$A$ person is sitting on a rotating table with their arms outstretched. If they suddenly pull their arms inward,then:

Two rods,each of mass $m$ and length $l$,are joined at their centers to form a cross. What is the moment of inertia of the cross about an axis passing through their common center and perpendicular to the plane of the rods?

Three point masses,each of mass $m$,are placed at the corners of an equilateral triangle of side $a$. The moment of inertia of this system about an axis passing along one side of the triangle is:

The moment of inertia $(M.I.)$ of four bodies,having the same mass $M$ and radius $R$,are reported as follows:
$I_{1} = M.I.$ of a thin circular ring about its diameter.
$I_{2} = M.I.$ of a circular disc about an axis perpendicular to the disc and passing through the centre.
$I_{3} = M.I.$ of a solid cylinder about its axis.
$I_{4} = M.I.$ of a solid sphere about its diameter.
Then:

We have two spheres,one solid and one hollow. Their moments of inertia about their respective diameters are equal. The ratio of their radii will be:

Three identical rods each of mass $M$ and length $L$ are joined to form a symbol $H$. The moment of inertia of the system about one of the sides of $H$ is