A uniform wire ($Young\text{'}s Modulus 2\times {10}^{11} N{m}^{-2}$) is subjected to a longitudinal tensile stress of $5\times {10}^7 N{m}^{-2}$. If the overall volume change in the wire is $0.02 \%$, the fractional decrease in the radius of the wire is

$\frac{\Delta L}{L}=\frac{Stress}{Y}=\frac{5\times {10}^7}{2\times {10}^{11}}=2.5\times {10}^{-4}$

$Given\frac{\Delta V}{V}=\frac{0.02}{100}=2\times {10}^{-4}$

Now,         $V=\pi r^{2}L$  . Therefore,

$$\begin{gathered} \frac{\Delta V}{V}=\frac{\Delta (\pi r^{2}L)}{\pi r^{2}L}=\frac{2\pi r\Delta rL+\pi r^{2}L\Delta L}{\pi r^{2}L} \\ =2\frac{\Delta r}{r}+\frac{\Delta L}{L} \\ or 2\frac{\Delta r}{r}=\frac{\Delta V}{V}-\frac{\Delta L}{L}=2\times {10}^{-4}-2.5\times {10}^{-4} \\ =-0.5\times {10}^{-4} \\ or \frac{\Delta r}{r}=- 0.25 X {10}^{-4} \end{gathered}$$

.·. Fractional decrease in $r=0.25\times {10}^{-4}$