Updated 19 Oct 2026

JEE Main 2025, 4 Apr Shift 1 question paper with solutions

JEE Main 2025, 4 Apr Shift 1 was held on 4 April 2025. The bank holds 74 of its questions (Physics 25, 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

74 Questions in bank for this shift

Try questions from JEE Main 2025, 4 Apr Shift 1

Q1PhysicsJEE Main 2025, 4 Apr Shift 1Average and RMS value of a function

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An alternating current is represented by the equation,

i = 1002 sin (100 π t)

ampere. The RMS value of current and the frequency of the given alternating current are

Answer key (C) & worked solution

Correct answer: Option C

i_{r} = \frac{i _{0}}{2} = 100 A

f = \frac{w}{2 π} = \frac{100 π}{2 π} = 50 H z

Q2PhysicsJEE Main 2025, 4 Apr Shift 1Energy, radius and speed of electron in hydrogen like elements

Considering the Bohr model of hydrogen like atoms, the ratio of the ratio of the radius 5

^{t h}

orbit of the electron in Li

^{2 +}

and He

^{+}

\frac{3}{2}

\frac{4}{9}

\frac{9}{4}

\frac{2}{3}

Answer key (D) & worked solution

Correct answer: Option D

r = r . \frac{n ^{2}}{2}

f or L i^{2 +}

r_{5} = r_{0} . \frac{25}{3}

f or H e^{+}

r_{5} = r_{0} . \frac{25}{2}

∴ \frac{r _{L i^{2 +}}}{r _{H e^{+}}} = \frac{2}{3}

Q3ChemistryJEE Main 2025, 4 Apr Shift 1Molecular Orbital Theory

Which of the following molecules(s) show/s paramagnetic behavior ? (A) O

_{2}

(B) N

_{2}

_{2}

(D) S

_{2}

(E) Cl

_{2}

Choose the correct answer from the options given below :

AB only

BA & C only

CA & E only

DA & D only

Answer key (D) & worked solution

Correct answer: Option D

No. of unpaired e− (A) O

_{2}

⇒ 2 (B) N

_{2}

⇒ 0 (C) F

_{2}

⇒ 0 (D) S

_{2}

⇒ 2 (E) Cl

_{2}

⇒ 0 If species contain unpaired electron than it is paramagnetic So A & D are paramagnetic

Q4ChemistryJEE Main 2025, 4 Apr Shift 1Rate Law

Rate law for a reaction between A and B is given by R = k [A]

^{n}

[B]

^{m}

If concentration of A is doubled and concentration of B is halved from their initial value, the ratio of new rate of reaction to the initial rate of reaction

(\frac{r _{2}}{r _{1}})

A2(n-m)

B(n − m)

C(m + n)

\frac{1}{2 ^{m + n}}

Answer key (A) & worked solution

Correct answer: Option A

_{1}

= k[A]

^{n}

[B]

^{m}

Now A is doubled & B is halved in concentration

\Rightarrow r_{2} = k 2^{n} [A]^{n} . \frac{[ B ] ^{m}}{2 ^{m}}

Now

\frac{r _{2}}{r _{1}} = 2^{(n - m)}

Q5MathematicsJEE Main 2025, 4 Apr Shift 1Shortest distance between two lines

Let the shortest distance between the lines

\frac{x - 3}{3} = \frac{y - α}{- 1} = \frac{z - 3}{1} an d \frac{x + 3}{- 3} = \frac{y + 7}{2} = \frac{z - β}{4}

330

. Then the positive value of 5α + β is

A42

B46

C48

D40

Answer key (B) & worked solution

Correct answer: Option B

A(3, α , 3) & B(–3, –7, β )

B A = 6 i^+ (α + 7) j^+ (3 - β) k^

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

\frac{B A . ( p \times q )}{∣ p \times q ∣} = 330

36 + 15 (α + 7) - 3 (3 - β) = (330)^{2}

36 + 15 α + 105 - 9 + 3 β = 270

15 α + 3 β = 138

5α + β = 46

Q6MathematicsJEE Main 2025, 4 Apr Shift 1General term of binomial expansion

In the expansion of

(32 + \frac{1}{3 3})^{n}, n \in N,

if the ratio of 15

^{t h}

term from the beginning to the 15

^{t h}

term from the end is

\frac{1}{6}

, then the value of

^{n} C_{3}

A4060

B1040

C2300

D4960

Answer key (C) & worked solution

Correct answer: Option C

T_{r + 1}^{n} C_{r} (2^{1/3})^{n - r} (\frac{1}{3 ^{1/3}})^{r}

r = 14

T_{15}^{n} C_{14} (2^{1/3})^{n - 14} (\frac{1}{3 ^{1/3}})^{14}

T_{15}^{'} =

^{t h}

term from last is

(n - 13)^{th}

term from beginning.

T_{15}^{1} =^{n} C_{n - 14} (2^{1/3})^{14} (\frac{1}{3 ^{1/3}})^{n - 14}

\Rightarrow \frac{T _{15}}{T _{15}} = \frac{^{n} C _{14} ( 2 ^{1/3} ) ^{n - 14} ( \frac{1}{3 ^{1/3}} ) ^{14}}{^{n} C _{n - 14} ( 2 ^{1/3} ) ^{14} ( \frac{1}{3 ^{1/3}} ) ^{n - 14}} = \frac{1}{6}

= (2^{1/3})^{n - 28} (3^{1/3})^{n - 28} = \frac{1}{6}

= 6^{\frac{n - 28}{3}} = 6^{- 1}

= n = 25

S o,^{n} C_{3} =^{25} C_{3} = 2300