A material has Poisson ratio 0.5. If a rod of material has a longitudinal strain $2 \times {10}^{-3}$, the percentage change in volume is:

Here Poisson's ratio, 
 $\sigma = \frac{{\displaystyle \left(\frac{∆\mathrm{r}}{\mathrm{r}}\right)}}{{\displaystyle \left(\frac{∆\mathrm{l}}{\mathrm{l}}\right)}}$

In this case $\frac{∆\mathrm{l}}{\mathrm{l}} = 2\times {10}^{-3} \mathrm{and} \mathrm{\sigma } = 0.5$

    $\frac{∆\mathrm{r}}{\mathrm{r}} = \mathrm{\sigma }\frac{∆\mathrm{l}}{\mathrm{l}}$ = $2\times {10}^{-3}\times 0.5 = {10}^{-3}$

Let V be the initial volume and (V + $∆\mathrm{V}$) be the final volume, then

          $V = \pi r^{2}l$

          $\mathrm{V} + ∆\mathrm{V} = \mathrm{\pi }{(\mathrm{r}-∆\mathrm{r})}^2(\mathrm{l}+∆\mathrm{l})$

          $∆\mathrm{V} = {\mathrm{\pi r}}^2∆\mathrm{l}-2\mathrm{\pi r} \mathrm{l} ∆\mathrm{r}$

(Subtracting and simplifying)

$\frac{∆\mathrm{V}}{\mathrm{V}} = \frac{∆\mathrm{l}}{\mathrm{l}}-\frac{2∆\mathrm{r}}{\mathrm{r}}$

         = 2 $\times {10}^{-3}-2\times {10}^{-3} = 0$