Four thin rods of same mass $M$ and same length $l$ form a square as shown in the figure. The moment of inertia of this system about an axis passing through the centre $O$ and perpendicular to its plane is

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 $I$)

Suppose there is a uniform circular disc of mass $M$ and radius $r$ shown in the figure. Two shaded circular regions, each of radius $r/4$, are cut out from the disc. The centers of these cut-out discs are at a distance of $3r/4$ from the center of the original disc. The moment of inertia of the remaining part about the axis $A$ (passing through the center of the disc and perpendicular to its plane) is given by $\frac{x}{256} Mr^2$. The value of $x$ is . . . . . . .

If the moment of inertia of a thin circular ring about an axis passing through its edge and perpendicular to its plane is $I$, then the moment of inertia of the ring about its diameter is

The moment of inertia of a rod about an axis passing through its center of mass and perpendicular to it is $\frac{1}{12}ML^2$. If the rod is bent at the middle such that the two halves make an angle of $90^{\circ}$ in the same plane,what will be the moment of inertia of the system about the same axis?

The rectangular block shown in the figure is rotated about the $x-x$,$y-y$,and $z-z$ axes passing through its center of mass $O$. Its moment of inertia is: