The stress along the length of a rod (with rectangular...

$\because$ Stress, $\frac{F}{\Delta A}=1 \%$ of $Y=\frac{Y}{100}$ But Young's modulus, $Y=\frac{\text { stress }}{\text { strain }}=\frac{\frac{F}{\Delta A}}{\frac{\Delta l}{l}}\therefore \quad Y=\frac{\frac{Y}{100}}{\frac{\Delta}{l}} \quad\left(\right.$ putting \left.\frac{F}{\Delta A}=\frac{Y}{100}\right)\therefore \quad \frac{\Delta l}{l}=\frac{1}{100}$Poisson's ratio,$\sigma=\frac{-\frac{\Delta r}{r}}{\frac{\Delta l}{l}}\therefore \quad \frac{\Delta r}{r}=-\sigma \cdot \frac{\Delta l}{l}=\frac{-0.3}{100}\frac{\Delta r}{r}=\frac{-0.3}{100}\therefore$Change in volume,$\frac{\Delta V}{V}=\frac{2 \Delta r}{r}+\frac{\Delta l}{l}$$\begin{array}{l} =\frac{2 \times(-0.3)}{100}+\frac{1}{100}\\ =\frac{1-0.6}{100}=\frac{0.4}{100} \end{array}$$\Delta V \%=0.4 \%