If the volume of a wire remains constant when subjected to tensile stress, the value of Poisson's ratio of the material of the wire is

Let L be the length, r be the radius of the wire.
Volume of the wire is
      $V = {\mathrm{\pi r}}^2\mathrm{L}$
Differentiating both sides, we get
$∆V = \mathrm{\pi }(2\mathrm{r}∆\mathrm{r})\mathrm{L}+{\mathrm{\pi r}}^2∆\mathrm{L}$

As the volume of the wire remains uncharged when it gets stretched, so $∆V=0$. Hence
$0 = 2\mathrm{\pi rL}∆\mathrm{r} + {\mathrm{\pi r}}^2∆\mathrm{L}$

$\therefore \frac{{\displaystyle \frac{∆\mathrm{r}}{\mathrm{r}}}}{{\displaystyle \frac{∆\mathrm{L}}{\mathrm{L}}}} = -\frac{1}{2}$

Poisson's ratio = $\frac{\mathrm{Lateral} \mathrm{strain}}{\mathrm{Longitudinal} \mathrm{strain}}$
 

                      = $-\frac{{\displaystyle \frac{∆\mathrm{r}}{\mathrm{r}}}}{{\displaystyle \frac{∆\mathrm{L}}{\mathrm{L}}}} = \frac{1}{2} = 0.5$