$A$ lamina is made by removing a small disc of diameter $2R$ from a bigger disc of uniform mass density and radius $2R$,as shown in the figure. The moment of inertia of this lamina about axes passing through $O$ and $P$ is $I_0$ and $I_P$,respectively. Both these axes are perpendicular to the plane of the lamina. The ratio $\frac{I_P}{I_0}$ to the nearest integer is:

The moment of inertia of a circular disc of radius $2 \,m$ and mass $1 \,kg$ about an axis passing through the centre of mass but perpendicular to the plane of the disc is $2 \,kg \,m^{2}$. Its moment of inertia about an axis parallel to this axis but passing through the edge of the disc is (see the given figure).

Four identical discs each of mass $M$ and diameter $a$ are arranged in a plane as shown in the figure. If the moment of inertia of the system about $OO'$ is $\frac{x}{4} Ma^2$,then the value of $x$ 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 $........$

$A$ 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:

The moment of inertia of a uniform semicircular wire of mass $m$ and radius $r$,about an axis passing through its centre of mass and perpendicular to its plane is ..........