A rod PQ of mass M and length L is hinged at end P. The rod is kept horizontal by a massless string tied to point Q as shown in figure. When string is cut, the initial angular acceleration of the rod is

When the string is cut, the rod will rotate about P. Let $\alpha$ be initial angular acceleration of the rod. Then

$$\begin{gathered} Torque,\tau =I\alpha =\frac{M{L}^2}{3}\alpha ---\left(i\right) \\ ( \because Moment of inertia of the rod about one end =\frac{M{L}^2}{3} ) \\ Also, \tau =Mg\frac{L}{2}---\left(ii\right) \\ Equating \left(i\right) and \left(ii\right), we get \\ Mg\frac{L}{2}=\frac{M{L}^2}{3}\alpha or\alpha =\frac{3g}{2L} \end{gathered}$$