The moment of inertia of a thin uniform rod about a perpendicular axis passing through one of its ends is $I$. Now,the rod is bent into a ring and its moment of inertia about its diameter is $I_{1}$. Then $\frac{I}{I_{1}}$ is:

The radius of gyration of a cylindrical rod about an axis of rotation perpendicular to its length and passing through its center is $k$. Given that the length of the rod is $10 \sqrt{3} \ m$,find the value of $k$ in meters.

If the radius of a solid sphere is doubled while keeping its mass constant,what is the ratio of its moment of inertia about any of its diameters?

$A$ disc of radius $R$ and thickness $\frac{R}{6}$ has moment of inertia $I$ about an axis passing through its centre and perpendicular to its plane. The disc is melted and recast into a solid sphere. The moment of inertia of the sphere about its diameter is

The moment of inertia of a uniform rod of mass $M$ and length $L$ about an axis passing through its center and perpendicular to it is $\frac{1}{12} ML^2$. The rod is bent at the center such that the two halves make an angle of $60^\circ$ with each other. Find the moment of inertia of the bent rod about the same axis passing through the center of the rod (the point of bending) and perpendicular to the plane containing the two halves.

$A$ solid sphere of mass $M$ and radius $R$ is divided into two unequal parts. The first part has a mass of $\frac{7M}{8}$ and is converted into a uniform disc of radius $2R$. The second part is converted into a uniform solid sphere. Let $I_1$ be the moment of inertia of the disc about its axis and $I_2$ be the moment of inertia of the new sphere about its axis. The ratio of $I_1/I_2$ is given by