30-08-2025

4602CJA101001250016 JA

PART-A-PHYSICS

SECTION-I(i)

1) The velocities of two steel balls before impact are shown. If after head on impact the velocity of
ball B is observed to be 3 m/s to the right, the coefficient of restitution is :-

(A) 7/18
(B) 6/7
(C) 5/18
(D) None of these

2) Block ‘ A ‘ is hanging from a vertical spring and is at rest. Block ‘ B ‘ strikes the block ‘A’ with
velocity ‘ v ‘ and sticks to it. Then the value of ‘ v ‘ for which the spring just attains natural length is

:-

(A)

(B)

(C)

(D) None of these

3) A 1kg sphere moving with velocity ms–1 and another sphere of same mass, but moving with
velocity , collide with each other. Then one of the spheres is found to have a velocity
. Then

(A) the speed of other is about 5 m/s.

(B) the collision between to sphere is inelastic.
(C) the collision between to sphere is elastic.
(D) final kinetic energy of system is 100 J.

4) A car of mass 'm' moves on a banked road having radius 'r' and banking angle θ. To avoid slipping
from banked road, the maximum permissible speed of the car is v0. The coefficient of friction µ
between the wheels of the car and the banked road is :-

(A)

(B)

(C)

(D)

5) A ball is projected with speed u at angle θ from horizontal. Neglected air resistance :-

List-I List-II

(I) Minimum radius of curvature (P) g

(II) Tangential acceleration at height point (Q) g cos θ

(III) Initial normal acceleration

(R)

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

6) A ball of mass m moving vertically down, collides with inclined surface of the wedge. After the
collision wedge starts moving in horizontal direction with velocity V0. If all the surfaces are smooth
then impulse provided by wedge to ball is given by :-

(A) MV0 sin θ

(B) MV0 cos θ

(C)

(D)

SECTION-I(ii)

1) Mass m is moving in a circle in horizontal plane with speed v. Choose the correct statement.

(A) Speed v is given by .

(B)
The angular speed is given by .
(C) Graph between tension (T) and angular speed is parabolic.
(D) Minimum tension in string is mg.

2) A ball tied to the end of a string performs (complete circle) in a vertical circle under the influence
of gravity. Choose the correct option(s)

When the string makes on an angle 90° with the vertical, the magnitude of tangential
(A) acceleration is zero and magnitude of radial acceleration is somewhere between maximum and
niinimiuii
When the string makes an angle 90° with the vertical, the magnitude of tangential acceleration
(B) is maximum and magnitude of radial acceleration is somewhere between maximum and
minimum
At no place in the circular motion, magnitude of tangential acceleration is equal to magnitude of
(C)
radial acceleration
Throughout the path whenever magnitude of radial acceleration has its extreme value, the
(D)
magnitude of tangential acceleration is zero

3) Two balls A and B of masses m and 2m respectively are lying on a smooth surface as shown in the
figure. Ball A hits the ball B (which is at rest) with a velocity v = 16 m/s. Such that B just reaches the
highest point of inclined plane. (g = 10 m/s2)
where e is the coefficient of restitution and vA is velocity of the ball A after collision.

(A) e = 3/4
(B) e = 7/8
(C) vA = −4 m/s
(D) vA = 8 m/s

4) A person, standing on a plank which is kept over a smooth surface, as shown. Person jumps
towards right, during the jump, plank also acquires some velocity :-

(A) Work done by friction, on person is negative.

(B) Work done by friction, on plank is negative.
Position of centre of mass of system (man + plank) will remain same, with respect to a fixed
(C)
point, before and after the jump.
Increase in mechanical energy of system (man + plank) will be equal to kinetic energy
(D)
of system in c-frame.

5) A block 'A' of mass m1 hits horizontally the rear side of a spring (ideal) attached to a block B of
mass m2 resting on a smooth horizontal surface. If v1 and v2 be the velocities of A and B at any

instant in opposite directions, then,

The kinetic energy of the system with respect to the centre of mass reference frame will be
(A)

The kinetic energy of the system with respect to the centre of mass reference frame will be
(B)

If block A hits the spring with a velocity 'v' and velocity of B is zero and the stiffness constant of
(C)
the spring be 'k' the maximum compression in the spring will be
If block A hits the spring with a velocity 'v' and the stiffness constant of the spring be 'k' the

(D)

maximum compression in the spring will be

6) Two blocks of equal mass m are connected by an unstretched spring and the system is kept at rest
on a smooth horizontal surface. A constant horizontal force (F) begins to act on block 1, at time t =
0, so as to pull it away from the other block. (If k = 10 N/m, m = 5 kg)

(A) Acceleration of COM is 10 m/s2.

(B) At maximum extension the velocity of both the blocks will be equal.
(C) Maximum extension of the spring is 10 m.
(D) Maximum extension of the spring is 5 m.

SECTION-III

1) A sphere A of mass 2kg moving with velocity v0 collide with bob of mass 1kg as shown in the
figure. value of tension at which string of pendulum break is 40N. If length of pendulum is 30 cm,

0
then find the maximum value of v in m/s so that string does not break. Coefficient of restitution is .

2) A marble bounces down a long flight of stairs in a regular manner, hitting each step vertically at
the same speed and distance from the edge, and bouncing up to the same height above each step, as
shown in figure. Each stair has the same height and width ℓ as shown. The horizontal component of
velocity Vh is unaffected, but the stairs have the property that e = 0.6 is a constant. Find the value Vi
(in m/s). Ignore the size of the marble and air resistance.Assume the trajectory of the marble lies in
plane of the paper.

[Given : ℓ = 0.8m]

3) Figure shows a bob moving in horizontal circular motion at angular speed where AB =
h. The ratio of tension in supporting strings is given by , find value of N. Strings are of equal

lengths and inextensible.

4) A uniform metallic chain in a form of circular loop of mass m = 3 kg with a length = 1 m rotates
at the rate of n = 5 revolutions per second. The tension T (in Newton) in the chain is 15n. Find the

value of n.

5) The figure shows the velocity and acceleration of a point like body at the initial moment of its
motion. The acceleration vector of the body remain constant. The minimum radius of curvature of

trajectory of the body is.

PART-B-CHEMISTRY

SECTION-I(i)

1) Hydroxylamine reduces iron (III) according to following equation :-

Which statement is correct ?

(A) n–factor for hydroxylamine is 1

(B)
Equivalent weight of Fe2(SO4)3 is
(C) 6 milliequivalent of Fe2(SO4)3 is required for complete oxidation of 6 millimoles of NH2OH.
(D) All of these
2) In 200 ml of 0.8 M H2SO4 solution 300 ml 0.4 M NaOH solution is added. Determine the nature &
Normality of final solution :-

(A) 0.08 & Acidic

(B) 0.4 & Acidic
(C) 0.56 & basic
(D) 0.4 , basic

3) The ratio of equivalent masses of NH3 in the reactions (i) and (ii) are :-
(i) 4NH3 + 5O2 → 4NO+ 6H2O,
(ii) 2NH3 → N2 + 3H2

(A) 5 : 6
(B) 6 : 5
(C) 5 : 3
(D) 3 : 5

4) Graph depicting correct behaviour of ideal gas & H2 gas will be (neglect a) :-

(A)

(B)

(C)
(D)

5) In case of CO and CH4 curve goes to minima then increases with increase in pressure but in case

of H2 and He the curve is linear because :-

(A) Intermolecular interactions for H2 and He are very low.

(B) Molecular size or atomic size for H2 and He is small.
(C) Both (A) and (B)
(D) Neither (A) nor (B)

6) Which of the following expression represents correctly the variation of density of an ideal gas with
change in temperature?

(A)

(B)

(C)

(D)

SECTION-I(ii)

1) A metal forms two oxides. The higher oxide contains 20% oxygen, while 4.29 g of the lower oxide
when converted to higher oxide, become 4.77 g. The equivalent weight of metal in :-

(A) lower oxide is 32.

(B) lower oxide is 64.4.
(C) higher oxide is 64.4.
(D) higher oxide is 32.
2) Natural H2 consists of 80% H2 and 20% D2 by mass. Which of the following options is/are correct.

(A) Rate of diffusion of H2 is times the rate of diffusion of D2.

(B) As the mixture effuses, the average molar mass of the mixture remaining increases with time.
(C) Both H2 and D2 will diffuse by same rate initially.
(D) Mole fraction of H2 in effusion mixture is higher than that of D2.

3) Select the correct option(s):

(A) Pressure in container-I is 3 atm before opening the valve.

(B) Pressure after opening the valve is 3.57 atm.
(C) Moles in each compartment are same after opening the valve.
(D) Pressure in each compartment are same after opening the valve.

4) 11.2 litre of a gas at 1 atm pressure and 273 K weighs 14 gm. The gas/gases would be :-

(A) CO
(B) N2
(C) C2H4
(D) N2O

5)

The crystalline form of borax has :-

2⊝
(A) Tetranuclear [B4O5(OH)4] unit
(B) All boron atoms in the same plane
(C) Equal number of sp2 and sp3 hybridized boron atoms
(D) One terminal hydroxide per boron atom

6) Select correct statement about B2H6 :-

(A) Bridging groups are electron-deficient with 12 valence electrons

(B) It has 2c-2e B–H bonds
(C) It has 3c-2e B-H-B bonds
(D) It has 3c-4e B-H-B bonds

SECTION-III
1) The maximum number of atoms lying in the same plane in B2H6 is :-

2) When the pressure of 5 L of N2 gas is doubled and its temperature is raised from 300 K to 600 K,
the final volume of N2 in litres will be :-

3) One mole of acidified K2Cr2O7 on reaction with sufficient KI will liberate how many moles of I2,
along with Cr3+?

4) The Vander Waal's constant for a gas are a = 3.6 atm L2 mol–2, b = 0.6 L mol–1, if R = 0.08 L atm

K–1mol–1 and Boyle's temperature is TB of this gas then what is the value of

5) A bubble of air is underwater at temperature 17oC and the pressure 2.9 bar. If the bubble rises to
the surface where the temperature is 27oC and the pressure is 1.0 bar. Ratio of final volume of
bubble to initial volume is …………

PART-C-MATHEMATICS

SECTION-I(i)

1) In a triangle ABC, if cos A + 2 cos B + cos C = 2 and the lengths of the sides opposite to the
angles A and C are 3 and 7 respectively, then cos A – cos C is equal to

(A)

(B)

(C)

(D)

2) In a ΔABC with usual notations if b + c = 3a, then cot . cot has the value equal to

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

3) The area enclosed by the graph of the following relation = 1, is

(A) 12
(B) 24
(C) 48
(D) 96

4) Two vertices of a rectangle are on the positive x-axis. The other two vertices lie on the lines y = 8x
and y = 5x + 4 such that rectangle lies inside the triangle formed by lines and x-axis. Then the
maximum area of the rectangle is

(A)

(B)

(C)

(D)

5) The vertices of a triangle are and where p,q,r are the roots of the
3 2
equation y – 3y + 6y + 1 = 0. The coordinates of its centroid are

(A) (1, 2)
(B) (2, –1)
(C) (1, –1)
(D) None of these

6) The area of triangle ABC is 20 cm2 . The co-ordinates of vertex A are (- 5, 0) and B are (3, 0) . The
vertex C lies on the line, x - y = 2 . The co-ordinates of C may be

(A) (5, 3)
(B) (- 3, - 5)
(C) (- 5, - 7)
(D) (3, 5)

SECTION-I(ii)

1) If curve C is the locus of the point which moves such that the ratio of its distances from (1, 0) and
(4, 0) is always 2 : 1. Then curve C

(A) meets x-axis at (–2, 0) and (2, 0)

(B) meets x-axis at (3, 0) and (7, 0)
(C) intersects x2 + y2 = 1 at two distinct points
(D) intersects y = 1 at two distinct points

2) An equilateral triangle ABC has centroid at origin and the base BC lies along the line x + y = 1,
then
(A)
Area of ΔABC is

(B)
Area of ΔABC is
(C) Gradient of other two lines are
(D) Gradient of other two lines are

3) The ΔABC is formed by the lines 2x – 3y = 6, 3x – y + 6 = 0 and 3x + 4y = 24. If the point P(α, 0)
and Q(0, β) always lie on or inside the ΔABC then (Where α, β ∈ R)

(A) α ∈ [–2, 3]
(B) β ∈ [–1, 6]
(C) α ∈ [–2, 2]
(D) β ∈ [–2, 6]

4) A triangle has two of its sides along the lines where are the roots of the
equation If be the orthocentre of triangle and equation of 3rd side is
(a,b,c and a,b,c are co-primes) then

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

5) Let straight line y = mx + 4 meets the curve at two points A and B

such that where m1 < m2 and 'O' is the origin. Identity which of the
following statement(s) is/are correct?

(A)

(B)

(C)
If m = 2, then area of units

(D)
If m = 2, then area of units

6) Let the equation of the pair of lines is given as, 2x2 + 5xy + 3y2 + 6x + 7y + 4 = 0. Which of the
following is/are true ?

(A) Orthocentre of the triangle formed by pair of lines and the line x – y = 7, is (3, –4)
(B) Acute angle between the pair of lines is tan–15
(C) Point of intersection of the pair of lines is (1, –2)

(D)
Circumcentre of the triangle formed by the pair of lines and the line x – y = 7 is

SECTION-III
1) Consider two points A(2, 4) & B(1, 2). Let P(a, a) be a point on the straight line L : y = x such that
AP + PB is minimum, then 'a' is equal to

2)

The lines given by

L1 = x + 3y – 5 = 0
L2 = 3x – ky – 1 = 0
L3 = 5x + 2y – 12 = 0
L1, L2, L3 are concurrent if k =

3) The vertex A of ΔABC is (3, –2) and the equations to two of its medians are 5x + 3y = 11 and 4x +
3y = 8. If the equation of side BC is ax + by = 7, then (a + b) is equal to

4) Two sides of a rhombus ABCD are parallel to the lines y = x + 2 & y = 7x + 3. If the diagonals of
the rhombus intersect at the point (1, 2) & the vertex A is on the y-axis, if the possible coordinates
of A are (a, b) and (c, d), then the value of (a + b + c + 2d) is

5) If line 2x – y – 1 = 0 intersect the curve ax2 + 2hxy + by2 + 2gx + 2ƒy + c = 0 in two points A and

B, such that ∠AOB = 90°, then, is equal to

 ANSWER KEYS

PART-A-PHYSICS

SECTION-I(i)

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

SECTION-I(ii)

Q. 7 8 9 10 11 12
A. B,C,D B,C,D B,C A,C,D A,C A,B,C

SECTION-III

Q. 13 14 15 16 17
A. 3 5 5 5 8

PART-B-CHEMISTRY

SECTION-I(i)

Q. 18 19 20 21 22 23
A. D B D A C A

SECTION-I(ii)

Q. 24 25 26 27 28 29
A. B,D B,D A,D A,B,C A,C,D A,B,C

SECTION-III

Q. 30 31 32 33 34
A. 6 5 3 5 3

PART-C-MATHEMATICS

SECTION-I(i)

Q. 35 36 37 38 39 40
A. C C C C B B

SECTION-I(ii)

Q. 41 42 43 44 45 46
A. B,D A,C A,D A,B,C B,C A,C,D
 SECTION-III

Q. 47 48 49 50 51
A. 2 5 5 5 2
 SOLUTIONS

PART-A-PHYSICS

1)

This is the problem based on head on collision.

v2 = (1 + e)vCM – eU2
3 = (1 + e) × 1.6 – e × (–2)
3 = 1.6 + 1.6 e + 2 e
3.6 e = 1.4

2)

The initial extension in spring is

Just after collision of B with A the speed of combined mass is

For the spring to just attain natural length the combined mass must rise up by
(sec fig.) and comes to rest.
Applying conservation of energy between initial and final states

Solving we get
3)

× 1 × [100 + 100] = –100J

: Note elastic

4)

Ncosθ – f sinθ = mg

Rgtanθ + µRg = v2 – v2µtanθ

Ans. (3)

5)

(P) Minimum radius of curvature = at highest point

(Q) Tangential acceleration at highest point = Zero {only vertical acceleration 'g' is present}
(R) Initial normal acceleration = g. cos θ
(S)
Here, 'Vx' will not change with time as no acceleration in Horizontal direction but 'Vy' will
change as gravitational acceleration is present in vertical.

So,

6) According to impulse momentum theorem

I sin θ = MV0

∴I=

7)
Tcosθ = mg

∴ Angular speed

Tsinθ = mω2ℓsinθ ⇒ T µω2

∴ graph between T v/s ω is parabolic

when θ = 0°,
and T cosθ = mg
Tmin = mg

8)
9)

10) (wfriction)net = 0
(wfriction)plank = +ive
(wfriction)boy = –ive

Also ΔEinternal =

11)

(KE) of system wrt to com =

By w.e.t. in center of mass frame

Work done by all = Δke

= µ(VRelf2 – VReli2)

12) ΣFext = F

At maximum extension velocity of both the blocks = v.

Using Work-Energy theorem

13) Velocity of B after collision will be

0
V= =v
Tension just after the collision is

T= = 40
⇒ = 30 × 0.3 ⇒ v0 = 3 m/s

14) and ℓ = 0.8 m; m/s

15)

16)

dmω2R = 2T sin

∴T= ω2R2 ....(1)

But ω = 2πn
 ∴

17) The acceleration vector shall change the component velocity u||along the acceleration

vector r = . Radius of curvature rmin means v is minimum and an is maximum. This is at point
P when component of velocity parallel to acceleration vector becomes zero, that is u|| = 0.

∴R= = 8 meters

PART-B-CHEMISTRY

18) n–factor of NH2OH = 1

n–factor of Fe2(SO4)3 = 2

19)

Nf =

=
Nf = 0.4 acidic

20) N3– —→ N2+ + 5e ∴ E = M/5

0
6e + 2N3– —→ (N )2 ∴ E = M/3

21) For H2 z > 1, So graph (A) is correct.

22) Due to small size of these species (H2 and He) intermolecular interactions (van der Waal
forces) are very low, therefore it is difficult to compress these .

23)
PMo = dRT

24) Higher oxide:

Lower oxide (4.29 g) → Higher oxide (4.77 g)

Mass of metal =
Mass of oxygen = 4.29 – 3.816 = 0.474 g

lower oxide :

25)
⇒ Effusion mixture has higher mole fraction of H2.

26)

no. of moles are equal before as well as after opening the valve.

27) PV = nRT

M = 28

28)

29)

30)
So maximum number of atoms in same plane = 6
 31)

32)

Cr2O72– + 14H+ + 6I 3I2 + 2Cr3+ + 7H2O

(v.f = 3×2 = 6) (v.f = 2)
Equivalents of K2Cr2O7 = Equivalent of I2.
1 × 6 = Moles of I2 × 2
Therefore, Moles of I2 = 3.

33)

34)

PART-C-MATHEMATICS

35)

as

so

2cos B/2 cos = 4sin B/2 cos B/2

Sin A + sin C = 2 sin B
a + c = 2b ⇒ a = 3, c = 7, b = 5

cos A – cos C =
36) ∵ cot cot = = =

= = =2 (Δb + c = 3a)

37) Required area = = 48 sq. units.

38) y = –5x + 4
A (h, 0)
B (h, 8h)

39)

40)

Let c ≡ α, β α–β=2 ; β=α–2

= 20 α = –3 , β = – 5

41)
curve C is the circle with diametric ends as A and B. These two points will divide P and Q in
the ratio 2 : 1 internally and externally respectively.
Hence A ≡ (3,0) and B ≡ (7,0)

42) Let side of triangle be 'a'

∴ a = 2pcot30°

Area =

Also tan60° =

∴m= and

43)

44) Pair of lines is

Third side

45) a = 3
(a)

(b)
(c) y = 2x + 4

Area

46) 2x2 + 5xy + 3y2 + 6x + 7y + 4 = 0

(2x + 3y + λ1)(x + y + λ2) = 0
2λ2 + λ1 = 6 and 3λ2 + λ1 = 7
λ2 = 1, λ1 = 4
Pair of lines are 2x + 3y + 4 = 0 and x + y + 1 = 0
Point of intersection of the pair of lines is (1, –2)

orthocentre : B(3, –4)

circumcentre of ΔABC =
47) As shown in figure, Let A' be
the image of point A in line L
∴ AP + PB = A'P + PB
For any point P on line L,
A'P + PB > A'B
∴ Minimum value of A'P + PB is A'B
when A', P, B are collinear
Now, coordinates of A' will be reflection of A(2, 4) in the line L : y = x
⇒ A' ≡ (4, 2)
Equation of A'B : y = 2
∴ Point P(a, a) will be (2, 2)
∴a=2

48)
k=5

49)
Point F lies on median CF

∴
⇒ α = 1 and point E lies on median BE

∴
⇒β=5
∴ B = (1,2) and C = (5,–4)
∴ equation of BC is 3x + 2y = 7

50)
Let lines to AB lines
y = x + 2 & y = 7x + 3
intersect on y-axis at (0, λ)
perpendicular distance from P(1, 2) to AD = perpendicular distance P to AB
(5 + λ) = ±5(λ – 1)

51)