Updated 19 Oct 2026

JEE Main 2026, 24 Jan Shift 1 question paper with solutions

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

73 Questions in bank for this shift

Try questions from JEE Main 2026, 24 Jan Shift 1

Q1PhysicsJEE Main 2026, 24 Jan Shift 1Energy, radius and speed of electron in hydrogen like elements

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Two electrons are moving in orbits of two hydrogen like atoms with speeds

3 \times 1 0^{5} m / s

and

2.5 \times 1 0^{5} m / s

respectively. If the radii of these orbits are nearly same then the possible order of energy states are

____

respectively.

Answer key (A) & worked solution

Correct answer: Option A

V \propto \frac{Z}{n}

r \propto \frac{n ^{2}}{Z}

Thus ;

r \propto \frac{n}{V}

Radii are same then

\frac{n _{1}}{V _{1}} = \frac{n _{2}}{V _{2}}

\frac{n _{1}}{n _{2}} = \frac{3 \times 1 0 ^{5}}{2.5 \times 1 0 ^{5}} = \frac{6}{5}

Possible order is 6 and 5

Q2PhysicsJEE Main 2026, 24 Jan Shift 1Electromagnetic spectrum

Match the List-I with List-II List – IList – IIA. Radio-waveI. is produced by Magnetron valueB. Micro-waveII. Due to change in the vibrational modes of atomsC. Infrared-waveIII. Due to inner shell electrons moving from higher energy level to lower energy levelD. X-rayIV. Due to rapid acceleration of electrons Choose the correct answer from the options given below:

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

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

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

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

Answer key (C) & worked solution

Correct answer: Option C

Radio wave ⇒ Produced by rapid acceleration of electrons Micro wave ⇒ By magnetron valve Infrared wave ⇒ Change in vibrational modes X ray ⇒ Transition of inner shell electrons from high energy level to low energy level. A - IV, B - I, C - II, D - III

Q3ChemistryJEE Main 2026, 24 Jan Shift 1Miscellaneous examples

Given below are statements about some molecules/ions. Identify the CORRECT statements. A. The dipole moment value of

NF_{3}

is higher than that of

NH_{3}

. B. The dipole moment value of

BeH_{2}

is zero. C. The bond order of

O_{2}^{2 -}

and

F_{2}

is same. D. The formal charge on the central oxygen atom of ozone is -1 . E. In

NO_{2}

, all the three atoms satisfy the octet rule, hence it is very stable. Choose the correct answer from the options given below :

AA, B, C, D & E

BB & C only

CB, C & D only

DA, C & D only

Answer key (B) & worked solution

Correct answer: Option B

(A) Dipole moment :

NF_{3} < NH_{3}

(B)

BeH_{2}

is 'sp' hybridized, linear molecule with zero dipole moment. (C)

O_{2}^{- 2} \Rightarrow

bond order

= 1

F_{2} \Rightarrow

bond order

= 1

(D) Formal charge on central oxygen atom in

O_{3}

is +1 . (E) In

NO_{2}

; nitrogen does not follow octet rule.

Q4ChemistryJEE Main 2026, 24 Jan Shift 1Arrhenius Equation

2 7^{\circ} C

in presence of a catalyst, activation energy of a reaction is lowered by

10 kJ mol^{- 1}

. The logarithm ratio of

\frac{k (catalysed)}{k (uncatalysed)}

is .... (Consider that the frequency factor for both the reactions is same)

A17.41

B1.741

C3.482

D0.1741

Answer key (B) & worked solution

Correct answer: Option B

\frac{K _{catalyst}}{K _{uncatalyst}} = e^{\frac{Δ E _{a}}{RT}}

ln \frac{K _{catalyst}}{K _{uncatalyst}} = \frac{Δ E _{a}}{RT}

lo g \frac{K _{catalyst}}{K _{uncatalyst}} = \frac{Δ E _{a}}{2.303 RT}

= \frac{10 \times 1000}{2.303 \times 8.314 \times 300}

lo g \frac{K _{catalyst}}{K _{uncatalyst}} = 1.741

Q5MathematicsJEE Main 2026, 24 Jan Shift 1Equation of lines in 3D and vector form

Let the lines

L_{1} : r = i^+ 2 j^+ 3 k^+ λ (2 i^+ 3 j^+ 4 k^)

λ \in R

and

L_{2} : r = (4 i^+ j^) + μ (5 i^+ 2 j^+ k^), μ \in R

, intersect at the point R . Let P and Q be the points lying on lines

L_{1}

and

L_{2}

, respectively, such that

∣ PR ∣ = 29

and

∣ PQ ∣ = \frac{47}{3}

. If the point P lies in the first octant, then

27 (QR)^{2}

is equal to

A340

B360

C320

D348

Answer key (B) & worked solution

Correct answer: Option B

For POI

2 λ + 1 = 5 μ + 4; 3 λ + 2 = 2 μ + 1; 4 λ + 3 = μ \Rightarrow λ = μ = - 1 R (- 1, - 1, - 1) P (2 λ + 1, 3 λ + 2, 4 λ + 3) PR^{2} = 29 \Rightarrow (2 λ + 2)^{2} + (3 λ + 3)^{2} + (4 λ + 4)^{2} = 29 \Rightarrow λ = 0 or λ = - 2 (Reject) \Rightarrow P (1, 2, 3) Q (5 μ + 4, 2 μ + 1, μ) ∣ PQ ∣ = \frac{47}{3} \Rightarrow PQ^{2} = \frac{47}{3} \Rightarrow (5 μ + 3)^{2} + (2 μ - 1)^{2} + (μ - 3)^{2} = \frac{47}{3} \Rightarrow μ = - \frac{1}{3} Q = (\frac{7}{3}, \frac{1}{3}, - \frac{1}{3}) (QR)^{2} = (\frac{7}{3} + 1)^{2} + (\frac{1}{3} + 1)^{2} + (- \frac{1}{3} + 1)^{2} = \frac{100 + 16 + 4}{9} = \frac{120}{9} \Rightarrow 27 \times (QR)^{2} = 27 \times \frac{120}{9} = 360

Q6MathematicsJEE Main 2026, 24 Jan Shift 1Binomial coefficient problems

Let

S = \frac{1}{25 !} + \frac{1}{3 ! 23 !} + \frac{1}{5 ! 21 !} + \dots

up to 13 terms. If

13 S = \frac{2 ^{k}}{n !}, k \in N

, then

n + k

is equal to

A51

B52

C49

D50

Answer key (C) & worked solution

Correct answer: Option C

\frac{1}{26 !} (\frac{26 !}{25 ! 1 !} + \frac{26 !}{3 ! 23 !} + \frac{26 !}{5 ! 21 !} + \dots .. + 13

terms

)

\frac{1}{26 !} (^{26} C_{1} +^{26} C_{3} +^{26} C_{5} + \dots . + 13

terms

)

\frac{1}{26 !} (^{26} C_{1} +^{26} C_{5} + \dots . +^{26} C_{25})

\Rightarrow S = \frac{1}{26 !} \times 2^{25}

\Rightarrow 13 S = \frac{2 ^{24}}{25 !}

n + k = 25 + 24 = 49