$A$ solid sphere and a thin-walled hollow sphere have the same mass and the same material. Which of them has a greater moment of inertia about its diameter? ($I_{h} =$ moment of inertia of a hollow sphere about an axis coinciding with its diameter,$I_{s} =$ moment of inertia of a solid sphere about an axis coinciding with its diameter)

From a circular ring of mass $M$ and radius $R$,an arc corresponding to a $90^{\circ}$ sector is removed. The moment of inertia of the remaining part of the ring about an axis passing through the centre of the ring and perpendicular to the plane of the ring is $K$ times $MR^{2}$. Then the value of $K$ 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 :

Why does a solid sphere have a smaller moment of inertia than a hollow cylinder of the same mass and radius,about an axis passing through their axes of symmetry?

The radius of gyration of a cylindrical rod about an axis of rotation perpendicular to its length and passing through its center is $k$. Given that the length of the rod is $10 \sqrt{3} \ m$,find the value of $k$ in meters.

Two identical thin rods of mass $M \ kg$ and length $L \ m$ are connected as shown in the figure. The moment of inertia of the combined rod system about an axis passing through point $P$ and perpendicular to the plane of the rods is $\frac{x}{12} M L^2 \ kg \ m^2$. The value of $x$ is . . . . . . .