(B) The elastic potential energy stored per unit volume is given by $u = \frac{1}{2} \times \text{Young's Modulus} \times (\text{strain})^2$.Since the

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(B) The elastic potential energy stored per unit volume is given by $u = \frac{1}{2} \times \text{Young's Modulus} \times (\text{strain})^2$.Since the track is constrained,the thermal strain is $\text{strain} = \alpha \Delta T$.Given $\alpha = 10^{-5} /^{\circ}C$ and $\Delta T = 10^{\circ}C$,the strain is $\text{strain} = 10^{-5} \times 10 = 10^{-4}$.The energy stored per unit length is $U = u \times \text{Area} = \frac{1}{2} Y (\text{strain})^2 \times A$.Substituting the values: $U = \frac{1}{2} \times 10^{11} \times (10^{-4})^2 \times 0.01$.$U = 0.5 \times 10^{11} \times 10^{-8} \times 10^{-2} = 0.5 \times 10^1 = 5\, J/m$.

Class 11 Physics 1 Thermal Properties of Matter: Thermometry, Expansion and Calorimetry Thermal Expansion for Solid

Medium

The area of cross-section of a railway track is $0.01\, m^2$. The temperature variation is $10^{\circ}C$. The coefficient of linear expansion of the material of the track is $10^{-5} /^{\circ}C$. The energy stored per meter in the track is ...... $J/m$. (Young's modulus of the material of the track is $10^{11}\, Nm^{-2}$)

JEE MAIN 2021 All JEE MAIN

A
$4$
B
$5$
C
$6$
D
$7$

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1 Thermal Properties of Matter: Thermometry, Expansion and Calorimetry

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Thermal Expansion for Solid

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$A$ clock with an iron pendulum keeps correct time at $15^{\circ} C$. If the room temperature is $20^{\circ} C$,the error in seconds per day will be nearly (coefficient of linear expansion of iron is $\alpha = 1.2 \times 10^{-5} /{ }^{\circ} C$) (in $s$)

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When the temperature of a rod increases from $t$ to $(t + \Delta t)$,its moment of inertia increases from $I$ to $(I + \Delta I)$. If $\alpha$ is the coefficient of linear expansion of the rod,then the value of $(\frac{\Delta I}{I})$ is

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$A$ cuboid $ABCDEFGH$ is anisotropic with $\alpha_x = 1 \times 10^{-5} /^{\circ}C$,$\alpha_y = 2 \times 10^{-5} /^{\circ}C$,$\alpha_z = 3 \times 10^{-5} /^{\circ}C$. The coefficient of superficial expansion of the faces can be:

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$A$ thin copper wire of length $l$ metre increases in length by $2\%$ when heated through $10^{\circ}C$. The percentage increase in area when a square copper sheet of side $l$ metre is heated through $10^{\circ}C$ is:

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$A$ clock which keeps correct time at $20\,^{\circ}C$ has a pendulum rod made of brass. How many seconds will it gain or lose per day when the temperature falls to $0\,^{\circ}C$? $(\alpha = 18 \times 10^{-6}/^{\circ}C)$

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$A$ clock with an iron pendulum keeps correct time at $15^{\circ} C$. If the room temperature is $20^{\circ} C$,the error in seconds per day will be nearly (coefficient of linear expansion of iron is $\alpha = 1.2 \times 10^{-5} /{ }^{\circ} C$) (in $s$)

When the temperature of a rod increases from $t$ to $(t + \Delta t)$,its moment of inertia increases from $I$ to $(I + \Delta I)$. If $\alpha$ is the coefficient of linear expansion of the rod,then the value of $(\frac{\Delta I}{I})$ is

$A$ cuboid $ABCDEFGH$ is anisotropic with $\alpha_x = 1 \times 10^{-5} /^{\circ}C$,$\alpha_y = 2 \times 10^{-5} /^{\circ}C$,$\alpha_z = 3 \times 10^{-5} /^{\circ}C$. The coefficient of superficial expansion of the faces can be:

$A$ thin copper wire of length $l$ metre increases in length by $2\%$ when heated through $10^{\circ}C$. The percentage increase in area when a square copper sheet of side $l$ metre is heated through $10^{\circ}C$ is:

$A$ clock which keeps correct time at $20\,^{\circ}C$ has a pendulum rod made of brass. How many seconds will it gain or lose per day when the temperature falls to $0\,^{\circ}C$? $(\alpha = 18 \times 10^{-6}/^{\circ}C)$

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