Updated 19 Oct 2026

JEE Main 2025, 23 Jan Shift 2 question paper with solutions

JEE Main 2025, 23 Jan Shift 2 was held on 23 January 2025. The bank holds 69 of its questions (Physics 24, Chemistry 20, 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

69 Questions in bank for this shift

Try questions from JEE Main 2025, 23 Jan Shift 2

Q1PhysicsJEE Main 2025, 23 Jan Shift 2Photoelectric effect and Einstein's photoelectric equation

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In photoelectric effect an em-wave is incident on a metal surface and electrons are ejected from the surface. If the work function of the metal is 2.14 eV and stopping potential is 2V, what is the wavelength of the em-wave? (Given hc = 1242 eVnm where h is the Planck's constant and c is the speed of light in vaccum.)

Answer key (D) & worked solution

Correct answer: Option D

e V_{s} = E - ϕ

2 e V = E - 2.14 e V

E = 4.14 eV

E = \frac{h c}{λ}

λ = \frac{1242}{4.14} = 300 nm

Q2PhysicsJEE Main 2025, 23 Jan Shift 2Properties of EM waves

A plane electromagnetic wave of frequency 20 MHz travels in free space along the +x direction. At a particular point in space and time, the electric field vector of the wave is E

_{y}

= 9.3 Vm–1. Then, the magnetic field vector of the wave at that point is-

B_{z} = 9.3 \times 1 0^{- 8} T

B_{z} = 1.55 \times 1 0^{- 8} T

B_{z} = 6.2 \times 1 0^{- 8} T

B_{z} = 3.1 \times 1 0^{- 8} T

Answer key (D) & worked solution

Correct answer: Option D

E = BC 9.3 = B × 3 × 10

^{8}

B = \frac{9.3}{3 \times 10 ^{8}} = 3.1 \times 1 0^{- 8} T

Q3ChemistryJEE Main 2025, 23 Jan Shift 2Physical Properties of Alcohols

Given below are two statements: Statement (I): The boiling points of alcohols and phenols increase with increase in the number of C-atoms. Statement (II): The boiling points of alcohols and phenols are higher in comparison to other class of compounds such as ethers, haloalkanes. In the light of the above statements, choose the correct answer from the options given below:

ABoth Statement I and Statement II are false

BStatement I is false but Statement II is true

CStatement I is true but Statement II is false

DBoth Statement I and Statement II are true

Answer key (D) & worked solution

Correct answer: Option D

B.P. ∝ M.W. B.P. ∝ Inter molecular hydrogen bonding Alcohol & Phenol have intermolecular H-bonding

Q4ChemistryJEE Main 2025, 23 Jan Shift 2Degree of Dissociation

Consider the reaction

X_{2} Y (g) ⇌ X_{2} (g) + \frac{1}{2} Y_{2} (g)

The equation representing correct relationship between the degree of dissociation (x) of X

_{2}

Y(g) with its equilibrium constant Kp is________. Assume x to be very very small.

x = 3 \frac{2 K p}{p}

x = 3 \frac{2 K p ^{2}}{p}

x = 3 \frac{K p}{2 p}

x = 3 \frac{K p}{p}

Answer key (B) & worked solution

Correct answer: Option B

1 mole

X_{2} Y_{(g)} ⇌ X_{2}_{(g)} + \frac{1}{2} Y_{2}_{(g)}

1-x mole x mole

\frac{x}{2}

mole

∴ P_{X_{2} Y} = \frac{1 - x}{1 + \frac{x}{2}} \times P

P_{X_{2}} = \frac{x}{1 + \frac{x}{2}} \times P

∴ K_{p} = \frac{( \frac{x}{1 + \frac{x}{2}} P ) ( \frac{x}{2 ( 1 + \frac{x}{2} )} P ) ^{\frac{1}{2}}}{( \frac{1 - x}{1 + \frac{x}{2}} ) \times P}

∴ K_{P} = (\frac{x}{1 - x}) (\frac{x}{2 ( 1 + \frac{x}{2} )})^{\frac{1}{2}} \times P^{\frac{1}{2}}

∵

x to be very very small

∴ K_{p} = \frac{x ^{3/2}}{( 2 ) ^{\frac{1}{2}}} \times P^{\frac{1}{2}}

∴ x^{\frac{3}{2}} = \frac{K _{P} \times 2 ^{\frac{1}{2}}}{P ^{\frac{1}{2}}}

∴ x^{3} = \frac{K _{p}^{2} \times 2}{P}

x = (\frac{K _{p}^{2} \times 2}{P})^{\frac{1}{3}}

Q5MathematicsJEE Main 2025, 23 Jan Shift 2Shortest distance between two lines

If the square of the shortest distance between the lines

\frac{x - 2}{1} = \frac{y - 1}{2} = \frac{z + 3}{- 3}

an d \frac{x + 1}{2} = \frac{y + 3}{4} = \frac{z + 5}{- 5}

\frac{m}{n}

, where m, n are coprime numbers, then m + n is equal to:

C21

D14

Answer key (B) & worked solution

Correct answer: Option B

a = (2, 1, - 3)

b = (- 1, - 3, - 5)

\vec p \times \vec q = \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k} \\ 1&2&{ - 3} \\ 2&4&{ - 5} \end{array}} \right|

= 2 i^- j^

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

S_{d} = \frac{∣ ( b - a ) \cdot ( p \times q ) ∣}{∣ p \times q ∣}

= \frac{2}{5}

(S_{d})^{2} = \frac{4}{5}

m = 4, n = 5 \Rightarrow m + n = 9

Q6MathematicsJEE Main 2025, 23 Jan Shift 2Shortest distance between two lines

The distance of the line

\frac{x - 2}{2} = \frac{y - 6}{3} = \frac{z - 3}{4}

from the point (1, 4, 0) along the line

\frac{x}{1} = \frac{y - 2}{2} = \frac{z + 3}{3}

17

14

15

13

Answer key (B) & worked solution

Correct answer: Option B

Let the parallel line is

\frac{x - 1}{1} = \frac{y - 4}{2} = \frac{z - 0}{3}

so their point of intersection is (λ + 1, 2λ + 4 3λ ) = (2t + 2, 3t + 6, 4t + 3) λ = 2t + 1 2λ + 4 = 3t + 6 ⇒ t = 0 so POI is (2,6,3) so distance =

(2 - 1)^{2} + (6 - 4)^{2} + (3 - 0)^{2} = 14