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

$A$ circular disc of mass $9M$ and radius $R$ has a smaller disc of radius $R/3$ cut from it. Calculate the moment of inertia of the remaining portion about an axis passing through the center of the original disc and perpendicular to its plane. (in $MR^2$)

The moment of inertia of a square loop made of four uniform solid cylinders, each having radius $R$ and length $L$ $(R < L)$, about an axis passing through the midpoints of opposite sides, is (Take the mass of the entire loop as $M$):

Two spheres each of mass $M$ and radius $\frac{R}{2}$ are connected at the ends of a massless rod of length $2R$. What will be the moment of inertia of the system about an axis passing through the centre of one of the spheres and perpendicular to the rod?

Three rods each of mass $m$ and length $L$ are joined to form an equilateral triangle as shown in the figure. What is the moment of inertia about an axis passing through the centre of mass of the system and perpendicular to the plane?

$M$ and $R$ are the mass and radius of a disc. $A$ small disc of radius $R/3$ is removed from the bigger disc as shown in the figure. The moment of inertia of the remaining part of the bigger disc about an axis $\text{AB}$ passing through the centre $O$ and perpendicular to the plane of the disc is $\frac{4}{x} MR^2$. The value of $x$ is . . . . . . .