 mol 1 respectively. The solubility of AgCl is

A) 1101 mol L1 B) 1102 mol L1 C) 1103 mol L1 D) 1.9 105 mol L1

48. 50 g of antifreeze (ethylene glycol) is added to 200 g water. What amount of ice will
separate out at 9.30 C ?  K f  1.86 K kg mol 1  .

A) 38.71 mg B) 42 g C) 38.71 g D) 42 mg

49. An energy of 24.6eV is required to remove the first electron from helium atom. The
energy required to remove both electrons from helium atom is

A) 54.4eV B) 79 eV C) 49.2 eV D) 51.8 eV

50. For the first order reaction 3 A  B concentration varies with time as shown in the
adjacent graph. The half – life of the reaction would be

A) 4 minutes B) 2 minutes C) 6 minutes D) 8 minutes

 SECTION-II
(NUMERICAL VALUE ANSWER TYPE)
This section contains 10 questions. The answer to each question is a Numerical value. If the Answer in the
decimals , Mark nearest Integer only. Have to Answer any 5 only out of 10 questions and question will be
evaluated according to the following marking scheme:
Marking scheme: +4 for correct answer, -1 in all other cases.
51. When the ionization energy Vs atomic number is plotted for the elements, of atomic
number 11 to 18, two peaks are observed for the element X and Y in between the curve.
What is the difference between atomic number of element X and element Y?
52. For a low spin , Cr 2  complex, the value of spin only magnetic moment is x B.M in an
octahedral field. Then the value of x2 is_____
53. The number of compounds among the following more reactive than acetic acid towards
decarboxylation by soda-lime is
HC  CCH 2COOH , H 2C  CH  COOH , PhCH 2COOH , CCl3COOH , Me3CCOOH

54. How many of the following groups activates benzene ring towards electrophilic aromatic
substitution?
O
||
 NHCOR,  OCOR,  C  O  R,  NR 2 ,  NH 2 ,  OH,  OR

55. .
The number of iodoform molecules produced per molecule of the reactant in above
reaction is ______
56. How many of the following are more reactive than acetaldehyde towards nucleophilic
addition?
FCH2CHO, O2N CH2CHO , CH3CH2CHO, CH3COCH3, PhCHO, PhCOCH3
57. At what pH the oxidation potential of hydrogen electrode will be 0.413 V?
 2.303RT 
  0.059V  (Given: PH 2  1 atm )
F 
58. If E Ag
0

/ Ag
 0.8V ; K sp  AgCl   10 10 M 2 and ECl0  / AgCl / Ag is ‘x’ volts, then the value of 10x is:
 2.303RT 
  0.06V 
F 
59. A buffer is 0.25 M in CH 3 COOH and 0.56 M in CH 3 COONa . What is the value of pH if
0.006 mol of HCl is added to 0.300 L of a buffer solution?
pK a  CH3COOH  4.7  log10 2  0.3 
0
60. NH 4 HS S  is added to a closed vessel containing H 2 S g  at 1 atm and 27 C. If the total
pressure at equilibrium is 3 atm at 270C, then, the value of K P [in (atm)2] is :
MATHEMATICS MAX.MARKS: 100
SECTION – I
(SINGLE CORRECT ANSWER TYPE)
This section contains 20 multiple choice questions. Each question has 4 options (1), (2), (3) and (4) for its answer,
out of which ONLY ONE option can be correct.
Marking scheme: +4 for correct answer, 0 if not attempted and -1 if not correct.
61. If ‘  ’ is a complex number satisfying the relation   1    3(1  i) . Then  equals to
1 1
A)   3i B)  3i
4 4
1
C)  3i D) No such complex number exists
4
62. The greatest integral value of k such that both the roots of the equation
 k  5 x 2  2kx  (k  4)  0 are positive, one root is less than 2 and the other root lies
between 2 and 3 is
A) 13 B) 14 C) 23 D) 24
 3 1
63. Let matrix A    then  I  A  (where I is a unit matrix of order 2) equals to
99

 6 2 

A) I  298 A B) I  299 A C) I   299  1 A D) I   299  1 A

64. In a bag there are three tickets with number 1, 2, 3. A ticket is drawn at random and the
number is noted and put back in the bag, this is repeated for four times. Then the chance
that the sum of those numbers is even is

39 40 41 42
A) B) C) D)
81 81 81 81

d
65. If g ( y )  y 5  2 y 3  3 y  4, then the value of 28
dy
 g 1 ( y )  at y  2 is

1
A) 2 B) 1 C) D) 2
14
x  4 y  5 z 1 x  2 y  1 z
66. Given lines   ,  
2 4 3 1 3 2
S1: The lines are intersecting
S2: The lines are not parallel
A) Both S1 and S2 are true B) S1 is true and S2 are false
C) S1 is false and S2 is true D) Both S1 and S2 are false
  cos   
67.    . sin   d 
  cos  

A)  .sin   c B)   .cos   c C) 2  .cos   c D) c  2  .cos 


x2
0 k  x dx
68. lim
  0   sin  
 1 where k  0 then the value of ‘k’ is

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

dy tan y
69. The solution of the differential equation   (1  t ) et sec y is, where ‘ c ’ is an
dt (1  t )

arbitrary constant

A) cos y   et  c   t  1 B) cos y   et  c   x  1

C) sin y   et  c   t  1 D) sin y   et  c   t  1

70. The minimum value of k for which the quadratic equation

 cot k  y   tan k  y  2  cot 1 k   0 has both positive roots is:  k  I 

1 2 1 3/ 2 2

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

71. If  3   6  2 then the maximum value of the independent term of x in the expansion of

 x   x 1/6    0,   0  is
1/3 9

A) 42 B) 68 C) 84 D) 148

72. The set of critical points of the function g ( x)  ( x  2)2 / 3 (2 x  1) is

1
A) {1} B) {1, 2} C) {1, 2} D)  , 1
2 

73. The mean and variance of the marks obtained by the students in a test are 10 and 4
respectively. It is known that one of the students got ‘12’ instead of 8. If the new mean
of the marks is 10.2 then the new variance is equal to

A) 4.04 B) 4.08 C) 3.96 D) 3.92

74. The area enclosed between the curves y  x 2 and y  x is
1 2 4
A) B) C) D) 2
3 3 3

75. Statement 1: f  x   x sin x is differentiable at x  0

Statement 2: If f  x  is not differentiable and g  x  is differentiable at x  a , then

f  x  .g  x  can still be differentiable at x  a

A) Statement 1 is true, Statement 2 is true

B) Statement 1 is true, Statement 2 is false
C) Statement 1 is false and Statement 2 is true
D) Statement 1 is false and Statement 2 is false
 1 ;   /2

76. If f ( )   sin cos   , where {.} represents fractional part function, then
 (   / 2) ;    / 2

A) f ( ) is continuous at    / 2
B) lim
 / 2
f ( ) exists but not continuous at    / 2

C) lim
 / 2
f ( ) does not exist

D) lim
 / 2
f ( )  1

77. Match the following

Column – I Column – II
1/3 1/3 2/3
A) g(x) = 2 – x and f(g(x)) = - x + 5x – x , the local
P) 0
maximum value of f(x) is
z 3 
B) No. of points of intersection of the curves arg  
  and
 z 1  4 Q) 1
z 1  i   z 1  i   4  0

C) If f(x) = ax3 + bx2 + cx + d, (a, b, c, d  Q) and two roots of

f(x)=0 are eccentricities of a parabola and a rectangular hyperbola, R) 2
then a + b + c + d =
D) Number of solution of equation 1x + 2x + 3x …. + nx = (n + 1)x
S) 3
are
A) A – Q ; B – S ; C – R ; D – P B) A – S ; B – Q ; C – P ; D – Q
C) A – S ; B – Q ; C – R ; D – Q D) A – S ; B – Q ; C – P ; D – R
 440

 cos r o

The value of 100 (k  1)  where [ x] represents the G.I.F. and k 

r 1
78. 440 is
 sin r
r 1
o

A) 144 B) 142 C) 141 D) 140

79. The sum of possible integral values of k for which the point P(0, k) lies on or inside the
triangle formed by the lines y  3x  2  0, 3 y  2 x  5  0 and 4 y  x  14  0 is

