Updated 19 Oct 2026

JEE Main 2025, 22 Jan Shift 2 question paper with solutions

JEE Main 2025, 22 Jan Shift 2 was held on 22 January 2025. The bank holds 70 of its questions (Physics 25, Chemistry 21, 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, 22 Jan Shift 2

Q1PhysicsJEE Main 2025, 22 Jan Shift 2Resonance in LCR circuits

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A series LCR circuit is connected to an alternating source of emf E. The current amplitude at resonant frequency is I

_{0}

. If the value of resistance R becomes twice of its initial value then amplitude of current at resonance will be

Answer key (B) & worked solution

Correct answer: Option B

Initially,

I_{0} = \frac{ε _{m}}{R}

Finally,

I_{0}^{1} = \frac{ε _{m}}{2 R} = \frac{I _{0}}{2}

Q2PhysicsJEE Main 2025, 22 Jan Shift 2Cross product of vectors

The torque due to the force

(2 i + j + 2 k)

about the origin, acting on a particle whose position vector is

(i + j + k)

, would be

i + j + k

i + k

i - k

j - k

Answer key (C) & worked solution

Correct answer: Option C

\overrightarrow \tau = \overrightarrow r \times \overrightarrow F = \left| {\begin{array}{*{20}{c}} {\overrightarrow i }&{\overrightarrow j }&{\overrightarrow k } \\ 1&1&1 \\ 2&1&2 \end{array}} \right| = \overrightarrow i - 0\overrightarrow j - \overrightarrow k

Q3ChemistryJEE Main 2025, 22 Jan Shift 2Dipole Moment

Arrange the following compounds in increasing order of their dipole moment : HBr, H

_{2}

S, NF

_{3}

and CHCl

_{3}

ANF

_{3}

< HBr < H

_{2}

S < CHCl

_{3}

BHBr < H

_{2}

S < NF

_{3}

< CHCl

_{3}

_{2}

S < HBr < NF

_{3}

< CHCl

_{3}

DCHCl

_{3}

< NF

_{3}

< HBr < H

_{2}

Answer key (A) & worked solution

Correct answer: Option A

Increasing order of Dipole moment NF

_{3}

< HBr < H

_{2}

S < CHCl

_{3}

μ

= 0.24D, 0.79D, 0.95D ,1.04D It is NCERT Data Based

Q4ChemistryJEE Main 2025, 22 Jan Shift 2Crystal Field Theory

Identify the homoleptic complex(es) that is/are low spin. (A) [Fe(CN)

_{5}

NO]2– (B) [CoF

_{6}

]3– (C) [Fe(CN)

_{6}

]4– (D) [Co(NH

_{3}

)

_{6}

]

^{3 +}

(E) [Cr(H

_{2}

_{6}

]

^{2 +}

Choose the correct answer from the options given below :

A(B) and (E) only

B(A) and (C) only

C(C) and (D) only

D(C) only

Answer key (C) & worked solution

Correct answer: Option C

(A) [Fe(CN)

_{5}

NO]–2 → Heteroleptic, Fe

^{+ 2}

-3d

^{6}

, t

_{2 g}

^{6}

_{g}

^{0}

, d

^{2}

^{3}

, Low spin (3d series + SFL) (B) [CoF

_{6}

]–3 → Homoleptic, sp

^{3}

^{2}

, High spin, Co+3 -3d

^{6}

,

t2g 4eg2 (3d series + WFL)(C) [Fe(CN)

_{6}

]–4 → Homoleptic Fe+2 -3d

^{6}

, d

^{2}

^{3}

, t

_{2 g}

^{6}

_{g}

^{0}

Low spin (3d series + SFL) (D) [Co(NH

_{3}

)

_{6}

]

^{+ 3}

→ Homoleptic, Co+3 -3d

^{6}

, d

^{2}

^{3}

, t

_{2 g}

^{6}

_{g}

^{0}

, Low spin (3d series + SFL) (E) [Cr(H

_{2}

_{6}

]

^{+ 2}

→ Homoleptic Cr

^{+ 2}

-3d

^{4}

, d

^{2}

^{3}

, High spin t

_{2 g}

^{3}

_{g}

^{1}

(3d series + WFL)

Q5MathematicsJEE Main 2025, 22 Jan Shift 2Equation of lines in 3D and vector form

Let a line pass through two distinct points P(–2, –1, 3) and Q, and be parallel to the vector

3 i^+ 2 j^+ 2 k

. If the distance of the point Q from the point R (1, 3, 3) is 5, then the square of the area of

Δ P QR

is equal to :

A136

B140

C144

D148

Answer key (A) & worked solution

Correct answer: Option A

P Q p a r a l l e l t o 3 i^+ 2 j^+ 2 k^, R (1, 3, 3)

\Rightarrow Q (3 λ - 2, 2 λ - 1, 2 λ + 3), λ \in R - {0}

QR = 5 = (3 λ - 3)^{2} + (2 λ - 4)^{2} + (2 λ)^{2}

∴ 17 λ^{2} - 34 λ + 25 = 25 \Rightarrow λ = 2 (∵ λ \neq = 0)

∴ Q (4, 3, 7), P (- 2, - 1, 3), R (1, 3, 3)

A r e a o f Δ P QR = [P QR] = \frac{1}{2} P Q \times P R

\left[ {PQR} \right] = \frac{1}{2}\left\| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k} \\ 6&4&4 \\ 3&4&0 \end{array}} \right\| = \left\| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k} \\ 3&2&2 \\ 3&4&0 \end{array}} \right\|

[P QR] = ∣ - 8 i^+ 6 j^+ 6 k^∣ = 136

∴ [P QR]^{2} = 136

Q6MathematicsJEE Main 2025, 22 Jan Shift 2General term of binomial expansion

Let

α, β, γ an d δ

be the coefficient of

x^{7}, x^{5}, x^{3}

and x respectively in the expansion of

(x + x^{3} - 1)^{5} + (x - x^{3} - 1)^{5}, x > 1.

If u and v satisfy the equations

α u + β v = 18, γ u + δ v = 20,

then u + v equals :

Answer key (A) & worked solution

Correct answer: Option A

(x + x^{3} - 1)^{5} + (x - x^{3} - 1)^{5}

= 2 {^{5} C_{0} . x^{5} +^{5} C_{2} . x^{3} (x^{3} - 1) +^{5} C_{4} . x (x^{3} - 1)^{2}}

= 2 {5 x^{7} + 10 x^{6} + x^{5} - 10 x^{4} - 10 x^{3} + 5 x}

\Rightarrow α = 10, β = 2, γ = - 20, δ = 10

Now, 10u + 2v=18 − 20u + 10v = 20

\Rightarrow u = 1, v = 4

u + v = 5