Updated 19 Oct 2026

JEE Main 2026, 24 Jan Shift 2 question paper with solutions

JEE Main 2026, 24 Jan Shift 2 was held on 24 January 2026. The bank holds 75 of its questions (Physics 25, Chemistry 25, 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

75 Questions in bank for this shift

Try questions from JEE Main 2026, 24 Jan Shift 2

Q1PhysicsJEE Main 2026, 24 Jan Shift 2Intensity of EM waves

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A point source is kept at the center of a spherically enclosed detector. If the volume of the detector increased by 8 times, the intensity will

Answer key (D) & worked solution

Correct answer: Option D

V \to 8 V \Rightarrow R \to 2 R

\Rightarrow A \to 4 A

\Rightarrow I \to \frac{I _{0}}{4}

Q2PhysicsJEE Main 2026, 24 Jan Shift 2Vernier calliper

In a vernier callipers, 50 vernier scale divisions are equal to 48 main scale divisions. If one main scale division

= 0.05 mm

, then the least count of the vernier callipers is

____

mm .

A0.002

B0.05

C0.02

D0.005

Answer key (A) & worked solution

Correct answer: Option A

LC = 1 MSD - 1 MSD = 1 MSD - \frac{48}{50} MSD

= \frac{2}{50} MSD = \frac{2}{50} \times .05 mm = 0.002 mm

Q3ChemistryJEE Main 2026, 24 Jan Shift 2Chemical Properties of Carbonyl Compounds

Given below are two statements : Statement I: Cross aldol condensation between two different aldehydes will always produce four different products. Statement II: When semicarbazide reacts with a mixture of benzaldehyde and acetophenone under optimum pH, it forms a condensation product with acetophenone only. 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 false but Statement II is true

CBoth Statement I and Statement II are true

DStatement I is true but Statement II is false

Answer key (A) & worked solution

Correct answer: Option A

Statement I : False Cross aldol can give 2 or 4 products Statement II : False Benzaldehyde & Acetone both react with semi carbazide.

Q4ChemistryJEE Main 2026, 24 Jan Shift 2Proteins

The number of possible tripeptides formed involving alanine (ala), glycine (gly) and valine (val), where no amino acid has been used more than once is :

Answer key (A) & worked solution

Correct answer: Option A

Gly ala val Gly val ala Val gly ala Val ala gly Ala val gly Ala gly val Total tri peptides

= 6

Q5MathematicsJEE Main 2026, 24 Jan Shift 2Shortest distance between two lines

The sum of all values of

α

, for which the shortest distance between the lines

\frac{x + 1}{α} = \frac{y - 2}{- 1} = \frac{z - 4}{- α}

and

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

2

, is

B-6

D-8

Answer key (B) & worked solution

Correct answer: Option B

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

2 = \frac{- 1 ( - 2 α + 2 α ) - 1 ( 2 α ^{2} + α ^{2} ) + 3 ( 2 α + α )}{i ^ ( - 2 α + 2 α ) - j ^ ( 2 α ^{2} + α ^{2} ) + k ^ ( 2 α + α )} 2 = \frac{- 3 α ^{2} + 9 α}{9 α ^{4} + 9 α ^{2}} 2 = \frac{- α + 3}{α ^{2} + 1} \Rightarrow 2 α^{2} + 2 = α^{2} + 9 - 6 α α^{2} + 6 α - 7 = 0 (α + 7) (α - 1) = 0 α = - 7, 1 sum = - 7 + 1 = - 6

option (b)

Q6MathematicsJEE Main 2026, 24 Jan Shift 2Intervals of monotonocity

Consider the following three statements for the function

f : (0, \infty) \to R

defined by

f (x) = ∣ lo g_{e} x ∣ - ∣ x - 1∣

: (I) f is differentiable at all

x > 0

. (II) f is increasing in

(0, 1)

. (III) f is decreasing in (

1, \infty

). Then.

AAll (I), (II) and (III) are TRUE.

BOnly (I) is TRUE.

COnly (II) and (III) are TRUE.

DOnly (I) and (III) are TRUE.

Answer key (D) & worked solution

Correct answer: Option D

$\mathrm{f}(\mathrm{x})=|\ell \mathrm{nx}|-|\mathrm{x}-1|$

$=\left\{\begin{array}{cc}\ln \mathrm{x}-(\mathrm{x}-1) & \mathrm{x} \geq 1 \\ -\ln \mathrm{x}+(\mathrm{x}-1) & 0<\mathrm{x}<1\end{array}\right.$

$=\left\{\begin{array}{cc}\ln \mathrm{x}-\mathrm{x}+1 & \mathrm{x} \geq 1 \\ -\ln \mathrm{x}+\mathrm{x}-1 & 0<\mathrm{x}<1\end{array}\right.$

$\mathrm{f}^{\prime}(\mathrm{x})=\left(\begin{array}{ll}\frac{1}{x}-1 & x \geq 1 \\ -\frac{1}{x}+1 & 0<x<1\end{array}\right.$

$\mathrm{f}^{\prime}\left(1^{+}\right)=\mathrm{f}^{\prime}\left(1^{-}\right)=0 \Rightarrow \mathrm{f}(\mathrm{x})

i s d i f f er e n t iab l e

\forall \mathrm{x}>0$

$\mathrm{f}^{\prime}(\mathrm{x})<0 \quad \forall \mathrm{x}>1$

$\mathrm{f}^{\prime}(\mathrm{x})<0 \quad \forall 0<\mathrm{x}<1$

$\Rightarrow \mathrm{f}(\mathrm{x})

i s d ecr e a s in g

\forall \mathrm{x} \in(0, \infty)$

Option (d)