24-08-2025

1001CJA101001250011 JA

PART-1 : PHYSICS

SECTION-I (i)

1) If a portion of uniform disc of radius 3R is removed and placed near its surface as shown, find x-

coordinate of COM of complete system (R = 9m).

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

2) In the figure shown the initial velocity of boat (30 kg) + person (15 kg) is 2 m/s. Find velocity of
person w.r.t. boat so that velocity of boat will be 1 m/s in right. (Neglect friction between boat and

water)

(A) 3 m/s towards right

(B) 3 m/s towards left
(C) 4 m/s towards right
(D) 4 m/s towards left

3) A block of mass 1 kg is resting on a rough horizontal surface with coefficient of friction 0.2. If an

impulse N-s is given to block then the distance travelled by the block till it stops (after

the impulse) is :
(A) 10 m
(B) 0 m
(C) 16 m
(D) 4 m

4) Find moment of inertia of a wire frame about an axis passing through point O & perpendicular to
plane of frame. (linear mass density of wire is 2 kg/m & its radius is 1m)

(A) 2π kg m2

(B)

(C) π kg m2
(D) (π + 1.5) kg m2

SECTION-I (ii)

1) In figure a block A of mass 2 kg is moving to the right with a speed 5 m/s on a horizontal
frictionless surface. Another block B of mass 3 kg with a massless spring of spring constant 120N/m
attached to it, is moving to the left on the same surface and with a speed 2 m/s. Let us take the
direction to the right as the positive X-direction. At some instant, block A collides with the spring
attached to block B. At some other instant, the spring has maximum compression and then, finally,

the blocks move with their final velocities.

(A) During the collision the momentum as well as mechanical energy of the system is conserved.

(B) When block A has a velocity of 4m/s in the positive x direction, the velocity of B is m/s in the
negative x direction.
(C) At the maximum compression, the blocks have a common speed of 0.8 m/s in positive direction.
(D) The maximum compression in the spring is 0.7 m.

2) If 3 point masses 1, 2, & 3 kg are rigidly fixed on a massless triangular frame and placed on
smooth frictionless table as shown below then choose correct statement base on given information.
(A) Acceleration of 3 kg must be m/s2.
(B) Acceleration of 2 kg must be m/s2.
(C) Acceleration of 1 kg must be m/s2.
(D) Acceleration of COM must be m/s2.

3) Choose the correct statement.

(A) Work done by kinetic friction can be zero.

(B) Work done by internal forces is always zero.
If a uniform disc is rotating about its axis with angular velocity ω then its linear momentum is
(C)
zero.
If a cylindrical bottle half filled with a liquid is projected at angle from ground then COM will
(D)
perform projectile motion.

SECTION-II (i)

Common Content for Question No. 1 to 2

2 kg and 3 kg blocks are placed on a smooth horizontal surface and connected by spring which is
unstretched initially. The blocks are imparted velocities as shown in the figure.

1) Maximum speed of 3 kg block in the subsequent motion will be n × v0, find the value of n.

2) The maximum energy stored in the spring in the subsequent motion will be n × , find the value
of n.

Common Content for Question No. 3 to 4

A gun crew observes a remotely controlled balloon launching an instrumented spy package in
enemy territory. When first noticed the balloon is at an altitude of 800 m and moving vertically
upward at a constant velocity of 5 m/s. It is 1600 m down range. Shells fired from the gun have an
initial velocity of 400 m/s at angle 37° from horizontal. The gun crew waits and fires so as to destroy
the balloon. Assume g = 10 m/s2. Neglect air resistance.

3)

What is flight time (in sec) of the shell before it strikes the balloon?

4) The altitude of collision in meter is :-

SECTION-II (ii)

1) In the shown arrangement all surfaces are smooth and system is released from rest. Mass of block
A and B are equal (m = 1kg), mass of block C is M = 3kg. After releasing from rest block B strikes
with the surfaces of C after travelling height h = 10 cm. Find the displacement (in cm) of the block

C, just before, the B strikes to C.

