Physics - Elasticity Question with Solution | TestHub

Let the rod length be $\ell$ (marked in the figure). Take torques about the lower end of the rod.

$mg\left(\frac{\ell }{2}\right)\cos \theta =T\ell \sin \theta$

$T=\frac{mg\cot \theta }{2}$

$\sigma =\frac{T}{A}=\frac{mg\cot \theta }{2A}$