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(B) The moment of inertia of a circular ring about an axis passing through its centre and perpendicular to its plane is $I_{Ring} = MR^2$.The moment o

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$2 : 1$

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(B) The moment of inertia of a circular ring about an axis passing through its centre and perpendicular to its plane is $I_{Ring} = MR^2$.The moment of inertia of a circular disc about an axis passing through its centre and perpendicular to its plane is $I_{Disc} = \frac{1}{2}MR^2$.Taking the ratio of the moment of inertia of the ring to that of the disc:$\frac{I_{Ring}}{I_{Disc}} = \frac{MR^2}{\frac{1}{2}MR^2} = \frac{1}{1/2} = 2:1$.

One circular ring and one circular disc,both having the same mass and radius. The ratio of their moments of inertia about the axes passing through their centres and perpendicular to their planes will be:

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