Calculate the moment of inertia of a thin uniform ring about an axis tangent to the ring and in a plane of the ring, if its moment of inertia about an axis passing through the centre and perpendicular to plane is $4 \mathrm{kg} {\mathrm{m}}^2$.

A solid sphere of radius $R$ has a moment of inertia $I$ about its geometrical axis. If it is melted into a disc of radius $r$ and thickness $t$. If it's moment of inertia about the tangential axis (which is perpendicular to the plane of the disc), is also equal to $I$, then the value of $r$ is equal to

A disc of mass $M$ and radius $R$ is lying on the $x\text{-}y$ plane. The locus of all points on the $x\text{-}y$ plane about which the moment of inertia of the rod is same as that about $0$ will be,

A thin uniform rectangular plate of mass $2 \mathrm{kg}$ is placed in $X-Y$ plane as shown in the figure. The moment of inertia about $x$-axis is $I_x=0.2 \mathrm{kg} {\mathrm{m}}^2$ and the moment of inertia about $y$ - axis is $I_y=0.3 \mathrm{kg} {\mathrm{m}}^2$. The radius of gyration of the plate about the axis passing through $O$ and perpendicular to the plane of the plate is