26-04-2025

1001CJA101029250016 JA

PART-1 : PHYSICS

SECTION-I

1) Moment of inertia of a uniform hollow hemisphere of mass m and radius r about various axes are

indicated in the figure. Which of the following is correct?

(A) I1 + I2 = I3
(B) I2 + I3 = I1
(C) I3 + I4 = I1
(D) I1 + I2 = 2I4

2)

System is in equilibrium. Find the hinge force (Ny) :-

(A)

(B)

(C)

(D) None

3) Assertion (A) : No work is required to be done to move a test charge between any two points on
same equipotential surface.
Reason (R) : Electric lines of force at the equipotential surfaces are mutually perpendicular to each
other.

(A) (A) is true, (R) is true, (R) is the correct explanation of (A).
(B) (A) is true, (R) is true, (R) is not correct explanation of (A).
(C) (A) is true, (R) is false.
(D) (A) is false, (R) is true.

4) A copper wire carries a current of 10 A. It has cross-sectional area of 0.05 cm2. Density of copper
is 8.9 g/cm3. Molar mass of copper is 63.5 × 10–3 kg/mol. The drift speed of the electrons is closest to

(A) 1.5 × 10–4 m/s

(B) 1.5 × 104 m/s
(C) 3.5 × 10–4 m/s
(D) 1.5 m/s

5) In a wire of cross-section radius r, free electrons travel with drift velocity v when a current I flows
through the wire. What is the current in another wire of half the radius and of the same material
when the drift velocity is 2v ?

(A) 2I
(B) I
(C) I/2
(D) I/4

6) Calculate energy needed for moving a mass of 4kg from the centre of the earth to its surface. If
radius of the earth is 6400 km and acceleration due to gravity at the surface of the earth is g = 10
m/sec2

(A) 1.28 × 108 J

(B) 1.28 × 106 J
(C) 2.56 × 108 J
(D) 2.56 × 1010 J

7) Two stars of mass m and 3m separated by distance r are moving in circular path about their
centre of mass due to mutual gravitation force. The angular velocity of any star will be

(A)

(B)

(C)

(D)
8) Three charges +q, –q, +q are placed randomly along a straight line at points P1, P2, P3. The
potential energy of this system of charges is u1 when –q charge is in the middle and is u2 when –q

charge is at one end. Then ratio is equal to :-

(A) 3
(B) –3
(C) 2
(D) –2

9) Variation of electric field along the line joining two charges is given in list-I. Relation between
charges is given in list-II. Match the proper variation. Take electric field along right as positive

List-I List-II

|q1| > |q2|

(P) (1)
q1 < 0; q2 > 0

|q1| = |q2|
(Q) (2)
q1 > 0; q2 > 0

|q1| = |q2|
(R) (3)
q1 < 0; q2 > 0

|q2| > |q1|

(S) (4)
q1 > 0; q2 > 0

(A) P→3, Q→1, R→2, S→4

(B) P→1, Q→4, R→2, S→3
(C) P→4, Q→3, R→2, S→1
(D) P→2, Q→3, R→4, S→1
10) In telescope, if the powers of an objective and eye lens are + 1.25 D and + 20 D respectively,
then for relaxed vision, the length of tube and magnification will be :-

(A) 85 cm and 25
(B) 85 cm and 16
(C) 21.25 cm and 16
(D) 21.25 cm and 25

11) When a thin convergent glass lens (µg = 1.5) and has power of +5 D is immersed in a liquid of
refractive index µℓ it acts as a divergent lens of focal length 100 cm. Then µℓ is :

(A) 4/3
(B) 5/3
(C) 5/4
(D) 6/5

12)

A graph between velocity and displacement of a particle performing SHM is shown. Find the
acceleration when the particle is at half of maximum displacement from mean position.

(A) 10 cm/s2
(B) 20 cm/s2
(C) 30 cm/s2
(D) 40 cm/s2

13) Two semi-infinite wires are joined with a quarter circular arc of radius R as shown in diagram.
The charge density is same on both the wires and the arc. The electric field intensity at point C, the

centre of arc is :

(A)

(B)

(C)
(D)

14) A segment of a circular wire of radius R, extending from θ = 0 to π/2 , carries a constant linear

charge density λ. The electric field at origin O is :-

(A)

(B)

(C)

(D) 0

15) Statement-1 : The centre of mass of an isolated system is having some angular momentum
which remains constant with time.
Statement-2 : The isolated system will not experience any net torque.

(A) Statement-1 is True, Statement-2 is True ; Statement-2 is a correct explanation for Statement-1.
Statement-1 is True, Statement-2 is True ; Statement-2 is not a correct explanation for
(B)
Statement-1.
(C) Statement-1 is True, Statement-2 is False.
(D) Statement-1 is False, Statement-2 is True.

16) A planet moving around sun sweeps area A1 in 2 days, A2 in 3 days and A3 in 6 days. Then the

relation between A1, A2 and A3 is :-

(A) 3A1 = 2A2 = A3

(B) 2A1 = 3A2 = 6A3
(C) 3A1 = 2A2 = 6A3
(D) 6A1 = 3A2 = 2A3

17) In free space, a particle A of charge 1 μC is held fixed at a point P. Another particle B of the
same charge and mass 4 μg is kept at a distance of 1 mm from P. if B is released, then its velocity at
a distance of 9 mm from P is :

(A) 2.0 × 103 m/s

(B) 3.0 × 104 m/s
(C) 1.5 × 102 m/s
(D) 1.0 m/s

18) In a certain region of space, the potential is given by : V = k[2x2 – y2 + z2]. The electric field at
the point (1, 1, 1) has magnitude =

(A)
(B)
2k
(C)

(D)

19)

When a copper wire is heated, its resistance :-

(A) decreases.
(B) remains the same.
(C) increases.
(D) any of the above, depending on the temperature.

20) A vessel of depth 2h is half filled with a liquid of refractive index and the upper half with
another liquid of refractive index . The liquids are immiscible. The apparent depth of the inner
surface of the bottom of vessel will be :

(A)

(B)

(C)

(D)

SECTION-II

1) In the given figure, if object is placed at a distance 9 cm from O1 then find the image distance (in
cm) from O2.

2) A thin prism P1 with angle 4° made of glass of refractive index 1.54 is combined with another thin
prism P2 made of glass of refractive index 1.72 to produce dispersion without deviation. The angle(
in degree) of the prism P2 is :

3) A particle starts from rest and performs SHM of amplitude A. Find the ratio of time taken by it

from mean to to time taken by it from A to :-

4) There are 8 drops of a conducting fluid. Each has radius r and they are charged to potential 1
volt. They are then combined to form a bigger drop. Find potential of big drop in volt.

5) A storage battery of emf 8.0 V and internal resistance 1.5 Ω is being charged by a 120 V dc supply
using a series resistor of 6.5 Ω. What is the terminal voltage (in V) of the storage battery during
charging ?

PART-2 : CHEMISTRY

SECTION-I

1) Product. , Product is :

(A) 3° alkyl bromide

(B) 1° alkyl bromide
(C) 2° alkyl bromide
(D) 1° alcohol

2)
Identify the reagent(s) 'A' and condition(s) for the reaction :
(A) A = HCl ; Anhydrous AlCl3
(B) A = HCl , ZnCl2
(C) A = Cl2 ; UV light
(D) A = Cl2 ; dark, Anhydrous AlCl3

3)
Identify major product of both respectively

(A)

(B)

(C)

(D)

4) Match the suitable mechanism given in list-II for the reaction given in list-I, in order to form major
product

List-I List-II

(P) (1) SN2

(Q) (2) SN1

 (R) (3) E2 elimination

(S) (4) E1 elimination

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

5) Statement-1 : Tertiary butoxide ion is example of strong base and weak nucleophile.
Statement-2 : Nucleophilicity is thermodynamic property but basicity in kinetic property.

(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.
Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for
(B)
statement-1.
(C) Statement-1 is true, statement-2 is false.
(D) Statement-1 is false, statement-2 is true.

6)
Above reaction is an example of 1, 4-elimination. Predict the product.

(A) CH3–CH=CH–CH2
(B) CH3–CH=CH–CH3
(C) CH3–CH2–CH≡CH
(D) H2C=CH–CH=CH2

7) Relation between (A) and (B) is:

(A) G.I.
(B) Positional isomer
(C) Enantiomer
(D) Chain isomer

8) Which of the following graphs most appropriately represents a zero order reaction ?
(A)

(B)

(C)

(D)

9) Rate constant for first order reaction is . What percentage of initial reactant will
react in 10 hours? (Take ℓn2 = 0.693)

(A) 12.5 %
(B) 25%
(C) 87.5 %
(D) 75%

10) Solution among the following which does not show significant deviation from ideal behaviour is :

(A) C6H6 + C6H5CH3

(B) C2H5OH + H2O
(C) CHCl3 + CH3COCH3
(D) C5H5N + H2O

11) Which of the following is isotonic with 15 % glucose solution (specific gravity = 1.2) at same
temperature :

(A) 1 M CaCl2
(B) 1 N CaCl2
(C) 0.5 M NaCl
(D) 0.5 M urea

12) Statement-I : The melting point of ice decreases with increase of pressure.
Because
Statement-II : Ice contracts on melting.

(A) Both Statement-I and Statement-II is true

(B) Statement-I is true and Statement-II is false
(C) Both Statement-I and Statement-II are false
(D) Statement-I is false and Statement-II is true

13) Number of square faces in truncated cube will be -

(A) 6
(B) 8
(C) 12
(D) zero

14) Which of the following molecule has intramolecular hydrogen bonding?

(A) Water
(B) o-nitro phenol
(C) HF
(D) p-nitro phenol

15)

Amongst the following the acid having – O – O – bond is

(A)
(B)
(C)
(D)

16) Among the following, the incorrect statement is :

(A) Al(CH3)3 has the three-centre two-electron bonds in its dimeric structure
(B) AlCl3 has the three-centre two-electron bonds in its dimeric structure
(C) BH3 has the three-centre two-electron bonds in its dimeric structure
(D) The Lewis acidity of BCl3 is greater than that of AlCl3

17) Which of the following species exhibits the diamagnetic behaviour?

(A)
(B)
(C) O2
(D) NO

18) The IUPAC name of [Pt(NH3)3Br(NO2)Cl]Cl is

(A) triamminechloridonitrito-Nplatinum(IV) Chloride

(B) triamminebromidochloridonitritoNplatinum(IV) Chloride
(C) triamminebromochloronitroplatinum(III) Chloride
(D) triamminenitrochlorobromoplatinum(III) Chloride

19) Which of the following complex shows ionization isomerism

(A) [Cr(NH3)6]Cl3
(B) [Cr(en)2 ]Cl2
(C) [Cr(en)3]Cl3
(D) [CoBr(NH3)5]SO4

20) Match the compounds given in List I with conductance (given in List II) and assign the correct
code.

List -II (Oxidation

List -I (Compound)
state C.A)

(P) [Pt(NH3)6]Cl4 (1) 97

(Q) [Pt(NH3)4Cl2]Cl2 (2) 523

(R) [Pt(NH3)4Cl3]Cl (3) 404

(S) [Pt(NH3)5Cl]Cl3 (4) 229

(A) P → 2 ; Q → 4 ; R → 1 ; S → 3
(B) P → 2 ; Q → 1 ; R → 3 ; S → 4
(C) P → 1 ; Q → 2 ; R → 4 ; S → 3
(D) P → 1 ; Q → 3 ; R → 4 ; S → 2

SECTION-II

1) Which of the following will give same Markovnikov and anti-Markovnikov product ?

, , , , ,

, , ,
2)

How many alkenes are possible on dehydrohalogenation of following compound :

3)

If in a simple cubic arrangement 'X' number of faces meet at a corner of a cube in crystal lattice then
what is the value of (X–3)

4) If there are 3 moles of atoms are present in the close packing of pattern ABC ABC ABC. Then
number of moles of tetrahedral voids in the packing is equal to.

5) Sum of P = O and P–O–P bonds is/are present in H6P6O18 ?

PART-3 : MATHEMATICS

SECTION-I

1) Number of values of m for which the lines x + y – 1 = 0, (m – 1)x + (m2 – 7)y – 5 = 0 & (m – 2)x +
(2m – 5)y = 0 are concurrent; are

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

2) Given A ≡ (1, 1) and Ab is any line through it cutting the x-axis at B. If AC is perpendicular to AB
and meets the y-axis in C, then the equation of the locus of midpoint P of BC is

(A) x + y = 1
(B) x + y = 2
(C) x + y = 2xy
(D) 2x + 2y = 1

3) is (where [.] represents greatest integer function)

(A) 0
(B) 1
(C) –1
(D) does not exist
4) Given 3x3 + y3 – xy2 – 3x2y = 0, then the value of at (1, 1) is

(A) 1
(B) 0
(C) 3
(D) –3

5) Let f(x) = (x + 1)2 – 1 & f : [–1, ∞] → R

Assertion (A) : The set {x : f(x) = f–1(x)} = {0, –1}
Reason (R) : f is a bijection.

(A) A is correct but R is not correct

(B) A is not correct but R is correct
(C) Both A and R are correct but R is NOT the correct explanation of A
(D) Both A and R are correct and R is the correct explanation of A

6) Let f : R → R be a function such that f(2) = 4 and f ′ (2) = 1. Then, the value
of is equal to

(A) 4
(B) 8
(C) 16
(D) 12

7) Let f : R → R be defined as f(x) = . If f ″ (0) exist value of λ is

(A) 5
(B) 6
(C) 7
(D) No value of λ possible

8) Let f : R → R be defined as f(x) = . Then, f is increasing function in

the interval

(A)

(B) (0, 2)

(C)
(D) (–3, –1)

9) Let f(x) = xcos–1(–sin|x|), , then which of the following is true ?

(A) f ′ is decreasing in and increasing in

(B) f ′ is increasing in and decreasing in

(C) f is not differentiable at x = 0

f ′ (0) =
(D)

10) If x = 3 tan t and y = 3 sec t, then the value of at , is

(A)

(B)

(C)

(D)

11) Let f : R → R, g : R → R and h : R → R be differentiable functions such that f(x) = x3 + 3x + 2,

g(f(x)) = x and h(g(g(x))) = x for all x ∈ R. Then match the lists

List-I List-II

(A) g ′ (2) (I) 666

(B) h ′ (1) (II)

(C) h(0) (III) 38

(D) h(g(3)) (IV) 16

(A) (A) – II, (B) – I, (C) – IV, (D) – III
(B) (A) – I, (B) – II, (C) – III, (D) – IV
(C) (A) – IV, (B) – III, (C) – II, (D) – I
(D) (A) – II, (B) – IV, (C) – I, (D) – III

12) Let f(x) = sin–1x and . If g(2) = g(x), then the domain of the function fog is
(A)

(B)

(C)

(D)

13) The inverse of y = 5logx is

(A) x = 5log y
(B) x = ylog 5

(C)

(D)

14) If y = y(x) is an implicit function of x such that loge(x + y) = 4xy, then at x = 0 is

(A) 10
(B) 20
(C) 30
(D) 40

15) Let f : R → R be defined as f(x) = x + cosx + 2 and g(x) be the inverse function of f(x) then (g ′ (3)
+ g ″ (3)) is

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

16) Let then

Statement-I : ƒ(x) is continuous at x = 0.
and

Statement-II : .

(A) Both Statement-I and Statement-II are true.

(B) Both Statement-I and Statement-II are false.
(C) Statement-I is true but Statement-II is false.
(D) Statement-I is false but Statement-II is true

17) The smallest integral value of p for which the function f(x) = 6px – p sin 4x – 5x – sin3x is
monotonic increasing and has no critical points on R.
(A) 4
(B) 5
(C) 6
(D) 1

18) A line L is perpendicular to the curve at its point P and passes through (10, –1). The
coordinates of the point P is

(A) (2, –1)

(B) (6, 7)
(C) (0, –2)
(D) (4, 2)

19) The function f : [a, ∞) → R where R denotes the range corresponding to the given domain, with
rule f(x) = 2x3 – 3x2 + 6, will have an inverse provided

(A) a ≥ 1
(B) a ≥ 0
(C) a ≤ 0
(D) a ≤ 1

20) Let C be the curve y = x3 (where x takes all real values). The tangent at A meets the curve again
at B. If the gradient at B is K times the gradient at A then K is equal to

(A) 4
(B) 2
(C) –2

(D)

SECTION-II

1) If f(x) = and g(x) is such that h(x) = f(x) + g(x) is an odd function
and identically zero for all x ∈ R, then is (where [.] represents
greatest integer function)

2) Let f(x) = , then absolute value of f(0) equals

3) is

4) If f(x) = g(x) |(x – 1) (x – 2) ... (x – 10)| – 2 is differentiable ∀ x ∈ R where g(x) = ax9 + bx6 + cx3 +

d; a, b, c, d ∈ R then at x = 0 is

5) Given f(x) = 3 + sinx number of integers in the range of f(x) are

 ANSWER KEYS

PART-1 : PHYSICS

SECTION-I

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

SECTION-II

Q. 21 22 23 24 25
A. 9 3 1 4 29

PART-2 : CHEMISTRY

SECTION-I

Q. 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45
A. B C C D C D B B C A C A D B D B A B D A

SECTION-II

Q. 46 47 48 49 50
A. 4 8 9 6 12

PART-3 : MATHEMATICS

SECTION-I

Q. 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70
A. A A C B A D A C A B A C C D C A B D A A

SECTION-II

Q. 71 72 73 74 75
A. 1 1 3 0 3
 SOLUTIONS

PART-1 : PHYSICS

1)
I2 = Icm + m(d)2 = I4

2)
(TP) = 0
200 × 2 = N × 7

N=

3) W.D. = qΔV
Electric lines of force are perpendicular to the equipotential surface.

4) I = neAVd

; ρ is density

5) I = nevdA

6)
ΔE =

7)

9) Apply the following tests :-

(1) Electric field will be zero in between the charges only if the charges are of same nature.
(2) E.F. will be zero in between the charges, closer to the smaller charge
(3) E.F. may be zero on outer side of charges (on the line joining them), if they are unequal,
closer to the smaller charge.

10)

f0 = 80 cm
fe = 5 cm
ℓ = f0 + fe = 85 cm

11) For lst case, f = = 20 cm

∴ (µg – 1) = ⇒
For 2nd case,

12)

Speed

10 = w×(2.5)
w=4
14)

where

16) = constant

17) WE = –[ΔU] = Ui – UF =

U=

v2 = 4 × 106
v = 2 × 103 m/s

18)
20)
For near normal incidence,

happ =

21)
Power of both refracting surfaces are same.

23)

(i)

(ii)

Ratio =

24) For small drops

For bigger drop

V' =
 25)
112 = I(8)
I = 14 Amp
T.p.d. = V + IR

= = 29 volt

PART-2 : CHEMISTRY

27)
A ⇒ Cl2/hv (allylic substitution)

31)

32)

33) For zero order reaction : A → P

Rate = k
[A]t = a0 – kt

35) (A) Benzene + toluene → ideal solution

 (B) C2H5OH + H2O → non ideal solution
(C) CHCl3 + CH3COCH3 → non ideal solution
(D) C5H5N + H2O → non ideal solution

36) [Glucose] = = 1M

37) Theory based.

40)

47)

8 products

PART-3 : MATHEMATICS
51)
m3 – 4m2 + 5m – 6 = 0
(m2 – m + 2)(m – 3) = 0
D < 0 of m2 – m + 2 = 0
so, m = 3, but
at m = 3 lines are parallel, so no. of values of m is zero.

