Updated 19 Oct 2026

JEE Main 2025, 23 Jan Shift 1 question paper with solutions

JEE Main 2025, 23 Jan Shift 1 was held on 23 January 2025. The bank holds 69 of its questions (Physics 25, 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

69 Questions in bank for this shift

Try questions from JEE Main 2025, 23 Jan Shift 1

Q1PhysicsJEE Main 2025, 23 Jan Shift 1Matter waves

Try right here · No sign-in

A sub-atomic particle of mass 10–30 kg is moving with a velocity 2.21 × 10

^{6}

m/s. Under the matter wave consideration, the particle will behave closely like ______. (h = 6.63 × 10–34 J.s)

Answer key (B) & worked solution

Correct answer: Option B

λ = \frac{h}{p} = \frac{6.63 \times 10 ^{- 34}}{10 ^{- 30} \times 2.21 \times 10 ^{6}}

= 3 \times 1 0^{- 10} m

Hence particle will behave as x-ray.

Q2PhysicsJEE Main 2025, 23 Jan Shift 1Electric flux

The electric flux is

ϕ = α σ + β λ

where

λ an d σ

are linear and surface charge density, respectively,

(\frac{α}{β})

represents

Acharge

Belectric field

Cdisplacement

Darea

Answer key (C) & worked solution

Correct answer: Option C

ϕ = α σ + β λ

[ϕ] = [α σ] + [β λ]

[α] = \frac{[ ϕ ]}{[ σ ]} [\frac{α}{β}] = \frac{[ λ ]}{[ σ ]}

[β] = \frac{[ ϕ ]}{[ λ ]} = \frac{[ Q / L ]}{[ Q / A r e a ]} = [\frac{A r e a}{L e n g t h}]

[\frac{α}{β}] = L

Q3ChemistryJEE Main 2025, 23 Jan Shift 1Preparation of Alkyl Halides

Match the LIST-I with LIST-II LIST-IName reactionLIST-IIProduct obtainableA. Swarts reactionI. Ethyl benzeneB. Sandmeyer’s reactionII. Ethyl iodideC. Wurtz Fittig reactionIII. CyanobenzeneD. Finkelstein reactionIV. Ethyl fluoride Choose the correct answer from the options given below:

AA-II, B-III, C-I, D-IV

BA-IV, B-I, C-III, D-II

CA-IV, B-III, C-I, D-II

DA-II, B-I, C-III, D-IV

Answer key (C) & worked solution

Correct answer: Option C

LIST-IName reactionLIST-IIProduct obtainableA.Swarts reactionI.

E t - I K F D M F E t - F

B.Sandmeyer’s reactionII.

P h N_{2}^{\oplus} C l^{-} C u C N / K C N P h C N + N_{2}

C.Wurtz Fittig reactionIII.

P h - C l + E tC l N a e t h er P h - E t + P h - P h + E t - E t

D.Finkelstein reactionIV.

E t - C l N a I a ce t o n e E t - I + N a C l

Q4ChemistryJEE Main 2025, 23 Jan Shift 1Lewis Dot Structures

Match the LIST-I with LIST-II LIST-I(Classification of moleculesbased on octet rule)LIST-II(Example)A. Molecules obeying octet ruleI. NO, NO

_{2}

B. Molecules with incomplete octetII. BCl

_{3}

, AlCl

_{3}

C. Molecules with incomplete octet with odd electronIII. H

_{2}

_{4}

, PCl

_{5}

D. Molecules with expanded octetIV. CCl

_{4}

, CO

_{2}

Choose the correct answer from the options given below :

AA-IV, B-II, C-I, D-III

BA-III, B-II, C-I, D-IV

CA-IV, B-I, C-III, D-II

DA-II, B-IV, C-III, D-I

Answer key (A) & worked solution

Correct answer: Option A

(A) A → IV (B) B → II (C) C → I (D) D → III

Q5MathematicsJEE Main 2025, 23 Jan Shift 1Modulus of complex number & its properties

Let

\frac{z - i}{2 z + i} = \frac{1}{3}, z \in C

, be the equation of a circle with center at C. If the area of the triangle, whose vertices are at the points (0, 0), C and

(α, 0)

is 11 square units, then

α^{2}

equals

A100

B50

\frac{121}{25}

\frac{81}{25}

Answer key (A) & worked solution

Correct answer: Option A

\frac{z - i}{2 z + i} = \frac{1}{3}

\frac{z - i}{z + \frac{i}{2}} = \frac{2}{3}

3 ∣ x - i y - i ∣ = 2 x - i y + \frac{i}{2}

9 (x^{2} + (y + 1)^{2}) = 4 (x^{2} + (y - 1/3)^{2})

9 x^{2} + 9 y^{2} + 18 y + 9 = 4 x^{2} + 4 y^{2} - 4 y + 1

5 x^{2} + 5 y^{2} + 22 y + 8 = 0

x^{2} + y^{2} + \frac{22}{5} y + \frac{8}{5} = 0

centre \Rightarrow (0, - \frac{11}{5})

\left| {\frac{1}{2}\left| {\begin{array}{*{20}{c}} 0&0&1 \\ 0&{ - 11/5}&1 \\ \alpha &0&1 \end{array}} \right|} \right| = 11

\Rightarrow (- \frac{11}{5} α)^{2} = (11 \times 2)^{2}

\Rightarrow α^{2} = 100

Q6MathematicsJEE Main 2025, 23 Jan Shift 1Continuity at a point

If the function

f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {\frac{2}{x}\left\{ {\sin \left( {{k_1} + 1} \right)x + \sin \left( {{k_2} - 1} \right)x} \right\}}&,&{x 0} \end{array}} \right.

is continuous at x = 0, then

k_{1}^{2} + k_{2}^{2}

is equal to

B20

D10

Answer key (D) & worked solution

Correct answer: Option D

x \to 0^{-} lim \frac{2}{x} {sin (k_{1} + 1) x + sin (k_{2} - 1) x} = 4

\Rightarrow 2 (k_{1} + 1) + 2 (k_{2} - 1) = 4

\Rightarrow k_{1} + k_{2} = 2

\Rightarrow x \to 0^{+} lim \frac{2}{x} ln (\frac{2 + k _{1} x}{2 + k _{2} x}) = 4

\Rightarrow x \to 0^{+} lim \frac{1}{x} ln (1 + \frac{( k _{1} - k _{2} ) x}{2 + k _{2} x}) = 2

\Rightarrow \frac{k _{1} - k _{2}}{2} = 2

\Rightarrow k_{1} - k_{2} = 4

∴ k_{1} = 3, k_{2} = - 1

k_{1}^{2} + k_{2}^{2} = 9 + 1 = 10