Updated 19 Oct 2026

JEE Main 2025, 2 Apr Shift 1 question paper with solutions

JEE Main 2025, 2 Apr Shift 1 was held on 2 April 2025. The bank holds 74 of its questions (Physics 25, Chemistry 25, 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

74 Questions in bank for this shift

Try questions from JEE Main 2025, 2 Apr Shift 1

Q1PhysicsJEE Main 2025, 2 Apr Shift 1Bohr's atomic model

Try right here · No sign-in

Considering Bohr's atomic model for the hydrogen atom : (A) The energy of the H-atom in the ground state is the same as the energy of the He

^{+}

ion in its first excited state.(B) The energy of the H-atom in the ground state is the same as that for the Li

^{++}

ion in its second excited state. (C) The energy of the H-atom in its ground state is the same as that of the He

^{+}

ion in its ground state.(D) The energy of He

^{+}

ion in its first excited state is same as that for Li

^{++}

ion in its ground state Choose the correct answer from the options given below:

Answer key (B) & worked solution

Correct answer: Option B

E \propto \frac{Z}{n ^{2}}

Z_{H} = 1 Z_{H e^{+}} = 2 Z_{L i^{+ 2}} = 3

^{s t}

excited state ⇒ n = 2 2

^{n d}

excited state ⇒ n = 3 From the given statements only A & B are correct.

Q2PhysicsJEE Main 2025, 2 Apr Shift 1Heating effects of current/joule's heating

The battery of a mobile phone is rated as 4.2 V, 5800 mAh. How much energy is stored in it when fully charged?

A43.8 kJ

B48.7 kJ

C87.7 kJ

D24.4 kJ

Answer key (C) & worked solution

Correct answer: Option C

Given V = 4.2 volt ∴ Energy supplied by battery = vq = 4.2 × 5800 × 3600 × 10–3 J = 87.696 kJ ∴ Energy stored in the battery when fully charged = 87.696 kJ ≈ 87.7 kJ

Q3ChemistryJEE Main 2025, 2 Apr Shift 1Chemical Properties of Amines

The correct order of basic nature on aqueous solution for the bases NH

_{3}

, H

_{2}

N – NH

_{2}

, CH

_{3}

_{2}

_{2}

, (CH

_{3}

_{2}

)

_{2}

NH and (CH

_{3}

_{2}

)

_{3}

N is:

N H_{3} < H_{2} N - N H_{2} < (C H_{3} C H_{2})_{3} N

< C H_{3} C H_{2} N H_{2} < (C H_{3} C H_{2})_{2} N H

N H_{3} < H_{2} N - N H_{2} < C H_{3} C H_{2} N H_{2}

< (C H_{3} C H_{2})_{2} N H < (C H_{3} C H_{2})_{3} N

H_{2} N - N H_{2} < N H_{3} < (C H_{3} C H_{2})_{3} N

< C H_{3} C H_{2} N H_{2} < (C H_{3} C H_{2})_{2} N H

N H_{2} - N H_{2} < N H_{3} < C H_{3} C H_{2} N H_{2}

< (C H_{3} C H_{2})_{3} N < (C H_{3} C H_{2})_{2} N H

Answer key (D) & worked solution

Correct answer: Option D

Basic strength of amine depends on hydrogen bonding and electronic inductive effect

N H (E t)_{2} > N (E t)_{3} > N H_{2} E t > N \cdot\cdot H_{3} > N H_{2} - N H_{2}

Q4ChemistryJEE Main 2025, 2 Apr Shift 1Amino Acids

Identify the correct statement among the following:

AAll naturally occurring amino acids except glycine contain one chiral centre.

BAll naturally occurring amino acids are optically active.

CGlutamic acid is the only amino acid that contains a –COOH group at the side chain.

DAmino acid, cysteine easily undergo dimerization due to the presence of free SH group.

Answer key (D) & worked solution

Correct answer: Option D

* Isoleucine has 2 chiral centre * Glycine is optically inactive * Aspartic acid also contain COOH group at the side chain. * Cysteine easily dimmerise due to free SH group

Q5MathematicsJEE Main 2025, 2 Apr Shift 1Maxima and Minima at a point

If the function

f (x) = 2 x^{3} - 9 a x^{2} + 12 a^{2} x + 1,

where a > 0, attains its local maximum and local minimum values at p and q, respectively, such that p

^{2}

= q, then f(3) is equal to:

A55

B10

C23

D37

Answer key (D) & worked solution

Correct answer: Option D

f'(x) = 6x

^{2}

– 18ax + 12a

^{2}

f'(x) = 6(x

^{2}

– 3ax + 2a

^{2}

) roots are a, 2a p

^{2}

= q ⇒ a

^{2}

= 2a a = 2 f(x) = 2x

^{3}

– 18x

^{2}

+ 48x + 1 f(3) = 37

Q6MathematicsJEE Main 2025, 2 Apr Shift 1Modulus of complex number & its properties

Let z be a complex number such that |z| = 1. If

\frac{2 + k ^{2} z}{k + z ˉ} = k z, k \in R

, then the maximum distance of k + ik

^{2}

from the circle |z – (1 + 2i)| = 1 is

5 + 1

3 + 1

Answer key (A) & worked solution

Correct answer: Option A

\frac{2 + k ^{2} z}{k + z ˉ} =

kz |z|

^{2}

k = 2 k = 2 point p(2, 4); center (1, 2) distance from circle (x – 1)

^{2}

+ (y – 2)

^{2}

= 1 is max. if (OP + r) =

1 + 4 + 1 = 5 + 1