52) Equation of AB is y – 1 = m(x – 1)

⇒ Equation of AC is y – 1 = (x – 1)

D is mid point of BC

⇒ 2h = 1 –

and 2k = 1 +
Eliminating 'm' we get locus of Δ is x + y = 1

53)

54) in case of homogeneous function

55)

56) f(2) = 4, f ′ (2) = 1

Now,

Applying L-Hospital Rule as form on

putting x = 2
So,
= 2 ⋅ 2 ⋅ f(2) – 4f ′ (2)
= 4 ⋅ 4 – 4 ⋅ 1 = 12

57) Given function,

f(x) =

⇒ f ′ (x) =

⇒ f ″ (x) =
∵ It is given that f ″ (0) exists.
So, f ″ (0+) = f ″ (0–)
⇒ 2λ = 10 ⇒ λ = 5

58) f(x) =

f ′ (x) =

f ′ (x) =

f ′ (x) > 0 ⇒

59) So, f ′ (x) =

⇒ f ″ (x) =
∴ f′ is decreasing in and increasing in .
Hence, option (A) is correct.

60) We have, x = 3 tan t and y = 3 sec t

Clearly,

and

Now,

61) Given
∵ g(f(x)) = x
⇒ g ′ (f(x)) . f ′ (x) = 1

put x = 0

* h(g(g(x))) = x ; given
x f(x)

⇒ h ′ (x) = f ′ (f(x)) . f ′ (x)

⇒ h ′ (x) = f ′ (x3 + 3x + 2) . (3x2 + 3)
put x = 1
⇒ h ′ (1) = f ′ (6) . 6

⇒ h ′ (1) = 111 × 6 ⇒

62) Given, g(x) = , f(x) = sin–1x

f(g(x)) = sin–1(g(x))
fog(x) = sin–1
For the domain of fog(x),
|g(x)| ≤ 1
[∵ Domain of f(x) is [–1, 1]]

63) y = 5log x
Taking log on both sides,
log y = log x . log 5

64) We have, ℓn(x + y) = 4xy

⇒ x + y = e4xy

If x = 0, then y = 1

At (0, 1),

At x = 0,

65) f(x) = x + cos x + 2

f ′ (x) = 1 – sin x and f ′ (0) = 1
f ″ (x) = – cos x and f ″ (0) = –1

Now,
When y = 3 then x = 0

∴ g ′ (3) = ....(1)
Again

or
Hence, g ′ (3) + g ″ (3) = 1 + 1 = 2.

66) Statement-2 :

Statement-1 :

{using statement-2}

67) We have f ′ (x) = 6p – 4p cos 4x – 5 – 3 cos 3x

= 4p(1 – cos 4x) + 2(p – 4) + 3(1 – cos 3x)
Hence f ′ (x) > 0 ⇒ p > 4 ∀ x ∈ R
Hence least integral value of p = 5. Ans.
Alternatively :
f ′ (x) = 6p – 4p cos 4x – 5 – 3 cos 3x
∴ 3p(3 – 2 cos 4x) – 5 – 3 cos 3x > 0

∴ 2p > max. value of

2p > when x = 2kπ k∈I

∴ 2p > 8 ∴ p > 4
∴ smallest p in 5

68) ⇒ slope of normal =

⇒ 3x1 + x1y1 = 20 ....(1)

also
⇒ 4y1 = x12 – 8 ....(2)
only (D) satisfies (1) and (2) both.
69) f(x) = 2x3 – 3x2 + 6
f ′ (x) = 6x2 – 6x = 6(x2 – x) = 0
gives x = 0 or x = 1
for inverse to exist function
must be one one onto
hence Domain is [1, ∞)
Hence a ≥ 1

70) = 3x2 = 3t2 at 'A'

∴ 3t2 = = T2 + Tt + t2
T2 + Tt – 2t2 = 0

71) f(x) =
Since h(x) is identically zero for all x
∴ g(x) = –f(x)

g(x) =
∴ f ′ (–ℓn 2) = 0
Since h(x) = 0
∴ h ′ (x) = 0 ⇒ h ′ (ℓn 2) = 0

Now, g ′ (x) =
⇒ g ′ (tan 1) = –cos 1.

72) Using L'Hospital's Rule, we get

f(x) =
Hence |f(0)| = |–1| = 1 Ans.

73) L = 3 as term

74) Since, f(x) = g(x) |(x – 1) (x – 2) ... (x – 10)| – 2 is differentiable ∀ x ∈ R

∴ g(x) = 0, hence f(x) = –2

∴ at x = 0 is zero.
75) Range of f(x) is [2, 4]
So 3 integers.