Updated 19 Oct 2026

JEE Main 2026, 22 Jan Shift 2 question paper with solutions

JEE Main 2026, 22 Jan Shift 2 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 2

Q1PhysicsJEE Main 2026, 22 Jan Shift 2Bohr's atomic model

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The smallest wavelength of Lyman series is 91 nm . The difference between the largest wavelengths of Paschen and Balmer series is nearly

____

nm .

Answer key (C) & worked solution

Correct answer: Option C

n = 4

n = 3

- paschen

n = 2

- Balmer

n = 1

Lyman Smallest wavelength of lyman

\frac{1}{λ} = R (\frac{1}{12} - \frac{1}{\infty ^{2}})

R = \frac{1}{λ} = \frac{1}{91} nm^{- 1}

λ_{max}

for balmer series

n_{1} = 2 \to n_{2} = 3

\frac{1}{λ _{B}} = R (\frac{1}{4} - \frac{1}{9})

\frac{1}{λ _{B}} = \frac{1}{91} (\frac{5}{36})

λ_{B} = (\frac{91 \times 36}{5}) = 655.2 nm

λ_{max}

paschen

n_{1} = 3 \to n_{2} = 4

\frac{1}{λ _{p}} = \frac{1}{91} (\frac{1}{3 ^{2}} - \frac{1}{4 ^{2}}) = \frac{1}{91} \times \frac{7}{144}

λ_{p} = (\frac{91 \times 144}{7}) = 1872 nm

Δ λ = λ_{P} - λ_{B} = 1872 - 655.2

Δ λ = 1216.8

Δ λ \approx 1217

Q2PhysicsJEE Main 2026, 22 Jan Shift 2Motion of centre of mass

Given below are two statements : Statement I: For a mechanical system of many particles total kinetic energy is the sum of kinetic energies of all the particles. Statement II : The total kinetic energy can be the sum of kinetic energy of the center of mass w.r.t. to the origin and the kinetic energy of all the particles w.r.t. the center of mass as the reference. In the light of the above statements, choose the correct answer from the options given below :

ABoth Statement I and Statement II are true

BStatement I is true but Statement II is false

CStatement I is false but Statement II is true

DBoth Statement I and Statement II are false

Answer key (A) & worked solution

Correct answer: Option A

K E = \frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2}

KE = \frac{1}{2} (m_{1} + m_{2}) v_{cm}^{2} + \frac{1}{2} \frac{m _{1} m _{2}}{m _{1} + m _{2}} v_{1} - v_{2}^{2}

Q3ChemistryJEE Main 2026, 22 Jan Shift 2Electrochemical Series and Its Applications

Consider the following reduction processes :

Al^{3 +} + 3 e^{-} \to Al (s), E^{\circ} = - 1.66 V

Fe^{3 +} + e^{-} \to Fe^{2 +}, E^{0} = + 0.77 V

Co^{3 +} + e^{-} \to Co^{2 +}, E^{\circ} = + 1.81 V

Cr^{3 +} + 3 e^{-} \to Cr (s), E^{\circ} = - 0.74 V

The tendency to act as reducing agent decreases in the order :

Al > Cr > Fe^{2 +} > Co^{2 +}

Al > Fe^{2 +} > Cr > Co^{2 +}

Al > Cr > Co^{2 +} > Fe^{2 +}

Cr > Fe^{2 +} > Al > Co^{2 +}

Answer key (A) & worked solution

Correct answer: Option A

Reducing power

\propto \frac{1}{Reduction potential}

Q4ChemistryJEE Main 2026, 22 Jan Shift 2Ionization Enthalpy

Given below are two statements : Statement-I: The first ionization enthalpy of Cr is lower than that of Mn. Statement-II: The second and third ionization enthalpies of Cr are higher than those of Mn. In the light of the above statements, choose the correct answer from the options given below :

ABoth Statement-I and Statement-II are false.

BStatement-I is true but Statement-II is false.

CBoth Statement-I and Statement-II are true.

DStatement-I is false but Statement-II is true.

Answer key (B) & worked solution

Correct answer: Option B

Cr = (Ar) 3 d^{5} 4 s^{1}

Mn = (Ar) 3 d^{5} 4 s^{2}

IE_{1} (Cr) IE_{2} (Mn)

IE_{3} (Cr) < IE_{3} (Mn)

Q5MathematicsJEE Main 2026, 22 Jan Shift 2Equation of lines in 3D and vector form

Let L be the line

\frac{x + 1}{2} = \frac{y + 1}{3} = \frac{z + 3}{6}

and let S be the set of all points

(a, b, c)

on L , whose distance from the line

\frac{x + 1}{2} = \frac{y + 1}{3} = \frac{z + 9}{0}

along the line L is 7 . Then

\sum_{(a, b, c) \in S} (a + b + c)

is equal to :

A34

B28

C40

Answer key (A) & worked solution

Correct answer: Option A

M

is the point of intersection of

L_{1} & L_{2}

\Rightarrow 2 λ - 1 = 2 μ - 1, 3 λ - 1 = 3 μ - 1, 6 λ - 3 = 9

\Rightarrow λ = 2 = μ

\Rightarrow M (3, 5, 9)

Now let point P be

(2 K - 1, 3 K - 1, 6 K - 3)

L_{2}

such that

PM = 7

\Rightarrow (2 K - 4)^{2} + (3 K - 6)^{2} + (6 K - 12)^{2} = 7

\Rightarrow 49 K^{2} + 196 - 196 K = 49

\Rightarrow K^{2} + 4 - 4 K = 1

\Rightarrow K^{2} - 4 K + 3 = 0

\Rightarrow K = 1, 3

So points

P & Q

are

(1, 2, 3) & (5, 8, 15)

So sum of all co-ordinates of

P & Q = 34

Q6MathematicsJEE Main 2026, 22 Jan Shift 2Basics of greatest integer function

Let

f (x) = [x]^{2} - [x + 3] - 3, x \in R

where

[∙]

is the greatest integer function. Then

f (x) > 0

only for

x \in [4, \infty)

f (x) < 0

only for

x \in [- 1, 3)

\int_{0}^{2} f (x) dx = - 6

f (x) = 0

for finitely many values of

x

Answer key (B) & worked solution

Correct answer: Option B

f (x) = [x]^{2} - [x] - 6 = ([x] + 2) ([x] - 3)

(a)

f (x) > 0 \Rightarrow [x] \in (- \infty, - 2) \cup (3, \infty)

\Rightarrow x \in (- \infty, - 2) \cup [4, \infty)

(b)

f (x) < 0 \Rightarrow [x] \in (- 2, 3)

\Rightarrow x \in [- 1, 3)

option (2) is correct (c)

\int_{0}^{2} f (x) d x = \int_{0}^{1} (0 - 0 - 6) d x + \int_{1}^{2} (1 - 1 - 6) d x

= - 6 - 6

= - 12

(d)

f (x) = 0 \Rightarrow [x] = 3

[x] = - 2

infinitely many solutions