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

Two rings of radius $R$ and $nR$ having different masses and made up of the same wire have the ratio of moment of inertia about an axis passing through the centre as $1 : 8$. The value of $n$ is

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

$A$ solid sphere $A$ of radius $R$ and mass $M$ is attached at a point to a smaller solid sphere $B$ of radius $r < R$ and mass $m < M$. Assume that the line joining their centres lies along the horizontal. The moment of inertia of the system calculated about a vertical axis passing through the centre of $A$ is $I_A$ and that calculated about a vertical axis passing through the centre of $B$ is $I_B$. The difference $I_A - I_B$ is :

If the length of a thin uniform rod is $L$ and the radius of gyration of the rod about an axis perpendicular to its length and passing through one end is $K$, then $K: L=$

The ratio of the radius of gyration of a hollow sphere to that of a solid cylinder of equal mass,for the moment of inertia about their diameter axis $AB$ as shown in the figure,is $\sqrt{\frac{8}{x}}$. The value of $x$ is: