23-02-2025

1001CJA101021250004 JA

PART-1 : PHYSICS

SECTION-I (i)

1) For a ground to ground projectile. Choose the CORRECT statement(s):

(A) Rate of change of speed is zero at only one instant.

(B) Rate of change of speed is always positive.
(C) Rate of change of speed is always negative.
(D) Rate of change of speed is first negative, then positive.

2) The motion of a body travelling along a straight line is given by the equation where
v is speed in m/s and t in second. If body was at rest at t = 0 :

(A) The terminal speed attained by the body is 2.0 m/s

(B) The speed varies with the time as v = 2(1 – e–3t) m/s
(C) The speed is 0.1 m/s when the acceleration is half the initial value
(D) The magnitude of the initial acceleration is 6.0 m/s2

3) A spring has natural length of 40 cm and spring constant 500 N/m. A block of mass 1 kg is
attached at one end of the spring and other end of the spring is attached to ceiling. The block is
released from the position, where the spring has length 45 cm. Then choose the correct
statement(s).

(A) the block will perform SHM of amplitude 5 cm.

(B) the block will have maximum velocity cm/sec.
(C) the block will have maximum acceleration 15 m/s2
(D) the minimum potential energy of the spring will be zero.

4) A particle slides back and forth between two inclined frictionless planes joined smoothly at the
bottom. The particle : (Consider that speed of particle won't change at the instant of change of

incline.)

(A) executes simple harmonic motion

(B) executes periodic motion

(C)
has time period
(D)
has time period

5) A fish is rising upward with 1 cm/sec in a container where liquid level is falling at 2 cm/sec. In the
sky, a bird is diving towards the container with 4 cm/sec along the same line, then choose the

correct statement(s). (Take µliquid = )

(A)
Speed of image of fish as seen by bird is cm/sec

(B)
Speed of image of bird as seen by fish is cm/sec
(C) Speed of fish with respect to bird is same as speed of bird with respect to fish
(D) Speed of image of fish as seen by bird is 7 cm/sec.

6) For the system shown below which of the statements is/are correct ? (Assume horizontal ground

to be smooth)

(A) Net force on all the blocks is same

(B) Net force on 4 kg block is more than 3 kg block.
(C) Net force on both 3 kg blocks is same
(D) Normal force between 3 kg blocks is equal to normal force between 3 kg and 4 kg blocks.

SECTION-I (ii)

1) List–I contains value of F and List–II contains corresponding value of acceleration. (Take : g = 10

m/s2)

List-I List-II
 (I) F = 20 N (P) 5.4 m/s2

(II) F = 100 N (Q) 0 m/s2

(III) F = 40 N (R) 3.2 m/s2

(IV) F = 10 N (S) 7.7 m/s2

(T) 2.5 m/s2

(A) I → P;II → Q;III → R;IV → S
(B) I → T;II → Q;III → Q;IV → S
(C) I → P;II → Q;III → Q;IV → S
(D) I → T;II → Q;III → R;IV → S

2) A block of mass m is connected to a spring and oscillating as shown. When the block is at one of
its extreme position, a wax piece of same mass m falls and sticks to the block. Match the lists.

List-I List-II

(I) (P) 1

(II) (Q)

(III) (R)

(IV) (S)

(T) 0
(A) I → S;II → R;III → P;IV → Q
(B) I → R;II → S;III → Q;IV → P
(C) I → S;II → R;III → Q;IV → P
(D) I → Q;II → S;III → P;IV → R

3) Both A & B are thrown simultaneously as shown from a very high tower.

List–I List–II

(I) (P) Distance between the two balls at two seconds is m.

(II) (Q) distance between two balls at 2 seconds is 40 m.

(III) (R) distance between two balls at 2 sec is m.

(IV) (S) Magnitude of relative velocity of B w.r.t A is m/s.

