A wire having a length L and cross-sectional area A is suspended at one of its ends from a ceiling.  Density and Young’s modulus of material of the wire are $\rho$ and $\gamma ,$  respectively.  Find its strain energy due to its own weight in $\mu J.$(Given: $\frac{{\rho }^2{g}^{2}A{L}^3}{\gamma }=12\times {10}^{-6}J)$

consider  an element as shown in the figure.

  Stress in the element =$\frac{Force}{Area}=\frac{xA\rho g}{A}=x\rho g$

Now, electric potential energy stored in the wire is

  $dU=\frac{1}{2}$(stress) (strain) (volume)                                                                  

$=\frac{1}{2}.\frac{{\left(stress\right)}^2}{Y}\left(volume\right)$

$dU=\frac{1}{2}.\frac{{\left(x\rho g\right)}^2}{Y}Adx=\frac{1}{2}.\frac{{\rho }^2{g}^{2}A}{Y}{x}^{2}dx$

    Total electric potential energy=$\frac{1}{2}.\frac{{\rho }^2{g}^{2}A}{Y}\underset{0}{\overset{L}{\int }}{x}^{2}dx=\frac{{\rho }^2{g}^{2}A{L}^3}{6Y}$

$=\frac{12\times {10}^{-6}}{6}=2\times {10}^{-6}J$