When the temperature of a wire of length L_0 | vedclass

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(D) The thermal strain produced in the wire due to a temperature change $T$ is given by $\frac{\Delta L}{L} = \alpha T$,where $\alpha$ is the linear e

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$\frac{1}{{18}}{\gamma ^2}{T^2}Y$

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(D) The thermal strain produced in the wire due to a temperature change $T$ is given by $\frac{\Delta L}{L} = \alpha T$,where $\alpha$ is the linear expansion coefficient.Since the volume expansion coefficient $\gamma = 3\alpha$,we have $\alpha = \frac{\gamma}{3}$.Thus,the strain is $\frac{\Delta L}{L} = \frac{\gamma T}{3}$.The energy density $u$ (energy per unit volume) stored in a stretched wire is given by $u = \frac{1}{2} 	imes 	ext{Stress} 	imes 	ext{Strain}$.Using Hooke's Law,$	ext{Stress} = Y 	imes 	ext{Strain}$,so $u = \frac{1}{2} Y (	ext{Strain})^2$.Substituting the value of strain: $u = \frac{1}{2} Y \left( \frac{\gamma T}{3} ight)^2$.$u = \frac{1}{2} Y \left( \frac{\gamma^2 T^2}{9} ight) = \frac{1}{18} \gamma^2 T^2 Y$.

When the temperature of a wire of length $L_0$ is increased by $T$,what is the energy density? The volume expansion coefficient of the wire is $\gamma$ and the Young's modulus is $Y$.

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