(T) magnitude of relative velocity of B with respect to A is m/s

(A) I → Q;II → R,T;III → P;IV → Q

(B) I → Q;II → S,T;III → P;IV → R
(C) I → S;II → R,T;III → T;IV → P
(D) I → T;II → R,T;III → S;IV → P

4) Consider the system shown in the figure. Here, the pulleys can move in a vertical plane, points A
and B are on the rope, point C is the centre of the indicated moving pulley while point D is a point on
the rope near the bottom of the middle pulley. The strings are taut.
 List-I List-II

(I) Speed of point A must be (P) 5 m/s

(II) Speed of point B must be (Q) 1.5 m/s

(III) Speed of point C must be (R) 3 m/s

(IV) Speed of point D must be (S) m/s

(T)
(A) I → R;II → P;III → Q;IV → S
(B) I → Q;II → P;III → R;IV → S
(C) I → R;II → Q;III → P;IV → S
(D) I → S;II → P;III → R;IV → Q

SECTION-II

1) A convex mirror of focal length 10 cm, produces a virtual image of size of the object.
Distance (in cm) of the object from the mirror is :-

2) Light is incident at point A on one of the faces of a diamond crystal If the maximum
allowed value of angle of incidence so that light suffers total internal reflection at point B is

then find n.

3) If the sphere in the figure shown has a mass of 6 kg, what is the tension in the spring (in N).
Assume that spring and all the ropes are massless and the pulley is ideal.
4) Two identical rods each of mass m and length ℓ, are rigidly joined and then suspended in a
vertical plane so as to oscillate freely about an axis normal to the plane of paper passing through ‘S’

(point of suspension). Time period of such small oscillations = , then find the value of α + β.

5) What can be the maximum amplitude (in m) of the system so that there is no slipping between any

of the blocks

6) There are two thin symmetrical lenses, one is converging, with refractive index n1 = 1.70, and the
other is diverging with refractive index n2 = 1.50. Both lenses have the same curvature radius of
their surface equal to R = 10 cm. The lenses are put close together and submerged into water. The

focal length of lens system is found to be in water. What is the value of x. (R.I. of water =
4/3)

7) An object of mass 0.2 kg executes simple harmonic oscillational along the x-axis with a frequency
of (25/π) Hz. At the position x = 0.04, the object has kinetic energy of 0.5 J and potential energy 0.4
J. The amplitude of oscillations is_____ m. (Assume mean position to be at x = 0)

8) A fluorescent lamp of length 1 m is placed horizontally at a depth of 1.2 m below a ceiling. A plane
mirror of length 0.6 m is placed below the lamp parallel to and symmetric to the lamp at a distance
2.4 m from it as shown in figure. Find the length in meters (distance between the extreme points of
the visible region along x-axis) of the reflected patch of light on the ceiling.

PART-2 : CHEMISTRY

SECTION-I (i)

1) Select correct order of leaving groups

(A)

(B)

(C)

(D)

2) For the above compound (A) the given newmann projection can be

represented
X & Y are respectively :

(A) CH3 and CHClBr

(B) CH3 & CH3

(C)
& Br

(D)
& –CH3

3) During titrations in aqueous solutions one mole of KMnO4 is used for complete oxidation of FeSO4,
FeC2O4 and H2C2O4 respectively and separately. Pick up the correct statement(s) :

(A) 5 mole of FeSO4 can be oxidised in acidic medium

(B) 3/5 mole of FeC2O4 can be oxidised in acidic medium
(C) 1 mole of FeC2O4 can be oxidised in neutral medium
(D) 0.5 mole of H2C2O4 can be oxidised in strong basic medium

4) Which of the following statement(s) is (are) correct :

(A) The coordination number of each type of ion in CsCl crystal is 8

(B) A metal that crystallizes as BCC bcc structure has a coordination number of 12
(C) A unit cell of an ionic crystal shares some of its ions with other unit cells
(D) The length of the unit cell in NaCl is 552 pm (Given ionic radii : )

5) When SiF4 undergoes hydrolysis then which of the following statement is correct?

(A) Reaction is redox

(B) H2[SiF6] is formed as one of the product
(C) Undergoes partial hydrolysis
(D) H2 is formed as one of the product

6) Which of the following species will have maximum solubility?