2) One end of a pendulum is attached to point O on the wedge of mass M = 4m. Mass of the
pendulum is m and its length is ℓ. Initially system is at rest on a smooth floor, now velocity v0 = 12
m/s is given to the pendulum in horizontal direction. v0 is sufficient for the bob to move in circle with
respect to O. If at highest point of the trajectory speed of bob is equal to speed of wedge then find

this speed in m/s.

3) A uniform rod of mass 120kg is at angle ϕ = 60° from horizontal. It is supported by a cable at an
angle 30° from horizontal. The rod is pivoted at the bottom and object of mass m = 200kg hangs

from it's top. If tension in the support cable is T, find (in SI unit). Consider system to be in
equilibrium.

4) Two masses m and m, that start simultaneously from the inter-section of two straight lines with
velocities 3 m/s and 4 m/s respectively on a horizontal surface as shown. Find the two times of

magnitude of the velocity of centre of mass (in m/s) when θ = 90°.

5) Spring is at natural length when force F1 and F2 start acting on the mass m1 and m2. If the
maximum elongation of spring during subsequent motion is x (in meter), fill the value of 5x . [m1 =

2kg, m2 = 3kg, F1 = 200 N, F2 = 100 N].

6) Find value of x if torque of force applied at point (x, 1, –1) about origin is
.

PART-2 : CHEMISTRY

SECTION-I (i)

1) Element A has successive ionization energies 8.2, 16.4, 102.5, 122.4 (in eV/atom). Element B has
corresponding values 10.4, 19.2, 35.6, 49.2, 166.5, 208.7 (eV/atom)
What would be the likely formula for a compound that may be formed from A and B

(A) A4B3
(B) A2B
(C) AB2
(D) A2B3

2)
Among these compounds, the correct order of C – N bond lengths is :

(A) IV > I > II > III

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

3)
Correct order of bond strength between carbon and oxygen is-

(A) a > b > c

(B) c > b > a
(C) a > b = c
(D) b = c > a

4) A mixture of FeO and Fe2O3 is reacted with acidified KMnO4 (0.2 M) whose 100 ml was used. The
solution was then titrated with zinc dust which converted Fe+3 in solution to Fe+2. In final solution
Fe+2 required 1000 ml, 0.1 M K2Cr2O7 solution for end point. Find millimoles of FeO & Fe2O3 in
original mixture respectively.

(A) 0.1, 0.25

(B) 100, 250
(C) 200, 100
(D) 150, 250

SECTION-I (ii)

1) Choose the correct statement(s)

(A) In Moseley's equation, square root of frequency is directly proportional to atomic number
(B) Silver and gold have nearly same atomic radius
(C) Sc <Y< La (correct order of atomic size)
(D) Atomic number of mercury (Hg) is 80

2) Correct order of basic strength-

(A)
> > >

(B)

(C) CH3CH2Na > CH2 = CHNa > HC ≡ CNa

(D) I– > Br– > Cl– > F⦵

3) A mixture of 20 ml of CO, CH4 and N2 was burnt in excess of O2 resulting in reduction of 13 ml of

volume. The residual gas then treated with KOH solution to show a reduction in volume of 14 ml.
Then the correct statement among the following is

(A) Volume of CO in mixture is 10 ml

(B) Volume of CH4 in mixture is 4 ml
(C) Volume of CH4 in mixture is 6 ml
(D) Volume of N2 in mixture is 6 ml

SECTION-II (i)

Common Content for Question No. 1 to 2

(I) Delocalisation of 'σ' electrons is known as hyper conjugation.
(II) During hyper conjugation 'C–H', 'C–D' and 'C–T' 'σ' orbital overlap with 'p' orbital of carbocation,
free radicals and π* orbital of double bond.
(III)

1) Find total number of hyperconjugative structure (involving C–H bond) for the following

carbocation.

