Find the moment of inertia of a ring about an axis passing through the centre and perpendicular to its plane.

The moment of inertia $(M.I.)$ of four bodies,having the same mass $M$ and radius $R$,are reported as follows:
$I_{1} = M.I.$ of a thin circular ring about its diameter.
$I_{2} = M.I.$ of a circular disc about an axis perpendicular to the disc and passing through the centre.
$I_{3} = M.I.$ of a solid cylinder about its axis.
$I_{4} = M.I.$ of a solid sphere about its diameter.
Then:

Two rings have their masses in ratio $1 : 2$ and their diameters are in the ratio $2 : 1$. The ratio of their moments of inertia is

Two half rings are joined as shown in the figure. Each half ring has a radius $R$ and mass $M$. The moment of inertia of the system about the axis $XX'$ is

$A$ system consists of two identical spheres,each of mass $1.5\, {kg}$ and radius $50\, {cm}$,attached to the ends of a light rod. The distance between the centers of the two spheres is $5\, {m}$. What is the moment of inertia of the system about an axis perpendicular to the rod and passing through its midpoint? (in ${kgm}^{2}$)

Match List-$I$ with List-$II$:

List-$I$	List-$II$
$(A)$ Moment of inertia of solid sphere of radius $R$ about any tangent	$(I)$ $\frac{5}{3} MR^2$
$(B)$ Moment of inertia of hollow sphere of radius $R$ about any tangent	$(II)$ $\frac{7}{5} MR^2$
$(C)$ Moment of inertia of circular ring of radius $R$ about its diameter	$(III)$ $\frac{1}{4} MR^2$
$(D)$ Moment of inertia of circular disc of radius $R$ about any diameter	$(IV)$ $\frac{1}{2} MR^2$

Choose the correct answer from the options given below: