There is no change in the volume of a wire due to change in its length on stretching. The Poisson’s ratio of the material of the wire is

The correct option is – 0.50.

Step-by-step derivation:

Consider a cylindrical wire of length $L$ and radius $r$.
The volume $V$ of the cylinder is given by the formula:
$V=\pi r^{2}L$

Since the volume of the wire does not change during stretching, the change in volume is zero:
$dV=0$

Differentiating the volume equation, we get:
$dV=d\left(\pi r^{2}L\right)=\pi r^{2}dL+2\pi rLdr=0$

To simplify, we divide the entire equation by the volume $V=\pi r^{2}L$:
$\frac{\pi r^{2}dL}{\pi r^{2}L}+\frac{2\pi rLdr}{\pi r^{2}L}=0$

This simplifies to:
$\frac{dL}{L}+2\frac{dr}{r}=0$

Rearranging the terms to relate the lateral strain and longitudinal strain:
$2\frac{dr}{r}=-\frac{dL}{L}$

$\frac{dr/r}{dL/L}=-\frac{1}{2}=-0.50$

By definition, Poisson's ratio is the ratio of lateral strain to longitudinal strain:
$\text{Poisson's ratio}=\frac{\text{Lateral strain}}{\text{Longitudinal strain}}=\frac{dr/r}{dL/L}=-0.50$

Therefore, the Poisson's ratio of the material of the wire is – 0.50.