Updated 19 Oct 2026

JEE Main 2026, 23 Jan Shift 1 question paper with solutions

JEE Main 2026, 23 Jan Shift 1 was held on 23 January 2026. The bank holds 73 of its questions (Physics 24, Chemistry 25, 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 2026 shifts Marks vs percentile table

2026 Exam session year

300 Maximum marks (+4 / −1)

180 min Official duration

73 Questions in bank for this shift

Try questions from JEE Main 2026, 23 Jan Shift 1

Q1PhysicsJEE Main 2026, 23 Jan Shift 1Bohr's atomic model

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In hydrogen atom spectrum, (

R \to

Rydberg's constant) A. the maximum wavelength of the radiation of Lyman series is

\frac{4}{3 R}

B. the Balmer series lies in the visible region of the spectrum C. the minimum wavelength of the radiation of Paschen series is

\frac{9}{R}

D. the minimum wavelength of Lyman series is

\frac{5}{4 R}

Choose the correct answer from the options given below:

Answer key (B) & worked solution

Correct answer: Option B

\frac{1}{λ} = R (1 - \frac{1}{4})

λ = \frac{4}{3 R}

\frac{1}{λ ^{'}} = R (\frac{1}{9})

λ^{'} = \frac{9}{R}

Q2PhysicsJEE Main 2026, 23 Jan Shift 1Matter waves

The de Broglie wavelength of an oxygen molecule at

2 7^{\circ} C

x \times 1 0^{- 12} m

. The value of x is (take Planck's constant

= 6.63 \times 1 0^{- 34} J . s

, Boltzmann constant

= 1.38 \times 1 0^{- 23} J / K

, mass of oxygen molecule

= 5.31 \times 1 0^{- 26} kg

A26

B24

C30

D20

Answer key (A) & worked solution

Correct answer: Option A

λ = \frac{h}{2 mK} = \frac{h}{2 m ( \frac{3}{2} kT )}

λ = \frac{h}{3 mkT} = \frac{6.63 \times 1 0 ^{- 34}}{3 \times 5.31 \times 1 0 ^{- 26} \times 1.38 \times 1 0 ^{- 23} \times 300} = 2.58 \times 1 0^{- 11} = 25.8 \times 1 0^{- 12}

So,

x = 26

Q3ChemistryJEE Main 2026, 23 Jan Shift 1Equilibrium Constants (Kp and Kc) and their Properties

Consider the general reaction given below at 400 K

xA (g) ⇌ yB (g)

The values of

K_{p}

and

K_{c}

are studied under the same condition of temperature but variation in x and y . (i)

K_{p} = 85.87

and

K_{c} = 2.586

appropriate units (ii)

K_{p} = 0.862

and

K_{c} = 28.62

appropriate units The value of

x

and

y

in (i) and (ii) respectively are:

A(i) 3,1 ; (ii) 3,1

B(i) 4, 1 ; (ii) 4, 1

C(i) 1,3 ; (ii) 2,1

D(i) 1, 2 ; (ii) 2, 1

Answer key (D) & worked solution

Correct answer: Option D

For reaction (i) :

K_{p} > K_{C}

Δ n_{g} > 0

y - x > 0

85.87 = 2.586 (0.0821 \times 400)^{y - x}

Solving

y - x ≃ 1

For reaction (ii) :

K_{p} < K_{C}

y - x < 0

Q4ChemistryJEE Main 2026, 23 Jan Shift 1Crystal Field Theory

Given below are two statements : Statement I:

[CoBr_{4}]^{2 -}

ion will absorb light of lower energy than

[CoCl_{4}]^{2 -}

ion. Statement II: In

[CoI_{4}]^{2 -}

ion, the energy separation between the two set of d-orbitals is more than

[CoCl_{4}]^{2 -}

ion. 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

CStatement I is false but Statement II is true

DBoth Statement I and Statement II are true

Answer key (B) & worked solution

Correct answer: Option B

Statement 1 (True) Strength of ligand :

Cl^{-} > Br^{-}

Δ_{t} : [CoCl_{4}]^{2 -} > [CoBr_{4}]^{2 -}

E_{absorbed} : [CoCl_{4}]^{2 -} > [CoBr_{4}]^{2 -}

Statement 2 (False) Strength of ligand :

I^{-} < Cl^{-}

Δ_{t} : [CoI_{4}]^{2 -} < [CoCl_{4}]^{2 -}

Q5MathematicsJEE Main 2026, 23 Jan Shift 1Equation reducible to quadratic

α

and

β (α < β)

are the roots of the equation

(- 2 + 3) (∣ x - 3∣) + (x - 6 x) + (9 - 23) = 0

x \geq 0

, then

\frac{β}{α} + α β

is equal to :

C10

D11

Answer key (C) & worked solution

Correct answer: Option C

(x - 6 x + 9) - (2 - 3) ∣ x - 3∣ - 23 = 0

\Rightarrow ∣ x - 3 ∣^{2} - (2 - 3) ∣ x - 3∣ - 23 = 0

\Rightarrow ∣ x - 3∣ = 2

∣ x - 3∣ = - 3

(not possible)

\Rightarrow x = 1

or 5

\Rightarrow x = 1

or 5

\Rightarrow α = 1

and

β = 25

Aliter: Let

x \geq 9

, let

x = t \Rightarrow t \geq 3

(3 - 2) (t - 3) + (t - 3)^{2} - 23 = 0

Let

t - 3 = u

u^{2} + (3 - 2) u - 23 = 0

u = 2

, or

u = - 3

\Rightarrow t - 3 = 2

t - 3 = - 3

\Rightarrow t = 5

t = 3 - 3

(rejected)

\Rightarrow x = 25

Now let

0 < x < 9

- (3 - 2) (t - 3) + (t - 3)^{2} - 23 = 0

let

t - 3 = u

u^{2} - (3 - 2) u - 23 = 0

\Rightarrow u = 3

u = - 2

\Rightarrow t = 3 + 3

(rejected) or

t - 3 = - 2

\Rightarrow t = 1 \Rightarrow x = 1

α = 1, β = 25

Now

\frac{β}{α} + α β = 25 + 25 = 10

Q6MathematicsJEE Main 2026, 23 Jan Shift 1Binomial coefficient problems

The value of

\frac{^{100} C _{50}}{51} + \frac{^{100} C _{51}}{52} + \dots . + \frac{^{100} C _{100}}{101}

is :

\frac{2 ^{101}}{100}

\frac{2 ^{100}}{100}

\frac{2 ^{101}}{101}

\frac{2 ^{100}}{101}

Answer key (D) & worked solution

Correct answer: Option D

S = \sum_{r = 50}^{100} \frac{^{100} C _{r}}{r + 1} = \sum_{r = 50}^{100} \frac{1}{r + 1} \cdot \frac{r + 1}{101} \cdot^{101} C_{r + 1}

S = \frac{1}{101} r = 50 \sum 100^{101} C_{r + 1} = \frac{1}{101} \times \frac{2 ^{101}}{2} = \frac{2 ^{100}}{101}

Option (4)