The position vector of a $1\,kg$ object is $\overrightarrow{r} = (3\hat{i} - \hat{j})\,m$ and its velocity is $\overrightarrow{v} = (3\hat{j} + \hat{k})\,m/s$. The magnitude of its angular momentum is $\sqrt{x}\,N\cdot m\cdot s$,where $x$ is:

An object of mass $m$ is projected from the origin in a vertical $xy$ plane at an angle $45^{\circ}$ with the $x$-axis with an initial velocity $v_0$. The magnitude and direction of the angular momentum of the object with respect to the origin,when it reaches the maximum height,will be [$g$ is the acceleration due to gravity].

If a particle of mass $m$ is moving with constant velocity $v$ parallel to the $X$-axis in the $x-y$ plane as shown in the figure,its angular momentum with respect to the origin at any time $t$ will be:

The angular momentum of a system of particles changes if:

$A$ particle of mass $m$ is projected from a point $P$ on the ground with an initial velocity $v_0$ at an angle of $45^{\circ}$ with the horizontal at $t = 0$. Find the magnitude of the angular momentum of the particle at time $t = \frac{v_0}{g}$.

$A$ 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: