A uniform rod of mass m and length l_0 is pivoted at one end and is...

Here, the rod is oscillating about an end point $O$. Hence, moment of inertia of rod about the point of oscillating is $\mathrm{I}=\frac{1}{3} \mathrm{~m} l_0^2$ Moreover, length $l$ of the pendulum = distance from the oscillation axis to centre of mass of rod $=l_0 / 2 \therefore$ Time period of oscillation, $$\begin{aligned} & \mathrm{T}=2 \pi \sqrt{\frac{\mathrm{I}}{\mathrm{mg} l}}=2 \pi \sqrt{\frac{\frac{\mathrm{l}}{3} \mathrm{~m} l_0^2}{m g\left(\frac{l_0}{2}\right)}} \\ & \Rightarrow \quad \mathrm{T}=2 \pi \sqrt{\frac{2 l_0}{3 \mathrm{~g}}} \\ & \end{aligned}$$