A) 4 B) 5 C) 6 D) 7

x 1 y  1 z x  1 y  3 z 1
80. The acute angle between the lines   and   where a  b  c
a b c b c a
and a, b, c are the roots of the equation t 3  t 2  4t  4  0 is

 63  4 2  3 
A) sin 1   B) cos 1 C) tan 1   D) cos 1  
 9  9 3  13 

SECTION-II
(NUMERICAL VALUE ANSWER TYPE)
This section contains 10 questions. The answer to each question is a Numerical value. If the Answer in the
decimals , Mark nearest Integer only. Have to Answer any 5 only out of 10 questions and question will be
evaluated according to the following marking scheme:
Marking scheme: +4 for correct answer, -1 in all other cases.
81. If ( p  iq) 2018  p  iq . Then the number of real ordered pairs ( p, q) that satisfy the given
N
equation is N. Then   where [ x] represents G.I.F. equals to _________
100 

82. An equilateral triangle has its centroid at the origin and one side is x  y  1 , then the sum
of the slopes of the other two sides is

If a  b  2, c  1,  a  c  .  b  c   0 . Then the value of a  b  2c .  a  b  is equal to

2
83.

84. If x1 , x2 are the roots of x 2  x  K  0 and x3 , x4 are the roots of x 2  4 x  L  0 such that
x1 , x2 , x3 , x4 are in G.P. Then the product of the integral values of K and L is ______
 t t2 1 t3
85. The total number of distinct real values of ‘t’ for which 2t 4t 2 1  8t 3  10 is _____
3t 9t 2 1  27t 3

86. If 
dt  1
 
 2 log 1  1    f (t )  c then 3 lim f (t )  _________
t  
2
t (1  t )  t t 

87. Let A  1, 2, 3, 4, 5 . The number of unordered pairs of subsets P and Q of A such that
P  Q   is ‘n’. Then the sum of digits of n is __________

88. If  ,  ,   is the foot of perpendicular drawn from the point P(1, 0, 3) to the line joining
the points A(4, 7, 1) and B(3, 5, 3) . Then the value of 3  6  9  ______

xy g ( x) . g ( y ) g (3)
89. If g    ; x, y  R, g (1)  g 1 (1) then 1  ________
 2  2 g (3)

(1  y )1/ y  e1
90. The value of real number ‘k’ for which the right hand limit lim is a non-
y  0 yk

zero real number is _________

 Date: 02-01-24
Time: 3 HRS Max. Marks: 300
KEY SHEET
PHYSICS
1 C 2 D 3 B 4 B 5 B
6 C 7 A 8 C 9 D 10 A
11 B 12 C 13 C 14 D 15 C
16 D 17 C 18 B 19 D 20 B
21 -13 22 9 23 8 24 701 25 25
26 10 27 79 28 1 29 20 30 328

CHEMISTRY
31 D 32 C 33 B 34 C 35 B
36 D 37 C 38 C 39 D 40 B
41 C 42 D 43 B 44 B 45 C
46 C 47 C 48 C 49 B 50 B
51 3 52 8 53 4 54 6 55 1
56 2 57 7 58 2 59 5 60 2

MATHEMATICS
61 D 62 C 63 D 64 C 65 D
66 C 67 C 68 D 69 D 70 B
71 C 72 B 73 C 74 B 75 A
76 C 77 B 78 C 79 B 80 B
81 20 82 4 83 10 84 64 85 2
86 3 87 5 88 46 89 3 90 1
 SOLUTIONS
PHYSICS
^ ^
 ^ ^

