The moment of inertia of a semicircular ring about an axis passing through the center and perpendicular to the plane of the ring is $\frac{1}{x} MR^2$,where $R$ is the radius and $M$ is the mass of the semicircular ring. The value of $x$ will be $...........$

On account of the melting of ice at the North Pole,the moment of inertia of the spinning Earth:

The moment of inertia of a body is $160 \ kg \ m^2$ and its mass is $10 \ kg$. What will be the radius of gyration in $m$?

One solid sphere $A$ and another hollow sphere $B$ have the same mass and the same outer radii. Their moments of inertia about their diameters are $I_{A}$ and $I_{B}$ respectively. Which of the following relations is correct?

$A$ 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 :

Moment of Inertia $(M.I.)$ of four bodies having same mass $M$ and radius $2R$ are as follows:
$I_{1} =$ $M.I.$ of solid sphere about its diameter
$I_{2} =$ $M.I.$ of solid cylinder about its axis
$I_{3} =$ $M.I.$ of solid circular disc about its diameter
$I_{4} =$ $M.I.$ of thin circular ring about its diameter
If $2(I_{2} + I_{3}) + I_{4} = x I_{1}$,then the value of $x$ will be...