2) , ,
Find out number of α-hydrogens in alkene among the following which release maximum heat of
hydrogenation on adding H2.

Common Content for Question No. 3 to 4

The strength of H2O2 is expressed in several ways like molarity, normality, % (w/V), volume strength,
etc. The strength of "10 V" means 1 volume of H2O2 on decomposition gives 10 volumes of oxygen at
STP or 1 litre of H2O2 gives 10 litre of O2 at STP. The decomposition of H2O2 is shown as under : H2O2

(aq) —→ H2O (ℓ) + O2 (g)

H2O2 can acts as oxidising as well as reducing agent. As oxidizing agent, H2O2 is converted into H2O
and as reducing agent, H2O2 is converted into O2.

3) What is the percentage strength (% w/V) of "11.35 V" H2O2 ?

4) 37.5 g Ba(MnO4)2 (mol. wt. = 375) sample containing some inert impurities in acidic medium is
completely reacted with 125 mL of "34.05 V" of H2O2. What is the percentage purity of the sample ?

SECTION-II (ii)

1) The atomic number of element “Unnilquadium” is X. The value of is-

2)

How many of the total number of processes given below are endothermic?
(i) (ii)
(iii) (iv)
(v) (vi)
(vii) (viii)
(ix)

3) How many unpaired electron(s) is/are present in tri-positive titanium ion?

4) A sample of iron weighing 20 g was heated with KClO3(s) in an evacuated container. The oxygen
generated from decomposition of KClO3 converted some of Fe to Fe2O3. If the combined mass of
Fe and Fe2O3 was 21.5 g. Calculated mass of Fe2O3 (in g) formed. (Atomic mass of Fe = 56)

5)

N2O4 dissociates into NO2, if % dissociation of N2O4 is 33.33 %. Calculate average molecular weight
of gaseous mixture formed .

6) How many compounds react with NaHCO3 and release CO2 gas ?
(i) Picric acid (ii) Squaric acid
(iii) Carbolic acid (Phenol) (iv) Acetic acid
(v) Formic acid (vi) Maleic acid

PART-3 : MATHEMATICS

SECTION-I (i)

1) Let p, q, r ∈ (0, ∞) & satisfying , then family of lines

passes through the point

(A) (1, 1)
(B) (1, –2)
(C) (–1, 2)
(D) (–1, 1)

2) A ray of light is sent along the line x – 2y + 5 = 0. Upon reaching the line 3x + 2y + 7 = 0, the ray
is reflected from it. Then the equation of the line containing the reflected ray is

(A) 19x – 22y + 79 = 0

(B) 22x + 19y + 11 = 0
(C) 11x – 13y + 9 = 0
(D) 19x + 22y + 11 = 0

3) The number of solution(s) of equation in (0, 7π)

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

4) A variable line L passes through the point (3, 5) and intersects the positive coordinate axes at the
points A and B. The minimum area of the triangle OAB, where O is the origin, is :

(A) 30
(B) 25
(C) 40
(D) 35

SECTION-I (ii)

1) If x and y are two real numbers satisfying and ,

then the values of siny is/are

(A)

(B)

(C)

(D)

2) For the straight lines 4x + 3y – 7 = 0 and 24x + 7y – 31 = 0, the equation of

(A) bisector of the obtuse angle between them is x – 2y + 1 = 0

(B) bisector of the acute angle between them is 2x + y – 3 = 0
(C) bisector of the angle containing origin is x – 2y + 1 = 0
(D) bisector of the angle containing the point (1, –2) is x – 2y + 1 = 0

3) If a, b and c are distinct real numbers, then the system

a2x + b2y + c2z = 0,
(a + 1)2x + (b + 1)2y + (c + 1)2z = 0,
(a – 1)2x + (b – 1)2y + (c – 1)2z = 0,

(A) is consistent
(B) has unique solution
(C) has a solution (x0, y0, z0) such that x0y0z0 > 0
(D) has some non-trivial solution

SECTION-II (i)
Common Content for Question No. 1 to 2
In triangle ABC given that A(2, 1), median through vertex B is y = x & altitude through vertex C is 2x
+ y = 1.

