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(C) The moment of inertia of a circular disc of mass $m$ and radius $R$ about an axis passing through its center of mass $(C.M.)$ and perpendicular to

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$1:3$

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(C) The moment of inertia of a circular disc of mass $m$ and radius $R$ about an axis passing through its center of mass $(C.M.)$ and perpendicular to its plane is given by:$I_z = \frac{1}{2} mR^2$The axis $z'$ is a tangential axis parallel to the $z$-axis. According to the parallel axis theorem,the moment of inertia about an axis parallel to the central axis is given by $I = I_{cm} + md^2$,where $d$ is the distance between the axes. Here,$d = R$.$I_{z'} = I_z + mR^2 = \frac{1}{2} mR^2 + mR^2 = \frac{3}{2} mR^2$The ratio of the moment of inertia about $z$ and $z'$ axes is:$\frac{I_z}{I_{z'}} = \frac{\frac{1}{2} mR^2}{\frac{3}{2} mR^2} = \frac{1}{3}$Thus,the ratio is $1:3$.

$A$ thin circular disc is in the $xy$ plane as shown in the figure. The ratio of its moment of inertia about $z$ and $z'$ axes will be

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