20-07-2025

1001CJA101023250009 JA

PART-1 : PHYSICS

SECTION-I (i)

1) A current of 10A exists in a wire of cross-sectional area of 5 mm2 with a drift velocity of 2 ×
10–3 ms–1. The number of free electrons in each cubic meter of the wire is :-

(A) 2 × 106
(B) 625 × 1025
(C) 2 × 1025
(D) 1 × 1023

2) A parallel plate capacitor with air as medium between the plates has a capacitance of 10μF. The
capacitor is divided into two halves and filled with two media as shown in the figure having
dielectric constant K1 = 2 and K2 = 4. The capacitance of the system will now be :-

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

3) Initial charge on conducting sphere of radius r is Q0 . If S is closed at t = 0 then charge on the

sphere at any time t is (switch S is in contact with earth)

(A)
(B)

(C)

(D)

4) A bullet of mass m and charge q is fired towards a solid uniformly fixed charged sphere of radius
R and total charge + q. If it strikes the surface of sphere with speed u, find the minimum speed u so
that it can penetrate through the sphere. (Neglect all resistive forces or friction acting on bullet

except electrostatic forces)

(A) Zero

(B)

(C)

(D) ∞

SECTION-I (ii)

1) A battery is of emf E is being charged from a charger such that positive terminal of the battery is
connected to terminal A of charger and negative terminal of the battery is connected to terminal B of
charger. The internal resistance of the battery is r.

(A) Potential difference across points A and B must be more than E.

(B) A must be at higher potential than B
(C) In battery, current flows from positive terminal to the negative terminal
(D) No current flows through battery

2) In a meter bridge the point D is a neutral point as shown in figure.

(A) The meter bridge can have no other neutral point for this set of resistances.
When the jockey contacts a point on meter wire left of D, current flows to B from the wire
(B)
through galvanometer.
When the jockey contacts a point on the meter wire to the right of D, current flows from B to
(C)
the wire through galvanometer.
(D) When R is increased, the neutral point shifts to left.

3) Young's modulus of a uniform cross section rod, linearly varries from Y0 (end A) to 2Y0 (end B).
Two equal forces are acting at ends of rod (area of cross-section = A, length = ℓ)

(A) Stress at end A = stress at mid point of rod

(B) Stress at end A ≠ stress at mid point of rod

(C)
Total extension in rod is

(D)
Total extension in rod is

SECTION-I (iii)

1) Match the List

List-I List-II

If distance between plates of an Potential difference

(P) (1)
isolated capacitor decreases across plate is decreased

If dielectric is inserted between

Capacitance of the
(Q) plates of capacitor whose plates (2)
capacitor will increase
are connected with battery
 If area of plates of an isolated Energy of capacitor will
(R) (3)
capacitor is increased increase

If distance between plates of

capacitor is decreased when the Force between the plates
(S) (4)
capacitor is connected with will decrease
battery

(5) None of these

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

2) Four identical thin, square metal sheets, S1, S2, S3 and S4, each of side a are kept parallel to each
other with equal distance d (<< a) between them, as shown in the figure. Let C0 = ε0a2/d, where ε0 is

the permittivity of free space.

List-I List-II

The capacitance between S1 and S4, with S2 and S3 not

(P) (1) 3C0
connected, is

The capacitance between S1 and S4, with S2 shorted to

(Q) (2) C0/2
S3, is

The capacitance between S1 and S3, with S2 shorted to

(R) (3) C0/3
S4, is

The capacitance between S1 and S2, with S3 shorted to

(S) (4) 2C0/3
S1, and S2 shorted to S4, is

(5) 2C0
(A) P → 3;Q → 2;R → 4;S → 5
(B) P → 2;Q → 3;R → 2;S → 1
(C) P → 3;Q → 2;R → 4;S → 1
(D) P → 3;Q → 2;R → 2;S → 5

3) In list-I, we have certain situations showing object and optical system. Match them with
description of image in list-II.

List-I List-II

(P) (1) Real & magnified

R=1m
Arc shows boundary separated
two medium

(Q) (2) Real & diminished

(R) (3) Virtual & magnified

(S) (4) Virtual & diminished

(5) Real and of same size

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

SECTION-II

1) The coil of a calorimeter C has a resistance of R1 = 60Ω. The coil R1 is connected to the circuit as
shown in figure. What is the rise in temperature (°C) of 240 grams of water poured into the
calorimeter when it is heated for 7 minutes during which a current flow through the coil and the
ammeter shows 3A? The resistance R2 = 30Ω. [Disregard the resistances of the battery and the
ammeter, and the heat losses and heat capacity of the calorimeter and the resistor and specific heat
of water = 4200 J/kg°C]

2) In an isolated parallel-plate capacitor of capacitance C, the two sheets have charges 7Q and –Q.

The potential difference between the plates is . Find N.

3) A conductor of resistance R = 10Ω and heat capacity C = 200 J/°C is connected to a voltage
source V = 20 V at time t = 0. It loses heat to surrounding at a rate Ploss = k(T – T0), where k = 2W/°C
& T0 = 20°C. Temperature T (in °C) of the conductor at time t = 100 sec is :- (e–1 = 0.632)

4) Find the current (in mA) in the wire between points A and B.

5) The rod connecting two reservoir and connected to a source is as shown in diagram. The rod is in
steady state. Find the temperature of source, so that rate of melting of ice is 16 times that rate of
vaporization. (Latent heat of vaporization is 540 Cal/gm and latent heat of fusion is 80 Cal/gm)
6) A body radiates energy 5W at a temperature of 127°C. If the temperature is increased to 927°C,
then it radiates energy (in W) at the rate of –

PART-2 : CHEMISTRY

SECTION-I (i)

1) Heat absorbed by the system going through a cyclic process as shown will be–

(A) 10π Joule

(B) 100π Joule
(C) 103π Joule
(D) 400π Joule

2) The rate of a certain biochemical reaction at physiological temperature (T) occurs 106 times faster
in presence of enzyme than without the enzyme. The magnitude of change in the activation energy
upon adding enzyme is : (ln 10 = 2.303)

(A) RT
(B) 6RT
(C) (2.303)RT
(D) 6(2.303)RT

3) Major product formed in the following reaction is: P

(A)
(B)

(C)

(D)

4) In the given following emperical formula of silicates shared and unshared oxygen corner are same
by each tetrahedron in.

(A)

(B)

(C)

(D)

SECTION-I (ii)

1) Which is/are correct information(s) about 0.1 M NaCl (aq) and 0.1 M Na2SO4 (aq) solutions at
given temperature : (Both solutions are highly diluted)

(A) the lowering in vapour pressure of Na2SO4 solution is less than NaCl solution
(B) the freezing point of Na2SO4 solution is less than NaCl solution
(C) the osmotic pressure of Na2SO4 is 1.5 times that of NaCl solution
(D) the boiling point of Na2SO4 solution is 1.5 times that of NaCl solution

2) Which of the following statements is correct?

(A) The order of splitting energy is

(B) is colourless whereas is coloured

(C) will exhibit geometrical isomerism
(D) The magnetic moment of is B.M
3) In which case incorrect product(s) is/are formed:

(A) Me3C–O–CH3 Me3C–OH + CH3I

(B) H3C–O–CH2–CH3 CH3OH + ICH2CH3

(C)

(D)
+ CH3I

SECTION-I (iii)

1) One mole of a monatomic ideal gas is taken along two cyclic processes E → F → G → E and E →

F → H → E as shown in the p-V diagram. (ln 2 = 0.7) The processes

involved are isochoric, isobaric, isothermal or adiabatic. Match the paths in List-I with magnitudes of
the work done in List-II and select the correct answer using the codes given below the lists.
(ln2 = 0.70)

List-I List-II

(P) G→E (1) 112 p0V0

(Q) G→H (2) 36 p0V0

(R) F→H (3) 24 p0V0

(S) F→G (4) 31 p0V0

(5) 96 p0V0
(A) P → 4;Q → 3;R → 5;S → 2
(B) P → 3;Q → 2;R → 4;S → 1
(C) P → 4;Q → 3;R → 2;S → 1
(D) P → 1;Q → 2;R → 3;S → 5

