The volume of a wire remains unchanged when the wire is subjected to a certain tension. The Poisson's ratio of the material of the wire is

Poisson's ratio $\sigma$ is defined as

               $\sigma =\frac{lateral strain}{longitudinal strain}=\frac{∆r/r}{∆l/l}$

where r is the radius of the wire and l its length and $∆r$ is the change in r and $∆l$ the change in $l$ when the wire is subjected to tension.

Volume of wire before elongation is

              $V_1=\pi r^{2}l$

Volume of wire after elongation is

$$\begin{gathered} V_2=\pi (r-\Delta r{)}^2(l+\Delta l) \\ Given V_1=V_2. Thus \\ \pi r^{2}l=\pi (r-\Delta r{)}^2(l+\Delta l) \\ =\pi [r^2-2r(\Delta r)+(\Delta r{)}^2\left]\right(l+\Delta l) \\ =\pi r^2(l+\Delta l)-2\pi r \Delta r(l+\Delta l)+\pi (\Delta r{)}^2(l+\Delta l) \end{gathered}$$

Since $∆r$ and $∆l$ are very small, terms of order $∆r∆l$ and ${\left(∆r\right)}^2$  

and higher can be ignored. Then, we have

$$\begin{gathered} \pi r^{2}l=\pi r^{2}l+\pi r^2\Delta l-2\pi rl\Delta r \\ or r\Delta l=2 l\Delta r or \frac{\Delta l}{l}=2\frac{\Delta r}{r} \\ \sigma =\frac{\Delta r/r}{\Delta l/l}=\frac{1}{2}=0.5 \end{gathered}$$