Updated 19 Oct 2026

JEE Main 2025, 24 Jan Shift 1 question paper with solutions

JEE Main 2025, 24 Jan Shift 1 was held on 24 January 2025. The bank holds 67 of its questions (Physics 23, 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

67 Questions in bank for this shift

Try questions from JEE Main 2025, 24 Jan Shift 1

Q1PhysicsJEE Main 2025, 24 Jan Shift 1Average and RMS value of a function

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An alternating current is given by

I = I_{A} sin ω t + I_{B} cos ω t .

The r.m.s. current will be:

Answer key (C) & worked solution

Correct answer: Option C

i_{r m s} = \frac{\int I ^{2} d t}{\int d t}

\frac{I _{A}^{2} + I _{B}^{2}}{2} = i_{r m s}

Q2PhysicsJEE Main 2025, 24 Jan Shift 1Energy of a capacitor

Consider a parallel plate capacitor of area A (of each plate) and separation 'd' between the plates. If E is the electric field and ε

_{0}

is the permittivity of free space between the plates, then potential energy stored in the capacitor is :-

\frac{1}{2} ε_{0} E^{2} A d

\frac{3}{4} ε_{0} E^{2} A d

\frac{1}{4} ε_{0} E^{2} A d

ε_{0} E^{2} A d

Answer key (A) & worked solution

Correct answer: Option A

\frac{U}{V} = \frac{1}{2} ε_{0} E^{2}

U = \frac{1}{2} ε_{0} E^{2} V

= \frac{1}{2} ε_{0} E^{2} (A d)

Q3ChemistryJEE Main 2025, 24 Jan Shift 1Potassium Permanganate

Preparation of potassium permanganate from MnO

_{2}

involves two step process in which the 1

^{s t}

step is a reaction with KOH and KNO

_{3}

to produce

_{4}

[Mn(OH)

_{6}

]

_{3}

MnO

_{4}

CKMnO

_{4}

_{2}

MnO

_{4}

Answer key (D) & worked solution

Correct answer: Option D

M n O_{2} K O H K N O_{3}, Δ K_{2} M n O_{4}

Q4ChemistryJEE Main 2025, 24 Jan Shift 1General Characteristics of f-block Elements

Which of the following ions is the strongest oxidizing agent ? (Atomic Number of Ce = 58, Eu = 63, Tb = 65, Lu = 71]

ALu

^{3 +}

BEu

^{2 +}

CTb

^{4 +}

DCe

^{3 +}

Answer key (C) & worked solution

Correct answer: Option C

^{4 +}

is strongest oxidising agent,

The common oxidation state of Terbium is +3.

indicating that it can also act as a good oxidizing agent.

Q5MathematicsJEE Main 2025, 24 Jan Shift 1Plotting region and area under curves

The area of the region {(x, y) : x

^{2}

+ 4x + 2 ≤ y ≤ |x + 2|} is equal to

B24/5

C20/3

Answer key (C) & worked solution

Correct answer: Option C

^{2}

+ 4x +2 ≤ y ≤ |x + 2| The area bounded between y = x

^{2}

+ 4x + 2 = (x + 2)

^{2}

–2 and y = |x + 2| is same as area bounded between y = x

^{2}

– 2 and y = |x| For P.O.I |x|

^{2}

– 2 = |x| ⇒ |x| = 2 ⇒ x = ± 2 ∴ Required area =

- - 2 \int 2 (x^{2} - 2) d x + - 2 \int 2 ∣ x ∣ d x

= - 2 - 2 \int 2 (x^{2} - 2) d x + - 2 \int 2 x . d x

= - 2 [\frac{x ^{3}}{3} - 2 x]_{0}^{2} + [\frac{x ^{2}}{2}]_{0}^{2}

= - 2 [\frac{8}{3} - 4] + 2 [\frac{4}{2}]

= - 2 \times (\frac{- 4}{3}) + 4

= \frac{20}{3}

Q6MathematicsJEE Main 2025, 24 Jan Shift 1Binomial coefficient problems

For some

n \neq = 10

, let the coefficients of the 5

^{t h}

, 6

^{t h}

and 7

^{t h}

terms in the binomial expansion of (1 + x)

^{n + 4}

be in A.P. Then the largest coefficient in the expansion of (1 + x)

^{n + 4}

is :

A70

B35

C20

D10

Answer key (B) & worked solution

Correct answer: Option B

(1 + x)

^{n + 4}

^{n + 4}

_{4}

^{n + 4}

_{5}

^{n + 4}

_{6}

, → A.P. ⇒ 2 ×

^{n + 4}

_{5}

^{n + 4}

_{4}

^{n + 4}

_{6}

⇒ 4 ×

^{n + 4}

_{5}

= (

^{n + 4}

_{4}

^{n + 4}

_{5}

) +(

^{n + 4}

_{5}

^{n + 4}

_{6}

) ⇒ 4 ×

^{n + 4}

_{5}

^{n + 5}

_{5}

^{n + 5}

_{6}

\Rightarrow 4 \times \frac{( n + 4 )!}{5 ! ( n - 1 )!} = \frac{( n + 6 )!}{6 ! . n !}

\Rightarrow 4 = \frac{( n + 6 ) ( n + 5 )}{6 n}

⇒ n

^{2}

+ 11n + 30 = 24 n ⇒ n

^{2}

– 13 n + 30 = 0 ⇒ n = 3, 10(rejected)

∵ n \neq = 10

∴ Largest binomial coefficient in expansion of (1 + x)

^{7}

(

∵

n + 4 = 7) is coeff. of middle term ⇒

^{7}

_{4}

^{7}

_{3}

= 35 N.T.A. Ans Option (2)