A uniform rod of mass m, length $l$ is freely suspended vertically from a fixed horizontal support.
( $\rho$=density of the rod, Y=Young’s modulus of the rod, A= area of cross section of the rod). In this situation mark out the correct statement(s)

Stress in the element  $=\frac{Force}{Area}=\frac{xA\rho g}{A}=x\rho g$
Now, elastic potential energy stored in the wire is
$$\begin{gathered} dU=\frac{1}{2}\left(Stress\right)\left(Strain\right)\left(Volume\right) \\ =\frac{1}{2}.\frac{{\left(Stress\right)}^2}{Y}\left(Volume\right) \\ dU=\frac{1}{2}.\frac{{\left(x\rho g\right)}^2}{Y}Adx=\frac{1}{2}.\frac{{\rho }^2{g}^{2}A}{Y}{x}^{2}dx \end{gathered}$$

Total elastic potential energy= $\frac{1}{2}.\frac{{\rho }^2{g}^{2}A}{Y}{\int }_0^L{x}^{2}dx$

$=\frac{{\rho }^2{g}^{2}AL}{6Y}$