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(A) Let the mass of both the ring and the disc be $M$ and the radius be $R$.For a ring,the moment of inertia about its center of mass is $I_{cm} = MR^

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$6$

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(A) Let the mass of both the ring and the disc be $M$ and the radius be $R$.For a ring,the moment of inertia about its center of mass is $I_{cm} = MR^2$. By the parallel axis theorem,the moment of inertia about a tangent in its plane is $I = I_{cm} + MR^2 = MR^2 + MR^2 = \frac{3}{2}MR^2$ is incorrect; the correct formula for a tangent in the plane is $I = I_{cm} + MR^2 = MR^2 + MR^2 = 2MR^2$ is also incorrect. Let's re-evaluate: The moment of inertia of a ring about its diameter is $\frac{1}{2}MR^2$. By the parallel axis theorem,the moment of inertia about a tangent in its plane is $I = I_{diameter} + MR^2 = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$.For a disc,the moment of inertia about its diameter is $I' = \frac{1}{4}MR^2$.The ratio is $\frac{I}{I'} = \frac{\frac{3}{2}MR^2}{\frac{1}{4}MR^2} = \frac{3}{2} 	imes 4 = 6$.Thus,the ratio is $6: 1$.

$A$ ring and a disc have the same mass and the same radius. The ratio of the moment of inertia of a ring about a tangent in its plane to that of the disc about its diameter is: (in $: 1$)

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