The Poisson’s ratio of a material is $0.1$. If the longitudinal strain of a rod of this material is ${10}^{-3}$, then the percentage change in the volume of the rod will be $\frac{\mathrm{x}}{100}$. Find $\mathrm{x}$.

(Round off to nearest integer)

Longitudinal strain, $\mathrm{\alpha }=\frac{{\mathrm{l}}_2-{\mathrm{l}}_1}{{\mathrm{l}}_1}={10}^{-3}$
Or   $\frac{{\mathrm{l}}_2}{{\mathrm{l}}_1}=1.001$
Poisson’s ratio, $\mathrm{\sigma }=\frac{\left(\text{lateral strain}\right)}{\left(\text{longitudinal strain}\right)}=\frac{\mathrm{\beta }}{\mathrm{\alpha }}$
$$\begin{gathered} \mathrm{or} \mathrm{\beta }=\mathrm{\sigma }\mathrm{\alpha }=0.1\times {10}^{-3}={10}^{-4}=\frac{{\mathrm{r}}_1-{\mathrm{r}}_2}{{\mathrm{r}}_1} \\ \mathrm{or} \frac{{\mathrm{r}}_2}{{\mathrm{r}}_1}=1-{10}^{-4}=0.9999 \\ \% \mathrm{increase} \mathrm{in} \mathrm{volume} =\left(\frac{{\mathrm{V}}_2-{\mathrm{V}}_1}{{\mathrm{V}}_1}\right)\times 100 \\ =[{(0.9999)}^2\times 1.001-1]\times 100=0.08\mathrm{\%} \end{gathered}$$