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:

$A$ mass $M$ hangs on a massless rod of length $l$ which rotates at a constant angular frequency $\omega$. The mass $M$ moves with steady speed in a circular path of constant radius $r$. The angular momentum of $M$ about point $A$ is $L_A$,which lies in the positive $z$-direction,and the angular momentum of $M$ about point $B$ is $L_B$. Which of the following statements is correct for this system?

$A$ particle of mass $m$ is moving with constant velocity $v$ parallel to the $x$-axis as shown in the figure. Its angular momentum about origin $O$ 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.

$A$ uniform rod $AB$ of mass $m$ and length $l$ is at rest on a smooth horizontal surface. An impulse $P$ is applied to the end $B$. The time taken by the rod to turn through a right angle is

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: