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(C) The moment of inertia $I$ of a ring about an axis passing through its rim and perpendicular to its plane is given by the parallel axis theorem: $I

Vedclass · Accepted Answer

$1.5$

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(C) The moment of inertia $I$ of a ring about an axis passing through its rim and perpendicular to its plane is given by the parallel axis theorem: $I = I_{cm} + mr^2 = mr^2 + mr^2 = 2mr^2$.However,the problem states the axis is horizontal and passes through the rim,and the ring oscillates such that its centre moves in a plane perpendicular to the ring's plane. This implies the axis of rotation is a diameter of the ring. The moment of inertia of a ring about its diameter is $I = \frac{1}{2}mr^2$.Using the parallel axis theorem,the moment of inertia about a parallel axis through the rim is $I = I_{cm} + mr^2 = \frac{1}{2}mr^2 + mr^2 = \frac{3}{2}mr^2$.The time period of a compound pendulum is $T = 2\pi \sqrt{\frac{I}{mgh}}$,where $h$ is the distance from the pivot to the centre of mass $(h = r)$.$T = 2\pi \sqrt{\frac{3/2 mr^2}{mgr}} = 2\pi \sqrt{\frac{3r}{2g}}$.Comparing this with the time period of a simple pendulum $T = 2\pi \sqrt{\frac{L_{eq}}{g}}$,we get $L_{eq} = \frac{3}{2}r$.Given diameter $d = 2 \ m$,so radius $r = 1 \ m$.Therefore,$L_{eq} = \frac{3}{2} 	imes 1 = 1.5 \ m$.

$A$ ring of diameter $2 \ m$ oscillates as a compound pendulum about a horizontal axis passing through a point at its rim. It oscillates such that its centre moves in a plane which is perpendicular to the plane of the ring. The equivalent length of the simple pendulum is .... $m$

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