13-07-2025

1001CJA101022250020 JA

PART-1 : PHYSICS

SECTION-I (i)

1) The flux linked with a coil at any instant ‘t’ is given by ϕ =10t2 – 50t + 250. The induced emf at t =
3 s is :-

(A) 190 V
(B) –190V
(C) –10 V
(D) 10V

2) Radiation from a black body at the thermodynamic temperature T1 is measured by a small

detector at distance d1 from it. When the temperature is increased to T2 and the distance to d2, the
power received by the detector is unchanged. What is the ratio d2/d1 ?

(A)

(B)

(C)

(D)

3) For a periodic motion represented by the equation Y = sin ωt + cos ωt. The amplitude of the
motion is :-

(A) 0.5
(B)
(C) 1
(D) 2

4) When x grams of steam at 100°C is mixed with y grams of ice at 0° C, We obtain (x + y) grams of
water at 100°C. What is the ratio y/x? (Lf of ice = 80 cal/gm, LV of steam = 540 cal/gm)

(A) 1
(B) 2
(C) 3
(D) 4
 SECTION-I (ii)

1) When the switch K is open, the equivalent resistance between A and B is Which of the

following statement(s) is/are correct?

(A)
(B) No current flows through K when it is closed
(C) The powers dissipated in R and in the resistor are equal
(D) The powers dissipated in the two resistors are unequal

2) In the circuit shown in figure, each capacitor has a capacitance C. The emf of the cell is E and

circuit is already in steady state. If the switch S is closed :-

(A) Some positive charge will flow out of the positive terminal of the cell
(B) Some positive charge will enter the positive terminal of the cell
(C) The amount of charge flowing through the cell will be CE

(D)
The amount of charge flowing through the cell will be CE

3) AB and CD are smooth parallel rails, separated by a distance ℓ, and inclined to the horizontal at
an angle θ. A uniform magnetic field of magnitude B, directed vertically upwards, exists in the
region. EF is a conductor of mass m, carrying a current i. For EF to be in equilibrium,

(A) i must flow from E to F

(B) Biℓ = mg tan θ
(C) Biℓ = mg sin θ
(D) Biℓ = mg

SECTION-I (iii)

1) Match the List.

List-I List-II [Value of magnetic field at

[Orientation of current carrying conductor] point P]

(P) (1)

(Q) (2)

(R) (3)

(S) (4)

(5)

(A) P → 3;Q → 4;R → 2;S → 1

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

2) You are given different situation of incident ray in List-I. Match them to the nature of image in
List-II.

List–I List–II

(P) (1) Real

(Q) (2) Erect

(R) (3) Magnified

(S) (4) Diminished

(5) None
(A) P → 1;Q → 1,2,4;R → 1,3;S → 1,2,3
(B) P → 1,3;Q → 1,2,4;R → 1,3;S → 5
(C) P → 1;Q → 1,4;R → 1,4;S → 1,2,3
(D) P → 1,4;Q → 1,2,4;R → 1,4;S → 5

3) Match the list.

List-I List-II

A point charge Q. Field at a point at a distance r

(P) (1) 0
from it.
 An infinite linear charge having charge per unit
(Q) (2)
length = λ. Field at a distance r from it.

An infinite sheet of charge having charge per unit

(R) (3)
area = σ. Field at a distance r from it.

A uniform hollow sphere of radius R and total

(S) charge Q. Field at a point inside the sphere at a (4)
distance r from centre.

(5)

(A) P → 1;Q → 5;R → 3;S → 4

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

SECTION-II

1)

Find charge in 4µF capacitor in µC. Assume capacitors are initially uncharged.

2) The EMF of a battery is determined using the following circuit with wire AB. The galvanometer
shows zero deflection when one of it’s terminals is connected to point C. If the internal resistance of
the battery is zero., then what is emf of battery (in volt) ? Length of wire AB is 100cm.

3) A particle having mass m, charge q enters a cylinder region having uniform magnetic field B in
the inward direction as shown. If the particle is deviated by 60° as it emerges out of the field then
the time spent by it in the field is . The value of N is : -

4) A steel wire of length 4.5 m and a copper wire of length 3.5 m are stretched same amount under a

given load. If ratio of Young's modulli of steel to that of copper is , then what is the ratio of cross
sectional area of steel wire to copper wire ?

5) A particle of mass m is moving in circular motion of radius r with constant speed v in influence of
two fixed particles of mass M. Plane of circular motion is perpendicular to line joining fixed particles

as shown. Speed of particle is given as , find value of n (a = r )

6) Three rods of same material, same area of cross-section but different lengths 10 cm, 20 cm and 30
cm are connected at a point as shown. What is temperature of junction O (in ℃)?

PART-2 : CHEMISTRY

SECTION-I (i)

1) At temperature T, compound AB2(g) dissociates as AB2(g) AB(g) + B2(g) having degree of

dissociation x (small compared to unity). The correct expression for x in terms of Kp and p is
(A)

(B)

(C)

(D)

2) For hydrogen like species, the shortest wavelength in the Lyman series is 911.6 Å. Then the
longest wavelength in the Lyman series is :-

(A) 1215 Å
(B) ∞
(C) 2430 Å
(D) 600 Å

3) Which of the following complex has maximum electrical conductance in aqueous solution.

(A) [CoCl3 (NH3)3]

(B) [Co (NH3)5Cl]Cl2
(C) [Co (NH3)4Cl2]Cl
(D) [Co (NH3)6]Cl3

4) Major product
Major product will be-

(A)

(B)

(C)

(D) CH4
 SECTION-I (ii)

1)

Which of the following compound can show isomerism ?

(A) [Cr(NH3)6] [Co(CN)6]

(B) [Co(NH3)5 Br] Cl2
(C) [Cr(NH3)3 Cl3]
+2
(D) [Co(NH3)5 NO2]

2)

Find correct statement for following complex compound [Pt(NH3)4 Cl2] [PtCl4]

(A) E.A.N of cationic part and anionic part is not equal

(B) Cationic part of given complex can show cis - trans isomerism
(C) Synergic bonding is involved in complex
(D) Oxidation state of platinum is +2 in both cationic and anionic part of complex.

3)
Correct options related to x and y is (are)

(A)

x=

(B)

y=

(C)

x=
(D)

y=

SECTION-I (iii)

1) Match the thermodynamic processes given Column I with the expressions given under Column II.

LIST-I LIST-II

(P) Freezing of water at 273 K and 1 atm (1) q=0

Expansion of 1 mol of an ideal gas into a vacuum under adiabatic

(Q) (2) w=0
conditions

Mixing of equal volumes of two ideal gases at constant temperature and

(R) (3) ΔSsys < 0
pressure in an isolated container

Heating of ideal gas H2(g) at 1 atm from 300 K to 600 K in closed rigid
(S) (4) ΔU = 0
container

(5) ΔSsys > 0

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

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

2) Match the list

List-I List-II

Bonding length is
(P) (1)
decreased during the process

Bond order is decreased

(Q) (2)
during the process

Bond length is increased

(R) (3)
during the process

Oxidation takes place

(S) (4)
during the process

Reduction takes place

(5)
during the process

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

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

3) Matching list type

 List-II
List-I
(Degree of unsaturation
(Reaction)
of major product)

(P) (1) DOU =1

Major product

(Q) (2) DOU = 2

Majo
r product

(R) (3) DOU = 3

Major
product

(S) (4) DOU = 4

Major product

(5) DOU = 5

(A) P → 3;Q → 1;R → 2;S → 4

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

SECTION-II

1) How many moles of photon would contain sufficient energy to raise the temperature of 245 g of
water from 18.5°C to 99.5 °C? The specific heat of water is 4.4 J/oC-g and frequency of light
radiation used is 2.45 × 1010 per second. (NA = 6 × 1023, h = 6.6 × 10–34 J sec)

2) The bond dissociation enthalpy of H2(g) and N2(g) are 436 kJmol–1 and 940 kJmol–1 and enthalpy of
formation of NH3 is – 45 kJ mol–1. The enthalpy of atomisation of NH3 is (kJ mol–1)

3) For the first order reaction A(g) → 2B(g) + C(g) the half life for decomposition of A is 3 minute at
300K. Calculate the time (in minute) in which partial pressure of A(g) will drop from 2 bar to 0.5 bar
at 400K. Given activation energy of reaction is (840 R) per mole of A(g). (Take ln2 = 0.7)

4) Total number of geometrical isomer that can exist for squar planar complex [Pt (Cl) (Py) (NH3)
(NH2OH)]+

5)

X is number of mole of Grignard required, find value of X.

6) P
Number of π bonds present in P.

PART-3 : MATHEMATICS

SECTION-I (i)

1) is
(Where C is constant of integration)

(A)

(B)

(C)

(D)

2) Let , where f(1) = ; then f(2) is :

(A)

(B)

(C)

(D)

3) Let y = x3 + 5x2 + x + 7, then slope of tangent at the point (1, 14) is

(A)

(B) –7
(C) 14
(D) 2

4) Let f : R → R ; f(x) = sin–1 then which of the following is not True ?

(A) f(x) is continuous for all x

(B) f(x) is many one function

(C)
f(x) has discontinuity at x =
(D) f(x) = f(|x|) for all x

SECTION-I (ii)

1) Let f : R → [0, ∞) ; f(x) = min. (|x|, ex), then which of following is/are Not true ?

(A) f(x) is continuous function

(B) f(x) is surjective function
(C) f(x) is differentiable everywhere
(D) f(x) is even function

2) Which of the following limit is greater than unity -

(A)

(B)

(C)

(D)
3) Let S1 = 0 & S2 = 0 are two circles & they are mirror image w.r.t. the line L = 0. If the centre of S1
= 0 is (2, 4) & the point of intersection of their internal common tangents is (6, 8), then which of the
following is true -

(A) Centre of S2 = 0 is (10,12)

(B) Sum of intercepts made by the line L = 0 on co-ordinate axes is 28
(C) Point of intersection of their external common tangent does not exist
(D) Radical axis of S1 = 0 & S2 = 0 is L = 0

SECTION-I (iii)

1) Match List-I with List-II and select the correct answer using the code given below the list.

List-I List-II

Sum of intercepts on co-ordinate axes made by tangent

(P) (1) 0

to the curve ℓny = at the point (0, 1) is

In adjacent figure, triangle is rotated along AM to form a

cone of maximum volume. If ℓ denotes its slant height, then

(Q) (2) 1

[3ℓ] is (where [.] denotes greatest integer

function)

Least positive integral value of k for which equation |x| =

(R) (3) 2
kex has exactly one solution, is

(S) , if f(x) is continuous at (4) 4

x= then the value of is (where b ∈ R)

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

2) Match List-I with List-II and select the correct answer using the code given below the list.

List-I List-II
 The value of definite integral

(P) (1) 1

is less than

(Q) (2) 2
then value of is greater than
(where [.] & {.} greatest integer function and fractional
part function respectively)

If ƒ(x) = (x3 + bx2 + cx + d)|(x – 1)(x – 2)(x – 3)3(x – 4)4| is

(R) derivable for all x ∈ R and ƒ'(3) + ƒ'''(4) = 0 then value of (3) 3
|b + c + d| is less than

(S) If L = (4) 4
where [.] greatest integer function
then value of 7 + L is greater than

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

3) Match the statement in List-I with given value in List-II

List-I List-II

If area bounded by x = (θ – sinθ); y = (1 – cosθ);

(P) (1) 6
0 ≤ θ ≤ 2π is λπ; then the value of λ can be equal to

If the least area of triangle formed by tangent to the

(Q) circle x2 + y2 = 1 and x = 0, y = 0 is A, then A is (2) 3
coprime with

Area of triangle formed by any tangent to the curve

(R) (3) 80
and pair of lines
xy – x + 2y = 2 is

If the maximum area bounded by the curves

(S) (4) 84
x2 = 4ay, y = ax, y = , 1 ≤ a ≤ 2 is α, then value of
(α + a) is

(5) 86
(A) P → 2;Q → 1,2;R → 1,4,5;S → 1,2
(B) P → 2;Q → 1,2,3,4,5;R → 1;S → 4
(C) P → 2;Q → 1,2,3,4,5;R → 2,5;S → 1,4
(D) P → 3;Q → 2;R → 4;S → 4

SECTION-II

1) is :

2) Let area bounded by y = x & y = x2 is λ then 3λ is

3) Let the area bounded by the curves y = ex . ℓn x and y = is then λ1 + λ2 is

(where λ1, λ2 ∈ R)

4) is

5) If f,g,h be continuous function on [0,a] such that f(a – x) = f(x), g(a – x) = –g(x) and 3h(x) – 4h(a –

x) = 5 then dx is equal to

6) Figure shows a circle x2 + y2 = a2 whose centre O is at origin. Line y = mx + c drawn parallel to

AB cuts circle at M and N & coordinate axes at P and Q as shown. If MP = & PN = 12, then

is
 ANSWER KEYS

PART-1 : PHYSICS

SECTION-I (i)

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

SECTION-I (ii)

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

SECTION-I (iii)

Q. 8 9 10
A. A A D

SECTION-II

Q. 11 12 13 14 15 16
A. 16.00 3.75 3.00 0.75 4.00 16.36

PART-2 : CHEMISTRY

SECTION-I (i)

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

SECTION-I (ii)

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

SECTION-I (iii)

Q. 24 25 26
A. C C A

SECTION-II

Q. 27 28 29 30 31 32
A. 9000.00 1169.00 3.00 3.00 4.00 3.00

PART-3 : MATHEMATICS
 SECTION-I (i)

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

SECTION-I (ii)

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

SECTION-I (iii)

Q. 40 41 42
A. C A B

SECTION-II

Q. 43 44 45 46 47 48
A. 0.25 0.50 1.50 0.00 0.00 3.00
 SOLUTIONS

PART-1 : PHYSICS

1)

E(t) = –

E(t) =
E(t) = – (20t – 50)
Here, Substitute t = 3s
E (C) = – (20 × 3 – 50)
E(C) = – (60–50)
E(C) = –10V

2) We use I1 =

I2 =
From (1) and (2), We get

3) y = sin ωt + cos ωt

y = sin ωt + sin (ωt + )

Δϕ =

Anet =
Anet =

4) Heat given out when x grams of steam at 100°C is converted to water at 100°C
H1= 540x ...............(1)
Heat gained by ice at 0°C to converted to water at 100°C

According to principle of calorimetry ,

5)

Bridge is balanced.
No current flows.
PR = I2R = I2 × 80
PS = (4I)2 × 5 = I2 × 80
P20 = I2 × 20 (upper line)
= (4I)2 × 20 (lower line)

6)

7)
mg sinθ = iℓB cosθ
Biℓ = mg tanθ

8)

(P)
due to wire = 0

due to half loop =

(Q)
due to wire = 0

due to half loop =

(R)

due to wire =

due to half loop =

(S)

due to wire =

due to half loop =

9) (P) Object is virtual and erect so image will be real & erect.
(Q) In this case concave mirror will form real, erect and diminished image.
(R) In this case concave mirror will form real and magnified image.
(S) In this case convex mirror will form real, erect and magnified image.

10)
(P)

(Q)

(R)
(S) E inside hollow sphere = 0

11) Asking about

The question asks for the charge stored on the 4 μF capacitor in a circuit with capacitors and
batteries.
Concept :
Capacitors in series when capacitors are in series, they share the same charge. the equivalent
capacitance is :

Capacitor charge, Voltage, and capacitance : The charge Q stored in a capacitor is related to
its capacitance C and voltage V by.
Q = CV
Kirchhoff's Voltage law (KVL): The sum of the voltages in a closed loop is zero
Calculation / Explanation
Step 1 : Calculate equivalent capacitance
The two capacitors are in series, so the equivalent capacitance is

Step 2 : Apply kirchhoff's Law

Let Q be the charge on each capacitor (since charge is same in series.)
Voltage equation for the first loop (12V battery and 4μF capacitor)

Voltage equation for the second loop (6V battery and 8μF capacitor):

Step 3 : solve for Charge Q

Adding the two equations eliminates Vjunction :

12V –

Final Answer
The charge on the 4 μF capacitor is 16 μC

12) Potential gradiant along wire AB is

=
So (VA – VB) =
So emf of battery E = 3.75 Volt

13)

14)

15)

16)

Let junction temperature = T

PART-2 : CHEMISTRY

17) AB2(g) AB(g) + B2(g)

teq.

⇒ x << 1 ⇒ 1and 1 – x 1

18)

For Lyman series, we have . Hence

λ1 = = 1215 Å

19) [Co(NH3)6]+3 + → 4 ions (Max ion)

20) The correct answer is option (D)

21) (A) → Coordination isomer

(B) → Ionisation isomer
(C) → Stereo isomer
(D) → Linkage isomer
22) (A) EAN is different
(B) Cationic part show G.I.
(C) No Synergic bonding
(D) Pt+4 & Pt+2

23) The correct answer is option (C), (D)

24) (P) 1 atm and 273 K.

∵ Exothermic reaction ∴ q < 0
also volume increase ∴ w < 0
H2O(s) has lesser entropy as comparted to H2O(l)
∴ ΔSsys < 0 ΔU < 0
Hence in (3) & (5)
(Q) ∵ isolated condition ∴ q = 0, also expansion against vacuum ∴ w = 0 ∴ ΔU = 0, also I.G.
∴ Temperature remains constant.
Also, entropy increases
Also for physical process
∴ Ans. is (1), (2), (4)
(R) ∵ isolated container ∴ q = 0 and w = 0
∵ Gases are getting mixed ∴ entropy of system is increasing
Also ΔU = 0 (∵ ΔU = q + w)
∴ Ans. is (1), (2), (4)
(S) q > 0
∴ ΔS > 0
at constant volume, w = 0
ΔU > 0
∴ Ans. (2), (5)

25) Molecular orbital theory based

26) (P) is DOU = 3

(Q) is DOU = 1
(R) is DOU = 2
(S) is DOU = 4

27)
or, n × 6.6 × 10–34 × 2.45 × 1010 = 245 × 4.4 × (99.5 – 18.5)
∴ Number of photons = 54 × 1026

∴ Moles of photon =

28)

= 1169
 29)

(t½) = 1.5 min. (at 400K)

PA = 2bar → 0.5 bar
We need (2t½) = 3 min.

30) [M abcd] ⇒ G.I = 3

31)

The value of X is 4.

32) 3 number of π bonds groups present in P.

PART-3 : MATHEMATICS

33)

I1 =

Let

Similarly I2 =

Let
34) Divide Nr & Dr by x2

Let

35)

36)

37)

38) By options A, B, C, D are Correct.

39) (A)
(6, 8) is middle point of (2, 4) & (x1, y1)
⇒ (x1, y1) ≡ (10, 12)
(B) The line L is passing through (6,8) & perpendicular to line joing their centres. Line joining
the centres have slope 1 then line L = 0 has slop –1. Equation of line L = 0 is y – 8 = (–1) (x –
6)
⇒ x + y = 14
sum of intercepts is equal to 28.
(C) External common tangents are parallel & their point of intersects does not exist.
(D) Radical axes is line perpendicular to line joining the centres & this time it is also
perpendicular bisector of line joining the centres because both circles will have the same
radius.
40) (P)

⇒ y ′ (0) = 1
tangent is (y – 1) = x ⇒ x – y + 1 = 0

(Q)

–x
(R) k =| x | e

(S)

a+b–2<b+2
⇒0<a<4

41) (P)
Apply king property and add

(Q)
(R) f(x) = (x3 + bx2 + cx + d)
|(x – 1) (x – 2) (x – 3)3(x – 4)4|
f(x) = (x – 1) (x – 2) (x – 3)
|(x – 1) (x – 2) (x – 3)3 (x – 4)4|
∴ |b + c + d| = 1
(S)

42) (P) 3

(Q) Let tangent x cosα + y sinα = 1

it's area is with coordinates axes

⇒ Amin = 1
(R) 6
Curve is rectangular Hyperbola
(x + 2)(y – 1) = –3
and pair of lines (x + 2)(y – 1) = 0 are pair of asymptodes.
∴ Area = 2c2 = 6.

(S)

Points of intersection are P(4a2, 4a3),

required area =

which is increasing
⇒ maximum value of area occures of a = 2
⇒ α = 84

43)
tan–1(2)

44)
45)

46) ....(1)

....(2)
(1) + (2)
⇒ 2I = 0 ⇒I=0

47) ...(1)

2I1 = 0 ⇒ I1 = 0

I=0

48) PM = QN & OP = OQ = c
PA.PC = PM.PN
(r – c) (r + c) = . 12
r2 – c2 = 12
PQ =

c2 + c2 =
r2 = 100
m = –1