$A$ solid sphere and a solid cylinder have the same mass and same radius. The ratio of the moment of inertia of the solid sphere about its diameter and the moment of inertia of the solid cylinder about its axis is

Assertion $(A)$: The moment of inertia of a steel sphere is larger than the moment of inertia of a wooden sphere of the same radius.
Reason $(R)$: Moment of inertia is independent of the mass of the body.
The correct one is:

Two discs have densities $d_1 = 7.2 \ g/cm^3$ and $d_2 = 8.9 \ g/cm^3$. Both discs have the same mass and the same thickness. Find the ratio of their moments of inertia.

The moment of inertia of a uniform thin rod of mass $m$ and length $\ell$ about a perpendicular axis passing through one end is $I_{1}$. The same rod is bent into a ring and its moment of inertia about a diameter is $I_{2}$. If $\frac{I_{1}}{I_{2}} = \frac{x \pi^{2}}{3}$,then the value of $x$ will be ...............

If $I_1$ is the moment of inertia of a thin rod of mass $M$ and length $l$ about an axis perpendicular to its length and passing through its centre of mass,and $I_2$ is the moment of inertia of the ring formed by bending the rod about an axis passing through its centre and perpendicular to its plane,then:

The mass per unit area of a circular disc of radius $a$ depends on the distance $r$ from its centre as $\sigma(r) = A + Br$. The moment of inertia of the disc about the axis perpendicular to the plane and passing through its centre is