A horizontal rod of mass m and length L is free to rotate about a...

The moment of inertia of the rod about the vertical axis passing through its centre is $I = \dfrac{mL^2}{12}$.

The torque acting on the rod due to the force $F$ applied at a distance of $L/4$ from the centre and always perpendicular to the rod is $\tau = F \times \dfrac{L}{4}$.

Using the relation $\tau = I\alpha$, the angular acceleration $\alpha$ is given by:

$\alpha = \dfrac{\tau}{I} = \dfrac{\dfrac{FL}{4}}{\dfrac{mL^2}{12}} = \dfrac{3F}{mL}$

Since the rod starts from rest, the angle rotated in time $t$ is:

$\theta = \dfrac{1}{2}\alpha t^2 = \dfrac{1}{2}\left(\dfrac{3F}{mL}\right)t^2 = \dfrac{3Ft^2}{2mL}$