$A$ circular disc of radius $10 \ cm$ and thickness $5 \ mm$ has a uniform density of $8 \ g/cc$. Find the moment of inertia of the disc about an axis passing through its center and perpendicular to its plane.

$A$ 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

The ratio of the radius of gyration of a ring to that of a disc (both circular) of the same radius and mass,about a tangential axis perpendicular to the plane is

$A$ semi-circular disc has mass $M$ and radius $r$. The moment of inertia about an axis passing through its center and perpendicular to the plane of the disc is:

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:

The ratio of the moments of inertia of two uniform rings of radii $R$ and $nR$ about an axis passing through their centers and perpendicular to their planes is $1 : 8$. What is the value of $n$?