What is the moment of inertia of a ring of mass $M$ and radius $R$ about the axis $PQ$ as shown in the figure?

If a solid sphere of mass $5 \, kg$ and a disc of mass $4 \, kg$ have the same radius $R$,then the ratio of the moment of inertia of the disc about a tangent in its plane to the moment of inertia of the sphere about its tangent is $\frac{x}{7}$. The value of $x$ is $.........$

$A$ disc and a ring both have the same mass and radius. The ratio of the moment of inertia of the disc about its diameter to that of a ring about a tangent in its plane is

$A$ thin uniform square lamina of side $a$ is placed in the $xy$-plane with its sides parallel to $X$ and $Y$-axes and with its centre coinciding with the origin. Its moment of inertia about an axis passing through a point on the $Y$-axis at a distance $y = 2a$ and parallel to the $X$-axis is equal to its moment of inertia about an axis passing through a point on the $X$-axis at a distance $x = d$ and perpendicular to the $xy$-plane. Then the value of $d$ is:

$A$ 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 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