A metal wire of length 2.5m and area of cross section 1.5 × 10^–6m2 is stretched through 2 mm. Calculate the work done during stretching. (Y=1.25×10¹¹Nm^–2) 

Given,

The length of the wire L = 2.5m

Area of cross-section = $\mathrm{A}=1.5\times {10}^{-6}{\mathrm{m}}^2$

change in length = 2mm

$e=2\times {10}^{-3}m$

$\mathrm{Y}=1.25\times {10}^{-3}\mathrm{N}/{\mathrm{m}}^2$

We know that,

$Workdone during stretching=Average force\times Extension$

$\mathrm{W}=\frac{1}{2}\mathrm{Fe}$

$\mathrm{W}=\frac{1}{2}\left(\frac{\mathrm{YAe}}{\mathrm{L}}\right)\mathrm{e}$

$\mathrm{W}=\frac{1}{2}\frac{{\mathrm{YAe}}^2}{\mathrm{L}}$

$\mathrm{W}=\frac{(1.25\times {10}^{11})(1.5\times {10}^{-6}){(2\times {10}^{-3})}^2}{2\times 2.5}$

$\mathrm{W}=0.15\mathrm{J}$

Hence the correct answer is 0.15J.