A tension of 20 ~N is applied to a copper wire of cross sectional...

Given, $\sigma=0.32, F=20 \mathrm{~N}$ A=0.01 $\mathrm{~cm}^2=0.01$\times$10^{-3}$\mathrm{~m}$and$Y=1-1 \times 10^{11} \mathrm{~N} / \mathrm{m}^2$We know that$\frac{\Delta l}{l}=\frac{F}{A Y}=\frac{20}{0.01 \times 10^{-3} \times 1.1 \times 10^{11}}=$18.1$\times$10^{-7} and we also known$\begin{gathered}\sigma=\frac{-\Delta r / r}{\Delta / /}\\ -\frac{\Delta r}{r}=0.32 \times18.1 \times10^{-7}=5.79 \times10^{-7} \end{gathered} Hence, decrease in cross reactional area of wire is $$\begin{aligned}\DeltaA=2 \frac{\Delta r}{r} \timesA & =2 \times5.79 \times10^{-7} \times0.01 \times10^{-3} \\ & =0.158 \times10^{-10} \mathrm{~m}^2 \\ & =1.26 \times10^{-6} \mathrm{~cm}^2 \end{aligned}$$