A steel wire of mass 3.16Kg is stretched to a tensile strain of 1 × 10^-3. The elastic deformation energy if density ρ =7.9 g/cc and Y = 2 × 10¹¹ N/m²

m = 3.16kg ; strain = 10^–3Nm^–2 
d = $\mathrm{\rho }$ = 7.9g / cc = 7.9 x 10³ kgm^-3
y = 2 x 10¹¹ N/m² ; U = 1/2 x (strain)² Y x V
$\mathrm{U}=\frac{1}{2} \mathrm{x} {\left(\mathrm{strain}\right)}^2 \mathrm{Y}\left(\frac{\mathrm{m}}{\mathrm{\rho }}\right)$

$\mathrm{U}=\frac{1}{2} \mathrm{x} {10}^{-6} \mathrm{x} 2 \mathrm{x} {10}^{11} \mathrm{x} \frac{3.16}{7.9 \mathrm{x} {10}^3}$
U = 10^-6  x 10¹¹ x  0.4 x 10^-3  = 0.4 x 10⁵ x  10^-3 = 0.04KJ