1. Initial velocity of stone w.r.t lift  20sin 300 j  20cos300 i  10 3 i  10 j  m / s
 

 ^ ^

Initial velocity of stone w.r.t ground  10 3 i  12 j  m / s
 
The initial position of stone and lift are same and when they again meet their final positions will also
be same. So both will have same displacement in vertical direction in same time
1 t2
Displacement of lift  2  t    1  t 2  2t 
2 2
1
Displacement of stone  12  t    10  t 2  12t  5t 2
2
2
t
So 2t   12t  5t 2
2
2
11t 20
 10t or t  sec
2 11
20
So time taken by stone to return to the floor of lift is sec
11
dW
2. For W to be maximum;  0;
dx
i.e., F  x   0  x  l , x  0
Clearly for d  l , the work done is maximum.
Alternate Solution:
External force and displacement are in the same direction
 work will be positive continuously so it will be maximum when displacement is maximum.
 I 4  107  18
3. B 0  T  18t
2 r 2  0.2
I I
Now, T  2 and T   2 d
MBH M  BH  B 
T BH T 24
Dividing  or  2
T BH  B T 24  18
T   2  0.1s  0.2 s
4. In the circular motion around the earth, the centripetal force on the satellite is a gravitational force.
Therefore, v 2  GM / R , where M is the mass of the Earth, R is the radius of the orbit of satellite and
G is the universal gravitational constant. Therefore, the kinetic energy increases with the decrease in
the radius of the orbit. The gravitational potential energy is negative and decreases with the decrease
in radius.
5. For an adiabatic process,
0 = dU + PdU
or d(a+bPV) + PdV=0
dV dP
or  b  1 b 0
V P
or (b+1) log V + b log P = constant
V b 1 p b  constant
b 1
or PV b
 constant
b 1
 
b
3RT
6. vrms  . According to problem T will becomes 2T and M will become M/2 so the value of vrms
M
will increase by 4  2 times i.e., new root mean square velocity will be 2v .
7. When sources are coherent, then I R  I1  I 2  2 I1 I 2 cos 
At middle point of the screen,   0 then
I R  I  I  2 II cos 0  4 I
When sources are in coherent, then I R '  I1  I 2  I  I  2I
I R 4I
 =2
I R' 2I
l
8. T 2  4 2
g
4 2l
g 2
T
g l T
  100   100  2  100
g l T
g
  y  2x
g
9. Using perpendicular axis theorem I = I1 + I2 and I = I3 + I4 also, I1 = I2 = I3 = I4 hence option 4 is
wrong.
10. We know that
1
PB  PA   2 a 2
2
PD  PA   ga
1 1
PC  PD   2 a 2  PA   ga   2 a 2
2 2
Therefore,
Pc  PA for all the values of  and PB  PD only
2g
If  
a
^
11. For p k it is equatorial point
1 P ^
 E1  k 
4 0 1  
P^
For k it is axial point
2
 P ^
 2 k 
1 2    1 p k^
 E2 
4 0 2 3
4 0 8
7p ^
 E  E1  E2   k
32 0

mv
12. Radius of circular orbit R 
qB
2mKE 2mT
 
qB qB
2
If T becomes double & ‘B” becomes tripled then radius becomes R
9
13. de  B  x  dx

3L
e  B  x dx
2L

5 B L2

2
14. I d  1mA  103 A
C  2 F  2  106 F
d dV
I D  I C   CV   V
dt dt
dV I D 103
Therefore,    500 Vs 1
dt C 2  106
Therefore, applying a varying potential difference of 500 Vs-1 would produce a displacement current
of desired value.
15. Radius of circular path described by a charged particle in a magnetic field is given
2mK q 2 B 2 r 2  e  eB 2 r 2
by r  ; Where K = Kinetic energy of electron  K   
qB 2m m 2
2
1  1 
 105   1  8  10 20 J  0.5eV
2
  1.7  1011  1.6 10 19  
2  17 
 12375 
By using  W0  E  K max    eV  0.5eV  4.5eV
 2475 
16.

1. As mutual repulsive force between the particles is internal for the system and as there is no other
external force on the system, linear momentum of the system is conserved in any direction.
2. As the forces on the particles due to one on the other are equal in magnitude. Opposite in direction
and act along the line joining them always, net torque on the system due to these forces about any
point in space in zero. Therefore angular momentum of the system remains constant about any point
in space.
3. As center of mass of the system lies on the line joining the particles always and force on any of
them is passing through C.M always, torque due to this force on any particle about C.M is zero.
Hence angular momentum of any particle about C.M is conserved individually.
4. About any other point except C.M, torque on any individual particle is not zero. Hence angular
momenta of individual particles change but total angular momentum of the system remains constant.
17.

