Updated 19 Oct 2026

JEE Main 2026, 28 Jan Shift 1 question paper with solutions

JEE Main 2026, 28 Jan Shift 1 was held on 28 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, 28 Jan Shift 1

Q1PhysicsJEE Main 2026, 28 Jan Shift 1Average and RMS value of a function

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The electric current in the circuit is given as

i = i_{o} (t / T)

. The r.m.s current for the period

t = 0

t = T

____

Answer key (D) & worked solution

Correct answer: Option D

i_{rms}^{2} = \frac{\int _{0}^{T} ( i _{0}^{2} t ^{2} / T ^{2} ) dt}{\int _{0}^{T} dt} = \frac{i _{0}^{2}}{T ^{3}} \cdot \frac{T ^{3}}{3} = \frac{i _{0}^{2}}{3}

i_{rms} = \frac{i _{0}}{3}

Q2PhysicsJEE Main 2026, 28 Jan Shift 1Parallel combination of a cell

For the two cells having same EMF E and internal resistance

r

, the current passing through the external resistor

6Ω

is same when both the cells are connected either in parallel or in series. The value of internal resistance

r

____

Ω

Answer key (D) & worked solution

Correct answer: Option D

In series,

i_{i} = \frac{2 E}{6 + 2 r}

In parallel,

i_{2} = \frac{E}{6 + \frac{r}{2}}

i_{1} = i_{2} \Rightarrow \frac{2 E}{6 + 2 r} = \frac{E}{6 + \frac{r}{2}}

12 + r = 6 + 2 r

r = 6Ω

Q3ChemistryJEE Main 2026, 28 Jan Shift 1Miscellaneous examples

Given below are two statements: Statement I: The number of species among

BF_{4}^{-}, SiF_{4}, XeF_{4}

and

SF_{4}

, that have unequal

E - F

bond lengths is two. Here, E is the central atom. Statement II: Among

O_{2}^{-}, O_{2}^{2 -}, F_{2}

and

O_{2}^{+}, O_{2}^{-}

has the highest bond order. In the light of the above statements, choose the correct answer from the options given below

ABoth Statement I and Statement II are false

BBoth Statement I and Statement II are true

CStatement I is true but Statement II is false

DStatement I is false but Statement II is true

Answer key (A) & worked solution

Correct answer: Option A

BF_{4}^{-}, SiF_{4}

and

XeF_{4}

all bond lengths are identical Molecules, B.O.

O_{2}^{+}

\to

2.5

O_{2}^{-}

\to

1.5

O_{2}^{2 -}

\to

F_{2}

\to

Q4ChemistryJEE Main 2026, 28 Jan Shift 1Differential and Integral Forms of Rate Law

An organic compound undergoes first order decomposition. The time taken for decomposition to

(\frac{1}{8})^{th}

and

(\frac{1}{10})^{th}

of its initial concentration are

t_{1/8}

and

t_{1/10}

respectively. What is the value of

\frac{t _{1/8}}{t _{1/10}} \times 10

(lo g 2 = 0.3)

B0.9

D30

Answer key (A) & worked solution

Correct answer: Option A

t = \frac{1}{k} ln \frac{A _{0}}{A _{t}}

t_{1/8} = \frac{1}{k} ln \frac{A _{0}}{A _{0} /8} = \frac{1}{k} ln 8 t_{1/10} = \frac{1}{k} ln \frac{A _{0}}{A _{0} /10} = \frac{1}{k} ln 10 \frac{t _{1/8}}{t _{1/10}} = \frac{ln 8}{ln 10} = \frac{lo g 8}{lo g 10} \frac{t _{1/8}}{t _{1/10}} = lo g 8 = 3 lo g 2 = 0.9 \frac{t _{1/8}}{t _{1/10}} \times 10 = 9

Q5MathematicsJEE Main 2026, 28 Jan Shift 1Reducible to Linear Differential Equation

Let

y = y (x)

be the solution of the differential equation

x \frac{d y}{d x} - sin 2 y = x^{3} (2 - x^{3}) cos^{2} y, x \neq = 0

. If

y (2) = x

, then

tan (y (1))

is equal to

\frac{3}{4}

\frac{7}{4}

- \frac{7}{4}

- \frac{3}{4}

Answer key (B) & worked solution

Correct answer: Option B

x \frac{d y}{d x} - sin 2 y = x^{3} (2 - x^{3}) cos^{2} y

sec^{2} y \frac{d y}{d x} - 2 tan y \cdot \frac{1}{x} = x^{2} (2 - x^{3})

tan y = t \Rightarrow sec^{2} y \frac{d y}{d x} = \frac{d t}{d x}

\frac{dt}{dx} - \frac{2 t}{x} = x^{2} (2 - x^{3})

(LDE) I.F.

= e^{\int - \frac{2}{x} dx} = e^{- 2 l n x} = \frac{1}{x ^{2}}

∴ \frac{t}{x ^{2}} = \int \frac{1}{x ^{2}} x^{2} (2 - x^{3}) dx + C

\frac{t a n y}{x ^{2}} = 2 x - \frac{x ^{4}}{4} + C

y (2) = 0 \Rightarrow 0 = 4 - 4 + C \Rightarrow C = 0

tany

= 2 x^{3} - \frac{1}{4} x^{6}

x = 1 \Rightarrow tan y = 2 - \frac{1}{4} = \frac{7}{4} \Rightarrow (2)

Q6MathematicsJEE Main 2026, 28 Jan Shift 1Standard trigonometric formats

\int (\frac{1 - 5 c o s ^{2} x}{s i n ^{5} x c o s ^{2} x}) d x = f (x) + C

where

C

is the constant of integration, then

f (\frac{π}{6}) - f (\frac{π}{4})

is equal to

\frac{1}{3} (26 + 3)

\frac{4}{3} (8 - 6)

\frac{1}{3} (26 - 3)

\frac{2}{3} (4 + 6)

Answer key (B) & worked solution

Correct answer: Option B

\int \frac{d x}{s i n ^{5} c o s ^{2} x} - 5 \int \frac{d x}{s i n ^{5} x}

\int \frac{s e c ^{2} x d x}{s i n ^{5} x} - 5 \int \frac{d x}{s i n ^{5} x}

By IBP

= \frac{tan x}{sin ^{5} x} - \int - \frac{5}{sin ^{6} x} \cdot cos x \cdot tan x d x - 5 \int \frac{d x}{sin ^{5} x} = \frac{tan x}{sin ^{5} x} + c

f (x) = \frac{tan x}{sin ^{5} x} f (\frac{π}{6}) - f (\frac{π}{4}) = \frac{2 ^{5}}{3} - (2)^{5} = 42 - \frac{32}{3} = \frac{32}{3} - 42 = \frac{4}{3} (8 - 6)