Updated 19 Oct 2026

JEE Main 2026, 28 Jan Shift 2 question paper with solutions

JEE Main 2026, 28 Jan Shift 2 was held on 28 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, 28 Jan Shift 2

Q1PhysicsJEE Main 2026, 28 Jan Shift 2Miscellaneous examples

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Identify the correct statements : A. Effective capacitance of a series combination of capacitors is always smaller than the smallest capacitance of the capacitor in the combination. B. When a dielectric medium is placed between the charged plates of a capacitor, displacement of charges cannot occur due to insulation property of dielectric. C. Increasing of area of capacitor plate or decreasing of thickness of dielectric is an alternate method to increase the capacitance. D. For a point charge, concentric spherical shells centered at the location of the charge are equipotential surfaces. Choose the correct answer from the options given below.

Answer key (C) & worked solution

Correct answer: Option C

For series combination

\frac{1}{C _{eq}} = \frac{1}{C _{1}} + \frac{1}{C _{2}}

∴ C_{eq}

is less than

C_{1}

C_{2}

. Note : In statement C, capacitor is assumed to be completely filled with dielectric then on decreasing thickness of dielectric capacitance will increase.

Q2PhysicsJEE Main 2026, 28 Jan Shift 2Photoelectric effect and Einstein's photoelectric equation

Number of photons of equal energy emitted per second by a 6 mW laser source operating at 663 nm is

____

. (Given :

h = 6.63 \times 1 0^{- 34}

J.s and

c = 3 \times 1 0^{8} m / s

)

5 \times 1 0^{16}

5 \times 1 0^{15}

10 \times 1 0^{15}

2 \times 1 0^{16}

Answer key (D) & worked solution

Correct answer: Option D

P = \frac{nhC}{λ}

6 \times 1 0^{- 3} = \frac{n \times 6.63 \times 1 0 ^{- 34} \times 3 \times 1 0 ^{8}}{663 \times 1 0 ^{- 9}}

n = 2 \times 1 0^{16}

photons

Q3ChemistryJEE Main 2026, 28 Jan Shift 2Test of Amines

Total number of alkali insoluble solid sulphonamides obtained by reaction of given amines with Hinsberg's reagent is Aniline, N-Methylaniline, Methanamine, N, N-Dimethylmethanamine, N-Methyl methanamine, Phenylmethanamine, N-propylaniline, N-phenylaniline, N, N-Dimethylaniline, Allyl amine, Isopropyl amine

Answer key (A) & worked solution

Correct answer: Option A

2^{\circ}

Amine are insoluble with hinsberg reagent.

Ph - NH - CH_{3}, Me - NH - Me

Ph - NH - CH_{2} - CH_{2} - CH_{3}, Ph - NH - Ph

Ans. (1) 4

Q4ChemistryJEE Main 2026, 28 Jan Shift 2VSEPR Theory

Match List-I with List-II according to shape. List-IList-IIA.

XeO_{3}

BrF_{5}

XeF_{2}

II.

NH_{3}

XeO_{2} F_{2}

III.

[I_{3}]^{-}

XeOF_{4}

IV.

SF_{4}

Choose the correct answer from the options given below :

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

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

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

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

Answer key (B) & worked solution

Correct answer: Option B

XeF_{2} & I_{3}^{-} : 2

bond pair 3 lone pair ; Linear

XeOF_{4} & BrF_{5} : 5

bond pair 1 lone pair ; Square pyramidal

XeO_{2} F_{2} & SF_{4} : 4

bond pair 1 lone pair ; See saw

XeO_{3} & NH_{3} : 3

bond pair 1 lone pair ; Pyramidal

Q5MathematicsJEE Main 2026, 28 Jan Shift 2Binomial coefficient problems

The sum of the coefficients of

x^{499}

and

x^{500}

(1 + x)^{1000} + x (1 + x)^{999} + x^{2} (1 + x)^{998} + \dots\dots + x^{1000}

^{1001} C_{501}

^{1002} C_{500}

^{1002} C_{501}

^{1000} C_{501}

Answer key (B) & worked solution

Correct answer: Option B

S = (1 + x)^{1000} + x (1 + x)^{999} + x^{2} (1 + x)^{998} + \dots . + x^{1000}

= (1 + x)^{1000} \frac{( 1 - ( \frac{x}{1 + x} ) ^{1001} )}{1 - \frac{x}{1 + x}}

= (1 + x)^{1001} - x^{1001}

Required sum

=^{1001} C_{499} +^{1001} C_{500} =^{1002} C_{500}

Q6MathematicsJEE Main 2026, 28 Jan Shift 2Miscellaneous Examples

Let the circle

x^{2} + y^{2} = 4

intersect

x

-axis at the points

A (a, 0), a > 0

and

B (b, 0)

. Let

P (2 cos α

2 sin α), 0 < α < \frac{π}{2}

and

Q (2 cos β, 2 sin β)

be two points such that

(α - β) = \frac{π}{2}

. Then the point of intersection of AQ and BP lies on :

x^{2} + y^{2} - 4 y - 4 = 0

x^{2} + y^{2} - 4 x - 4 = 0

x^{2} + y^{2} - 4 x - 4 y = 0

x^{2} + y^{2} - 4 x - 4 y - 4 = 0

Answer key (A) & worked solution

Correct answer: Option A

Let point of intersection

R (h, k)

m_{B R} = m_{B P} \Rightarrow \frac{k}{h + 2} = \frac{2 sin α}{2 cos α + 2} \Rightarrow \frac{k}{h + 2} = tan \frac{α}{2} m_{A R} = m_{A Q} \Rightarrow \frac{k}{h - 2} = \frac{2 sin β}{2 cos β - 2} = \frac{sin β}{cos β - 1} = - cot \frac{β}{2} \frac{α}{2} - \frac{β}{2} = \frac{π}{4} tan (\frac{α}{2} - \frac{β}{2}) = tan \frac{π}{4} = 1 \frac{tan \frac{α}{2} - tan \frac{β}{2}}{1 + tan \frac{α}{2} tan \frac{β}{2}} = 1 \frac{\frac{k}{h + 2} + \frac{h - 2}{k}}{1 + ( \frac{k}{h + 2} ) ( \frac{2 - h}{k} )} = 1 \Rightarrow \frac{k ^{2} + h ^{2} - 4}{\frac{k ( h + 2 )}{4}} (h + 2) = 1 \frac{h ^{2} + k ^{2} - 4}{4 k} = 1 x^{2} + y^{2} - 4 y - 4 = 0