Explanation

(c) Elastic potential energy(U) is given by
$U = \frac{1}{2}F \times l$
$= \frac{1}{2} \times \frac{F}{A} \times \frac{l}{L} \times AL\ \ \ \ldots(i)$
where, L is length of wire, A is area of cross-section of wire, F is stretching force and $l\ $is increase in length.
Eq. (i) may be written as
$U = \frac{1}{2} \times stress \times strain \times volume\ of\ the\ wire$
$\therefore\ $Elastic potential energy per unit volume of the wire
$u = \frac{U}{AL} = \frac{1}{2} \times stress \times srain$
$= \frac{1}{2} \times ({Young}^{'}s\ modulus\ \times strain) \times strain$
$= \frac{1}{2} \times (Y) \times (strain)^{2}$
Hence,
$u = \frac{1}{2} \times 1.1 \times 10^{11} \times \left( \frac{0.1}{100} \right)^{2}$
$= 5.5 \times 10^{4}{Jm}^{- 3}$