2)
 List-I List-II
(Reaction) (Major product)

(P) (1)

(Q) (2)

(R) (3)

(S) (4)

(5)

(A) P → 1;Q → 3;R → 2;S → 4

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

3)

List-I List-II

(P) [FeCl6]4– (1) Central atom uses orbtial in hybridisation

(Q) [Cr(ox)3]3– (2) Shows optical activity

(R) [Co(en)2(H2O)2]3+ (3) Complex is inner orbital complex

(S) [Pt(gly)3]+1 (4) Shows geometrical isomerism

(5) Complex contains symmetrical bidentate ligand

The correct option is
(A) P → 1;Q → 2,5;R → 3,4;S → 2,4
(B) P → 2;Q → 3,5;R → 2,4;S → 2,5
(C) P → 1;Q → 4,5;R → 2,3;S → 4,5
(D) P → 4;Q → 3,4;R → 4,5;S → 3,5
 SECTION-II

1) Solid AB has NaCl type structure. If the radius of A+ and B– are 1.5 Å and 2.5Å respectively and
formula mass of AB is 384 g/mole, what is the density of AB solid (in gm/cm3).
Take : Avogadro's number = 6 × 1023 atoms/mole

2) How many of the following can be prepared by Williamson's ether synthesis in good yield

(a) (b) Ph – O – Ph (c)

(d) (e) (f)

(g) (h) CH3–O–CH3

3) Number of correct statements :

(i) (Rate of SN1)

(ii) ( Rate of SN2)

(iii) (Rate of SN2 in DMSO)

(iv) (Rate of dehydration)

(v) (Stability of alkene)

4) The total number of species which have bond order greater than two
O2, O2+, O2–, N2, N2+, CN–, B2, C2

5) Let, a = the total number of ligands which can act as didentate / polydentate ligands with
identical donor sites.
DMG, en, Gly, acac, pyridine, PPh3 , CN⊖ , bn , EDTA–4
And
b = (number of Geometrical isomers of [Pt(H2NCH(CH3)COO–)2] complex)
Then the value of a + b is:

6) How many are sulphide ores in the following?

(i) Azurite (ii) Chalcocite
(iii) Iron Pyrites (iv) Limonite
(v) Malachite (vi) Siderite
(vii) Pyrolusite (viii) Argentite
(ix) Cinnabar.

PART-3 : MATHEMATICS

SECTION-I (i)

1) Let f(x) = , then

(A) f(x) is one-one function

(B) f(x) is an odd function

(C)
Range of f(x) is
(D) f(x) is invertible function

2) The value of limit is

(A) –2
(B) 0
(C) 1

(D)

3) If ,then the value of is equal to -

(Where C is constant of integration)

(A) 2010
(B) 2011
(C) 2012
(D) 2013
4) If , then at is

(A)

(B)

(C) –1
(D) –2

SECTION-I (ii)

1) Let two circles x2 + y2 + 2(cos θ)x + 2(sin θ)y + λ = 0 and x2 + y2 + 4(cosθ)x – (6 sinθ)y – 7 = 0
intersect each other orthogonally, then (for all θ ∈ R)

(A) maximum value of λ is 11

(B) maximum value of λ is 3
(C) minimum value of λ is 0
(D) minimum value of λ is 1

2) Which of the following limits is/are unity ?

(A)

(B)

(C)

(D)

3) is equal to-
(where C is constant of integration)

(A) x secx + C
(B) x secx – ℓn|secx + tanx| + C

(C)

(D) x tanx – x ℓn|secx| + C

SECTION-I (iii)

1) If ax2 – 2y2 + bxy + 4x – 6y = 0 (a, b ∈ R) represents the equation of a circle with center (x0, y0)
and radius 'r', then
Match each entry in List-I to the correct entries in List-II.
 List-I List-II

(P) a (1) 17

(Q) b (2) –2

(R) x0 + 4y0 (3) 0

(S) 4r2 (4) 13

(5) –5
The correct option is :
(A) P → 1;Q → 3;R → 5;S → 4
(B) P → 2;Q → 1;R → 3;S → 5
(C) P → 2;Q → 4;R → 5;S → 1
(D) P → 2;Q → 3;R → 5;S → 4

