The moment of inertia of a ring of mass $M$ and radius $R$ about an axis passing through its center and perpendicular to its plane is:

The moments of inertia of a non-uniform circular disc (of mass $M$ and radius $R$) about four mutually perpendicular tangents $AB, BC, CD, DA$ are $I_1, I_2, I_3$ and $I_4$,respectively (the square $ABCD$ circumscribes the circle). The distance of the centre of mass of the disc from its geometrical centre is given by

The ratio of the radius of gyration of a solid sphere of mass $M$ and radius $R$ about its own axis to the radius of gyration of a thin hollow sphere of the same mass and radius about its axis is:

$A$ circular disc is made using iron and aluminum such that its moment of inertia about its geometric axis is maximum. For this to happen:

The radius of gyration of a disc of mass $50 \, g$ and radius $2.5 \, cm$ about an axis passing through its center of gravity and perpendicular to its plane is ....... $cm$.

From a circular ring of mass $M$ and radius $R$,an arc corresponding to a $90^{\circ}$ sector is removed. The moment of inertia of the remaining part of the ring about an axis passing through the centre of the ring and perpendicular to the plane of the ring is $K$ times $MR^{2}$. Then the value of $K$ is