Updated 19 Oct 2026

JEE Main 2026, 22 Jan Shift 1 question paper with solutions

JEE Main 2026, 22 Jan Shift 1 was held on 22 January 2026. 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 2026 shifts Marks vs percentile table

2026 Exam session year

300 Maximum marks (+4 / −1)

180 min Official duration

74 Questions in bank for this shift

Try questions from JEE Main 2026, 22 Jan Shift 1

Q1PhysicsJEE Main 2026, 22 Jan Shift 1Relation between electric field and electric potential

Try right here · No sign-in

Electric field in a region is given by

E = A x i^+ B y j^

, where

A = 10 V / m^{2}

and

B = 5 V / m^{2}

. If the electric potential at a point

(10, 20)

is 500 V , then the electric potential at origin is

____

Answer key (C) & worked solution

Correct answer: Option C

E = 10 x i^+ 5 y j^

V_{at (10, 20)} = 500 V Δ V - \int E \cdot d r

500 - V_{0} = - \int_{(0, 0)}^{(10, 20)} (10 x i^+ 5 y j^) \cdot (d x i^+ d y j^)

500 - V_{0} = - [5 x^{2} + \frac{5 y ^{2}}{2}]_{(0, 0)}^{(10, 20)}

V_{0} - 500 = (500 + 5 \times \frac{400}{2}) - (0 - 0)

V_{0} - 500 = 500 + 1000

V_{0} = 2000 V

Q2PhysicsJEE Main 2026, 22 Jan Shift 1Surface tension

Given below are two statements: Statement I: Pressure of fluid is exerted only on a solid surface in contact as the fluid-pressure does not exist everywhere in a still fluid. Statement II: Excess potential energy of the molecules on the surface of a liquid, when compared to interior, results in surface tension. In the light of the above statements, choose the correct answer from the options given below.

AStatement I is true but Statement II is false

BBoth Statement I and Statement II are false

CBoth Statement I and Statement II are true

DStatement I is false but Statement II is true

Answer key (D) & worked solution

Correct answer: Option D

According to pascal's law pressure at any point in liquid at rest is same in all direction. It exist at every point in the liquid not just at boundaries. So statement (1) is false. For interior molecule net cohesive forces are zero statement (2) is correct.

Q3ChemistryJEE Main 2026, 22 Jan Shift 1Chemical Properties of Carbonyl Compounds

Match the LIST-I with LIST-II List-I (Reagents)List-II (Name of Reaction involving carbonyl compound)A.

NH_{2} - NH_{2}, KOH

I. Tollen's TestB.

Ag (NH_{3})_{2} OH

II. Clemmensen ReductionC. Aq. CuSO

_{4}

, Sodium Potassium tartarate, KOHIII. Wolff-Kishner ReductionD.

Zn - Hg, HCl

IV. Fehling's Test Choose the correct answer from the options given below

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

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

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

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

Answer key (A) & worked solution

Correct answer: Option A

Theoretical (NCERT Based)

Q4ChemistryJEE Main 2026, 22 Jan Shift 1Nucleophiles

The correct order of reactivity of

CH_{3} Br

in methanol with the following nucleophiles is

F^{-}, I^{-}, C_{2} H_{5} O^{-}

and

C_{6} H_{5} O^{-}

I^{-} > C_{6} H_{5} O^{-} > F^{-} > C_{2} H_{5} O^{-}

I^{-} > C_{2} H_{5} O^{-} > C_{6} H_{5} O^{-} > F^{-}

I^{-} > C_{2} H_{5} O^{-} > F^{-} > C_{6} H_{5} O^{-}

I^{-} > F^{-} > C_{6} H_{5} O^{-} > C_{2} H_{5} O^{-}

Answer key (B) & worked solution

Correct answer: Option B

Order of nucleophilicity :

I^{⊖} > C_{2} H_{5} O^{⊖} > PhO^{⊖} > F^{⊖}

Q5MathematicsJEE Main 2026, 22 Jan Shift 1Shortest distance between two lines

Let

P (α, β, γ)

be the point on the line

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

at a distance

414

from the point

(1, - 1, 0)

and nearer to the origin. Then the shortest distance, between the lines

\frac{x - α}{1} = \frac{y - β}{2} = \frac{z - γ}{3}

and

\frac{x + 5}{2} = \frac{y - 10}{1} = \frac{z - 3}{1}

, is equal to

7 \frac{5}{4}

4 \frac{7}{5}

4 \frac{5}{7}

2 \frac{7}{4}

Answer key (B) & worked solution

Correct answer: Option B

Let

P (2 λ + 1, - 3 λ - 1, λ)

Then

4 λ^{2} + 9 λ^{2} + λ^{2} = 16 \cdot 14 \Rightarrow λ = \pm 4 \Rightarrow - 4

(nearer to origin)

∴ P (- 7, 11, - 4)

∴

Shortest distance

= \frac{2 1 2 - 1 2 1 7 3 1}{i ^ 1 2 j ^ 2 1 k 3 1}

= \frac{28}{1 + 25 + 9} = \frac{4 7}{5}

Q6MathematicsJEE Main 2026, 22 Jan Shift 1global extreme values in an interval

Let

f (x) = x^{2025} - x^{2000}, x \in [0, 1]

and the minimum value of the function

f (x)

in the interval

[0, 1]

(80)^{80} (n)^{- 81}

. Then n is equal to

A-81

B-40

C-41

D-80

Answer key (A) & worked solution

Correct answer: Option A

f (x) = x^{2025} - x^{2000}

f^{'} (x) = 0 \Rightarrow x = (\frac{2000}{2025})^{1/25} = α (

say

)

∴ f (0) = 0, f (1) = 0, f (α) = (\frac{80}{81})^{80} \cdot \frac{- 1}{81} = 8 0^{80} \cdot (- 81)^{- 81}