Thermal Expansion – JEE Main Previous Year Questions with Solutions
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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]

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Sol. (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]

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Sol. (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}$$\therefore The ring is pressing the wheelfrom both sides, thus\mathrm{F}_{\mathrm{net}}=2 \mathrm{F}=2 \mathrm{YS} \alpha \Delta \mathrm{T} Q. The pressure that has to be applied to the ends of a steel wire of length 10 cm to keep its length constant when its temperature is raised by 100^{\circ} \mathrm{C} is :(For steel Young’s modulus is 2 \times 10^{11} \mathrm{Nm}^{-2} and coefficient of thermal expansion is 1.1 \times 10^{-5} \mathrm{K}^{-1} )(1) 2.2 \times 10^{7} \mathrm{Pa}(2) 2.2 \times 10^{6} \mathrm{Pa}(3) 2.2 \times 10^{8} \mathrm{Pa}(4) 2.2 \times 10^{9} \mathrm{Pa} [JEE-Main-2014] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (3)Thermal strain =\alpha \Delta \mathrm{T}(by \left.\ell=\ell_{0}(1+\alpha \Delta \mathrm{T})\right)$$\Rightarrow$ Thermal stress in Rod (Pressure due toThermal 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]

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Sol. (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]

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Sol. (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]

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Sol. (3)Due to thermal expansion$\frac{\Delta \mathrm{v}}{\mathrm{v}}=3 \alpha \Delta \mathrm{T}$Due to External pressure$\frac{\Delta \mathrm{v}}{\mathrm{v}}=\frac{\mathrm{P}}{\mathrm{K}}$

• May 17, 2022 at 12:08 pm

New questions please update

• May 17, 2022 at 12:08 pm

New questions????

• January 1, 2021 at 7:50 am

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• December 22, 2020 at 11:32 am

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• December 21, 2020 at 11:52 pm

Only 3 questions in all the years?

• December 18, 2020 at 2:46 pm