$A$ thin metal wire of length $L$ and mass $M$ is bent to form a semicircular ring as shown. The moment of inertia about the axis $XX^1$ passing through its ends is:

$A$ solid sphere of radius $10 \ cm$ is rotating about an axis which is at a distance $15 \ cm$ from its centre. The radius of gyration about this axis is $\sqrt{n} \ cm$. The value of $n$ is

Point masses $1, 2, 3$ and $4 \text{ kg}$ are located at the points $(0,0,0), (2,0,0), (0,3,0)$ and $(-2,-2,0)$ respectively. The moment of inertia of this system about the $x$-axis will be: (in $\text{ kg m}^2$)

Four particles,each of mass $1 \ kg$,are placed at the four corners of a square of side $2 \ m$. The moment of inertia of the system about an axis perpendicular to its plane and passing through one of its vertices is . . . . . . $kg \ m^2$.

$A$ solid sphere $A$ of radius $R$ and mass $M$ is attached at a point to a smaller solid sphere $B$ of radius $r < R$ and mass $m < M$. Assume that the line joining their centres lies along the horizontal. The moment of inertia of the system calculated about a vertical axis passing through the centre of $A$ is $I_A$ and that calculated about a vertical axis passing through the centre of $B$ is $I_B$. The difference $I_A - I_B$ is :

Match the following columns ($R=$ radius,$k=$ radius of gyration):

Column $I$	Column $II$
$(A)$ 'k' for a solid sphere rotating about its tangent	$(P)$ $\sqrt{2}R$
$(B)$ 'k' for a ring rotating about its tangent perpendicular to its plane	$(Q)$ $\frac{R}{2}$
$(C)$ 'k' for a uniform solid right circular cone rotating about its central axis	$(R)$ $\frac{\sqrt{7}}{\sqrt{5}}R$
$(D)$ 'k' for a uniform disc rotating about its diameter	$(S)$ $\frac{\sqrt{3}}{\sqrt{10}}R$