Question

A wire of length $L_{0}$ is supplied heat to raise its temperature by $T$. If $\gamma$ is the coefficient of volume expansion of the wire and $Y$ is Young's modulus of the wire then the energy density stored in the wire is

• A. $\frac{1}{2} \gamma^{2} T^{2} Y$
• B. $\frac{1}{3} \gamma^{2} T^{2} Y^{3}$
• C. $\frac{1}{18} \frac{\gamma^{2} T^{2}}{Y}$
• D. $\frac{1}{18} \gamma^{2} T^{2} Y$

Answer 1

Answer: (D)

Due to heating the length of the wire increases.
$\therefore $ Longitudinal strain is produced
$\Rightarrow \frac{\Delta L}{L}=\alpha \times \Delta T$
Elastic potential energy per unit volume
$E =\frac{1}{2} \times \text { Stress } \times \text { Strain }=\frac{1}{2} \times Y \times(\text { Strain })^{2} $
$\Rightarrow E =\frac{1}{2} \times Y \times\left(\frac{\Delta L}{L}\right)^{2}=\frac{1}{2} \times Y \times \alpha^{2} \times \Delta T^{2}$
or $E=\frac{1}{2} \times Y \times\left(\frac{\gamma}{3}\right)^{2} \times T^{2}=\frac{1}{18} \gamma^{2} Y T^{2}$
[As $\gamma=3 \alpha$ and $\Delta T=T$ (given)]