Explanation

(a) The Poisson’s ratio of the material of the wire is
$\sigma = \frac{\frac{\mathrm{\Delta}D}{D}}{\frac{\mathrm{\Delta}l}{l}}$
The relation for volume of wire is
$V = \pi r^{2}l\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( But,\ r = \frac{D}{2} \right)$
$V = \pi\left( \frac{D}{2} \right)^{2}\ \ l = \frac{\pi D^{2}l}{4}\ \ \ \ \ \ \ \ \ \ \ \ldots(i)$
Differentiating both sides of Eq. (i)
$dV = \frac{\pi l}{4} \bullet 2D \bullet dD + \pi D^{2} \times \frac{l}{4}dl$
As volume remains constant hence, we get
$0 = \pi\frac{l}{2}DdD + \pi\frac{D^{2}}{4}dl$
or $- \pi\frac{l}{2}D\ dD = \ \frac{\pi D^{2}}{4}dl$
or $- \frac{dD}{D}\frac{l}{dl} = \frac{2}{4} = 0.5$
Therefore, Poisson’s ratio, $\sigma = 0.5$