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

$A$ thin wire of length $L$ and linear mass density $m$ is bent into a circular ring (in $x-y$ plane) with centre $C$ as shown in the figure. The moment of inertia of the ring about an axis $yy'$ will be:

$A$ thin uniform rod has mass $M$ and length $L$. The moment of inertia about an axis perpendicular to it and passing through the point at a distance $\frac{L}{3}$ from one of its ends,will be

$I_1$ is the moment of inertia of a circular disc about an axis passing through its centre and perpendicular to the plane of the disc. $I_2$ is its moment of inertia about an axis $AB$ perpendicular to the plane and parallel to the axis $CM$ at a distance $\frac{2R}{3}$ from the centre. The ratio of $I_2$ to $I_1$ is $\frac{I_2}{I_1} = \frac{x}{9}$. The value of $x$ is ($R =$ radius of the disc).

Consider a uniform square plate of side '$a$' and mass '$m$'. The moment of inertia of this plate about an axis perpendicular to its plane and passing through one of its corners is

$A$ sphere of radius $R$ is cut from a larger solid sphere of radius $2R$ as shown in the figure. The ratio of the moment of inertia of the smaller sphere to that of the rest part of the sphere about the $Y$-axis is