The moment of linear momentum is called:

$A$ particle starts from the point $(0, 8) \, m$ and moves with a uniform velocity of $\vec{v} = 3 \hat{i} \, m/s$. What is the angular momentum of the particle about the origin after $5 \, s$ (mass of the particle is $1 \, kg$)?

$A$ cylinder of mass $5 \ kg$ and radius $30 \ cm$ is free to rotate about its axis. It receives an initial angular impulse of $3 \ kg \ m^2 s^{-1}$ and subsequently receives the same impulse every $4 \ s$. What will be the angular speed of the cylinder $30 \ s$ after the initial impulse? The cylinder is initially at rest.

$A$ 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 length of the string is now halved, keeping the angular velocity the same, then the angular momentum will be:

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 position vector of a particle of mass $m$ moving with a constant velocity $\vec{v} = v \hat{i}$ is given by $\vec{r} = x(t) \hat{i} + b \hat{j}$, where $b$ is a constant. At an instant, $\vec{r}$ makes an angle $\theta$ with the $x$-axis. The variation of the magnitude of the angular momentum $|\vec{L}|$ of the particle about the origin with $\theta$ will be: