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Thermal Expansion - JEE Main Previous Year Questions with Solutions

Master Thermal Expansion for JEE Main through PYQs covering linear and volume expansion, thermal stress and strain, Young’s modulus, bulk modulus, pendulum clocks, and temperature-dependent dimensional changes.

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Thermal Expansion - JEE Main Previous Year Questions with Solutions

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What Is Thermal Expansion? Key Concepts & Formulae 

Thermal expansion refers to the tendency of matter to change its dimensions — length, area, or volume — when temperature changes. For JEE Main, this topic sits under Properties of Matter (Class 11 Physics, Chapter 11) and is tested alongside elasticity, Young's modulus, and bulk modulus.

Thermal Expansion JEE Main PYQs with Solutions 

JEE Main Previous Year Question of Physics with Solutions are available here. Practicing JEE Main Previous Year Papers Questions of Physics will help all the JEE aspirants in realizing the question pattern as well as help in analyzing their weak & strong areas. Get detailed Class 11th &12th Physics Notes to prepare for Boards as well as competitive exams like IIT JEE, NEET etc. eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects. Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more. Download eSaral app for free study material and video tutorials.  

Q. Two wires are made of the same material and have the same volume. However wire 1 has cross-sectional area A and wire 2 has cross-sectional area 3A. If the length of wire 1 increases by Dx on applying force F, how much force is needed to stretch wire 2 by the same amount ? (1) 6F                (2) 9F                (3) F                 (4) 4F [AIEEE-2009]
Ans. (2) . $\mathrm{A} \ell_{1}=3 \mathrm{A} \ell_{2}$ $\Rightarrow \ell_{2}=\frac{\ell_{1}}{3}$ For wire $1: Y=\frac{F / A}{\Delta x / \ell_{1}}$ For wire $2: \mathrm{Y}=\frac{F^{\prime} / 3 A}{\Delta x / \ell_{2}}$ $\Rightarrow \mathrm{F}^{\prime}=9 \mathrm{F}$
Q. A wooden wheel of radius R is made of two semicircular parts (see figure). The two parts are held together by a ring made of a metal strip of cross sectional area S and Length L. L is slightly less than $2 \pi \mathrm{R}$. To fit the ring on the wheel, it is heated so that its temperature rises by $\Delta \mathrm{T}$ and it just steps over the wheel. As it cools down to surrounding temperature, it presses the semicircular parts together. If the coefficient of linear expansion of the metal is , and its Young's modulus is Y, the force that one part of the wheel applies on the other part is : [AIEEE 2012]
Ans. (1) $Y=\frac{F / S}{\Delta L / L}$ $\frac{\Delta L}{L}=\frac{\Delta R}{R}=\alpha \Delta \mathrm{T} \quad[\mathrm{L}=2 \pi \mathrm{R}]$ $\Rightarrow \mathrm{F}=\mathrm{YS} \alpha \Delta \mathrm{T}$ $\therefore$ The ring is pressing the wheel from both sides, thus $\mathrm{F}_{\mathrm{net}}=2 \mathrm{F}=2 \mathrm{YS} \alpha \Delta \mathrm{T}$
Q. The pressure that has to be applied to the ends of a steel wire of length 10 cm to keep its length constant when its temperature is raised by $100^{\circ} \mathrm{C}$ is : (For steel Young's modulus is $2 \times 10^{11} \mathrm{Nm}^{-2}$ and coefficient of thermal expansion is $1.1 \times 10^{-5} \mathrm{K}^{-1}$ ) (1) $2.2 \times 10^{7} \mathrm{Pa}$ (2) $2.2 \times 10^{6} \mathrm{Pa}$ (3) $2.2 \times 10^{8} \mathrm{Pa}$ (4) $2.2 \times 10^{9} \mathrm{Pa}$ [JEE-Main-2014]
Ans. (3) Thermal strain $=\alpha \Delta \mathrm{T}$ (by $\left.\ell=\ell_{0}(1+\alpha \Delta \mathrm{T})\right)$ $\Rightarrow$ Thermal stress in Rod (Pressure due to Thermal strain $)=\mathrm{Y} \alpha \Delta \mathrm{T}$ $=2 \times 10^{11} \times 1.1 \times 10^{-5} \times 100$ $=2.2 \times 10^{8} \mathrm{Pa}$
Q. A pendulum made of a uniform wire of cross sectional area A has time period T. When an additional mass M is added to its bob, the time period changes to $\mathrm{T}_{\mathrm{M}}$. If the Young's modulus of the material of the wire is Y then $\frac{1}{\mathrm{Y}}$ is equal to :- (g = gravitational acceleration) ( 1)$\left[1-\left(\frac{\mathrm{T}_{\mathrm{M}}}{\mathrm{T}}\right)^{2}\right] \frac{\mathrm{A}}{\mathrm{Mg}}$ ( 2)$\left[1-\left(\frac{\mathrm{T}}{\mathrm{T}_{\mathrm{M}}}\right)^{2}\right] \frac{\mathrm{A}}{\mathrm{Mg}}$ ( 3)$\left[\left(\frac{\mathrm{T}_{\mathrm{M}}}{\mathrm{T}}\right)^{2}-1\right] \frac{\mathrm{A}}{\mathrm{Mg}}$ ( 4)$\left[\left(\frac{\mathrm{T}_{\mathrm{M}}}{\mathrm{T}}\right)^{2}-1\right] \frac{\mathrm{Mg}}{\mathrm{A}}$ [JEE-Main-2015]
Ans. (3)
Q. A pendulume clock loses 12s a day if the temperature is $40^{\circ} \mathrm{C}$ and gains 4s a day if the temperature is $20^{\circ} \mathrm{C}$. The temperature at which the clock will show correct time, and the coeffecient of linear expansion $(\alpha)$ of the metal of the pendulum shaft are respectively :- (1) $55^{\circ} \mathrm{C} ; \alpha=1.85 \times 10^{-2} /^{\circ} \mathrm{C}$ (2) $25^{\circ} \mathrm{C} ; \alpha=1.85 \times 10^{-5} /^{\circ} \mathrm{C}$ (3) $60^{\circ} \mathrm{C} ; \alpha=1.85 \times 10^{-4} /^{\circ} \mathrm{C}$ (4) $30^{\circ} \mathrm{C} ; \alpha=1.85 \times 10^{-3} /^{\circ} \mathrm{C}$ [JEE-Main-2016]
Ans. (2) $\mathrm{T}=2 \pi \sqrt{\frac{\ell}{\mathrm{g}}}$ $\frac{\Delta \mathrm{T}}{\mathrm{T}}=\frac{1}{2} \frac{\Delta \ell}{\ell}$ When clock gain 12 sec $\frac{12}{\mathrm{T}}=\frac{1}{2} \alpha(40-\theta)$ …. (1) When clock lose 4 sec. $\frac{4}{\mathrm{T}}=\frac{1}{2} \alpha(\theta-20) \quad \ldots(2)$ From equation (1) & (2) $3=\frac{40-\theta}{\theta-20}$ $3 \theta-60=40-\theta$ $4 \theta=100$ $\theta=25^{\circ} \mathrm{C}$ from equation (1) $\frac{12}{\mathrm{T}}=\frac{1}{2} \alpha(40-25)$ $\frac{12}{24 \times 3600}=\frac{1}{2} \alpha \times 15$ $\alpha=\frac{24}{24 \times 3600 \times 15}$ $\alpha=1.85 \times 10^{-15} /^{\circ} \mathrm{C}$
Q. An external pressure P is applied on a cube at $0^{\circ} \mathrm{C}$ so that it is equally compressed from all sides. K is the bulk modulus of the material of the cube and $\alpha$ is its coefficient of linear expansion. Suppose we want to bring the cube to its original size by heating. The temperature should be raised by : (1) $\frac{3 \alpha}{\mathrm{PK}}$ (2) $3 \mathrm{PK} \alpha$ (3) $\frac{\mathrm{P}}{3 \alpha \mathrm{K}}$ (4) $\frac{\mathrm{P}}{\alpha \mathrm{K}}$ [JEE-Main-2017]
Ans. (3) Due to thermal expansion $\frac{\Delta \mathrm{v}}{\mathrm{v}}=3 \alpha \Delta \mathrm{T}$ Due to External pressure $\frac{\Delta \mathrm{v}}{\mathrm{v}}=\frac{\mathrm{P}}{\mathrm{K}}$

Frequently Asked Questions

Find answers to common questions.

How many questions from Thermal Expansion appear in JEE Main?

Typically 1–2 questions appear from Thermal Expansion in JEE Main each year. The topic is clubbed under Thermal Properties of Matter and Properties of Solids & Liquids in the NTA syllabus. Based on PYQ analysis from 2009–2024, thermal stress and pendulum clock problems are the most repeated question types.

What is the formula for thermal stress in JEE Main problems?

Thermal stress = Y × α × ΔT, where Y is the Young's modulus of the material, α is its coefficient of linear expansion, and ΔT is the temperature change. This applies when a rod or wire is heated but not allowed to expand. The pressure required to maintain constant length equals this thermal stress value.

Why does a pendulum clock lose time when the temperature increases?

A pendulum clock loses time when temperature rises because the length of the pendulum shaft increases due to linear thermal expansion. Since time period T = 2π√(ℓ/g), a longer pendulum means a longer time period — the clock ticks more slowly and loses seconds each day. This is tested directly in JEE Main 2016.

What is the difference between linear, area, and volumetric expansion coefficients?

Linear expansion coefficient α relates to change in one dimension. Area expansion coefficient β = 2α, and volumetric (cubical) expansion coefficient γ = 3α. For JEE Main, the relation γ = 3α is used in bulk modulus + thermal expansion combination problems, such as the JEE Main 2017 question on restoring a compressed cube by heating.

How do I find the temperature at which a pendulum clock shows correct time?

Set up two equations using ΔT/T = ½α(Δθ) — one for each given temperature and time gain/loss per day. Divide the two equations to eliminate α and solve for the correct temperature θ. Then substitute back to find α. This method is shown step-by-step in Q5 above (JEE Main 2016), yielding θ = 25°C and α = 1.85×10⁻⁵/°C.

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Comments

Yojith
May 17, 2022, 12:08 p.m.
New questions please update
Kuldeep
Sept. 7, 2023, 6:35 a.m.
I tried this question and solved but I nees some or questions and I am enjoying
Yojith
May 17, 2022, 12:08 p.m.
New questions????
Karthi
Jan. 1, 2021, 7:50 a.m.
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Bhumi Tiwari
Sept. 2, 2024, 6:35 a.m.
Question chahiye
Dipika Pawar
April 21, 2025, 6:35 a.m.
Pyq thermal expansion physics jee
Saransh Gupta
Dec. 22, 2020, 11:32 a.m.
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Ananymously1
Dec. 21, 2020, 11:52 p.m.
Only 3 questions in all the years?
sai
Dec. 18, 2020, 2:46 p.m.
very helpful
asgard
April 25, 2020, 12:09 a.m.
this is fucking bullshit!!!!!!!
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