$A$ 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.

The moment of inertia of a cylinder of mass $M$,length $L$,and radius $R$ about an axis passing through its centre and perpendicular to the axis of the cylinder is $I = M \left(\frac{R^2}{4} + \frac{L^2}{12}\right)$. If such a cylinder is to be made for a given mass of material,the ratio $L/R$ for it to have the minimum possible $I$ is

$A$ long slender rod is welded to a thin circular disc of diameter $0.5 \ m$ at a point on its circumference. The rod is in the same plane as that of the disc and forms a tangent to the disc. The radius of gyration of the disc about the rod (in $m$) is

$A$ one-quarter sector is cut from a uniform circular disc of radius $R$. This sector has mass $M$. It is made to rotate about a line perpendicular to its plane and passing through the centre of the original disc. Its moment of inertia about the axis of rotation is

Two solid spheres ($A$ and $B$) are made of metals of different densities $\rho_A$ and $\rho_B$ respectively. If their masses are equal, the ratio of their moments of inertia $(I_B / I_A)$ about their respective diameter is

The given figure shows a disc of mass $M$ and radius $R$ lying in the $x-y$ plane with its center on the $x$-axis at a distance $a$ from the origin. Then,the moment of inertia of the disc about the $x$-axis is: