If the mass and radius of a hollow sphere are $M$ and $R$ respectively,then its moment of inertia with respect to its diameter is:

Two identical rods each of mass $M$ and length $l$ are joined in a crossed position as shown in the figure. The moment of inertia of this system about a bisector ($B_1$ or $B_2$) is:

Four spheres of diameter $2a$ and mass $M$ are placed with their centers on the four corners of a square of side $b$. The moment of inertia of the system about an axis along one of the sides of the square is

Two circular discs $A$ and $B$ are of equal masses and thickness but made of metals with densities ${d_A}$ and ${d_B}$ $({d_A} > {d_B})$. If their moments of inertia about an axis passing through their centres and normal to the circular faces are ${I_A}$ and ${I_B}$ respectively,then:

$A$ circular disc $A$ of radius $r$ is made from an iron plate of thickness $t$ and another circular disc $B$ of radius $4r$ is made from an iron plate of thickness $t/4$. The relation between the moments of inertia $I_A$ and $I_B$ is:

Let $M$ be the mass and $L$ be the length of a thin uniform rod. In the first case,the axis of rotation passes through the center and is perpendicular to the length of the rod. In the second case,the axis of rotation passes through one end and is perpendicular to the length of the rod. The ratio of the radius of gyration in the first case to the second case is