A ring with mass m and radius r is suspended from a point on its...

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 from the point of suspension to the center of mass.

For a ring of mass $m$ and radius $r$, the moment of inertia about its center of mass is $I_{cm} = mr^2$.

Using the parallel axis theorem, the moment of inertia about a point on its periphery is $I = I_{cm} + md^2 = mr^2 + mr^2 = 2mr^2$.

The distance from the point of suspension to the center of mass is $d = r$.

Substituting these values into the time period formula, we get $T = 2\pi \sqrt{\dfrac{2mr^2}{mgr}} = 2\pi \sqrt{\dfrac{2r}{g}}$.