Updated 19 Oct 2026

JEE Main 2026, 23 Jan Shift 2 question paper with solutions

JEE Main 2026, 23 Jan Shift 2 was held on 23 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, 23 Jan Shift 2

Q1PhysicsJEE Main 2026, 23 Jan Shift 2Faraday's laws of electromagnetic induction

Try right here · No sign-in

A circular loop of radius 7 cm is placed in uniform magnetic field of 0.2 T directed perpendicular to plane of loop. The loop is converted into a square loop in 0.5 s . The EMF induced in the loop is

____

mV .

Answer key (D) & worked solution

Correct answer: Option D

induced EMF

= \frac{d ϕ}{dt}

circumference

≃ 14 π

side length of square loop

= \frac{14 π}{4} = \frac{7 π}{2}

Δ ϕ = B (A_{1} - A_{2})

= (.2) ((\frac{7 π}{2}) \times \frac{7 π}{2} - 49 π) \times 1 0^{- 4}

= .2 (\frac{49 π ^{2}}{4} - 49 π) \times 1 0^{- 4}

Δ ϕ = .2 \times 33.07 \times 1 0^{- 4}

Δ ϕ = 6.614 \times 1 0^{- 4}

EMF = \frac{6.614 \times 1 0 ^{- 4}}{\frac{1}{2}} V = 13.23 \times 1 0^{- 4} V

EMF = 1.32 mV

Q2PhysicsJEE Main 2026, 23 Jan Shift 2Time varying magnetic fields

Suppose a long solenoid of 100 cm length, radius 2 cm having 500 turns per unit length, carries a current

I = 10 sin (ω t) A

, where

ω = 1000 rad ./ s

. A circular conducting loop (B) of radius 1 cm coaxially slided through the solenoid at a speed

v = 1 cm / s

. The r.m.s. current through the loop when the coil B is inserted 10 cm inside the solenoid is

α / 2 μ A

. The value of

α

____

. [Resistance of the loop

= 10Ω

]

A197

B80

C280

D100

Answer key (A) & worked solution

Correct answer: Option A

EMF induced

ε = A \frac{dB}{dt} = A μ_{0} n \frac{di}{dt} ε = A μ_{0} n i_{0} ω cos ω t

current induced

i = \frac{ε}{R} = \frac{π r ^{2} μ _{0} ni _{0} ω}{R} cos ω t

i = \frac{π r ^{2} μ _{0} n i _{0} ω}{2 R}

= \frac{π \times 1 0 ^{- 4} \times 4 π \times 1 0 ^{- 7} \times 500 \times 10 \times 1 0 ^{3}}{2 \times 10}

= \frac{20 π ^{2}}{2} \times 1 0^{- 6}

≃ \frac{197}{2} μ A

Q3ChemistryJEE Main 2026, 23 Jan Shift 2Nucleic Acids

Both human DNA and RNA are chiral molecules. The chirality in DNA and RNA arises due to the presence of

ABase unit

BChiral phosphate unit

CD-sugar component

DL-sugar component

Answer key (C) & worked solution

Correct answer: Option C

DNA & RNA are chiral molecules due to presence of chiral 2-deoxy-Ribose & Ribose sugar unit respectively.

Q4ChemistryJEE Main 2026, 23 Jan Shift 2Rate of a Chemical Reaction

Observe the following reactions at

T (K)

I. A → products. II.

5 Br^{-} (aq) + BrO_{3}^{-} (aq) + 6 H^{+} (aq) \to 3 Br_{2} (aq) + 3 H_{2} O (l)

Both the reactions are started at 10.00 am . The rates of these reactions at 10.10 am are same. The value of

- \frac{Δ [ Br ^{-} ]}{Δ t}

at 10.10 am is

2 \times 1 0^{- 4} mol L^{- 1} Min^{- 1}

. The concentration of A at 10.10 am is

1 0^{- 2} mol L^{- 1}

. What is the first order rate constant (in

min^{- 1}

) of reaction I?

2 \times 1 0^{- 3}

1 0^{- 3}

1 0^{- 2}

4 \times 1 0^{- 3}

Answer key (D) & worked solution

Correct answer: Option D

t = 10

minutes Rate of reaction

= - \frac{1}{5} \frac{Δ [ Br ^{-} ]}{Δ t} = \frac{1}{5} \times (2 \times 1 0^{- 4}) = 4 \times 1 0^{- 5}

For reaction

A \to P

t = 10

minutes Rate of reaction

= 4 \times 1 0^{- 5} = k [A]

k = 4 \times 1 0^{- 3} min^{- 1}

Q5MathematicsJEE Main 2026, 23 Jan Shift 2Log & its basic properties

The sum of all the real solutions of the equation

lo g_{(x + 3)} (6 x^{2} + 28 x + 30) = 5 - 2 lo g_{(6 x + 10)} (x^{2} + 6 x + 9)

is equal to :

Answer key (C) & worked solution

Correct answer: Option C

lo g_{x + 3} [(x + 3) (6 x + 10)] = 5 - 2 lo g_{6 x + 10} (x + 3)^{2}

1 + lo g_{x + 3} (6 x + 10) = 5 - 4 lo g_{6 x + 10} (x + 3)

Let

lo g_{(x + 3)} (6 x + 10) = A

\Rightarrow A + \frac{4}{A} = 4

A = 2

\Rightarrow lo g_{(x + 3)} (6 x + 10) = 2

\Rightarrow 6 x + 10 = (x + 3)^{2}

\Rightarrow 6 x + 10 = x^{2} + 9 + 6 x

\Rightarrow x^{2} = 1, x = \pm 1

So sum of roots

= 0

Q6MathematicsJEE Main 2026, 23 Jan Shift 2Family of circle

If the points of intersection of the ellipses

x^{2} + 2 y^{2} - 6 x - 12 y + 23 = 0

and

4 x + 2 y^{2} - 20 x - 12 y + 35 = 0

lie on a circle of radius r and centre

(a, b)

, then the value of

ab + 18 r^{2}

A53

B51

C52

D55

Answer key (D) & worked solution

Correct answer: Option D

By family of curve equation of circle will be

\Rightarrow S_{1} + λ S_{2} = 0

\Rightarrow (x^{2} + 2 y^{2} - 6 x - 12 y + 23)

+ λ (4 x^{2} + 2 y^{2} - 20 x - 12 y + 35) = 0

\Rightarrow

for circle coeff of

x^{2} =

coeff. of

y^{2}

\Rightarrow λ = \frac{1}{2}

So equation of circle is

\Rightarrow x^{2} + y^{2} - \frac{16}{3} x - 6 y + \frac{27}{2} = 0

Centre

(\frac{8}{3}, 3) :

Radius

r = \frac{47}{18} = r

∴ ab + 18 r^{2} = 8 + 47 = 55