Explanation

(b) Poisson’s ratio, $\sigma = 0.4 = \frac{\mathrm{\Delta}d}{d}/\frac{\mathrm{\Delta}l}{l}$
Area $A = \ \pi r^{2} = \frac{\pi dA^{2}}{4}\ \ \ \ \ or\ d^{2} = \frac{4A}{\pi}\ $
Differentiating
$2d\ \mathrm{\Delta}d = \frac{4}{\pi}\mathrm{\Delta}A$
As $A = \frac{\pi d^{2}}{4},\ so\ \mathrm{\Delta}A = \ \frac{2\pi d\mathrm{\Delta}d}{4}$
$\frac{\mathrm{\Delta}A}{A} = \ \frac{\pi\frac{d}{2}\mathrm{\Delta}d}{\pi d^{2}/4}$= 2$\frac{\mathrm{\Delta}d}{d}$
Given $\frac{\mathrm{\Delta}A}{A}\ \times 100 = 2\%\ $
= 2=2 $\frac{\mathrm{\Delta}d}{d}\ \ \ \ or\ \ \ \frac{\mathrm{\Delta}d}{d} = 1\%$
Given $\sigma = \ \frac{\mathrm{\Delta}d/d}{\mathrm{\Delta}l/l} =$0.4
Or $\frac{\mathrm{\Delta}d}{d} = 0.4\frac{\mathrm{\Delta}l}{l}\ $
$\frac{\mathrm{\Delta}l}{l} = \frac{1}{0.4}\ \frac{\mathrm{\Delta}l}{l}$
= 2.5 $\times 1\%$
= 2.5%