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]

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]

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 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]

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 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]

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]

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]

Sol. (3)

Due to thermal expansion

$\frac{\Delta \mathrm{v}}{\mathrm{v}}=3 \alpha \Delta \mathrm{T}$

Due to External pressure

$\frac{\Delta \mathrm{v}}{\mathrm{v}}=\frac{\mathrm{P}}{\mathrm{K}}$


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Comments
  • April 25, 2020 at 12:09 am

    this is fucking bullshit!!!!!!!