Updated 19 Oct 2026

JEE Advanced 2024 question paper with solutions

JEE Advanced 2024 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 96 questions from the 2024 papers (Paper 1: 51, 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 2024 questions and worked solutions

Q1PhysicsJEE Advanced 2024 Paper 1Work energy theorem

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A block of mass 5 kg moves along the x-direction subject to the force F = (−20x + 10) N, with the value of x in metre. At time t = 0 s, it is at rest at position x = 1 m. The position and momentum of the block at t = (π /4) s are

Answer key (C) & worked solution

Correct answer: Option C

F = - 20 (x - \frac{1}{2}) = - 20 X (X = x - \frac{1}{2})

∴ Particle will perform SHM about

x = \frac{1}{2}

with ω = 2 rad/sec ⇒ T = π sec. ∴ Phase covered in

t = \frac{π}{4}

second = 90°. Given particle is at rest at x = 1m ⇒ x = 1 is extreme position.

∴

\frac{π}{4}

sec, it will be at equilibrium

∴

x = 0.5 m and momentum = mω A = 5 × 2 × 0.5 = 5 kg m/s Direction will be towards –ve x.

Q2ChemistryJEE Advanced 2024 Paper 1VSEPR Theory

Based on VSEPR model, match the xenon compounds given in List-I with the corresponding geometries and the number of lone pairs on xenon given in List-II and choose the correct option. List-IList-II(P)XeF

_{2}

(1)Trigonal bipyramidal and two lone pair of electrons(Q)XeF

_{4}

(2)Tetrahedral and one lone pair of electrons(R)XeO

_{3}

(3)Octahedral and two lone pair of electrons(S)XeO

_{3}

_{2}

(4)Trigonal bipyramidal and no lone pair of electrons(5)Trigonal bipyramidal and three lone pair of electrons

\to

5, Q

\to

2, R

\to

3, S

\to

\to

5, Q

\to

3, R

\to

2, S

\to

\to

4, Q

\to

3, R

\to

2, S

\to

\to

4, Q

\to

2, R

\to

5, S

\to

Answer key (B) & worked solution

Correct answer: Option B

XeF

_{2}

\Rightarrow

2 sigma bonds and 3 lone pairs on Xe, number of hybrid orbitals = 5, sp

^{3}

d hybridisation, geometry will be trigonal bipyramidal. P–5 XeF

_{4}

\Rightarrow

4 sigma bonds and 2 lone pairs on Xe, number of hybrid orbitals = 6, sp

^{3}

^{2}

hybridisation, geometry will be octahedral. Q–3 XeO

_{3}

\Rightarrow

3 sigma bonds and 1 lone pairs on Xe, number of hybrid orbitals = 4, sp

^{3}

hybridisation, geometry will be tetrahedral. R–2 XeO

_{3}

_{2}

\Rightarrow

5 sigma bonds and 0 lone pairs on Xe, number of hybrid orbitals = 5, sp

^{3}

d hybridisation, geometry will be trigonal bipyramidal. S–4

Q3ChemistryJEE Advanced 2024 Paper 1Isomerism in Coordination Compounds

Among the following options, select the option in which each complex in Set-I shows geometrical isomerism and the two complexes in Set-II are ionization isomers of each other. [en = H

_{2}

NCH

_{2}

_{2}

_{2}

]

ASet-I : [Ni(CO)

_{4}

] and [PdCl

_{2}

(PPh

_{3}

)

_{2}

], Set-II : [Co(NH

_{3}

)

_{5}

Cl] SO

_{4}

and [Co(NH

_{3}

)

_{5}

(SO

_{4}

)]Cl

BSet-I : [Co(en)(NH

_{3}

)

_{2}

_{2}

] and [PdCl

_{2}

(PPh

_{3}

)

_{2}

], Set-II : [Co(NH

_{3}

)

_{6}

] [Cr(CN)

_{6}

] and [Cr(NH

_{3}

)

_{6}

] [Co(CN)

_{6}

]

CSet-I : [Co(NH

_{3}

)

_{3}

(NO

_{2}

)

_{3}

] and [Co(en)

_{2}

_{2}

], Set-II : [Co(NH

_{3}

)

_{5}

Cl]SO

_{4}

and [Co(NH

_{3}

)

_{5}

(SO

_{4}

)]Cl

DSet-I : [Cr(NH

_{3}

)

_{5}

Cl]Cl

_{2}

and [Co(en)(NH

_{3}

)

_{2}

_{2}

], Set-II : [Cr(H

_{2}

_{6}

]Cl

_{3}

and [Cr(H

_{2}

_{5}

Cl]Cl

_{2}

· H

_{2}

Answer key (C) & worked solution

Correct answer: Option C

Set-I : [Co(NH

_{3}

)

_{3}

(NO

_{2}

)

_{3}

] shows two geometrical isomers :- facial and meridional [Co(en)

_{2}

_{2}

] shows two geometrical isomers :- cis and trans Set-II : [Co(NH

_{3}

)

_{5}

Cl]SO

_{4}

and [Co(NH

_{3}

)

_{5}

_{4}

]Cl are ionization isomers of each other.

Q4MathematicsJEE Advanced 2024 Paper 1Miscellaneous examples

Let f(x) be a continuously differentiable function on the interval (0, ∞ ) such that f(1) = 2 and

t \to x lim \frac{t ^{10} f ( x ) - x ^{10} f ( t )}{t ^{9} - x ^{9}} = 1

for each x > 0. Then, for all x > 0, f(x) is equal to:

\frac{31}{11 x} - \frac{9}{11} x^{10}

\frac{9}{11 x} + \frac{13}{11} x^{10}

\frac{- 9}{11 x} + \frac{31}{11} x^{10}

\frac{13}{11 x} + \frac{9}{11} x^{10}

Answer key (B) & worked solution

Correct answer: Option B

t \to x lim \frac{t ^{10} f ( x ) - x ^{10} f ( t )}{t ^{9} - x ^{9}} = 1

\Rightarrow t \to x lim \frac{10 t ^{9} f ( x ) - x ^{10} f ^{'} ( t )}{9 t ^{8}} = 1

\Rightarrow 10 xf (x) - x^{2} f’ (x) = 9

\Rightarrow x^{2} f’ (x) = 10 xf (x) - 9

\Rightarrow f’ (x) = \frac{10 f ( x )}{x} - \frac{9}{x ^{2}}

\Rightarrow \frac{dy}{dx} - \frac{10}{x} y = - \frac{9}{x ^{2}}

\Rightarrow y . \frac{1}{x ^{10}} = \int - \frac{9}{x ^{2}} . \frac{1}{x ^{10}} dx

\Rightarrow \frac{y}{x ^{10}} = \frac{9}{11 x ^{11}} + c

… (1)

∵ f (1) = 2 \Rightarrow \frac{2}{1} = \frac{9}{11} + c \Rightarrow c = \frac{13}{11}

∴ f (x) = \frac{9}{11 x} + \frac{13}{11} x^{10}

⇒ Option (B) is correct.

Q5MathematicsJEE Advanced 2024 Paper 1Conditional probability

A student appears for a quiz consisting of only true-false type questions and answers all the questions. The student knows the answers of some questions and guesses the answers for the remaining questions. Whenever the student knows the answer of a question, he gives the correct answer. Assume that the probability of the student giving the correct answer for a question, given that he has guessed it, is

\frac{1}{2}

. Also assume that the probability of the answer for a question being guessed, given that the student's answer is correct, is

\frac{1}{6}

. Then the probability that the student knows the answer of a randomly chosen question is :

\frac{1}{12}

\frac{1}{7}

\frac{5}{7}

\frac{5}{12}

Answer key (C) & worked solution

Correct answer: Option C

C → Correct G → Guess K → Knows

P (\frac{C}{G}) = \frac{1}{2}

P (\frac{C}{K}) = 1

P (\frac{G}{C}) = \frac{1}{6}

Let required probability = x

∴ P (\frac{G}{C}) = \frac{( 1 - x ) P ( \frac{C}{G} )}{( 1 - x ) P ( \frac{C}{G} ) + x . P ( \frac{C}{K} )}

\frac{1}{6} = \frac{( 1 - x ) ( \frac{1}{2} )}{( 1 - x ) ( \frac{1}{2} ) + ( x ) ( 1 )}

⇒

x = \frac{5}{7}

⇒ Option (C) is correct.

Q6PhysicsJEE Advanced 2024 Paper 2X-rays and its production

A metal target with atomic number Z = 46 is bombarded with a high energy electron beam. The emission of X-rays from the target is analyzed. The ratio 𝑟 of the wavelengths of the Kα

_{-}

line and the cut-off is found to be r = 2. If the same electron beam bombards another metal target with Z = 41, the value of r will be

A2.53

B1.27

C2.24

D1.58

Answer key (A) & worked solution

Correct answer: Option A

r = \frac{λ k _{α}}{λ _{0}} \propto \frac{1}{( Z - 1 ) ^{2}}

\frac{r _{2}}{r _{1}} = \frac{( Z _{1} - 1 ) ^{2}}{( Z _{2} - 1 ) ^{2}} = \frac{( 45 ) ^{2}}{( 40 ) ^{2}}

r_{2} = 1.265 \times r_{1} = 2.53