F  mg sin   mg tan   is small  .

dy 2x
i.e, F  mg   mg 
dx 40
x 1
 a    
2 2
18. Potential at any point inside the shell = potential at any point on the surface
 potential at A = potential at C due to ‘q’ and induced charges =
1 q 1 q 1  q  1 q
   
4 0 r 4 0 R 4 0  R  4 0 r
19. For A  B  C   ,  is the positive. This is because Eb for D and E is greater then Eb for F.

20. It is a case of resonance

X L  X C
Z R
Vrms 200
 I rms    2A
Z 100
2
 Pav  I rms R  4  100  400W
21. Change in linear momentum P   Fdt
15
 
15  v f  u    40cos   t dt
0  10 
15
40  sin  / 10  t  400
v f  4     4   1  12.5 m / s
15   / 10  0 15
22. ‘O’ is off axis for axis for both the parts. Size of object for upper and lower parts is
2mm and 1 mm respectively.
1 1 1
 
f v u
1 1 1
   v  60Cm
20 v 30
v
m 2
u
 Distance between the two image points is 4 + 2 + 1 + 2 = 9mm
23. Because the collision is perfectly inelastic, the two blocks stick together. By conservation of linear
momentum.
v
2 mV  mv orV 
2
By conservation of energy,
1 2 v2
2mgh   2mv or h 
2 8.g
 x8
T TL L
24. V & Y &   
 AL L
T Y L Y  
V   
A L 
1.3  1011  1.7  105  20
V  70 m / s
9  103
Cdl
25. dR 
l
l 1
dl dl
0 l l l
C  C

1
2 l l0  2 l
l

2 l  22 l
4 l 2
1
 l   0.25m
4
26. The forward biased resistance of a diode is
V 0.7  06
R 
I 15  5   103
01
R  10
10  10 3
27. Energy required to remove first electron is 24.6 eV. After removing first electrons from this atom, it
will become He+
E1   13.6  2 
2
 as EZ 2
and Z  2 
= - 54.4 eV
 Energy required to remove this second electrons will be 54.4 eV.
 Total energy required to remove both electrons
= 24.6 + 54.4
= 79 eV
 1 12
28. Req = 0.3   0.3   1.5
1 1 1 10
 
2 4 12
Applying ohm’s law,
V  IR
1.5
I  1A
1.5
29. FBD of the spherical ball

For translation equilibrium

T cos150  Mg   N
T sin150  N
Alsot  L
   NR  t  I0
2
MR 20
t  5
 NR
Solving we get t = 20 s
1
30.  I 0  11
4
2
 I 0  27
4
2  1
 16 cm
4
 1 1 
V     0.64 m
 256 512 
 V  512  0.64 m / s
 328 m / s

CHEMISTRY
2
31. A) SO3 & CO3 ; Both are sp & planar triangular

B) SO32  & NH 3 ; Both are sp3 & pyramidal

C) PCl5 : sp 3 d & trigonal bipyramidal
POCl5 : sp 3 & tetra hedral
D) XeF2 : sp 3 d & linear
ClF3 : sp 3 d & T–shape
As per M.O.T
32. B2 : By distributing 10 electrons only two electrons in   B .M .O are extra left without cancelling
with A.B.M.O electrons
C2 : By distributing 12 electrons, only 4 electrons in  B.M .O are extra left without cancelling with A
B M O electrons
N2 : By distributing 14 electrons only 4 electrons & 2  electrons in B M O & extra left without
cancelling with A B M O electrons.
 22.44
33. No of moles of M  g  formed =
374
 0.06 moles  0.06  6.023  10 23

