$A$ constant torque of $200 \, N-m$ turns a flywheel, which is at rest, about an axis through its centre and perpendicular to its plane. If its moment of inertia is $50 \, kg-m^{2}$, then in $4 \, s$, what will be the change in its angular momentum?

$A$ mass $m$ is moving with a constant velocity $v$ along a line parallel to the $X$-axis. Its angular momentum with respect to the origin or the $Z$-axis is:

$A$ particle of mass $2\, kg$ is on a smooth horizontal table and moves in a circular path of radius $0.6\, m$. The height of the table from the ground is $0.8\, m$. If the angular speed of the particle is $12\, rad\, s^{-1}$,the magnitude of its angular momentum about a point on the ground right under the centre of the circle is ........ $kg\, m^2\, s^{-1}$.

Two bodies rotate with kinetic energies $E_1$ and $E_2$. Their moments of inertia about their axes of rotation are $I_1$ and $I_2$. If $I_1 = \frac{I_2}{3}$ and $E_1 = 27 E_2$,then the ratio of their angular momenta $L_1$ to $L_2$ is:

$A$ particle having mass $4 \ kg$ is moving with velocity $(4 \hat{i} + 2 \hat{j}) \ m/s$. Find the angular momentum of the particle about the origin when it is at position $(1, 1, 0) \ m$.

The angular momentum of a particle is: