Updated 19 Oct 2026

JEE Main 2025, 4 Apr Shift 2 question paper with solutions

JEE Main 2025, 4 Apr Shift 2 was held on 4 April 2025. The bank holds 71 of its questions (Physics 24, Chemistry 23, Mathematics 24). A sample with answer keys and worked solutions is below, or sit the complete shift paper in the official NTA CBT interface (180 minutes, 300 marks, +4 / −1). Real PYQ papers are never filtered by your covered syllabus.

Sit this paper in the exam screen All JEE Main 2025 shifts Marks vs percentile table

2025 Exam session year

300 Maximum marks (+4 / −1)

180 min Official duration

71 Questions in bank for this shift

Try questions from JEE Main 2025, 4 Apr Shift 2

Q1PhysicsJEE Main 2025, 4 Apr Shift 2Vernier calliper

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For the determination of refractive index of glass slab, a travelling microscope is used whose main scale contains 300 equal divisions equals to 15 cm. The vernier scale attached to the microscope has 25 divisions equals to 24 divisions of main scale. The least count (LC) of the travelling microscope is (in cm) :

Answer key (B) & worked solution

Correct answer: Option B

300 msd = 15 cm

1 m s d = \frac{15}{300} c m = 0.05 c m

25 vsd = 24 msd

1 v s d = \frac{24}{25} m s d

L C = 1 m s d - 1 v s d

L C = 1 m s d - \frac{24}{25} m s d = \frac{1}{25} m s d

L C = \frac{1}{25} \times 0.05 = 0.002 c m

Q2PhysicsJEE Main 2025, 4 Apr Shift 2Areal expansion

Consider a rectangular sheet of solid material of length

ℓ

= 9 cm and width d = 4 cm. The coefficient of linear expansion is

α

= 3.1 × 10–5 K–1 at room temperature and one atmospheric pressure. The mass of sheet m = 0.1 kg and the specific heat capacity C

_{v}

= 900 J kg–1K–1. If the amount of heat supplied to the material is 8.1 × 10

^{2}

J then change in area of the rectangular sheet is :-

A2.0 × 10–6 m

^{2}

B3.0 × 10–7 m

^{2}

C6.0 × 10–7 m

^{2}

D4.0 × 10–7 m

^{2}

Answer key (A) & worked solution

Correct answer: Option A

Δ Q = m s Δ T

8.1 \times 1 0^{2} = 0.1 \times 900 \times Δ T

Δ A = A_{0} 2 α Δ T = 2.0 \times 1 0^{- 6} m^{2}

Q3ChemistryJEE Main 2025, 4 Apr Shift 2Physical Properties of d Block Elements

The incorrect relationship in the following pairs in relation to ionisation enthalpies is :

AMn

^{+}

< Cr

^{+}

BMn

^{+}

< Mn

^{2 +}

CFe

^{2 +}

< Fe

^{3 +}

DMn

^{2 +}

< Fe

^{2 +}

Answer key (D) & worked solution

Correct answer: Option D

^{2 +}

: [Ar]3d

^{5}

Half filled stability More IE than Fe

^{2 +}

^{2 +}

: [Ar]3d

^{6}

Q4ChemistryJEE Main 2025, 4 Apr Shift 2Physical Properties of Group 13 Elements

The elements of Group 13 with highest and lowest first ionisation enthalpies are respectively:

AB & Ga

BB & Tl

CTl & B

DB & In

Answer key (D) & worked solution

Correct answer: Option D

IE order B > T

ℓ

> Ga > Al > In

Q5MathematicsJEE Main 2025, 4 Apr Shift 2Shortest distance between two lines

Let the values of p, for which the shortest distance between the lines

\frac{x + 1}{3} = \frac{y}{4} = \frac{z}{5}

and

r = (p i + 2 j + k) + λ (2 i + 3 j + 4 k)

\frac{1}{6}

, be a, b, (a < b). Then the length of the latus rectum of the ellipse

\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1

is:-

\frac{3}{2}

\frac{2}{3}

D18

Answer key (C) & worked solution

Correct answer: Option C

Shortest distance =

\frac{( a ˉ - b ˉ ) . ( p ˉ \times q ˉ )}{∣ p ˉ \times q ˉ ∣}

Where

a = - i + 0 j + 0 k

b = p i + 2 j + k => a - b = (- 1 - p) i - 2 j - k

p = 3 i + 4 j + 5 k

q = 2 i + 3 j + 4 k

\frac{1}{6} = \frac{∣ - 1 - p + 4 - 1 ∣}{6}

∣ - p + 2 ∣ = 1

p = 3 & p = 1

\frac{x ^{2}}{1 ^{2}} + \frac{y ^{2}}{3 ^{3}} = 1

L . R = \frac{2 a ^{2}}{b} = \frac{2 \times 1}{3} = \frac{2}{3}

Q6MathematicsJEE Main 2025, 4 Apr Shift 2Maxima and Minima at a point

Let a > 0. If the function f(x) = 6x

^{3}

– 45ax

^{2}

+ 108a

^{2}

x + 1 attains its local maximum and minimum values at the points x

_{1}

and x

_{2}

respectively such that x

_{1}

_{2}

= 54, then a + x

_{1}

+ x

_{2}

is equal to :-

A15

B18

C24

D13

Answer key (B) & worked solution

Correct answer: Option B

f^{'} (x) = 18 x^{2} - 90 a x + 108 a^{2} = 0

x = 2 a & x = 3 a

x_{1} = 2 a x_{2} = 3 a

x_{1} x_{2} = 54

6 a^{2} = 54

a = 3

3 + 2 \times 3 + 3 \times 3 = 18

Option (b)