$A$ particle is moving in a circular path of radius $a$ with a constant speed $v$ as shown in the figure. The centre of the circle is marked by $C$. The angular momentum about the origin $O$ can be written as:

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:

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:

When the position vector $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ changes sign to $-\vec{r}$, which one of the following vectors will not flip its sign?

$A$ 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 is moving uniformly along a straight line as shown in the figure. During the motion of the particle from $A$ to $B$,the angular momentum of the particle about '$O$' is: