Updated 19 Oct 2026

JEE Advanced 2025 question paper with solutions

JEE Advanced 2025 consists of two compulsory 3-hour papers (Paper 1 and Paper 2, 180 marks each, 360 marks total) across Physics, Chemistry and Mathematics. The bank holds 93 questions from the 2025 papers (Paper 1: 48, Paper 2: 45). A sample with worked solutions is below; sit the full papers in the TCS iON section screen.

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Sample JEE Advanced 2025 questions and worked solutions

Q1ChemistryJEE Advanced 2025 Paper 1Crystal Field Theory

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The correct order of the wavelength maxima of the absorption band in the ultraviolet-visible region for the given complexes is

Answer key (A) & worked solution

Correct answer: Option A

Δ_{0} \propto \frac{1}{λ}

According to the spectrochemical series, the strength of ligands is

C l^{-} < H_{2} O < N H_{3} < C N^{-}

. Since

Δ_{0}

is inversely proportional to wavelength, the order of

λ_{ma x}

is:

[C o (C N)_{6}]^{3 -} < [C o (N H_{3})_{6}]^{3 +} < [C o (N H_{3})_{5} H_{2} O]^{3 +} < [C o (N H_{3})_{5} C l]^{2 +}

Q2ChemistryJEE Advanced 2025 Paper 1Potassium Permanganate

One of the products formed from the reaction of permanganate ion with iodide ion in neutral aqueous medium is

I_{2}

I O_{3}^{-}

I O_{4}^{-}

I O_{2}^{-}

Answer key (B) & worked solution

Correct answer: Option B

In neutral or faintly alkaline medium, permanganate ion oxidizes iodide to iodate:

2 M n O_{4}^{-} + H_{2} O + I^{-} \to 2 M n O_{2} + 2 O H^{-} + I O_{3}^{-}

Q3MathematicsJEE Advanced 2025 Paper 1Maxima and Minima at a point

Let

R

denote the set of all real numbers. Define the function

f : R \to R

f (x) = {2 - 2 x^{2} - x^{2} sin \frac{1}{x} 2 if x \neq = 0 if x = 0

Then which one of the following statements is TRUE?

AThe function

f

is NOT differentiable at

x = 0

BThere is a positive real number

δ

, such that

f

is a decreasing function on the interval

(0, δ)

CFor any positive real number

δ

, the function

f

is NOT an increasing function on the interval

(- δ, 0)

x = 0

is a point of local minima of

f

Answer key (C) & worked solution

Correct answer: Option C

(a) RHD at

x = 0 : lim_{h \to 0} \frac{( 2 - 2 h ^{2} - h ^{2} s i n \frac{1}{h} ) - 2}{h} = 0

. LHD is also 0. So

f

is differentiable at

x = 0

. (b)

f^{'} (x) = - 4 x - 2 x sin \frac{1}{x} - x^{2} (cos \frac{1}{x}) (- \frac{1}{x ^{2}}) = - (2 x (4 - sin \frac{1}{x})) + cos \frac{1}{x}

. As

cos \frac{1}{x}

oscillates, we can't say it's decreasing on

(0, δ)

. (d)

f (0) = 2, f (0 + h) 0

Q4MathematicsJEE Advanced 2025 Paper 1Some Special type of Matrices

Consider the matrix

P = 200020003

. Let the transpose of a matrix

X

be denoted by

X^{T}

. Then the number of

3 \times 3

invertible matrices

Q

with integer entries, such that

P Q = QP

and

Q^{T} Q = I

is:

A32

C16

D24

Answer key (C) & worked solution

Correct answer: Option C

Let

Q = a_{1} a_{2} a_{3} b_{1} b_{2} b_{3} c_{1} c_{2} c_{3}

. From

P Q = QP

, we find

c_{1} = 0, c_{2} = 0, a_{3} = 0, b_{3} = 0

Q = a_{1} a_{2} 0 b_{1} b_{2} 0 00 c_{3}

. From

Q^{T} Q = I

and integer entries:

a_{1}^{2} + b_{1}^{2} = 1, a_{2}^{2} + b_{2}^{2} = 1, c_{3}^{2} = 1

and

a_{1} a_{2} + b_{1} b_{2} = 0

. Possible values:

c_{3} \in {1, - 1}

. The

2 \times 2

block yields 8 integer combinations for

(a_{1}, b_{1}, a_{2}, b_{2})

. Total matrices

= 8 \times 2 = 16

Q5PhysicsJEE Advanced 2025 Paper 2Other problems involving circular orbits

Consider a star of mass

m_{2}

kg revolving in a circular orbit around another start of mass

m_{1} k g

with

m_{1} >> m_{2} .

The heavier start slowly acquires mass from the lighter star at a constant rate of

γ

kg/s. In this transfer process, there is no other loss of mass. If the separation between the centers of the stars is r, then its relative rate of change

\frac{1}{r} \frac{d r}{d t} (in s^{- 1})

is given by:

- \frac{3 γ}{2 m _{2}}

- \frac{2 γ}{m _{2}}

- \frac{2 γ}{m _{1}}

- \frac{3 γ}{2 m _{1}}

Answer key (B) & worked solution

Correct answer: Option B

m_{2} ω^{2} r = \frac{G m _{1} m _{2}}{r ^{2}}

ω = \frac{G m _{1}}{r ^{3}}

L = m_{2} ω r^{2}

= m_{2} \frac{G m _{1}}{r ^{3}} r^{2}

L = m_{2} G m_{1} r

= const.

ℓ n L = ℓ n m_{2} + ℓ n G + \frac{1}{2} ℓ n m_{1} + \frac{1}{2} ℓ n r

0 = \frac{d m _{2}}{m _{2}} + \frac{1}{2} \frac{d m _{1}}{m _{1}} + \frac{1}{2} \frac{d r}{r}

\frac{d r}{r d t} = - \frac{2 d m _{2}}{m _{2} d t} - \frac{d m _{1}}{m _{1} d t} \approx - \frac{2 γ}{m _{2}}

Here

(\frac{d m}{d t} = γ)

Q6PhysicsJEE Advanced 2025 Paper 2Dimensional analysis

A temperature difference can generate e.m.f. in some materials. Let S be the e.m.f. produced per unit temperature difference between the ends of a wire,

σ

the electrical conductivity and

k

the thermal conductivity of the material of the wire. Taking

M, L, T, I

and

K

as dimensions of mass, length, time, current and temperature, respectively, the dimensional formula of the quantity

Z = \frac{S ^{2} σ}{k}

is:

[M^{0} L^{0} T^{0} I^{0} K^{0}]

[M^{0} L^{0} T^{0} I^{0} K^{- 1}]

[M^{1} L^{2} T^{- 2} I^{- 1} K^{- 1}]

[M^{1} L^{2} T^{- 4} I^{- 1} K^{- 1}]

Answer key (B) & worked solution

Correct answer: Option B

S = emf per unit temperature difference

σ =

Electrical conductivity k = Thermal conductivity

[S] = [M L^{2} T^{- 3} I^{- 1} K^{- 1}]

[σ] = [M^{- 1} L^{- 3} T^{3} I^{2}]

[K] = [M^{1} L^{1} T^{- 3} K^{- 1}]

[Z] = \frac{S ^{2} σ}{K} = \frac{[ M ^{1} L ^{1} T ^{- 3} K ^{- 2} ]}{[ M ^{1} L ^{1} T ^{- 3} K ^{- 1} ]}

[Z] = [K^{- 1}]