$I_1$ is the moment of inertia of a circular disc about an axis passing through its centre and perpendicular to the plane of the disc. $I_2$ is its moment of inertia about an axis $AB$ perpendicular to the plane and parallel to the axis $CM$ at a distance $\frac{2R}{3}$ from the centre. The ratio of $I_2$ to $I_1$ is $\frac{I_2}{I_1} = \frac{x}{9}$. The value of $x$ is ($R =$ radius of the disc).

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

$A$ solid sphere has mass $M$ and radius $R$. Its moment of inertia about a parallel axis passing through a point at a distance $R/3$ from its centre is:

The moment of inertia of a uniform circular disc is maximum about an axis perpendicular to the disc and passing through which of the following points?

$A$ square plate of mass $m$ and side $a$ has a moment of inertia about an axis perpendicular to its plane and passing through one of its corners equal to $\alpha \,ma^2$. Then $\alpha$ is:

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 . . . . . . .