A uniform rod of length l is suspended from one end and undergoes...

The time period of a physical pendulum is given by $T = 2\pi \sqrt{\dfrac{I}{mgd}}$, where $I$ is the moment of inertia about the point of suspension and $d$ is the distance of the center of mass from the point of suspension.

For a uniform rod of mass $m$ and length $l$ suspended from one end, the moment of inertia is $I = \dfrac{ml^2}{3}$ and the distance to the center of mass is $d = \dfrac{l}{2}$.

Substituting these values into the time period formula:

$T = 2\pi \sqrt{\dfrac{\dfrac{ml^2}{3}}{mg\left(\dfrac{l}{2}\right)}} = 2\pi \sqrt{\dfrac{2l}{3g}}$

The time period of a simple pendulum of length $L$ is given by $T = 2\pi \sqrt{\dfrac{L}{g}}$.

Equating the time periods of the physical pendulum and the simple pendulum:

$2\pi \sqrt{\dfrac{L}{g}} = 2\pi \sqrt{\dfrac{2l}{3g}}$

Squaring both sides and solving for $L$, we get:

$L = \dfrac{2l}{3}$