The moment of inertia of a straight thin rod of mass $M$ and length $l$ about an axis perpendicular to its length and passing through its one end,is

$A$ 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 :

Four particles each of mass $M$ are placed at the corners of a square of side $L$. The radius of gyration of the system about an axis perpendicular to the square and passing through its centre is

The moment of inertia of a regular circular disc of mass $0.4\, kg$ and radius $100\, cm$ about an axis perpendicular to the plane of the disc and passing through its centre is ...... $kg\, m^2$.

The moment of inertia of a sphere of mass $M$ and radius $R$ is $I.$ If $M$ is kept constant and a graph is plotted between $I$ and $R,$ then its form would be:

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	$(S)$ $\frac{\sqrt{3}}{\sqrt{10}}R$