A thin rod AB of length ‘l’ is hinged at end A and the other end B is attached to a string hanging from a fixed ceiling as shown in the figure. Just after the string is gently burnt

Just after the string is burnt, reaction at the hinge will be vertical and the rod will start rotating clockwise about end A. Angular acceleration $\mathrm{\alpha }=\frac{\mathrm{mg}\mathrm{l}/2}{\frac{1}{3}{\mathrm{ml}}^2}=\frac{3\mathrm{g}}{2\mathrm{l}}\text{.}$

If we apply tangential pseudo force on the rod, the rod will be in dynamic equilibrium as shown in figure.

$\therefore \mathrm{N}+\mathrm{m}\mathrm{\alpha }\frac{\mathrm{l}}{2}=\mathrm{mg}\Rightarrow \mathrm{N}=\mathrm{mg}-\mathrm{m}\left(\frac{3\mathrm{g}}{2\mathrm{l}}\right)\frac{\mathrm{l}}{2}=\frac{1}{4}\mathrm{mg}.$

So the reaction at the hinge is upward.

Now, as viewed from ground frame net force acting on the rod $=\mathrm{mg}-\mathrm{N}=\mathrm{mg}-\frac{\mathrm{mg}}{4}=\frac{3\mathrm{mg}}{4}.$  Hence the rod will not be in translational equilibrium.

Again since $\mathrm{\alpha }\ne 0,$ the rod will not be in rotational equilibrium.