$A$ square lamina of side $b$ has the same mass as a disc of radius $R$. The moment of inertia of the two objects about an axis perpendicular to the plane and passing through the centre is equal. The ratio $\frac{b}{R}$ is

The moment of inertia of a semicircular ring of radius $R$ and mass $M$ about an axis passing through point $A$ (one end of the ring) and perpendicular to the plane of the paper is:

Moment of inertia of a disc of mass $M$ and radius $R$ about any of its diameter is $MR^2/4$. The moment of inertia of this disc about an axis normal to the disc and passing through a point on its edge will be $(x/2)MR^2$. The value of $x$ is

State and prove the theorem of perpendicular axes and the theorem of parallel axes.

$A$ 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

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