$A$ particle of mass $2\, kg$ is moving such that at time $t$,its position,in meters,is given by $\vec{r}(t) = 5\hat{i} - 2t^2\hat{j}$. The angular momentum of the particle at $t = 2\, s$ about the origin in $kg\, m^2\, s^{-1}$ is:

$A$ time-varying force $F = 2t$ is applied on a spool rolling as shown in the figure. The angular momentum of the spool at time $t$ about the bottommost point is:

$A$ thin uniform rod of length $L$ and mass $m$ is kept on a frictionless horizontal table with a massless string of length $L$ fixed to one end (top view is shown in the figure). The other end of the string is pivoted to a point $O$. If a horizontal impulse $P$ is imparted to the rod at a distance $x = L/n$ from the mid-point of the rod (see figure), then the rod and string revolve together around the point $O$, with the rod remaining aligned with the string. In such a case, the value of $n$ is:

$A$ 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?

Angular momentum is

$A$ particle of mass $0.5\, kg$ is rotating in a circular path of radius $2\, m$ and centripetal force on it is $9\, N$. Its angular momentum (in $J\cdot s$) is: