From a circular disc of radius $R$ and mass $9M$,a small disc of radius $\frac{R}{3}$ is removed concentrically. If the mass of the removed disc is $M$,what is the moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through its centre?

$A$ uniform rod of mass $M$ and length $L$ has a moment of inertia $I$ about an axis passing through its center and perpendicular to the rod. What will be its moment of inertia about an axis passing through one of its ends and perpendicular to the rod?

The moment of inertia of a square plate of side $l$ and mass $M$ about an axis passing through one of its corners and perpendicular to the plane of the square plate is given by:

The moment of inertia of a thin uniform rod of length $L$ and mass $M$ about an axis passing through a point at a distance of $\frac{L}{3}$ from one of its ends and perpendicular to the rod is

$A$ rigid body of mass $m$ is rotating with angular velocity $\omega$ about an axis at a distance '$a$' from the center of mass $G$. The radius of gyration about an axis passing through $G$ is $K$. The rotational kinetic energy of this body about the new parallel axis will be:

Suppose there is a uniform circular disc of mass $M$ and radius $r$ shown in the figure. Two shaded circular regions, each of radius $r/4$, are cut out from the disc. The centers of these cut-out discs are at a distance of $3r/4$ from the center of the original disc. The moment of inertia of the remaining part about the axis $A$ (passing through the center of the disc and perpendicular to its plane) is given by $\frac{x}{256} Mr^2$. The value of $x$ is . . . . . . .