A copper wire of cross-sectional area 0.01 ~cm^2 is under a tension...

Young's modulus, $$\begin{aligned}Y & =\frac{F / A}{\Delta l / l}\\ \frac{\Delta l}{l}& =\frac{F}{Y A} \end{aligned}$$where, $\frac{\Delta l}{l}=$ longitudinal strain Given, $F=22 \mathrm{~N}, Y=1.1 \times 10^{11} \mathrm{~N}-\mathrm{m}^2$, $$\begin{aligned}A & =0.01 \mathrm{~cm}^2=10^{-6} \mathrm{~m}^2 \\ \frac{\Delta l}{l}& =\frac{22}{1.1 \times 10^{11} \times 10^{-6}}=2 \times10^{-4} \end{aligned}$$Now, Poisson ratio $$\begin{aligned}\sigma& =\frac{\text { Lateral strain }}{\text { Longitudinal strain }}=\frac{\Delta d / d}{\Delta l / l}\\ \frac{\Delta d}{d}& =\sigma \cdot \frac{\Delta l}{l}=0.32 \times2 \times10^{-4} \end{aligned}$$Change in diameter, $\frac{\Delta d}{d}=6.4 \times 10^{-5}$ or change (decrease) in radius, $\frac{\Delta r}{r}=6.4 \times 10^{-5}$ Area, A=$\pi$r^2 Fractional change in area, $\frac{\Delta A}{A}=$2 $\cdot \frac{\Delta r}{r}$$\begin{aligned}& \frac{\Delta A}{A}=2 \times6.4 \times10^{-5} \\ & \DeltaA= \left(12.8 \times10^{-5}\right)\mathrm{A} \end{aligned}$$$Decrease in area, \begin{gathered}\DeltaA= \left(12.8 \times10^{-5}\right)\times(0.01) \mathrm{cm}^2 \\ \DeltaA=1.28 \times10^{-6} \mathrm{~cm}^2 \end{gathered}