1) If orthocentre of triangle ABC is (α, β), then is equal to

2) If (α, α2) lie inside the triangle ABC, then largest interval in which α lie is (a, b) then value of (a +
b) is :

Common Content for Question No. 3 to 4

Consider a triangle ABC with sides AB and AC having the equations L1 = 0 and L2 = 0. Let the
centroid, orthocentre and circumcentre of the ΔABC are G, H and S respectively. L = 0 denotes the
equation of side BC.

3) If L1 : x – y + 1 = 0 and L2 : 2x – y = 0 and H(1, 3) if equation of line L = 0 is ax + by + c = 0, then

value of |a + b + c| is

4) If L1 : x + 3 = 0 and L2 : x – y = 0 and S(0, 0) if co-ordinates of orthocentre is (α, β),

then is equal to

SECTION-II (ii)

1) A rod of unit length slides along co-ordinate axes such that its two ends A and B always lie on
positive X and Y axes respectively. Then, the locus of a point P (lying in the first quadrant) such that

ΔPAB is equilateral, is given by ; where k = .......

2) The radius of the circle touching incircle of an equilateral triangle of circumradius 6 unit,
externally and its two sides, is

3) If α, β, γ are roots of x3 + 2x2 – x + 3 = 0, then the value of

is

4) Let N be the number of triplets (α, β, γ) where α, β, γ ∈ [0, 2π] satisfying the inequality (4 +

sin4α)(2 + cot2β)(1 + sin4γ) ≤ 12sin2γ then the value of is

5) If an = ; (n ≥ 1), then value of is

6) Let A(–2, –1), B(1, 0), C(α, β) and D(γ, δ) be the vertices of a parallelogram ABCD. If the point C
lies on 2x – y = 5 and the point D lies on 3x – 2y = 6, then the value of |α + β + γ + δ| is equal to
 ANSWER KEYS

PART-1 : PHYSICS

SECTION-I (i)

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

SECTION-I (ii)

Q. 5 6 7
A. A,B,C,D D A,C,D

SECTION-II (i)

Q. 8 9 10 11
A. 3.00 15.00 5.00 1075.00

SECTION-II (ii)

Q. 12 13 14 15 16 17
A. 2 4 40 5 8 2

PART-2 : CHEMISTRY

SECTION-I (i)

Q. 18 19 20 21
A. B C A B

SECTION-I (ii)

Q. 22 23 24
A. A,B,C,D A,B,C A,B,D

SECTION-II (i)

Q. 25 26 27 28
A. 4.00 4.00 3.40 75.00

SECTION-II (ii)

Q. 29 30 31 32 33 34
A. 4 7 1 5 69 5

PART-3 : MATHEMATICS
 SECTION-I (i)

Q. 35 36 37 38
A. D A D A

SECTION-I (ii)

Q. 39 40 41
A. B,D A,B,C,D A,B

SECTION-II (i)

Q. 42 43 44 45
A. 2.50 1.50 5.00 4.50

SECTION-II (ii)

Q. 46 47 48 49 50 51
A. 4 1 5 8 7 32
 SOLUTIONS

PART-1 : PHYSICS

1) Let 1 be part moved and 2 remaining

scom =

scom = {s2 = 0}

scom =

2)

Correct Answer is (A)

3) Ndt = 10 Ns
fdt = µ Ndt = 0.2 × 10 = 2Ns
10 – 2 = 1v
⇒ v = 8 m/s
s = 16 m

4) I1(semicircular) = (2 × (π × 1)) × (1)2 = 2π

I2(Rod) =

5) (A) Momentum will be conserved at all instants.

(B) Pi = Pf = (2 × 5) + (3 × –2) = (2 × 4) + 3vB
(C) Compression will be maximum, when both blocks have same velocity.
Use momentum and energy conservation.

6)

Fexternal = m × acom

7)

Theoretical question

8) Velocity of centre of mass is v0 towards left and both the blocks 2kg & 3 kg will perform
simple harmonic motion with speed at mean position 3v0 and 2v0 respectively. So, maximum
speed of 2 kg block will be 3v0 + v0 = 4v0 and maximum speed of 3 kg block will be 2v0 + v0 =
3v0.
9) Maximum energy will be stored in the spring when blocks will be moving with same
velocity.

towards left

So, =

10)

along x direction

11)

= 1075 m

12)

Displacement of mass A is 10 cm
m × 10 = (m + m + 3m) × x
10m = 5xm
x = 2cm

13)

According to momentum conservation principle, we get

mv0 = 4mv – mV
3V = V0

V= = 4 m/s

14)
 T = (30g + 100g)

15)

So,

16) In COM frame

aCM =

(F1)cm =

(F1)cmx1 + (F2)cmx2 =

(x1 + x2) = xmax =

17) τ = r × F =

PART-2 : CHEMISTRY

18)

Ans is (B)
A ⇢ 8.2, 16.4 , 102.5, 122.4 …….
Valence e– = 2
B ⇢ 10.4, 19.2 , 35.6, 49.2, 166.5, 208.7
Valence e– = 4
∴ Formula is A2B1
19)
III > II > I > IV

20) a > b > c

21) FeO = x millimoles

Fe2O3 = y millimoles

FeO + KMnO4 → Fe+3 + Mn+2

x meq. 100 meq.
Zn + Fe+3 → Fe+2 + Zn+2
(x+2y)mmol (x+2y)mmol
m.equivalent of Fe+2 = m.equivalent of K2Cr2O7
1(x+2y) = 6 × 100
x + 2y = 600
y = 250
x = 100

22)

The correct answer is option (A),(B),(C),(D)

23) The correct answer option is (A), (B), (C)

24)

CO + CH4 + N2 = 20
x y z

CO + O2 → CO2
x x/2 x
CH4 + 2O2 → CO2 + 2H2O(ℓ)
y 2y y
Volume reduction due to

Reaction =

+ 2y = 13 ....... (i)
x + y = 14 ...... (ii)
y = 4 ml
x = 10 ml
z = 6 ml.

25)

The correct answer is (4)

26)
This alkene is least stable and release maximum heat of hydrogenation.

27) Strength in percentage mean how many g H2O2 present per 100 mL
∵ M ⇒ 1 and mol. wt. of H2O2 = 34
∴ 34 H2O2 present per litre of solution or 3.4 H2O2 present per 100 mL of solution.

28) m-eq. of Ba(MnO4)2 = m. eq. of H2O2

× 10 × 1000 = 3 × 125 × 2 ; w = 28.125

% purity = = = 75

29)

30)

(ii), (iv), (v), (vi), (vii), (viii), (ix)

31)

Ti3+ = [Ar] 3d1

32) Mass of oxygen added to Fe = (21.5 – 20)g = 1.5g

2 mol mol gives 1mol

 112g 48g gives 160 g

O2 is L.R.

33) Meff =

1 + 1/3 =
Mav = 69

34) (i), (ii), (iv), (v), (vi)

PART-3 : MATHEMATICS

35)

⇒
⇒
⇒
∴ family passes through (–1, 1)

36)
First we will find the intersection points of these two lines,
x – 2y + 5 = 0 ...(1)
3x + 2y + 7 = 0 ...(2)
from eqn (1) & eqn (2)

⇒
by taking ⊕ sign by taking –ve sign →

This is not possible as incident & reflected rays cannot be parallel

So, the eqn of the reflected ray having point
P(–3, 1) & slope

37)

cos2x = 1 ⇒ x = nπ, n ∈ I

38)

⇒ ,a>3

39)
= (cosy + cos3y)2 + (siny – sin3y)2
⇒ 2 = cos2y + 2cos4y + cos6y + sin2y + sin6y – 2sin4y
= 1 + 2(cos2y – sin2y) + (cos6y + sin6y)
= 1 + 2(cos2y – sin2y) + cos4y + sin4y – cos2y.sin2y
= 1 + 2(cos2y – sin2y) + (cos2y + sin2y)2 – 3cos2y.sin2y
⇒ 0 = 2 – 4sin2y – 3(1 – sin2y)sin2y
= 2 – 4sin2y – 3sin2y + 3sin4y
⇒ 3sin4y – 7sin2y + 2 = 0

⇒ sin2y =

40)

4x + 3y – 7 = 0
24x + 7y – 31 = 0
a1a2 + b1b2 > 0
obtuse angle bisector
20x + 15y – 35 = 24x + 7y – 31
4x – 8y + 4 = 0 ⇒ x – 2y + 1 = 0
acute angle bisector
20x + 15y – 35 + 24x + 7y – 31 = 0
44x + 22y – 66 = 0
2x + y – 3 = 0
for origin L1 < 0 L2 < 0
same sign so origin lies in x – 2y + 1 = 0
for point (1, –2) L1 < 0 L2 < 0
so same sign so point (1, –2) lies in x – 2y + 1 = 0

41)

= –4(a – b)(b – c)(c – a) ≠ 0

(as a, b and c are distinct)
so, the system has only trivial solution.

42)
Line perpendicular to CN is x – 2y = λ satisfy (2, 1) ⇒ λ = 0
∴ equation of line AB is x = 2y ⇒ B(0, 0)
midpoint of A and C lie on line
y = x ⇒ α = 0 ⇒ C(0, 1)
therefore ABC is right angle triangle whose 90° at vertex C. (α, α2) lie inside the triangle
⇒ α > 0, α2 < 1 and 2α2 – α > 0

⇒ <α<1
43)
Line perpendicular to CN is x – 2y = λ satisfy (2, 1) ⇒ λ = 0
∴ equation of line AB is x = 2y ⇒ B(0, 0)
midpoint of A and C lie on line
y = x ⇒ α = 0 ⇒ C(0, 1)
therefore ABC is right angle triangle whose 90° at vertex C. (α, α2) lie inside the triangle
⇒ α > 0, α2 < 1 and 2α2 – α > 0

⇒ <α<1

44)
Equation of line HB :

⇒ 2y + x = 7 ...(1)
On solving with equation of line AB :

Co-ordinate of B ≡

Similarly co-ordinates of C ≡

⇒ Equation of L :

45)
46)

D≡

Also,

Now, ...(i)

and ...(ii)
solving (i) and (ii) for sinθ and cosθ

and

Hence,

47)

Now,

⇒ r = 3r1 ⇒ 4R × = 3r1
⇒ r1 = 1

48)

α3 + 2α2 – α + 3 = 0
⇒ α3 + α2 + 3 = α – α2 = α(1 – α)

= = (1 – α)(1 – β)(1 – γ)
3 2
Now x + 2x – x + 3 = (x – α)(x – β)(x – γ)
put (x = 1) ⇒ 1 + 2 – 1 + 3 = (1 – α)(1 – β)(1 – γ)
So (1 – α)(1 – β)(1 – γ) = 5

49)

As, sin2γ + cosec2γ ≥ 2

2 + cot2β ≥ 2
4 + sin4α ≥3
⇒ sin2γ = 1, cot2β = 0, sin4α = –1
⇒ ,

50)

51)

and
⇒α–γ=3 ...(1)
β–δ=1 ...(2)
Also, (γ, δ) lies on 3x – 2y = 6
3γ – 2δ = 6 ...(3)
and (α, β) lies on 2x – y = 5
⇒ 2α – β = 5 ...(4)
Solving (1), (2), (3), (4)
α = –3, β = –11, γ = –6, δ = –12
|α + β + γ + δ| = 32