Four thin rods of same mass M and same length L form a square as shown in figure. Moment of inertia of this system about an axis passing through point O and perpendicular to its plane is

A solid sphere and a ring have equal mass and equal radius of gyration. If sphere is rotating about its diameter and ring about an axis passing through centre and perpendicular to its plane, then the ratio of radius of sphere to that of ring is $\sqrt{\frac{x}{2}}$ then the value of ' $x$ ' is

Two bodies A and B have their moments of inertia ' I ' and ' 2 I ' respectively about their axis of rotation. If their kinetic energy of rotation are equal then angular momentum of body A to that of body $B$ will be in the ratio

A thin rod of length ' 4 L ' and mass ' 4 m ' is bent at the points as shown in the figure. The moment of inertia of the rod about an axis passing through point ' $O$ ' and perpendicular to plane of the paper is