Updated 19 Oct 2026

JEE Main 2025, 3 Apr Shift 1 question paper with solutions

JEE Main 2025, 3 Apr Shift 1 was held on 3 April 2025. The bank holds 73 of its questions (Physics 23, 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

73 Questions in bank for this shift

Try questions from JEE Main 2025, 3 Apr Shift 1

Q1PhysicsJEE Main 2025, 3 Apr Shift 1Photoelectric effect and Einstein's photoelectric equation

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The work function of a metal is 3 eV. The color of the visible light that is required to cause emission of photoelectrons is

Answer key (B) & worked solution

Correct answer: Option B

(KE)_{m a x} = \frac{hc}{λ} - ϕ

\frac{hc}{λ} > ϕ [for emission]

λ < \frac{h c}{ϕ} \Rightarrow λ < \frac{1242}{3} nm

So blue light option (B)

Q2PhysicsJEE Main 2025, 3 Apr Shift 1Intensity of EM waves

The radiation pressure exerted by a 450 W light source on a perfectly reflecting surface placed at 2m away from it, is :

A1.5 × 10–8 Pascals

C6 × 10–8 Pascals

D3 × 10–8 Pascals

Answer key (C) & worked solution

Correct answer: Option C

P_{rad} = \frac{2 I}{C}

Where I = intensity at surface C = Speed of light

I = \frac{P o w er}{A r e a} = \frac{450}{4 π r ^{2}}

= \frac{450}{4 π \times 4} = \frac{450}{16 π}

P_{rad} = \frac{2 \times 450}{16 π \times 3 \times 10 ^{8}} = \frac{150}{8 π \times 10 ^{8}}

= 5.97 × 10–8 ≈ 6 × 10–8 Pascals

Q3ChemistryJEE Main 2025, 3 Apr Shift 1Chemical Properties of Carbonyl Compounds

Number of molecules from below which cannot give iodoform reaction is : Ethanol, Isopropyl alcohol, Bromoacetone, 2-Butanol, 2-Butanone, Butanal, 2-Pentanone, 3-Pentanone, Pentanal and 3-Pentanol

Answer key (B) & worked solution

Correct answer: Option B

Following will not give iodoform reaction/test. (1) Butanal (2) 3-Pentanone (3) Pentanal (4) 3- Pentanol

Q4ChemistryJEE Main 2025, 3 Apr Shift 1Introduction to Equilibrium

Given below are two statements Statement I : A catalyst cannot alter the equilibrium constant (K

_{C}

) of the reaction, temperature remaining constant Statement II : A homogenous catalyst can change the equilibrium composition of a system temperature remaining constant In the light of the above statements, choose the correct answer from the options given below

AStatement I is false but Statement II is true

BBoth Statement I and Statement II are true

CBoth Statement I and Statement II is false

DStatement I is true but Statement II is false

Answer key (D) & worked solution

Correct answer: Option D

A catalyst can change equilibrium composition if it is added at constant pressure, but it can not change equilibrium constant.

Q5MathematicsJEE Main 2025, 3 Apr Shift 1Shortest distance between two lines

Line L

_{1}

passes through the point (1, 2, 3) and is parallel to z-axis. Line L

_{2}

passes through the point (λ , 5, 6) and is parallel to y-axis. Let for λ = λ

_{1}

, λ

_{2}

, λ

_{2}

< λ

_{1}

, the shortest distance between the two lines be 3. Then the square of the distance of the point (λ

_{1}

, λ

_{2}

, 7) from the line L

_{1}

A40

B32

C25

D37

Answer key (C) & worked solution

Correct answer: Option C

L_{1} \equiv \frac{x - 1}{0} = \frac{y - 2}{0} = \frac{z - 3}{1}

L_{2} \equiv \frac{x - λ}{0} = \frac{y - 5}{1} = \frac{z - 6}{0}

{\text{SD}} = \frac{{\left| {\begin{array}{*{20}{c}} {\lambda - 1}&3&3 \\ 0&0&1 \\ 0&1&0 \end{array}} \right|}}{{\left| {\begin{array}{*{20}{c}} {\widehat {\text{i}}}&{\widehat {\text{j}}}&{\widehat {\text{k}}} \\ 0&0&1 \\ 0&1&0 \end{array}} \right|}}

= ∣ λ - 1 ∣ = 3

λ = 4, - 2

λ_{1} = 4

λ_{2} = - 2

Let foot of perpendicular from P(4, –2, 7) is Q(1, 2, t + 3) So (3, –4, 4 – t) . (0, 0, 1) = 0

t = 4

So Q(1, 2, 7) PQ

^{2}

= 9 + 16

P Q^{2} = 25

Q6MathematicsJEE Main 2025, 3 Apr Shift 1General term of binomial expansion

The sum of all rational terms in the expansion of

(2 + 3)^{8}

A16923

B3763

C33845

D18817

Answer key (D) & worked solution

Correct answer: Option D

S =

(2 + 3)^{8}

For sum of rational terms

=^{8} C_{0} (2)^{8} +^{8} C_{2} (2)^{6} . (3)^{2} +^{8} C_{4} (3)^{4} +^{8} C_{6} (2)^{2} (3)^{6} +^{8} C_{8} (3)^{8}

= 2^{8} + 2^{8} \times 2^{6} .3 + 70. 2^{4} .9 + 28. 2^{2} .27 + 81

= 256 + 5376 + 10080 + 3024 + 81 = 18817