Question

The work done per unit volume to stretch the length
of area of cross-section 2 mm2 by 2% will be
[Y = 8 x 1010 N/m2
(1) 40 MJ/m3
(2) 16 MJ/m3
(3) 64 MJ/m3
14 32 MJ/m3​

Answer 1

$\sf \: \frac{U}{V} = \frac{1}{2} \times Y \times ( \frac{ \triangleℓ}{ℓ} ) {}^{2}$

$\sf \: \frac{U}{V} = \frac{1}{2} \times 8 \times {10}^{10} \times {( \frac{2}{100} )}^{2}$

$\sf \frac{U}{V}= 16 \times {10}^{6} = 16MJ/ {m}^{3}$