20-07-2025

1402CJA101021250044 JA

PART-1 : PHYSICS

SECTION-I(i)

1) Current in R3 in the shown circuit

(A) just after closing the switch is zero

(B) long time after closing the switch is zero

(C)
just after closing the switch is

(D)
long time after closing the switch is

2) In the circuit shown switch S is closed at t = 0. At time (t) which of the following is/are correct:

(A)

(B)

At t = 0,
(C)

(D)

3) A variable voltage V = 2t applied across an inductor of inductance L = 2H is shown in figure.

Then,
(A) Current versus time graph is a parabola
(B) Energy stored in magnetic field at t = 2s is 4J
(C) Potential energy at time t = 1s in magnetic field is increasing at a rate of 1 J/s
(D) Energy stored in magnetic field is zero all the time

4) Two lenses L1 and L2 are arranged as shown in the figure. An object of length 1 cm is kept at a
distance of 10 cm in front of the combination. Select the correct options.

(A) The equivalent focal length of combination is +20 cm

(B) The size of image is 2 cm
(C) The magnification produced by combination is –2.
(D) The image formed by combination is real.

5) For the resistance network shown in the figure, choose the correct option(s).

(A) the current through PQ is zero

(B) I1 = 3A
(C) The potential at S is less than that at Q
(D) I2 = 2A

6) In Fig., a body A of mass m slides on a plane inclined at angle θ1 to the horizontal and μ1 is the
coefficient of friction between A and the plane. A is connected by a light string passing over a
frictionless pulley to another body B, also of mass m, sliding on a frictionless plane inclined at angle
θ2 to the horizontal. Which of the following statement(s) is (are) true ?

(A) A will never move up the plane

A will just start moving up the plane when
(B)

(C) For A to move up the plane, θ2 must always be greater than θ1

(D) B will always slide down with constant speed

SECTION-I(ii)

Common Content for Question No. 1 to 2

A massless square loop of side a is kept in xz plane as shown. Magnetic field in space is non uniform

given by . The loop is rotated about x-axis with constant angular velocity ω.

1)

Torque required to rotate the loop with constant angular velocity (as a function of time). Take
resistance of loop = R.

(A)

(B)

(C)

(D)

2)

e.m.f. induced in the loop as function of time is equal to :-

(A)
(B)
2
(C) 2B0a ω cos ωt
2 2
(D) B0a ω cos ωt

Common Content for Question No. 3 to 4

A non conducting sphere having relative permittivity unity (εr = 1) has radius R and have a non
uniform charge distribution. The volume charge density ρ(r) varies with radial distance r as per the

following graph. Here ρ is a constant.

3)

Consider an isolated sphere (as explained in paragraph). If is electric field at any point because of

sphere and if C is centre of sphere and S is surface of sphere find the value of

(A)

(B)

(C)

(D)

4) If a small dipole of dipole moment P is placed at point A which is at large distance x from centre of
sphere as shown in figure. The force exerted by dipole on sphere would be :

(A)

(B)
(C)

(D)

SECTION-II

1) A rectangular loop with a sliding connector of length ℓ = 1.0 m is situated in a uniform magnetic
field B = 2 T perpendicular to the plane of loop. Resistance of connector is r = 2Ω. Two resistances
of 6 Ω and 3 Ω are connected as shown in figure. The external force (in N) required to keep the

connector moving with a constant velocity v = 2 m/s, is :-

2) In the figure, a long thin wire carrying a varying current i = 2sin 5t lies at a distance y above one
edge of a rectangular wire loop of length L and width W lying in the x-z plane. Magnitude of emf (in

V) induced in the loop at t = sec is α × 10–7 . Then the value of is (y = 2m, L = 3m,

W = 1m).

3) A triangular wire frame (each side = 2m) is placed in a region of time variant magnetic field
having dB/dt = T/s. The magnetic field is perpendicular to the plane of the triangle. The base of
the triangle AB has a resistance 1Ω while the other two sides have resistance 2Ω each. The
magnitude of potential difference between the points A and B will be V0. Report 10V0.

4) The given figure shows an inductor and resistor fixed on a conducting wire. A movable conducting
wire PQ starts moving on the fixed rails from t = 0 with constant velocity 1 m/s. The work done (in
Joule) by the external force on the wire PQ in 2 seconds is :-

5) A uniform rope of mass 1.0 kg is connected with a box of mass 2.0 kg, which is placed on a
smooth horizontal surface. The free end of the rope is pulled horizontally by a force 6 N. Find the

tension in N at the midpoint of the rope.

6) Two blocks, each of mass 3kg, are connected by a spring, whose spring constant is 200 N/m. They
are placed onto an inclined plane of angle 37°. The coefficient of friction between the upper block
and the inclined plane is 0.6, while between the lower block and the inclined plane it is 0.1. After a
while, the two blocks move together with the same acceleration. Use g = 10 m/s2. Find the extension

(in cm) of the spring.

7) The gravitational field in a region is given by . Find the gravitational

potential at (1, 1) in J kg–1 [consider zero potential at origin] :-

8) A body of mass m is projected horizontally just above the earth's surface with velocity

. The maximum height attained by the body is . Find n.

 PART-2 : CHEMISTRY

SECTION-I(i)

1) Which are true statements for the following equilibrium H2O(ℓ)⇌ H2O(g)

(A) Increase in pressure will result in the formation of more liquid water.
(B) Increase in pressure will increase boiling point
(C) Decrease in pressure will vaporize H2O(l) to a greater extent
(D) Increase in pressure will liquefy steam

2) In the depression of freezing point experiment, it is found that the

(A) vapour pressure of the solution is less than that of pure solvent
(B) vapour pressure of the solution is more than that of pure solvent
(C) only solute molecules solidify at the freezing point
(D) only solvent molecules solididy at the freezing point

3)

The correct statement/s about the following reaction sequence is/are

(A) 'R' gives aldol condensation reaction on heating with NaOH solution
(B) The compound 'Q' gives orange or orange red precipitate upon being heated with acetone
(C) Step '4' is aromatic nucleophilic substitution reaction
(D) The end product exists as diastereomeric pair

4) In which of the following first is having higher resonance energy than second ?

(A)
(B)

(C)

(D) H2C = HC – F, H2C = HC – Br

5)

Among the following pair of compounds, identify the pair(s) which can be differentiated by Fehling
as well as Tollen's reagent ?

(A)

(B)

(C)

(D)

6) The compound which give addition product with sodium bisulphite

(A) Ph–CHO

(B)

(C) Cyclohexanone

(D)

SECTION-I(ii)
Common Content for Question No. 1 to 2
The rate law for reaction A + B C is :
rate = K[A] [B]. Given : K = 6.93 × 10–4 M–1 sec–1

1) Find the time taken (sec.) when concentration of [A] changes from 10–4 M to 5 × 10–5 M. The [B] =
1M remains constant (excess) :

(A) 10
(B) 100
(C) 1000
(D) 10000

2) Starting with [A] = 1M and [B] = 2M. What is the rate in (M sec–1). When [A] changes to 0.25 M :

(A) 2.16 × 10–3

(B) 2.16 × 10–4
(C) 8.64 × 10–4
(D) 8.64 × 10–3

Common Content for Question No. 3 to 4

3) Structure of E is

(A)

(B)

(C)
(D)

4) Structure of A is

(A) H2C = CH – CHO

(B) Ph–CH=CH–CH3

(C)

(D)

SECTION-II

1) Under the equilibrium conditions for the reaction,

, the total pressure is 12 atm. The value of KP is

2) Equilibrium constant Kp for the reaction

CaCO3(s) CaO(s) + CO2 (g) is 8.21 atm at 727°C,
if 10 mole of CaCO3(s) is placed in a 10 L container, what is the weight (in gm) of CaO formed at
equilibrium.

3)
Moles of CH3MgCl required to react with 1 mole of the above compound.

4) Number of reactions in which aldehyde can be obtained as one of the product.

(i) CH3–CH2OH (ii) CH3–CH2–CH2–OH

(iii) (iv)

(v) (vi) CH3–CH=NH

(vii) (viii) CH3–CH=CH–CH3

(ix)

5) AB2 is 10% dissociated in water to A2+ and B–. The boiling point of a 10.0 molal aqueous solution of
AB2 is _______ºC. (Round off to the Nearest Integer).
[Given : Molal elevation constant of water Kb = 0.5 K kg mol–1 boiling point of pure water = 100ºC]

6) Consider reaction

Product X is having insecticidal properties. What is degree of unsaturation (DOU) of X ?

7) Number of compounds that give Cannizaro reaction are :

8)
Number of SP2 carbon in final product ?
 PART-3 : MATHEMATICS

SECTION-I(i)

1)

The equation of the curve satisfying the differential equation can be a

(A) Circle
(B) Straight line
(C) Parabola
(D) Ellipse

2) A function f : R → R satisfies the differential equation 2xy + (1 + x2)y' = 1, where f(0) = 0, then

(A) f(x) is odd

(B) f(x) = 1 has no real solution

(C)
has exactly two distinct solution
(D) f(x) = –1 has no real solution

3)

Area bounded by & y < 0 is less than

(A)

(B)

(C)

(D)

4) If where & , then the

possible value(s) of α is/are

(A)

(B)

(C)

(D)
5) If = y2 + 4y + 5, then y can be equal to-

(A) 1
(B)
(C)
(D)

6) If are the roots of ax2 – 4x + 1 = 0 and β, δ are the roots of bx2 – 6x + 1 = 0 and
are in H.P. then which of the following is correct ?

(A)

(B)

(C) b = 8
(D) a = 3

SECTION-I(ii)

Common Content for Question No. 1 to 2

A curve is defined by the parametric equations x = cos2t, y = sin2t where

On the basis of above information, answer the following questions :

1) If , then k is equal to -

(A) –1
(B) 0
(C) 1
(D) 2

2) The equation of the tangent when is -

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

Common Content for Question No. 3 to 4

Let A(z1), B(z2), C(z3) and D(z4) be the vertices of a trapezium in an Argand place such that AB || CD.
Let |z1 – z2| = 4, |z3 – z4| = 10 and the diagonals AC and BD intersect at P. It is given that

and
3) Which of the following option(s) is/are INCORRECT ?

(A)
value of |CP – DP| is equal to
(B) PB : PD = 2 : 5
(C) PC : AC = 5 : 7
(D) PC : AC = 2 : 5

4) Which of the following option(s) is/are CORRECT ?

(A)
Area of the trapezium ABCD is equal to sq. units.

(B)
Area of the trapezium ABCD is equal to sq. units.

(C)
Area of the triangle BCP is equal to sq. units.

(D)
Area of the triangle BCP is equal to sq. units.

SECTION-II

1) Consider the functions ƒ, g : → defined by

ƒ(x) = x2 + and g(x) =

If α is the area of the region

then the value of 9α is ______.

2) A tangent is drawn to the curve x2 + 2x – 4ky + 3 = 0 at a point whose abscissa is 3. This tangent
is perpendicular to x + 3 = 2y. Find the area bounded by the curve, this tangent and ordinate x = – 1

3) The value of y(x) obtained from the differential equation = y – y2, where y(0) = 2 is

4) If , then

5) Let then the natural number n for which

is ________.
6) If differential equation of family of curves y ℓn|cx| = x, where c is an arbitrary constant, is y’ =

+ ϕ , for some function ϕ, then , for some function f, then is equal tom :

7) If (where k ∈ N), then the value of k is

8) If the sum upto 20 terms is , then is

 ANSWER KEYS

PART-1 : PHYSICS

SECTION-I(i)

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

SECTION-I(ii)

Q. 7 8 9 10
A. B A A A

SECTION-II

Q. 11 12 13 14 15 16 17 18
A. 2.00 9.00 4.00 32.00 5.00 3.00 40.00 0.67

PART-2 : CHEMISTRY

SECTION-I(i)

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

SECTION-I(ii)

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

SECTION-II

Q. 29 30 31 32 33 34 35 36
A. 16.00 56.00 3.00 9.00 106.00 8.00 6 8

PART-3 : MATHEMATICS

SECTION-I(i)

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

SECTION-I(ii)

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

Q. 47 48 49 50 51 52 53 54
A. 6.00 5.33 1.00 0.00 7.00 4.00 5 1.20
 SOLUTIONS

PART-1 : PHYSICS

1)

Just after switch is closed, L is open circuit and C is short circuit

Just after

⇒ ⇒i=0
Long time

⇒i=0

2)

(A)

(C)
at t = 0 ⇒ I = 0
Vab = 0

3) parabola

at t = 2 sec U = 4J
4)

5)
Since resitances in two arms upper and lower are in same ratio.

Current through PQ is zero.

, and Potential of S is less than that at Q.

6)

The block A will just starts moving up if,

mg sin θ1 + f = mg sin θ2
mg sin θ1 + μ mg cos θ1 = mg sin θ2

When block A moves upwards

f = mg sin θ2 − mg sin θ1 > 0
sin θ2 > sin θ1 θ2 > θ1

7)

Input power = Out put power (ω is constant)

τω =

τ=

8) From the diagram flux through the given loop is same as the flux through a rectangular loop
having width a sin ωt in xy frame.

9)

Electric field inside sphere.

10)
we can assume, sphere is placed at equitorial position of dipole, F = Q (Eeq)

Q = ρ0pR3

11)

E = Bvℓ
F = Biℓ

12)

B=
dϕ = B(dA)cosθ

dϕ =
ϕ= ;e=

13)

14)

P = U2B2ℓ2

15) Acceleration a =

NLM on half portion of rope 6 – T = ×2⇒T=5N

16) m1g sinα + m2gsinα – µ1m1g cosα – µ2m2g cosα = (m1 + m2) . acentre.
2mg sin α – (µ1 + µ2) mg cos α = 2macentre,

acentre = g(sinα – )
mg sin α – µ2mg cos α – kΔℓ = macentre,
mg sin α + kΔℓ – µ1mg cos α = macentre

17) . (dx + dy )
v = 40 J/Kg
 18)
L = mv0R = mvr

on solving, h = R

So ⇒

PART-2 : CHEMISTRY

19) As pressure increases equilibrium shift towards left, boiling point also increases and
stream gets easily liquefy. Decrease in pressure shifts the equilibrium towards right.

20)

On addition of non volatile solute in to volatile solvent

- Vapour pressure of solution decreases
- boiling point of solution increases
- freezing point of solution decreases. At freezing point only solvent get freezed not solute
molecule

21)
22)

Option 1,, 2 & 4 are correct

23)

Comp. Fehling Test Tollen's test

HCHO Y Y

N N

Y Y

N N

Y Y

N Y

N N
24) Aldehydes and Methyl ketones give NaHSO3 test
Acetophenone doesn’t give product on reaction with NaHSO3 due to steric crowding.

25) Question Explanation:

Find time

Concept:
This question is based on Pseudo first order reaction.

Solution:
A+B→C
r = K[A] [B]
when B is in excess
r = K' [A] (first order reaction)
here K' = K[B]
= 6.93 × 10–4 × 1 = 6.93 × 10–4 sec–1
for first order

t = 1000 sec

Final Answer:
The correct option is (3)

26) Question Explanation:

Calculate rate of reaction.

