Updated 19 Oct 2026

JEE Main 2025, 3 Apr Shift 2 question paper with solutions

JEE Main 2025, 3 Apr Shift 2 was held on 3 April 2025. The bank holds 74 of its questions (Physics 24, Chemistry 25, Mathematics 25). 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

74 Questions in bank for this shift

Try questions from JEE Main 2025, 3 Apr Shift 2

Q1PhysicsJEE Main 2025, 3 Apr Shift 2Average and RMS value of a function

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An electric bulb rated as 100 W-220 V is connected to an ac source of rms voltage 220 V. The peak value of current through the bulb is :

Answer key (A) & worked solution

Correct answer: Option A

P = v_{r m s} i_{r m s} i_{r m s} = \frac{100}{220} i_{0} = 2 i_{r m s} = 0.64 A

Q2PhysicsJEE Main 2025, 3 Apr Shift 2Heating effects of current/joule's heating

A motor operating on 100 V draws a current of 1 A. If the efficiency of the motor is 91.6%, then the loss of power in units of cal/s is

B8.4

D6.2

Answer key (C) & worked solution

Correct answer: Option C

P_{in p u t} = V i = 100 W

η = \frac{P _{o u t}}{P _{in p u t}} = 0.916 P_{o u t} = 91.6 W

Loss = 100 – 91.6 = 8.4 J/s = 2 cal/s

Q3ChemistryJEE Main 2025, 3 Apr Shift 2Vitamins

Fat soluble vitamins are : A. Vitamin B

_{1}

B. Vitamin C C. Vitamin E D. Vitamin B

_{12}

E. Vitamin K Choose the correct answer from the options given below :

AC & D Only

BA & B Only

CB & C Only

DC & E Only

Answer key (D) & worked solution

Correct answer: Option D

Vit D, E, K. A are fat soluble vitamins.

Q4ChemistryJEE Main 2025, 3 Apr Shift 2Arrhenius Equation

Consider the following statements related to temperature dependence of rate constants. Identify the correct statements, A. The Arrhenius equation holds true only for an elementary homogenous reaction. B. The unit of A is same as that of k in Arrhenius equation. C. At a given temperature, a low activation energy means a fast reaction. D. A and Ea as used in Arrhenius equation depend on temperature. E. When Ea >> RT. A and Ea become interdependent. Choose the correct answer from the options given below :

AA, C and D Only

BB, D and E Only

CB and C Only

DA and B Only

Answer key (C) & worked solution

Correct answer: Option C

Arrhenious equation hold true for elementary as well as complex reactions. Unit of A is same as unit of k. Rate of reaction is high if activation energy is low, A and Ea are temperature independent.

Q5MathematicsJEE Main 2025, 3 Apr Shift 2Concept of DC and DR

Each of the angles β and γ that a given line makes with the positive y– and z–axes, respectively, is half of the angle that this line makes with the positive x-axes. Then the sum of all possible values of the angle β is

\frac{3 π}{4}

π

\frac{π}{2}

\frac{3 π}{2}

Answer key (A) & worked solution

Correct answer: Option A

β = \frac{α}{2}, γ = \frac{α}{2} cos^{2} α + cos^{2} β + cos^{2} γ = 1

cos^{2} α + 2 cos^{2} \frac{α}{2} = 1 cos^{2} α + cos α = 0

cos α (cos α + 1) = 0 cos α = 0, - 1 α = \frac{π}{2}, π

N o w β = \frac{α}{2} \Rightarrow \frac{π}{4}, \frac{π}{2} so s u m i s \frac{3 π}{4}

Q6MathematicsJEE Main 2025, 3 Apr Shift 2Describing Complex number and its Algebra

If z

_{1}

, z

_{2}

, z

_{3}

∈ C are the vertices of an equilateral triangle, whose centroid is z

_{0}

, then

k = 1 \sum 3 (z_{k} - z_{0})^{2}

is equal to

B(2)

D–i

Answer key (A) & worked solution

Correct answer: Option A

z_{1} + z_{2} + z_{3} = 3 z_{0} (z_{1} + z_{2} + z_{3})^{2} = 9 z_{0}^{2}

\Rightarrow z_{1}^{2} + z_{2}^{2} + z_{3}^{2} + 2 (z_{1}^{2} + z_{2}^{2} + z_{3}^{2}) = 9 z_{0}^{2} \Rightarrow z_{1}^{2} + z_{2}^{2} + z_{3}^{2} = 3 z_{0}^{2}

k = 1 \sum 3 (z_{k} - z_{0})^{2} = (z_{1} - z_{0})^{2} + (z_{2} - z_{0})^{2} + (z_{3} - z_{0})^{2}

= z_{1}^{2} + z_{2}^{2} + z_{3}^{2} + 3 z_{0}^{2} - 2 (z_{1} + z_{2} + z_{3}) z_{0} = 6 z_{0}^{2} - 6 z_{0}^{2} = 0