Solution

$$V = \pi {r^2}I$$
$$\frac{{\Delta V}}{V} = \frac{{\Delta \left( {\pi {r^2}I} \right)}}{{\pi {r^2}I}}$$
$$\frac{{\Delta V}}{V} = \frac{{{r^2}\Delta I + 2rI\Delta r}}{{{r^2}I}}$$
$$\frac{{\Delta V}}{V} = \frac{{\Delta I}}{I} + \frac{{2\Delta r}}{r}$$
But $$\frac{{\Delta V}}{V} = 0$$
therefore$$,$$ $$\frac{{\Delta I}}{I} = \frac{{ - 2\Delta r}}{r}$$
Now$$,$$ Poisson's ratio$$,$$ $$\sigma = \frac{{\frac{{ - \Delta r}}{r}}}{{\frac{{\Delta I}}{I}}}$$-------------------$$(1)$$
from equation $$(1),$$
$$\sigma = - \left( {\frac{{\frac{{\Delta r}}{r}}}{{\frac{{ - 2\Delta r}}{r}}}} \right) = \frac{1}{2} = 0.5$$
Hence,
option $$(C)$$ is correct answer.