A ring of radius `r` is suspended from a point on its circumference. Determine its angular frequency of small oscillations.

Answer

B

Explanation

It is a physical pendulum, the time period of which is,
`T = 2pi sqrt((I)/(mgl))`
Here, `I` = moment of inertia of the ring about point of suspension
` = mr^(2) + mr^(2) = 2mr^(2)`
and `l =` distance of point of suspension from center of gravity `= r`
`:. T = 2pi sqrt ((2mr^(2))/(mgr))`
`= 2pi sqrt ((2 r)/(g))`
`:.` Angular frequency
`omega = (2pi)/(T)`
or `omega = sqrt ((g)/(2r))`.