2) Match each entry in List-I to the correct entry in List-II. (here C is constant of integration)

List-I List-II

(P) (1)

(Q) (2)

(R) (3)

(S) (4)

(5)

The correct option is

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

3)

List-I List-II

(P) The function f(x) = –x4e–x is strictly decreasing in (1)

The centre of circle passing through the points (0,

(Q) (2) (0, 4)
0), (1, 0) and touching the circle x2 + y2 = 9 can be

(R) (3)
If then x ∈
 (S) (4)

, then (q, p) is

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

SECTION-II

1) The value of is equal to :

2) Rolle's theorem holds for the function ƒ(x) = x3 + αx2 + βx + 2, in x ∈ [0,2] at the point , then
α + 2β is equal to

3) If y = a ℓn|x| + bx2 + x has its extremum values at x = –1 and x = 2, then value of (a + b) is

4)

Let where ƒ(0) = –ℓn2 & , then is equal to

5) Let then number of points where function is non-

differentiable is

6) If ax + by – 13 = 0 is tangent to the circle x2 + y2 – 4x + 6y – 13 = 0 at (3, 2), then value of (b – 2a)

is
 ANSWER KEYS

PART-1 : PHYSICS

SECTION-I (i)

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

SECTION-I (ii)

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

SECTION-I (iii)

Q. 8 9 10
A. C C B

SECTION-II

Q. 11 12 13 14 15 16
A. 25.00 4.00 32.60 to 32.64 7.50 640.00 405.00

PART-2 : CHEMISTRY

SECTION-I (i)

Q. 17 18 19 20
A. B D A C

SECTION-I (ii)

Q. 21 22 23
A. B,C A,C,D A,B,C

SECTION-I (iii)

Q. 24 25 26
A. C B A

SECTION-II

Q. 27 28 29 30 31 32
A. 5.00 4.00 5.00 4.00 8.00 4.00

PART-3 : MATHEMATICS
 SECTION-I (i)

Q. 33 34 35 36
A. C A D A

SECTION-I (ii)

Q. 37 38 39
A. A,D A,B,C B,C

SECTION-I (iii)

Q. 40 41 42
A. D C A

SECTION-II

Q. 43 44 45 46 47 48
A. 3.00 1.75 1.50 0.60 3.00 3.00
 SOLUTIONS

PART-1 : PHYSICS

1)

i = 10 A, A = 5 mm2 = 5 × 10–6 m2
and vd = 2 × 10–3 m/s
We know, i = neAvd
∴ 10 = n × 1.6 × 10–19 × 5 × 10–6 × 2 × 10–3
⇒ n = 0.625 × 1028 = 625 × 1025

2) Both are in parallel

Ceq. = C1 + C2 =

=
Ceq. = 3 × 10 = 30 µF

3) This is discharging of a capacitor where

4) (TE)surface = (PE)centre
Velocity at centre is nearly zero then it can cross the sphere

5)
While charging, current goes into positive terminal of battery and comes out of negative
terminal.
Apply KVL
E + i –VAB = 0
VAB = E + i
VAB > E
VA > VB

6)
7)

stress is , ∴ it will be same at all points.

Young's modulus on element dx will be

let extension in dx be dy

then

ΔL : total alongation in rod.

8) (P) For an isolated capacitor

Q = constant
So, if distance between plate decreases, then capacitance will increase.
Since Q = CV

⇒V∝
So as C increases V decreases.

Since Energy
So as C increases energy decreases.

Since Force F =

9)

(P) ⇒

(Q)

(R)
(S)

⇒ ⇒ 3C

10)

We know,
Since v = –ve & |m| < 1
Hence image is virtual & diminished.

11)

Current in 60Ω resistance = 1A

∴ Q = i2Rt = (1)2 × 60 × 7 × 60 J
∴ Q = mSΔT ⇒ ΔT = 25°C
12) Plane conducting surfaces facing each other must have equal and opposite charge
densities.
Potential difference between the plates = charge on the capacitor ÷ capacitance.

