If a spherical planet of mass $M$ and radius $R$ suddenly shrinks to half its size,and its mass reduces to half,what is the new moment of inertia of the planet about its diameter?

$A$ quarter circular sector is cut from a uniform circular disc. The mass of this sector is $M$. It is rotated about an axis passing through the center of the original disc and perpendicular to its plane. The moment of inertia of this sector about the axis of rotation is:

$A$ circular disc of radius $10 \ cm$ and thickness $5 \ mm$ has a uniform density of $8 \ g/cc$. Find the moment of inertia of the disc about an axis passing through its center and perpendicular to its plane.

$A$ circular disc of radius $b$ has a hole of radius $a$ at its centre (see figure). If the mass per unit area of the disc varies as $\sigma(r) = \frac{\sigma_0}{r}$,then the radius of gyration of the disc about its axis passing through the centre is

Four point masses,each of value $m$,are placed at the corners of a square $ABCD$ of side $l$. The moment of inertia of this system about an axis passing through $A$ and parallel to $BD$ is

Two circular discs $A$ and $B$ are of equal masses and thickness but made of metals with densities ${d_A}$ and ${d_B}$ $({d_A} > {d_B})$. If their moments of inertia about an axis passing through their centres and normal to the circular faces are ${I_A}$ and ${I_B}$ respectively,then: