Two identical rings A and B of same mass and radius are revolving, ring A arounds its own diameter and ring B about tangential axis in its own plane. Both the rings A and B have same rotational kinetic energy. The ratio of the angular velocity of ring B to that of ring A is

A body is rotating about its own axis. Its rotational kinetic energy is '$x$' and its angular momentum is '$y$'. Hence its moment of inertia about its own axis is

A force of $3\hat{i} + 2\hat{j} - \hat{k}$ N acts on a particle with position vector $\hat{i} + \hat{j} - \hat{k}$ m. The magnitude of torque of given force is

A thin uniform rod AB of mass '$m$' and length '$l$' is hinged at one end A to the ground level. Initially the rod stands vertically and is allowed to fall freely to the ground in the vertical plane. The angular velocity of the rod when its B end strikes the ground is
($g$ = acceleration due to gravity)