The moment of inertia of a sphere about its diameter is $I$. Four such spheres are arranged as shown in the figure. Calculate the moment of inertia of the system about the axis $XX'$. (in $I$)

The moment of inertia of a rod of mass $M$ and length $L$ about an axis passing through its center and normal to its length is $\alpha$. Now,the rod is cut into two equal parts,and these parts are joined symmetrically to form a cross shape. The moment of inertia of the cross about an axis passing through its center and normal to the plane containing the cross is:

Four identical uniform solid spheres,each of mass $M$ and radius $R$,are placed touching each other in a line as shown in the figure,with centers $A, B, C, D$. Let $I_A, I_B, I_C,$ and $I_D$ be the moment of inertia of the entire system of four spheres about an axis passing through the center of the respective sphere and perpendicular to the plane containing the centers. The difference $|I_A - I_B|$ is: (in $MR^2$)

$A$ circular thin disc of mass $2 \ kg$ has a diameter $0.2 \ m$. Calculate its moment of inertia about an axis passing through the edge and perpendicular to the plane of the disc (in $kg \cdot m^2$).

If an object lies in the $X-Y$ plane,then according to the Perpendicular Axis Theorem:

Four identical solid spheres each of mass $m$ and radius $a$ are placed with their centres on the four corners of a square of side $b$. The moment of inertia of the system about one side of the square is: