Updated 19 Oct 2026

JEE Main 2025, 28 Jan Shift 2 question paper with solutions

JEE Main 2025, 28 Jan Shift 2 was held on 28 January 2025. The bank holds 62 of its questions (Physics 21, Chemistry 18, Mathematics 23). 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 2025 shifts Marks vs percentile table

2025 Exam session year

300 Maximum marks (+4 / −1)

180 min Official duration

62 Questions in bank for this shift

Try questions from JEE Main 2025, 28 Jan Shift 2

Q1PhysicsJEE Main 2025, 28 Jan Shift 2Energy, radius and speed of electron in hydrogen like elements

Try right here · No sign-in

The frequency of revolution of the electron in Bohr's orbit varies with n, the principal quantum number as

Answer key (B) & worked solution

Correct answer: Option B

Frequency of revolution

\propto \frac{1}{n ^{3}}

Q2PhysicsJEE Main 2025, 28 Jan Shift 2Energy of a capacitor

A parallel plate capacitor of capacitance 1 μF is charged to a potential difference of 20 V. The distance between plates is 1 μm. The energy density between plates of capacitor is :

A1.8 × 10

^{3}

J/m

^{3}

B2 × 10–4 J/m

^{3}

C2 × 10

^{2}

J/m

^{3}

D1.8 × 10

^{5}

J/m

^{3}

Answer key (A) & worked solution

Correct answer: Option A

C = 1 μm V = 20 V d = 1 μm Energy density

= \frac{1}{2} ε_{0} E^{2}

E = \frac{V}{d} = 20 \times 1 0^{6} v / m

U = 1.77 × 10

^{3}

J/m

^{3}

Q3ChemistryJEE Main 2025, 28 Jan Shift 2Carbohydrates

Identify correct conversion during acidic hydrolysis from the following : (A) starch gives galactose. (B) cane sugar gives equal amount of glucose and fructose. (C) milk sugar gives glucose and galactose. (D) amylopectin gives glucose and fructose. (E) amylose gives only glucose. Choose the correct answer from the options given below :

A(C), (D) and (E) only

B(A), (B) and (C) only

C(B), (C) and (E) only

D(B), (C) and (D) only

Answer key (C) & worked solution

Correct answer: Option C

(A) Starch

H^{+} / H_{2} O

Glucose (B) Cane sugar

H^{+} / H_{2} O

glucose + fructose (Sucrose) 50% 50% (C) Milk sugar

H^{+} / H_{2} O

glucose + galactose (Lactose) (D) Amylopectin

H^{+} / H_{2} O

Glucose (E) Amylose

H^{+} / H_{2} O

Glucose

Q4ChemistryJEE Main 2025, 28 Jan Shift 2Carbohydrates

Match List-I with List-II List-I (Saccharides) List-II (Glycosidic-linkages found)(A) Sucrose (I) α 1 – 4(B) Maltose(II) α 1 – 4 and α 1 – 6(C) Lactose(III) α 1 – β 2(D) Amylopectin(IV) β 1 – 4 Choose the correct answer from the options given below :

A(A)-(III), (B)-(I), (C)-(IV), (D)-(II)

B(A)-(IV), (B)-(II), (C)-(I), (D)-(III)

C(A)-(II), (B)-(IV), (C)-(III), (D)-(I)

D(A)-(I), (B)-(II), (C)-(III), (D)-(IV)

Answer key (A) & worked solution

Correct answer: Option A

(A) Sucrose → α

_{1}

–β

_{2}

Glycosidic linkage (B) Maltose → α1 – 4 Glycosidic linkage (C) Lactose → β 1 – 4 Glycosidic linkage (D) Amylopectin → α 1–4 and α 1–6 Glycosidic linkage A–III, B–I, C–IV, D–II

Q5MathematicsJEE Main 2025, 28 Jan Shift 2Miscellaneous Examples

If A and B are the points of intersection of the circle x

^{2}

+ y

^{2}

– 8x = 0 and the hyperbola

\frac{x ^{2}}{9} - \frac{y ^{2}}{4} = 1

and a point P moves on the line 2x – 3y + 4 = 0, then the centroid of

Δ P A B

lies on the line :

4 x - 9 y = 12

x + 9 y = 36

9 x - 9 y = 32

6 x - 9 y = 20

Answer key (D) & worked solution

Correct answer: Option D

x^{2} + y^{2} - 8 x = 0, \frac{x ^{2}}{9} - \frac{y ^{2}}{4} = 1 .... (1)

4 x^{2} - 9 y^{2} = 36 .... (2)

Solve (1) & (2)

4 x^{2} - 9 (8 x - x^{2}) = 36

13 x^{2} - 72 x - 36 = 0

(13 x + 6) (x = 6) = 0

x = \frac{- 6}{13}, x = 6

x = \frac{- 6}{13} (rejected)

y \to Imaginary

n = 6, \frac{36}{9} - \frac{y ^{2}}{4} = 1

y^{2} = 12, y = I 12

A (6, 12), B (6, - 12)

P (α, \frac{2 α + 4}{3}) P

lies on Centroid (h, k)

2 x - 3 y + y = 0

h = \frac{12 + α}{3}, α = 3 h - 12

k = \frac{\frac{2 α + 4}{3}}{3} \Rightarrow 2 α + 4 = 9 k

α = \frac{9 k - 4}{2}

6 h - 2 y = 9 k - 4

6 x - 6 y = 20

Q6MathematicsJEE Main 2025, 28 Jan Shift 2Definite integration by method of indefinite integration

Let f be a real valued continuous function defined on the positive real axis such that

g (x) = 0 \int x t f (t) d t

. If

g (x^{3}) = x^{6} + x^{7}

, then value of

r = 1 \sum 15 f (r^{3})

is:

A320

B340

C270

D310

Answer key (D) & worked solution

Correct answer: Option D

g (x) = x^{2} + x^{\frac{7}{3}}

g^{'} (x) = 2 x + \frac{7}{3} x^{\frac{4}{3}}

’

f (x) = \frac{g ^{'} ( x )}{x}

f (x) = 2 + \frac{7}{3} x^{\frac{1}{3}}

f (r^{3}) = 2 + \frac{7 r}{3}

r = 1 \sum 15 (1 + \frac{7}{3} r) = 310