(A) BeSO4
(B) CaSO4
(C) SrSO4
(D) BaSO4

SECTION-I (ii)

1)

List-I List-II

Structural
(P) (1)
isomers

Conformational
(Q) (2)
isomers

Geometrical
(R) (3)
isomers
 (S) (4) Diastereomers

(5) Identical
(A) P → 2;Q → 1;R → 2,3,5;S → 4
(B) P → 3,4;Q → 1;R → 5;S → 2
(C) P → 2;Q → 3;R → 5;S → 4
(D) P → 3,4;Q → 4;R → 2;S → 1

2) Liquids A and B form an ideal solution . The correct match for

the given equilibrium compositions in liquid (mole-fraction = X) and vapour (mole-fraction = Y) in
List-I with the partial and total vapours pressure in List-II, is.

List-I List-II

(P) XA = 0.50 (1) PA = 16 cm Hg

(Q) XA = 0.40 (2) PA = 40 cm Hg

(R) YA = 0.50 (3) Ptotal = 53.33 cm Hg

(S) YA = 0.40 (4) Ptotal = 57.14 cm Hg

(5) Ptotal = 60 cm Hg
(A) P → 5;Q → 1;R → 3;S → 4
(B) P → 2;Q → 1;R → 4;S → 3
(C) P → 5;Q → 1;R → 3;S → 2
(D) P → 5;Q → 4;R → 3;S → 2

3)

List-I List-II

(P) pn (1) bidentate ligand with same donor atoms

bidentate ligand with different donor

(Q) dmg–1 (2)
atoms

(R) gly–1 (3) ligand contain chiral center

five member ring having one central atom,

2–
(S) ox (4) two carbon and two nitrogen atoms is
formed

five member ring having one central atom,

(5) two carbon and two oxygen atoms is
formed
The CORRECT option is :
(A) P → 1,3,4;Q → 1,4;R → 2;S → 1,4
(B) P → 2,3;Q → 1,2;R → 1;S → 3,4
(C) P → 1,3,5;Q → 1,2;R → 2;S → 1,4
(D) P → 1,3,4;Q → 1,4;R → 2;S → 1,5

4) Match the orbital overlap figures shown in List-I with the description given in List-II and select
the correct answer using the code given below the lists.

List - I List - II

P. 1. p – p π ABMO

Q. 2. d – d π BMO

R. 3. d – d π ABMO

S. 4. p – p σ BMO

5. Gerade MO is formed
(A) P → 3,5;Q → 1;R → 1,5;S → 4,5
(B) P → 2,3;Q → 2;R → 1,5;S → 3,5
(C) P → 3,5;Q → 2;R → 1,5;S → 4,5
(D) P → 1,4;Q → 2;R → 1,5;S → 3,5

SECTION-II

1) How many groups has more priority (or higher priority) than –C≡CH according to C.I.P. sequence
rule.

(i) (ii) (iii) –CH2OH (iv) –CH = CH2

(v) –CH2CCl3 (vi) –CH = CH – Cl (vii) –CH2 – F (viii)

2) How many total 1,2-shift are possible in following carbocation

3) How many sets are correctly matched :
(i) (Sets of ambidentate Nucleophiles)

(ii) , , (Sets of Z isomers)

(iii) , , , (Sets of Leaving groups)

(iv) CH3OH, HF, CH3COOH, Liq. NH3, H2O (Sets of Polar protic solvent)

(v) DMF, DMSO, THF, acetone (Sets of Polar aprotic solvent)

4) A complex of potassium, iron and cyanide ions Kx [Fe(CN)y ] is 100% ionized in its 1 molal aqueous
solution. If elevation in the boiling point of the solution is 2.08°C, find out the value of 'x' in the
complex: (Given : Kb = 0.52 K-kg mol-1)

5) A compound has cubical unit cell in which 'A' atoms are found at six corners and 'B' atoms are
present at remaining corners and at face centres which are not opposite to each other and 'C' atoms
are present at body centres and remaining face centres. Calculate : Density of the compound in
amu/Å3.
[Given : Edge length of unit cell = 2Å, Atomic mass : A = 20 u, B = 60 u, C = 80 u]

6) Rate of formation of SO3 in the following reaction :

2SO2 + O2 → 2SO3
is 100 g min–1, hence rate of disappearance (in g min–1) of O2 is :
[Given atomic mass : S = 32]

7) Water molecule which is required for complete hydrolysis of following compounds.

a) For PCl5 water molecule required are x
b) For XeF6 water molecule required are y
c) For IF7 water molecule required are z
Find x + y – z = ?

8) How many of the following statement is/are correct?

(i) Complex compounds are additional salt which loose their identity in aqueous solution.
(ii) Potash alum is complex compounds.
(iii) Chlorophyll of plants and vitamin B12 are the coordination compound of Mg and Co respectively.
(iv) [Cr(en)3]Cl3 compound has coordination number 3.
(v) Py, CO and H2O are neutral monodentate ligands.
(vi) bn ligand form a 5-member ring with central metal atom and has two chiral center.

PART-3 : MATHEMATICS

SECTION-I (i)

1) f(x) = sin–1x + cos–1x + cot–1x and g(x) = sec–1x + cos–1x + tan–1x, then

(A)
Range of f(x) is

(B)
Range of f(x) is

(C)
Range of g(x) is

(D)
Range of g(x) is

2)
where a, b, c, ℓ ∈ R then

(A)
a=
b=
(B)

(C)
c=

(D)

3) If cos–1x + sin–12x = and

sin–1y + cos–12y = then

(A)

(B)

(C)

(D)
4) Let , (where [x] denotes Greatest Integer function), then

(A)

(B)

(C)

(D)

5) f(x) = x2 – 2x + 3, f:A→ [2, ∞) where f is bijective function. If g(x) is inverse of f(x) then

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

6) If then

(A)

(B)

(C)

(D)

SECTION-I (ii)

1) List-I contains function and List-II contains their points of discontinuity. Match the List-I and List-
II. (where [.] denotes Greatest integer function and {.} denotes Fractional part function)

List-I List-II

(I) f(x) = [x2] (P)

(II) g(x) = [ex] (Q) x=

(III) (R)

(IV) h(x) = [x] + {2x} (S) x = ℓn2

(T)

(A) I → P;II → Q;III → R;IV → S

(B) I → Q;II → S;III → T;IV → P
(C) I → P;II → T;III → Q;IV → S
(D) I → R;II → S;III → P;IV → Q

2) Match the List-I and List-II.

List-I List-II

(I) Characteristics of log3300 is = (P) 10

(II) log932 log581.log425 = (Q) 6

If log2(x–1) + log2(x+5) = 4 ,
(III) (R) 5
then x =

If 3 lies between roots of equation

(IV) x2 – 2x + k2 – 6k + 2 = 0 then (S) 3
maximum integral value of k is

(T) 4
(A) I → Q;II → R;III → S;IV → T
(B) I → P;II → Q;III → R;IV → S
(C) I → R;II → P;III → S;IV → T
(D) I → T;II → P;III → Q;IV → R

3) List-I contains equation and List-II contains its number of real solutions, then match the List-I and
List-II.

List-I List-II

(I) x3 + 4x2 + 4x + 1 = 0 (P) 4

(II) x4 – 3x2 – 10 = 0 (Q) 3

(III) ||x2 – 2| – 5| = 2 (R) 2

If f(x) = x + sinx, then number of solutions

(IV) (S) 1
of equation f(x) = f–1(x), for x ∈ [0, 2π]

(T) 0
(A) I → P;II → Q;III → R;IV → S
(B) I → S;II → T;III → P;IV → Q
(C) I → T;II → P;III → Q;IV → R
(D) I → Q;II → R;III → P;IV → Q

4) Match the List-I and List-II. (where [.] = Greatest integer function and {.} = Fractional part
function.)

List-I List-II

(I) (P) 0
 (II) (Q) 1

(III) (R) 2

(IV) (S) –2

(T) 3
(A) I → R;II → R;III → P;IV → S
(B) I → P;II → Q;III → R;IV → S
(C) I → T;II → S;III → P;IV → R
(D) I → S;II → P;III → Q;IV → T

SECTION-II

1) If f(x) = [3 sinx] , f : [0, π] → R and g(x) = {2x} , g : [0, 3] → R

then sum of number of points of discontinuity of f(x) and g(x) is equal to ______
(where [.] = Greatest integer function , {.} = Fractional part function)

2)

3) If , where , then number of

values of 'x' is equal to

4) Let f(x) is an even periodic function with period = 4 and f(x) = x2, x ∈ [0, 2] then number of
solutions of f(x) = 1 for x ∈ [0, 20] is equal to

5) If [x] denotes greatest integer function of x , then value of [sin–1(sin10) + cos–1(cos10) +

tan–1(tan10)] is equal to

6) If , then the value of a2 + b2 = ______

7) Let , where f(x) is continuous at x = 0

and then value of is equal to ________

8) The number of integral values of 'a' for which the equation (x2 + x + 2)2 – (a – 3)(x2 + x + 2)(x2 + x
+ 1) + (a – 4)(x2 + x + 1)2 = 0 has atleast one real root is
 ANSWER KEYS

PART-1 : PHYSICS

SECTION-I (i)

Q. 1 2 3 4 5 6
A. A,D A,B,D B,C,D B,D A,B B,C

SECTION-I (ii)

Q. 7 8 9 10
A. C C A A

SECTION-II

Q. 11 12 13 14 15 16 17 18
A. 50.00 2.00 34.64 19.00 1.00 3.00 0.06 3.00

PART-2 : CHEMISTRY

SECTION-I (i)

Q. 19 20 21 22 23 24
A. A,D A,C A,C,D A,C,D B,C A

SECTION-I (ii)

Q. 25 26 27 28
A. B A D C

SECTION-II

Q. 29 30 31 32 33 34 35 36
A. 4.00 4.00 5.00 3.00 40.00 20.00 3.00 3.00

PART-3 : MATHEMATICS

SECTION-I (i)

Q. 37 38 39 40 41 42
A. B,D A,B,C,D A,C A,B,C A,D A,B,C

SECTION-I (ii)

Q. 43 44 45 46
A. B C D C
 SECTION-II

Q. 47 48 49 50 51 52 53 54
A. 11.00 2.50 3.00 10.00 2.00 17.00 20.00 1.00
 SOLUTIONS

PART-1 : PHYSICS

1) is perpendicular to at highest point.

2)

n6 – n(6 – 3v) = 3t

v = 2(1 – e–3t)
at t = ∞, v = 2 m/s
initial acceleration, a = 6
a=3
6 – 3v = 3

3) kx0 = mg

∴
= 0.02 m = 2 cm
So equilibrium is obtained after an extension of 2 cm of at a length of 42 cm. But it is releases
from a length of 45 cm.
∴ A = 3 cm = 0.03 m

(b) = 30 cm/s

(c) m/s2
(d) Mean position is at 42 cm length and amplitude is 3 cm. Hence, block oscillates between 45
cm and length 39 cm. Natural length 40 cm lies in between these two, where elastic potential
energy = 0.

4)
T = 4t Motion is periodic
5)
For direct image

(–4 + 2) + (–1–2)

–2 + =
As seen by fish

–8 – 3 =

6)

Net force on 4 kg block is more than 3 kg block

Net force on both 3 kg blocks is same

7)

fK = 0.4 × 16 = 6.4 N
frequired = 28 N
f = fK = 6.4
fK = 0.4 × 80 = 32 N
frequired = 20 N
∴ a = 0, f = 20N

fK = 0.4 × 32 = 12.8 N
frequired = 16 N
∴ a = 0, f = 16 N

fK = 0.4 × 8 = 3.2 N
frequired = 34N
f = 3.2 N

8)

New amplitude will be same as old amplitude.

Since wax fell on it when block was at extreme position.

New maximum speed

9)

(P) VBA = 10 + 10 = 20
so distance b/w B and A in 2sec. = 2 × 20 = 40 m

(Q) ⇒ | VBA| =
distance between A and B in 2 sec. = m

(R) :- so
so distance between A and B in 2 sec. = 2 ×

(S) :- so = 20
so distance between A and B in 2 sec. = 2 × 20 = 40 m.

10)

11) As image is virtual, so magnification is positive.

