Updated 19 Oct 2026

JEE Main 2025, 8 Apr Shift 2 question paper with solutions

JEE Main 2025, 8 Apr Shift 2 was held on 8 April 2025. The bank holds 70 of its questions (Physics 22, Chemistry 24, Mathematics 24). 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

70 Questions in bank for this shift

Try questions from JEE Main 2025, 8 Apr Shift 2

Q1PhysicsJEE Main 2025, 8 Apr Shift 2Electric potential energy of a system of charges

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Two metal spheres of radius R and 3R have same surface charge density σ . If they are brought in contact and then separated, the surface charge density on smaller and bigger sphere becomes σ

_{1}

andσ

_{2}

, respectively. The ratio

\frac{σ _{1}}{σ _{2}}

Answer key (D) & worked solution

Correct answer: Option D

For conducting sphere,

V = \frac{σ r}{ε _{0}}

After contact, V

_{1}

= V

_{2}

_{1}

_{1}

= σ

_{2}

_{2}

\frac{σ _{1}}{σ _{2}} = \frac{r _{2}}{r _{1}}

\frac{σ _{1}}{σ _{2}} = 3

Q2PhysicsJEE Main 2025, 8 Apr Shift 2Mechanical equivalent of heat

Water falls from a height of 200 m into a pool. Calculate the rise in temperature of the water assuming no heat dissipation from the water in the pool. (Take g = 10 m/s

^{2}

, specific heat of water = 4200 J/(kg K))

A0 . 2 3 Κ

B0 . 3 6 Κ

C0 . 1 4 Κ

D0.48 K

Answer key (D) & worked solution

Correct answer: Option D

mgh = ms∆ T

Δ T = \frac{g h}{s} = \frac{10 \times 200}{4200} K = \frac{10}{21} K

Q3ChemistryJEE Main 2025, 8 Apr Shift 2Some Important First Order Reactions

In a first order decomposition reaction, the time taken for the decomposition of reactant to one fourth and one eighth of its initial concentration are t

_{1}

and t

_{2}

(s), respectively. The ratio t

_{1}

_{2}

will :

\frac{4}{3}

\frac{3}{2}

\frac{3}{4}

\frac{2}{3}

Answer key (D) & worked solution

Correct answer: Option D

For I

^{s t}

order reaction When C

_{t}

= C

_{o}

/4 t

_{1}

= 2t50%. when C

_{t}

= C

_{o}

/8 t

_{2}

= 3t50% so,

\frac{t _{1}}{t _{2}} = \frac{2}{3}

Q4ChemistryJEE Main 2025, 8 Apr Shift 2VBT in Coordination Compounds

The number of species from the following that are involved in sp

^{3}

^{2}

hybridization is [Co(NH

_{3}

)

_{6}

]

^{3 +}

, SF

_{6}

, [CrF

_{6}

]3–, [CoF

_{6}

]

^{3}

–,[Mn(CN)

_{6}

]3– and [MnCl

_{6}

]3–

Answer key (D) & worked solution

Correct answer: Option D

In [Co(NH3]

_{6}

]

^{3 +}

, Co

^{+ 3}

: [Ar]3d

^{6}

, NH

_{3}

is S.F.L Hybridisation state of Co

^{3 +}

is d

^{2}

^{3}

In SF

_{6}

, Hybridisation state of sulphur is sp

_{3}

^{2}

In [CrF

_{6}

]3– , Cr

^{+ 3}

: [Ar]3d

^{3}

Hybridisation state of Cr

^{3 +}

is d

^{2}

^{3}

[CoF

_{6}

]3–, Co

^{+ 3}

: [Ar]3d

^{6}

F– is W.F.L Hybridisation state of Co

^{3 +}

is sp

^{3}

^{2}

[Mn(CN)

_{6}

]3–, Mn

^{+ 3}

: [Ar]3d

^{4}

CN– is S.F.L Hybridisation state of Mn

^{3 +}

is d

^{2}

^{3}

[MnCl

_{6}

]3–, Mn

^{+ 3}

: [Ar]3d

^{4}

Cl– is W.F.L Hybridisation state of Cl– is sp

^{3}

^{2}

‘ Total number of sp

^{3}

^{2}

hybridized molecules is 3

Q5MathematicsJEE Main 2025, 8 Apr Shift 2Shortest distance between two lines

Let the values of λ for which the shortest distance between the lines

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

and

\frac{x - λ}{3} = \frac{y - 4}{4} = \frac{z - 5}{5}

\frac{1}{6}

be λ

_{1}

and λ

_{2}

. Then the radius of the circle passing through the points (0, 0), (λ

_{1}

, λ

_{2}

) and (λ

_{2}

, λ

_{1}

) is

\frac{5 2}{3}

\frac{2}{3}

Answer key (A) & worked solution

Correct answer: Option A

p = 2 i^+ 3 j^+ 4 k^, q = 3 i^+ 4 j^+ 5 k^

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

A \equiv (1, 2, 3) B \equiv (λ, 4, 5)

Shortest Distance

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

= \frac{1}{6} = \frac{( ( λ - 1 ) i ^ + 2 j ^ + 2 k ^ ) . ( - i ^ + 2 j ^ - k ^ )}{6}

\Rightarrow ∣ - λ + 1 + 4 - 2 ∣ = 1 \Rightarrow ∣ λ - 3 ∣ = 1

\Rightarrow λ = 3 \pm 1 = 4, 2

Radius of circle passing through points (0, 0), (4, 2) & (2, 4) $ = \frac{{abc}}{{4\Delta }} = \frac{{\sqrt {20} \times \sqrt {20} \times \sqrt 8 }}{{4 \times \frac{1}{2}\left| {\begin{array}{*{20}{c}} 1&1&1 \\ 0&4&2 \\ 0&2&4 \end{array}} \right|}} = \frac{{20 \times 2\sqrt 2 }}{{2 \times 12}}$

= \frac{5 2}{3}

Q6MathematicsJEE Main 2025, 8 Apr Shift 2Intervals of monotonocity

Let the function

f (x) = \frac{x}{3} + \frac{3}{x} + 3, x \neq = 0

be strictly increasing in (–∞ , α

_{1}

) U (α

_{2}

, ∞ ) and strictly decreasing in (α

_{3}

, α

_{4}

) U (α

_{4}

,α

_{5}

). Then

i = 1 \sum 5 α_{i}^{2}

is equal to

A48

B28

C40

D36

Answer key (D) & worked solution

Correct answer: Option D

f (x) = \frac{x}{3} + \frac{3}{x} + 3, x \neq = 0

f’ (x) = \frac{1}{3} - \frac{3}{x ^{2}} = 0 \Rightarrow x = \pm 3

f’ (x) = \frac{x ^{2} - 9}{3 x ^{2}}

f (x) > 0 \forall (- \infty, - 3) \cup (3, \infty) \to increasing

f'(x) < 0 ∀ (–3, 0) ∪ (0, 3) → decreasing

i = 1 \sum 5 α_{i}^{2} = (- 3)^{2} + (3)^{2} + (- 3)^{2} + (0)^{2} + (3)^{2}

= 36