Updated 19 Oct 2026

JEE Main 2025, 7 Apr Shift 1 question paper with solutions

JEE Main 2025, 7 Apr Shift 1 was held on 7 April 2025. The bank holds 72 of its questions (Physics 23, Chemistry 24, 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

72 Questions in bank for this shift

Try questions from JEE Main 2025, 7 Apr Shift 1

Q1PhysicsJEE Main 2025, 7 Apr Shift 1Average and RMS value of a function

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An ac current is represented as

i = 52 + 10 cos (650 π t + \frac{π}{6}) A m p

The r.m.s value of the current is

Answer key (C) & worked solution

Correct answer: Option C

i = 52 + 10 cos (650 π t + \frac{π}{6})

i^{2} = 50 + 100 cos^{2} (650 π t + \frac{π}{6})

+ (2) (52) (10) cos (650 π t + \frac{π}{6})

= 50 + \frac{100}{2} + 0

= 100

= 10 Amp.

Q2PhysicsJEE Main 2025, 7 Apr Shift 1Energy, radius and speed of electron in hydrogen like elements

In a hydrogen like ion, the energy difference between the 2

^{n d}

excitation energy state and ground is 108.8 eV. The atomic number of the ion is

Answer key (D) & worked solution

Correct answer: Option D

Δ E = 13.6 z^{2} [\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}}]

(13.6) z^{2} [\frac{1}{1} - \frac{1}{9}] = 108.8

\frac{( 13.6 ) ( 8 )}{9} (z^{2}) = 108.8

z = 3

Q3ChemistryJEE Main 2025, 7 Apr Shift 1Physical Properties of Ethers

Given below are two statements : Statement I : Dimethyl ether is completely soluble in water. However, diethyl ether is soluble in water to a very small extent. Statement II : Sodium metal can be used to dry diethyl ether and not ethyl alcohol. In the light of given statements, choose the correct answer from the options given below

AStatement I is false but Statement II are true

BBoth Statement I and Statement II are false

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

St-I – Dimethyl ether is more soluble in water because it can form stronger hydrogen bonds with water molecules due to its smaller size compared to diethyl ether, which has larger ethyl groups St-II : - St. II is also correct because sodium metal is not used with ethyl alcohol as H2 gas release with ethyl a below

Q4ChemistryJEE Main 2025, 7 Apr Shift 1Carbohydrates

Given below are two statements : Statement I: D-(+)-glucose + D-(+) fructose

- H_{2} O

sucrose sucrose

H y d r o l y s i s

D-(+)-glucose + D-(+) fructose Statement II : Invert sugar is formed during sucrose hydrolysis. In the light of the above statements, choose the correct answer from the options given below -

ABoth Statement I and Statement II are true.

BStatement I is false but Statement II are true.

CStatement I is true but Statement II is false.

DBoth Statement I and Statement II are false.

Answer key (B) & worked solution

Correct answer: Option B

On hydrolysis of sucrose gives D-(+)-glucose and D-(–)-fructose while in St. (1) D-(+)-fructose is given hence St-(1) is incorrect. St. II – It is correct because sucrose on hydrolysis gives invert sugar

Q5MathematicsJEE Main 2025, 7 Apr Shift 1Maxima and Minima at a point

Let x = –1 and x = 2 be the critical points of the function f(x) = x

^{3}

+ ax

^{2}

+ b log

_{e}

|x| + 1, x

\neq =

0. Let m and M respectively be the absolute minimum and the absolute maximum values of f in the interval

[- 2, - \frac{1}{2}]

. Then |M + m| is equal to (Take log

_{e}

2 = 0.7):

A21.1

B19.8

C22.1

D20.9

Answer key (A) & worked solution

Correct answer: Option A

f (x) = x^{3} + a x^{2} + b ℓ n ∣ x ∣ + 1,

x \neq = 0

f^{'} (x) = 3 x^{2} + 2 a x + \frac{b}{x}

f^{'} (- 1) = 3 - 2 a - b = 0

f^{'} (- 2) = 12 + 4 a - \frac{b}{2} = 0

a = \frac{- 9}{2}, b = 12

f^{'} (x) = 3 x^{2} - 9 x + \frac{12}{x} = \frac{3 ( x + 1 ) ( x + 2 ) ^{2}}{x}

Max. at x = –1

f (x) = x^{3} - \frac{9}{2} x^{2} + 12 ℓ n ∣ x ∣ + 1

f (- 1) = - 1 - \frac{9}{2} + 1 = \frac{9}{2}

M = –4.5 Min. value at x = –2

f (- 2) = - 8 - 18 + 12 ℓ n 2 + 1

m = - 25 + 12 ℓ n 2 = - 16.6

∣ M + m ∣ = 21.1

Q6MathematicsJEE Main 2025, 7 Apr Shift 1Application of binomial theorem

The remainder when

((64)^{(64)})^{(64)}

is divided by 7 is equal to

Answer key (B) & worked solution

Correct answer: Option B

Let N =

((64)^{(64)})^{(64)}

N = (64)^{64^{2}}

N = (1 + 63)^{64^{2}}, l e t 6 4^{2} = n

Expanding by binomial N = (1 + 63)

^{n}

= 1 +

^{n}

_{1}

63 +

^{n}

_{2}

(63)

^{2}

+ …..

= 1 + 63 λ = 1 + 7 (9 λ)

Remainder when divided by 7 is 1