Thus,

u = (6 – 1)10 = 50 cm

12)
13)
T1 sin 60° = 60
T1 cos 60° = kx

14)

15) a = ω2A
f1 = (1)ω2A ≤ 6.....(1)
f2 = 3ω2A ≤12
ω2A ≤ 4 ....(2)
2
⇒ωA=4

16)
 Now ,

⇒x=3

17) mv2 = mw2(a2–x2) = 0.5 and mω2x2 = 0.4

= 0.06 m

PART-2 : CHEMISTRY

19)

Leaving ability of any leaving

group

Note : good leaving groups are weak bases

(A) (size)

(B) >
Correct order :

>
(Resonance stablize)
(C)
Correct order : NH3 > NMe3 >
(Strong base)

(D)
Correct order

21)
(i) Fe+2 → Fe+3 + e

(ii)

(iii)
(A) 5 mole of FeSO4 can be oxidised
(C) 5 mole of FeC2O4 can be oxidised
(D) 0.5 mole of H2C2O4 can be oxidised

22) A metal that crystallizes in bcc structure has a coordination number of 8

26) (P) PA = XA . = 0.50 × 40 = 20 cm Hg

and Ptotal = XA . + XB . = 0.50 × 40 + 0.5 × 80 = 60 cm Hg
(Q) PA = XA . = 0.40 × 40 = 16 cm Hg
and Ptotal = XA . + XB . = 0.4 × 40 + 0.6 × 80 = 64 cm Hg

(R) ⇒ Ptotal = 53.33 cm Hg

and PA = YA.Ptotal = 0.50 × 53.33 = 26.67 cm Hg

(S) ⇒ Ptotal = 57.14 cm Hg

and PA = YA.Ptotal = 0.40 × 57.14 = 22.86 cm Hg

29) i, ii, iii, vii

31) i, ii, iii, iv, v

32)

i=4

33) A → 6 × =
 B→2× +3× = + =

C→1+3× =

Formula of unit cell =

∴ Formula of the compound = A3 B7 C10

amu/Å3
= 40 amu/Å3.

34)

⇒ = 20 g min–1

PART-3 : MATHEMATICS

37)

Domain of f(x) is [–1,1]

so range of f is to
Domain of g(x) is x = ±1 only

38)

then a + c – 1 = 0
a+b=0

39) ⇒

⇒ 3x2 = 1 – x2
40)

41)

y = x2 – 2x + 3
⇒ x = y2 – 2y + 3 ⇒ (y – 1)2 = x – 2

if
if

42)

43)

[x2] is discontinuous at x =
[ex] is discontinous at x = ℓn2

{2 sinx} is discontinuous at

[x] + 2{x} is discontinuous at

44)

(I) 35 < 300 < 36

5 < log3300 < 6
(II) log932. log581. log425
(III) (x – 1) (x + 5) = 16
⇒ x2 + 4x – 21 = 0 ⇒ x = –7,3
(IV) f(x) = x2 – 2x + k2 – 6k + 2
f(3) = 9 – 6 + k2 – 6k + 2 < 0
k2 – 6k + 5 < 0
1<k<5⇒k=4

45)

(I) (x3 + 1) + 4x(x + 1) = 0

⇒ (x + 1) (x2 + 3x + 1) = 0 ⇒ 3 real solution
(II) x2 = 5 or x2 = –2 ⇒ two real solution
(III) |x2 – 2| = 7,3 ⇒ x2 = 9, –5,5,–1 ⇒ 4 real solution
(IV) f(x) = f–1(x) ⇒ f(x) = x ⇒ x + sinx = x
x = xπ ⇒ x = 0,π,2π

46)

(I) ,

(II)

(III)

(IV)

47)

0 < 3sinx < 3, for x ∈ [0,π)

Discontinuous for 3 sinx = 1,2,3
⇒ 5 points
0 < 2x < 6 for x ∈ [0,π]
Discontinuous for 6 points

48)
49)

x2 = 1 ⇒ x = ±1
solution are x = 0,1,–1

50)
two solutions in [0,4] ⇒ 10 solution

51)

[3π –10 + 4π – 10 + 10 – 3π]

= [4π – 10] = 2

52)
a = 1, –b = 4 ⇒ b = –4
a2 + 52 = 17

53)
using sandwich

54)
 t = 1, a –4

so
a=6
only 1 integral value