A uniform rod of length l and mass m is suspended by two vertical inextensible string as shown in the fig. The tension in the left string when right string snaps, is

Given that a rod's length is l and its mass is m
Let T be the force equal to the tension in the string. Before to snapping

$\mathrm{T}+\mathrm{T}=\mathrm{mg} \Rightarrow \mathrm{T}=\frac{\mathrm{mg}}{2}$

The torque acting on the rod is

Moment of inertia I of the rod,

$\mathrm{\tau }=\mathrm{T}\times l = \frac{\mathrm{mg}}{2}\times l\mathit{ }$

$\Rightarrow \mathrm{I\alpha }=\frac{\mathrm{mg}l\mathit{}}{2}$

$\frac{\mathrm{m}{l}^2}{3}\mathrm{\alpha }=\frac{\mathrm{mg}l}{2}\Rightarrow \mathrm{\alpha }=\frac{3\mathrm{g}}{2l}$

When a string snaps, it will be moving radially, hence acceleration will be

$\mathrm{a}=\frac{\mathrm{l}\times \mathrm{\alpha }}{2}=\frac{\mathrm{l}\times {\displaystyle \frac{3\mathrm{g}}{2l}}}{2}=\frac{3\mathrm{g}}{4}$

Force acting after snapping

$\mathrm{mg}-\mathrm{T}\text{'}=\mathrm{ma}=\mathrm{m}\times \frac{3\mathrm{g}}{4}$

$\Rightarrow \mathrm{mg}-\frac{3\mathrm{mg}}{4}=\mathrm{T}\text{'}$

$\Rightarrow \mathrm{T}\text{'}=\frac{\mathrm{mg}}{4}$

Hence the correct answer is $\frac{mg}{4}.$