A tension of 22 ~N is applied to a copper wire of cross-sectional...

Young's modulus of materials $Y=\frac{F \times l}{A \Delta l}, \frac{\Delta l}{l}=\frac{F}{A Y}=\frac{22}{0.02 \times 10^{-4} \times 1.1 \times 10^{11}}=10^{-4}$ Poisson's ratio, $\sigma=\frac{\frac{\Delta l}{l}}{\frac{\Delta r}{r}}$$\begin{aligned} \frac{\Delta l}{l} & =\sigma \frac{\Delta r}{r}=0.32 \times \frac{\Delta r}{r} \\ & =0.32 \times 10^{-4}=32 \times 10^{-6}\end{aligned}$$$The decrease in cross-sectional area,$$\begin{aligned} & \frac{\Delta A}{A}=\frac{\Delta A}{0.02}=0.32 \times \frac{\Delta l}{l} \\ & \Delta A=0.64 \times 10^{-6} \mathrm{~cm}^2\end{aligned}$$