A copper wire of cross sectional area A is under a tension F. Young's modulus of copper is $\gamma$ and Poisson's ratio $\sigma .$ Then find the fractional change in its cross sectional area.

$\begin{array}{l}StressS=F/A\\ Longitudinalstrain\epsilon =\frac{S}{\gamma }=\frac{F}{A\gamma }\\ Now,\frac{\Delta r/r}{\Delta l/l}=-\sigma \Rightarrow \frac{\Delta r}{r}=-\sigma .\frac{\Delta l}{l}=-\sigma \epsilon \\ Now,A=\pi r^2\\ \therefore \frac{\Delta A}{A}=2.\left(\frac{\Delta r}{r}\right)=-2\sigma \epsilon \\ \therefore Fractionaldecreaseinitscross\sec tionalarea\\ =2\sigma \epsilon =\frac{2\sigma F}{A\gamma }.\end{array}$