Updated 19 Oct 2026

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

JEE Main 2025, 2 Apr Shift 2 was held on 2 April 2025. The bank holds 75 of its questions (Physics 24, Chemistry 26, 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

75 Questions in bank for this shift

Try questions from JEE Main 2025, 2 Apr Shift 2

Q1PhysicsJEE Main 2025, 2 Apr Shift 2Matter waves

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An electron with mass 'm' with an initial velocity

(t = 0) v = v_{0} i^(v_{0} > 0)

enters a magnetic field

B = B_{0} j^.

If the initial de-Broglie wavelength at t = 0 is

λ_{0}

then its value after time 't' would be :

Answer key (D) & worked solution

Correct answer: Option D

Magnetic field does not work

∴

Speed will not change, so De-Broglie wavelength remains same.

Q2PhysicsJEE Main 2025, 2 Apr Shift 2Introduction to dielectrics

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : Net dipole moment of a polar linear isotropic dielectric substance is not zero even in the absence of an external electric field. Reason (R) : In absence of an external electric field, the different permanent dipoles of a polar dielectric substance are oriented in random directions. In the light of the above statements, choose the most appropriate answer from the options given below :

A(A) is correct but (R) is not correct

BBoth (A) and (R) are correct but (R) is not the correct explanation of (A)

CBoth (A) and (R) are correct and (R) is the correct explanation of (A)

D(A) is not correct but (R) is correct

Answer key (D) & worked solution

Correct answer: Option D

A : Since polar dielectrics are randomly oriented

P_{n e t} = 0 .

R : If

E

is absent, polar dielectric remain polar & are randomly oriented.

Q3ChemistryJEE Main 2025, 2 Apr Shift 2Proteins

A tetrapeptide "x" on complete hydrolysis produced glycine (Gly), alanine (Ala), valine (Val), leucine (Leu) in equimolar proportion each. The number of tetrapeptides (sequences) possible involving each of these amino acids is

A16

B32

D24

Answer key (D) & worked solution

Correct answer: Option D

The number of tetrapeptides (sequences) possible involving each of these amino acids (glycine, alanine, valine, leucine) ; It has three (3) peptides linkage the number of permutations in which they can be arranged

= 4 \times 3 \times 2 \times 1

= 24

Q4ChemistryJEE Main 2025, 2 Apr Shift 2Degree of Dissociation

Consider the following chemical equilibrium of the gas phase reaction at a constant temperature :

A (g) ⇌ B (g) + C (g)

If p being the total pressure, Kp is the pressure equilibrium constant and

α

is the degree of dissociation, then which of the following is true at quilibrium ?

AIf p value is extremely high compared to

K_{p}, α \approx 1

BWhen p increases

α

decreases

CIf kp value is extremely high compared to

p, α

becomes much less than unity

DWhen p increases

α

increases

Answer key (B) & worked solution

Correct answer: Option B

A (g) ⇌ B (g) + C (g)

t = 0: a, 0, 0 t = t:a (1 –

α

), a

α

, a

α

a moles of A(g) taken initially and at time Now moles fraction of A(g), B(g) and C(g) are

X_{A} = \frac{a - a α}{a + a α} = \frac{1 - α}{1 + α}

X_{B} = \frac{a α}{a + a α} = \frac{α}{1 + α}

X_{C} = \frac{a α}{a + a α} = \frac{α}{1 + α}

Now if P is total pressure then partial pressure of A(g), B(g) and C(g) are

P_{A} = (\frac{1 - α}{1 + α}) P

P_{B} = (\frac{α}{1 + α}) P

P_{C} = (\frac{α}{1 + α}) P

K_{P} = \frac{( \frac{α}{1 + α} ) P ( \frac{α}{1 + α} ) P}{( \frac{1 - α}{1 + α} ) P}

K_{P} = \frac{α ^{2} P}{1 - α ^{2}}

As KP is only function of temperature. So as P

↑ α ↓

Q5MathematicsJEE Main 2025, 2 Apr Shift 2Equation of lines in 3D and vector form

If the image of the point P(1, 0, 3) in the line joining the points A(4, 7, 1) and B(3, 5, 3) is Q

(α, β, γ)

, then

α + β + γ

is equal to

\frac{47}{3}

\frac{46}{3}

18

13

Answer key (B) & worked solution

Correct answer: Option B

P (1, 0, 3) A(4, 7, 1), B(3, 5, 3) Line AB

\Rightarrow \frac{x - 3}{1} = \frac{y - 5}{2} = \frac{z - 3}{- 2} = λ

Let foot of perpendicular of P on AB be

R \equiv (λ + 3, 2 λ + 5, - 2 λ + 3)

\Rightarrow (λ + 3 - 1) (1) + (2 λ + 5 - 0) (2) + (- 2 λ + 3 - 3)

(- 2) = 0

\Rightarrow λ + 2 + 4 λ + 10 + 4 λ = 0

\Rightarrow λ = - \frac{4}{3}

\Rightarrow R \equiv (\frac{5}{3}, \frac{7}{3}, \frac{17}{3})

Q \equiv (\frac{10}{3} - 1, \frac{14}{3} - 0, \frac{34}{3} - 3) \equiv (\frac{7}{3}, \frac{14}{3}, \frac{25}{3})

\Rightarrow α + β + γ = \frac{7 + 14 + 25}{3} = \frac{46}{3}

Q6MathematicsJEE Main 2025, 2 Apr Shift 2Shortest distance between two lines

The line L

_{1}

is parallel to the vector

a = - 3 i^+ 2 j^+ 4 k^

and passes through the point (7, 6, 2) and the line L

_{2}

is parallel to the vector

b = 2 i^+ j^+ 3 k^

and passes through the point (5, 3, 4). The shortest distance between the lines L

_{1}

and L

_{2}

is :

\frac{23}{38}

\frac{21}{57}

\frac{23}{57}

\frac{21}{38}

Answer key (A) & worked solution

Correct answer: Option A

L_{1} : (7 i^+ 6 j^+ 2 k) + λ (- 3 i^+ 2 j^+ 4 k)

L_{2} : (5 i^+ 3 j^+ 4 k) + λ (2 i^+ j^+ 3 k)

Distance between skew lines

= \frac{( 2 i ^ + 3 j ^ - 2 k ^ ) . ( 2 i ^ + 17 j ^ - 7 k ^ )}{342}

= \frac{69}{342} = \frac{69}{3 38} = \frac{23}{38}