$A$ mass tied to a string is whirled in a horizontal circular path with a constant angular velocity and its angular momentum is $L$. If the string is now halved,keeping the angular velocity the same,then the new angular momentum will be:

$A$ 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 $(1, 1, 0) \ m$.

$A$ fan of moment of inertia $0.6\,kg \cdot m^2$ is turned up to a working speed of $0.5$ revolutions per second. The angular momentum of the fan is:

$A$ particle of mass $m$ is moving in time $t$ on a trajectory given by $\overrightarrow{r} = 10 \alpha t^2 \hat{i} + 5 \beta (t - 5) \hat{j}$. Where $\alpha$ and $\beta$ are dimensional constants. The angular momentum of the particle becomes the same as it was for $t = 0$ at time $t = \dots$ seconds.

$A$ 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:

The units of angular momentum are