Explanation

(b) Let L be length of body, A the area of cross-section and $l$ the increase in length.
$Stress = \frac{F}{A^{'}}\ \ strain = \frac{l}{L}$
Force necessary to deform the body is
$F = \frac{YA}{L}l$
If body is deformed by a distance, then
$Work\ done = F \times dl = \frac{YA}{L}ldl$
$W = \int_{0}^{1}\frac{YA}{L}ldl = \frac{YA}{L}\left\lbrack \frac{l^{2}}{2} \right\rbrack_{0}^{l} = \frac{1}{2}YA\frac{l^{2}}{L}$
$= \frac{1}{2}\left( Y\frac{l}{L} \right)\left( \frac{l}{L} \right)(AL)$
$= \frac{1}{2}(stress \times strain) \times volume$
Hence, work done for unit volume is
$W = \frac{1}{2}stress \times srain.$