Updated 19 Oct 2026

JEE Main 2025, 24 Jan Shift 2 question paper with solutions

JEE Main 2025, 24 Jan Shift 2 was held on 24 January 2025. The bank holds 68 of its questions (Physics 24, Chemistry 19, 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

68 Questions in bank for this shift

Try questions from JEE Main 2025, 24 Jan Shift 2

Q1PhysicsJEE Main 2025, 24 Jan Shift 2Photoelectric effect and Einstein's photoelectric equation

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The energy E and momentum p of a moving body of mass m are related by some equation. Given that c represents the speed of light, identify the correct equation.

Answer key (D) & worked solution

Correct answer: Option D

[E] = M^{1} L^{2} T^{- 2}

[P c] = M^{1} L^{1} T^{- 1} . L^{1} T^{- 1} = M^{1} L^{2} T^{- 2}

[m c^{2}] = M^{1} L^{2} T^{- 2}

E^{2} = P^{2} c^{2} + m^{2} c^{4}

Q2PhysicsJEE Main 2025, 24 Jan Shift 2Graphs in photoelectric effect

In photoelectric effect, the stopping potential (V

_{0}

) v/s frequency (v) curve is plotted. (h is the Planck's constant and

ϕ_{0}

is work function of metal) (A) V

_{0}

v/s ν is linear (B) The slope of V

_{0}

v/s ν curve

\frac{ϕ _{0}}{h}

_{0}

v/s ν line (D) The value of electric charge of electron is not required to determine h using the V

_{0}

v/s ν curve. (E) The work function can be estimated without knowing the value of h. Choose the correct answer from the options given below:

A(A), (B) and (C) only

B(C) and (D) only

C(A), (C) and (E) only

D(D) and (E) only

Answer key (C) & worked solution

Correct answer: Option C

h v = ϕ + K E_{m a x}

K E_{m a x} = e V_{0}

V_{0} = \frac{h v - ϕ}{e}

(A) V

_{0}

v/s is linear correct (B) Slope

v_{0} = (\frac{h}{e}) v - \frac{ϕ}{e}

Wrong Slope

\frac{h}{e}

Q3ChemistryJEE Main 2025, 24 Jan Shift 2Resonance

Given below are two statements : Statement(I): Experimentally determined oxygen-oxygen bond lengths in the O

_{3}

are found to be same and the bond length is greater than that of a O = O (double bond) but less than that of a single (O – O) bond. Statement (II) : The strong lone pair-lone pair repulsion between oxygen atoms is solely responsible for the fact that the bond length in ozone is smaller than that of a double bond (O=O) but more than that of a single bond (O – O). In the light of the above statements, choose the correct answer from the options given below:

AStatement I is true but Statement II is false

BBoth Statement I and Statement II are true

CBoth Statement I and Statement II are false

DStatement I is false but Statement II is true

Answer key (A) & worked solution

Correct answer: Option A

Due to resonance bond length is identical in ozone. Because of resonance, both the O-O bond lengths have partial double bond character. In other words, both the O-O bond lengths are equal (128 pm) and lie in between those O = O double bond length of 121 pm and O-O single bond length of 148 pm.Therefore statement I is true and statement II is false.

Q4ChemistryJEE Main 2025, 24 Jan Shift 2Physical Properties of d Block Elements

Match List-I with List-II. List-I (Transition metal ion)List-II (Spin only magnetic moment (B.M.))(A) Ti

^{3 +}

(I) 3.87(B) V

^{2 +}

(II) 0.00(C) Ni

^{2 +}

(III) 1.73(D) Sc

^{3 +}

(IV) 2.84 Choose the correct answer from the options given below :

A(A)-(III), (B)-(I), (C)-(II), (D)-(IV)

B(A)-(III), (B)-(I), (C)-(IV), (D)-(II)

C(A)-(IV), (B)-(II), (C)-(III), (D)-(I)

D(A)-(II), (B)-(IV), (C)-(I), (D)-(III)

Answer key (B) & worked solution

Correct answer: Option B

^{+ 3}

= 3dº ∴ μ

_{s p in}

= 0 V

^{+ 2}

= 3d

^{3}

∴ μ

_{s p in}

= 3.87 B.M. Ni

^{+ 2}

= 3d

^{8}

∴ μ

_{s p in}

= 2.84 B.M. Ti

^{+ 3}

= 3d

^{1}

∴ μ

_{s p in}

= 1.73 B.M.

Q5MathematicsJEE Main 2025, 24 Jan Shift 2Intervals of monotonocity

Let (2, 3) be the largest open interval in which the function f(x) = 2 log

_{e}

(x – 2) – x

^{2}

+ ax + 1 is strictly increasing and (b, c) be the largest open interval, in which the function g(x) = (x – 1)

^{3}

(x + 2 – a)

^{2}

is strictly decreasing. Then 100(a + b – c) is equal to:

A280

B360

C420

D160

Answer key (B) & worked solution

Correct answer: Option B

f^{'} (x) = \frac{2}{x - 2} - 2 x + a ⩾ 0

f^{''} (x) = \frac{- 2}{( x - 2 ) ^{2}} - 2 < 0

f^{'} (x) ↓

f^{'} (3) ⩾ 0

2 - 6 + a ⩾ 0

a ⩾ 4

a_{m i n} = 4

g (x) = (x - 1)^{3} (x + 2 - a)^{2}

g (x) = (x - 1)^{3} (x - 2)^{2}

g^{'} (x) = (x - 1)^{3} 2 (x - 2) + (x - 2)^{2} 3 (x - 1)^{2}

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

= (x - 1)^{2} (x - 2) (5 x - 8) < 0

x \in (\frac{8}{5}, 2)

100 (a + b - c) = 100 (4 + \frac{8}{5} - 2) = 360

Q6MathematicsJEE Main 2025, 24 Jan Shift 2General term of binomial expansion

Suppose A and B are the coefficients of 30

^{t h}

and 12

^{t h}

terms respectively in the binomial expansion of

(1 + x)^{2 n - 1} .

2 A = 5 B,

then n is equal to:

A22

B21

C20

D19

Answer key (B) & worked solution

Correct answer: Option B

A =^{2 n - 1} C_{29}

B =^{2 n - 1} C_{11}

2^{2 n - 1} C_{29} = 5^{2 n - 1} C_{11}

2 \frac{( 2 n - 1 ) !}{29 ! ( 2 n - 30 ) !} = 5 \frac{( 2 n - 1 ) !}{( 2 n - 12 ) ! 11 !}

\frac{1}{29....12 \cdot 5} = \frac{1}{( 2 n - 12 ) ( 2 n - 13 ) .... ( 2 n - 29 ) 2}

\frac{1}{30 \cdot 29...12} = \frac{1}{( 2 n - 12 ) ( 2 n - 13 ) ... ( 2 n - 29 ) 12}

2 n - 12 = 30

n = 21