Updated 19 Oct 2026

JEE Main 2026, 21 Jan Shift 2 question paper with solutions

JEE Main 2026, 21 Jan Shift 2 was held on 21 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, 21 Jan Shift 2

Q1PhysicsJEE Main 2026, 21 Jan Shift 2Energy, radius and speed of electron in hydrogen like elements

Try right here · No sign-in

The energy of an electron in an orbit of the Bohr's atom is

- 0.04 E_{0} eV

where

E_{0}

is the ground state energy. If L is the angular momentum of the electron in this orbit and h is the Planck's constant, then

\frac{2 π L}{h}

____

Answer key (C) & worked solution

Correct answer: Option C

Angular momentum

L = \frac{nh}{2 π}

n = \frac{2 π L}{h}

Energy

E = - \frac{13.6}{n ^{2}} . z^{2}

E \Rightarrow - \frac{E _{0}}{n ^{2}} = - 0.04 E_{0} n^{2} = 25, n = 5

Q2PhysicsJEE Main 2026, 21 Jan Shift 2Maximum power transfer theorem

A battery with EMF E and internal resistance

r

is connected across a resistance

R

. The power consumption in R will be maximum when :

R = 2 r

R = \frac{r}{2}

R = 2 r

R = r

Answer key (D) & worked solution

Correct answer: Option D

For maximum power drawn across load Resistance

R_{Load} = R_{internal}

R = r

Q3ChemistryJEE Main 2026, 21 Jan Shift 2Preparation of Carbonyl Compounds

Match List-I with List-II. List-I (Reagents)List-II Reaction Name (Involving aldehydes)A.

H_{2}, Pd - BaSO_{4}

I. Etard ReactionB.

SnCl_{2}, HCl

II. Rosenmund ReductionC.

CrO_{2} Cl_{2}, CS_{2}

III. Gatterman-KochReactionD.

CO, HCl, A nh y d . A l C l_{3}

IV. Stephen Reaction Choose the correct answer from the options given below:

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

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

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

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

Answer key (D) & worked solution

Correct answer: Option D

NCERT Name reaction theory based

Q4ChemistryJEE Main 2026, 21 Jan Shift 2Proteins

The correct statements are: A. Activation energy for enzyme catalysed hydrolysis of sucrose is lower than that of acid catalysed hydrolysis. B. During denaturation, secondary and tertiary structures of a protein are destroyed but primary structure remains intact. C. Nucleotides are joined together by glycosidic linkage between

C_{1}

and

C_{4}

carbons of the pentose sugar D. Quaternary structure of proteins represents overall folding of the polypeptide chain. Choose the correct answer from the options given below:

AA, C and D Only

BA, B and D Only

CA and B Only

DB and C Only

Answer key (C) & worked solution

Correct answer: Option C

Activation energy for enzyme catalysed hydrolysis of sucrose is lower than that of acid catalysed hydrolysis. During denaturation secondary and tertiary structures of a protein are destroyed but primary structure remains intact.

Q5MathematicsJEE Main 2026, 21 Jan Shift 2Shortest distance between two lines

Let the line

L_{1}

be parallel to the vector

- 3 i^+ 2 j^+ 4 k^

and pass through the point

(2, 6, 7)

and the line

L_{2}

be parallel to the vector

2 i^+ j^+ 3 k^

and pass through the point

(4, 3, 5)

. If the line

L_{3}

is parallel to the vector

- 3 i^+ 5 j^+ 16 k^

and intersects the lines

L_{1}

and

L_{2}

at the points C and D , respectively, then

∣ CD ∣^{2}

is equal to :

A171

B290

C312

D89

Answer key (B) & worked solution

Correct answer: Option B

L_{1} : \frac{x - 2}{- 3} = \frac{y - 6}{2} = \frac{z - 7}{4}

Point C on

L_{1} : (- 3 λ_{1} + 2, 2 λ_{1} + 6, 4 λ_{1} + 7)

L_{2} : \frac{x - 4}{2} = \frac{y - 3}{1} = \frac{z - 5}{3}

Point D on

L_{2} : (2 λ_{2} + 4, λ_{2} + 3, 3 λ_{2} + 5)

Dr's of line

L_{3}

L_{3} : \frac{2 λ _{2} + 3 λ _{1} + 2}{- 3} = \frac{λ _{2} - 2 λ _{1} - 3}{5} = \frac{3 λ _{2} - 4 λ _{1} - 2}{16}

λ_{1} = - 3, λ_{2} = 2

C (11, 0, -5) D (8, 5, 11)

∣ CD ∣^{2} = 3^{2} + 5^{2} + 1 6^{2} = 290

Q6MathematicsJEE Main 2026, 21 Jan Shift 2Monotonocity of Composite functions

Let

f : R \to R

be a twice differentiable function such that

f^{''} (x) > 0

for all

x \in R

and

f^{'} (a - 1) = 0

, where a is real number. Let

g (x) = f (tan^{2} x - 2 tan x + a)

0 < x < \frac{π}{2}

. Consider the following two statements : (I) g is increasing in

(0, \frac{π}{4})

(II) g is decreasing in

(\frac{π}{4}, \frac{π}{2})

Then,

ANeither (I) nor (II) is True

BOnly (II) is True

COnly (I) is True

DBoth (I) and (II) are True

Answer key (A) & worked solution

Correct answer: Option A

g (x) = f ((tan x - 1)^{2} + a - 1)

g^{'} (x) = f^{'} ((tan x - 1)^{2} + a - 1) \cdot 2 (tan x - 1) sec^{2} x

∵ f^{'} (a - 1) = 0

and

f^{''} (x) > 0

∴ f^{'} ((tan x - 1)^{2} + a - 1) > 0

g^{'} (x) > 0

(tan x - 1) > 0

g is increasing in

x \in (\frac{π}{4}, \frac{π}{2})

g^{'} (x) < 0

tan x - 1 < 0

g is decreasing in

x \in (0, \frac{π}{4})