Question

A wire of length $L_0$ is supplied heat to to raise its temperature by T. If $\gamma$ is the coefficient of volume ex the wire and Y is the 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

The correct option is D $\frac{1}{18}{\gamma }^2{T}^{2}Y$

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