If the rotational kinetic energy of a body is $10 \ J$,and its angular momentum vector coincides with the axis of rotation,and its moment of inertia about this axis is $8 \times 10^{-7} \ kg \ m^2$,then the angular momentum of the body will be:

$A$ ring of mass $10 \ kg$ and diameter $0.4 \ m$ is rotating about its axis. If it completes $2100$ revolutions per minute,what is its angular momentum in $kg \cdot m^2/s$?

$A$ particle is moving uniformly along a straight line as shown in the figure. During the motion of the particle from $A$ to $B$,the angular momentum of the particle about '$O$' is:

Explain the angular momentum of a particle and show that it is the moment of linear momentum about the reference point.

The moment of inertia of a uniform rod of mass $M$ and length $L$ about an axis through its centre and perpendicular to its length is given by $\frac{ML^2}{12}$. Now,consider one such rod pivoted at its centre,free to rotate in a vertical plane. The rod is at rest in the vertical position. $A$ bullet of mass $M$ moving horizontally at a speed $v$ strikes and embeds itself in one end of the rod. The angular velocity of the rod just after the collision will be

If the Earth is considered a sphere of radius $R$ and mass $M$,then its angular momentum about its axis of rotation in terms of the time period $T$ will be: