A thin uniform rod of mass $m$ and length $L$ is hinged at its upper end and released from rest in horizontal position. The tension at a point located at a distance $L/3$ from the hinge point , when the rod becomes vertical, will be -

The correct option is D $2mg$
When the rod becomes vertical, apply conservation of energy.

$\frac{mgL}{2}=\frac{1}{2}\frac{m{L}^2}{3}{\omega }^2$
$\omega =\sqrt{3g/L}$
Centre of mass of the lower two-third part moves in a circle of radius $2L/3$. We apply Newton’s second law on this part
$T-\frac{2mg}{3}=\left(\frac{2m}{3}\right)\left(\frac{2L}{3}\right){\omega }^2\left(\text{or}\right)T=2mg$