Moment of inertia of a ring about a tangent perpendicular to the plane of the ring is $I$ . Then moment of inertia of the same ring about a tangent lying in the plane of the ring is

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$\frac{3 I}{4}$

$\frac{I}{4}$

$\frac{2 I}{3}$

$\frac{3 I}{2}$

answer is D.

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Detailed Solution

By parallel axis theorem, $I = {mR}^{2} + {mR}^{2} = 2 {mR}^{2}$

Now moment of inertia about a diameter $= \frac{1}{2} {mR}^{2}$

Moment of inertia about tangent AB

$I_{AB} = \frac{1}{2} {mR}^{2} + {mR}^{2} = \frac{3}{2} {mR}^{2} = \frac{3}{2} x \frac{I}{2} = \frac{3 I}{4}$

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