 3.613 1022 atoms

At.wt.
 22.44  374
8
 At .Wt .  133.33 g .
34. Due to small B.L of N–N bond lp–lp repulsions on ‘N’ weaken the bond.
35. 0.1 moles of the complex – 28.7 g of AgCl
1 mole gives of complex – 287 g of AgCl
– 2 moles of AgCl
-
 2 Cl ions should be ionisable.
36. The complex cannot show hydration isomerism as no H 2O ligands are present.
37. The colour of KMnO4 is due to charge transfer phenomenon
38. nm.eq NH 3  nm.eq H 2 SO4
 10  1  2  20 meq of NH3 = 20 m mol of NH3
1400  neq NH 3
%N 
wt.of organic compound
1400  20  10 3
  56% .
0.5
39.

40. Reactivity order IV > I > III > II > V on the basis of R and I effect of associated groups.
41.
CH2 CH=CH2 Friedal- craft's aklylation
A=

CH2 CH2 CH2 OH Hydroboration -oxidation

B=

C=

42.

43. Cleavage of the double bond by Ozonolysis, iodoform Rxn, dry distillation of calcium salts to give
cyclopentanone, followed by wolf–kishner reduction to give cyclohexane.
44. Benzyllic oxidation to give potassium salt of Benzoic acid, followed by acidification to give Benzoic
acid.
45. Gabriel pthalamide synthesis

46. Keratin and myosin are fibrous proteins and insoluble in H 2O .

 1.382  104  1000
47.  61.9  76.3 
S
 S  10 3 M .
48. T f  K f  m
50 1000
9.3  1.86  
62 x
 x  161.29 g
 Amount of ice separated  200  161.29
 38.71g
49. Required energy = I1  I 2

I1  24.6eV

I 2  I H  Z 2  13.6  22  54.4eV

 E  24.6  54.4  79 eV

50. 3A B
t  4 min; a  3 x  x
a
 4x  a  x 
4
 At 4 min 75% of first order is completed.
2t1 t
 t75%   1  2 min .
2 2
51. X  12  Mg  ; Y  15  P 
52. conceptual
53. Greater the stability of carbanion, greater is the rate of decarboxylation.
Except  CH 3 3 C  COOH remaining are more reactive than CH 3COOH .
O
||
54. Except  C  O  R , remaining are ring activating groups.
55.

56. FCH2CHO, O2N CH2CHO are more reactive than acetaldehyde.

57. H 2 g   2 Haq   2e
0.059 2
0.413   log  H  
2
 pH  7 .
58. E 0  E Ag
0

/ Ag
 0.06 log K SP

59. Number of moles of CH 3COOH  0.25  0.3  0.075 moles

Number of moles of CH 3COO   0.56  0.3  0.168
 Number of moles of CH 3COO  left = 0.168  0.006  0.162
Final number of moles of CH 3COOH  0.075  0.006  0.081
0.162
 pH  4.7  log .
0.081
60. NH 4 HS S   NH 3 g   H 2 S g 
x 1  x 
At equilibrium
1  x  x  3  x  1atm

 K P  2  1  2 atm 2 .
MATHS
2 2
61. Clearly ( x  1)  y  ( x  3)  i ( y  3)
 y  3 and ( x  1) 2  9  ( x  3) 2
1
 x
4
1
But    3i does not satisfy the given equation
4
 2k  k 4
62. Let f ( x)  x 2   x   0 by the given data, f (0)  0, f (2)  0, f (3)  0
 k 5  k 5
k 4
 0 ………. (1)
(k  5)
k  24
 0 …….. (2)
(k  5)
4k  49
 0 …….. (3)
(k  5)
 49 
From (1), (2) and (3), k   , 24 
 4 
2
63. Here A  A  A is an Idempotent matrix
 A  A2  A3  ..........  A99
Hence ( I  A)99  I   299  1 A
64. We can take 3 cases namely four odd numbers, two odd numbers and zero odd numbers.
Let X be the number of odd numbers chosen
 P(sum is even)  P ( X  4)  P ( X  2)  P ( X  0)
4 2 2 4
2  2 1 1 41
    4 C2        
3  3   3   3  81
65. Let f ( y ) be the inverse of g ( y )
 f '  g ( y )  g '( y )  1
1 1
 f '  g (2)  g '(2)  1   g 1  y    
y 2 f '  1 14
66. S1 and S2 are skew lines
cos x  x sin x
67. I  dx
x.cos x
Put x.cos x  t 2  I  2 x.cos x  c
 