∴ potential difference =

13)

T = 32.64

14)
2 i1 = 1 (i – i1)
1 i2 = 2(i – i2)
+ 30 – 103(i – i1) – 2 × 103 (i – i2) = 0
Solving equations i1 – i2 = 7.5 mA

15)

; ;

16) Rate of energy = P = AεσT4 ⇒ P ∝ T4

 ⇒ P2 = 5 × 81 = 405 W

PART-2 : CHEMISTRY

17) For a cyclic process

W=–q
= Area of circle

= = 100πJ

18)
E'a = Ea – RT ln106
E'a – Ea = – RT ln106 = – 6RT × 2.303

19)
It follows SN2 mechanism
Hence inversion in configuration takes place

20)

In
No of shared O-atom = 2
No of unshared O-atom = 2

21) ∴ Osmotic pressure of Na2SO4 will be more than NaCl.

∴ Osmotic pressure of Na2SO4 will be times that of NaCl.

∴ (B) & (C).

22) C is incorrect it does not show geometrical isomerism.

 is coloured due to d-d transition.

23) (A) Me3C–O–CH3 Me3C–I + CH3OH

(B) H3C–O–CH2–CH3 CH3–I + HO–CH2–CH3

(C)

24) Process :
FG : Isothermal expension
FH : Adiabatic expension
For process FH :
p1V1r = P2V2r
32p0 × V05/3 = p0 × VH5/3
VH = 8 V0
∴ WGE = – p0[32V0 – V0] = 31p0V0
∴ WGH = – p0 [32V0 – 8V0] = – 24p0V0

0 0

∴ WFH = = – 36p V

0 0 0 0
∴ WFG = – 32 p V ℓn = – 160 p V ℓn 2

25)

(P) is (Q) is

(R) is (R) is

26)

P→1
Q → 1, 2, 3, 5
R → 1, 2, 3, 4, 5
S → 1, 2, 3

27) d = = 5 gm/cc

28) (a, c, f, h)
 29)

The correct answer is 5.00

30)

O2+, N2, N2+, CN–

31)

a=4
b=4
a+b=8

32) Iron Pyrites, Argentite, Cinnabar, Chalcocite

PART-3 : MATHEMATICS

33)

Now since sin2x ∈ [0, 1] so f(x)

also f(x) is an even, many-one function

34)

35) Put 1 – x2013 = t2

⇒ –2013x2012dx = 2tdt

=
36)

at

37) Use 2(g1g2 + ƒ1ƒ2) = C1 + C2

2(cos θ)(2 cosθ) + 2(sinθ) (–3 sinθ) = λ – 7
λ = 7 + 4 cos2θ – 6sin2θ
λ = 7 + 2 (1 + cos2θ) – 3(1 – cos2θ)
λ = 6 + 5 cos2θ

38) (A)

(B)

(C)

(D) L.H.L. = –1, D.N.E.

39)

40)

coefficient of x2 = coefficient of y2 and coefficient of xy = 0

∴ a = –2, b = 0
Equation of circle : –2x2 – 2y2 + 4x – 6y = 0
x2 + y2 – 2x + 3y = 0

0 0
(x , y ) ≡ ,

41) (P)

(Q)

(R)

(S)

42) (P) f'(x) = –4x3e–x + x4e–x < 0

e–x(x3)(x – 4) < 0

(Q) (x – 0)(x – 1) + y2 + λy = 0
x2 + y2 – x + λy = 0
C1C2 = |r1 – r2|

Centre :

(R)

when 3x > 1

(S) &

43)
= tan–1(r + 2) – tan–1(r + 1)

tan(tan–1(n + 2) – tan–12) =
44) By ƒ(0) = ƒ(2) 2α + β = –4

By

45) y = a ℓn|x| + bx2 + x

46)

Let x + 1 =

⇒ ...(i)
Put t = 1

⇒ ƒ(0) =

In eq. (1), put t = 2

=
⇒ k=3

47)
Total points are 3.

48) T = 0
3x + 2y – 2(x + 3) + 3(y + 2) – 13 = 0
x + 5y – 13 = 0.