Concept:
This question is based on rate law r = K[A] [B].

Solution:
A + B → C
t=0 1M 2M
1–.75 2–.75
t=t .25 M = 1.25 M .75 M
r = K[A] [B]
= 6.93 × 10–4 [.25] [1.25]
= 2.165625 × 10–4 Ms–1
Final Answer:
The correct option is (2)

27)
28)

29) At equilibrium, the pressure of CO2 is p, In the reaction

or p + 2p = 12
p=4
Partial pressure of CO2 = 4
Partial pressure of CO = 2 × 4 = 8

30) KP = = 8.21 atm

8.21 × 10 = × 0.0821 × 1000
=1
= =1
weight of CaO in gm = 1 × 56 = 56 gm

31)

Answer is 3
 32) Theory based.

33) AB2 → A2+ + 2B–

t=0 a 0 0
t = t a –aα aα 2aα
nT = a –aα+aα+2aα
= a (1 + 2α)
so i = 1 + 2α
Now ΔTb = i × m × Kb
ΔTb = (1 + 2a) × m × Kb
a = 0.1 m = 10 Kb = 0.5
ΔTb = 1.2 × 10 × 0.5
=6
So boiling point = 106

34)
Hence its DOU is 8

35) A, B, E, H, I & L

36)

PART-3 : MATHEMATICS

37)

Straight line
38)
d((1 + x2)y) = 1
(1 + x2)y = x + C

39)

Domain x + y + 1 > 0,
y–x+2>0
Squaring x + y + 1 < y – x + 2

i.e.
Shaded region indicated in the figure

and its area

40)

As

as

41)

y2 + 4y + 5 =

=
=

42) From the given equations and ,

Also given α, β, γ, δ are in H.P. ⇒ and

Hence and
Substituting the values of roots in the respective equations gives b = 8 and a = 3.

43) k = sin2t

= – cosec22t + cot22t
= –(1 + cot22t) + cot22t = –1

44) Here

Now the equation of tangent is

45)

ΔABP and ΔCDP are similar if AP = 2x,

BP = 2y then CP = 5x, DP = 5y
Area of trapezium ABCD = xy

tanα = , tanβ = also α + β = 45°

⇒ xy =

Also AB2 = AP2 + BP2 ⇒ x2 + y2 = 4 ⇒ xy =

Ar(ΔPCD) = 5xy

(x – y)2 = 4 – ⇒ |x – y| =

46)

ΔABP and ΔCDP are similar if AP = 2x,

BP = 2y then CP = 5x, DP = 5y

Area of trapezium ABCD = xy

tanα = , tanβ = also α + β = 45°

⇒ xy =

Also AB2 = AP2 + BP2 ⇒ x2 + y2 = 4 ⇒ xy =

Ar(ΔPCD) = 5xy

(x – y)2 = 4 – ⇒ |x – y| =

47)

48) x2 + 2x – 4ky + 3 = 0
2x + 2 – 4k

put x = 3,
6 + 2 + 8k = 0
k=–1

Tangent is 4x + 2y – 3 = 0

49) = y – y2 ⇒ =

+ dy = x + c ⇒ ℓn =x+c

= kex ⇒ y = kex – kyex

⇒y=

x = 0, y = 2 ; 2 =
⇒ 2 + 2k = k

⇒ k = –2 , y =

⇒y=

(y(x)) = =1

50)
Differentiating both sides w.r.t. x, we get

So,
⇒

51)

52) y ℓn|cx| = x
yℓln|c| + ℓn|x|) = x
y(ℓn|c|+ ℓn|x|) = x ...(i)

k + ℓn|x| + =1

k=

Now y. + yℓn|x| = x

y– – y ℓn|x| + yℓn|x| =x

x =y–

= –

So =–

ϕ(2) = –( 2)–2 = –

53)

∴
∴ k=5

54) General terms of above is

Sn = T1 + T2 + T3 + ... + Tn