x2 d x2
0 k  x d
d 0 k  x
d
68. I  lim  lim
 0    sin   0 (1  cos )
(By. L. Hospitals Rule)
  2

 
k  2
 lim  2   1  k  4

 0 2sin  / 2
 k
69. Given equation can be reduced to
dy 1
cos y    sin y   et (1  t )
dt 1  t 
dy dv
Put sin y  v  cos y 
dt dt
sin y
Hence the solution is   et  c 
(t  1)
 cot k  y   tan k  y  2  cot 1 k   0 has real roots
1 2 1 3/ 2 2
70.
 D  0  tan 1 k   / 3  k  3 and sum of roots  0, product of roots  0
 k  0 and k  3
71. Let Tr 1 be the Independent term of x then r  6,
 t61  84 3  6
72. Check for what values of x , g '( x)  0 or does not exist
2

73. Mean
x i
 10 and Variance
x 2
i   xi
 

  4
n n  n 


x 2
i
 104 
x i

4
 102
n n n
 n  20
Hence new Variance = 3.96
1
2
74. From the graph of the functions, the required area = 2   
x  x 2 dx 
3
0

75. f  x   x sin x is differentiable at x  0

76. L.H.L.  lim  f ( )  0 and R.H.L.  lim  f ( )  
   
2 2
 lim f ( ) does not exist
  / 2
77.
x1/ 3  2  g  x 
 f  x  52  x  2  x  2  x 
2 3

 x3  7 x 2  11x  2  f max  3
Clearly 1 is the root  a + b + c + = 0
x x x
 1   2   n 
     ....   1
 n 1   n 1   n 1 

0
1
78. Clearly k  cot 22  2 1
2
Hence 100 (k  1)   141
79. Use L11 . L22  0 for three sides of the triangle
80. Clearly a  2; b  1 and c  2 and

a1a2  b1b2  c1c2

Use   cos1
a a2
1
2
2

81. p 2  q 2  1 or p 2  q 2  0
1
 z 2018  z   z 2019  1
z
So there are total 2020 solutions
82. The inclination of the line L  x  y  1 is 1350 . So the slopes of the other two sides will be
tan 1350  600 

83. E  a  b  2c. a  c . b 
 a  b  2  c . a  c .b  a .b 
2 2

But  a  c  .  b  c   0
 a .c b . c  a .b 1
 E  8  2  10
84. Let the common ratio of G.P. be ‘ r ’ then x2  x1r , x3  x1r 2 , x4  x1r 3
And x1  x2  1, x1 x2  K , x3  x4  4, x3 . x4  L
 r  2 and x1  1
 K  2, L  32
t t2 1 t3
85. 2t 4t 2 1  8t 3  10
3t 9t 2 1  27t 3
1 1 1 1 1 1
3 6
 x 2 4 1  x 2 4 8  10
1 9 1 3 9 27
 6 x 6  x3  5  0  x 3  5 / 6 or x3  1
dt
86. I  2
dt
 1 
t 2 1  1  
 t
1
Put 1   U
t
 1 2
Then I  2 log 1  1    c
 t  1
1  1  
 t
2
 f (t ) 
 1
1  1  
 t
35  1
87. The number of un-ordered pairs of subsets of A is  122
2
88. If ‘Q’ is the foot of perpendicular and it divide AB in the ratio  :1 then
 3  4 5  7 3  1 
Q , , 
  1  1  1 
Now PQ perpendicular AB    7 / 4
 5 7 17 
  ,  ,     , , 
3 3 3 
 3  6   9  46
g (2 x) g ( x)
89. Using the derivative from the first principle g 1 ( x)  g (1) 
2x x
g 1 ( x) 1
 
g ( x) x
n (1 y )

90. Write (1  y )1/ y as e y

and evaluate the limit