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:

The angular momentum of a system of particles changes if:

$A$ particle of mass $2\, kg$ is moving such that at time $t$,its position,in meters,is given by $\vec{r}(t) = 5\hat{i} - 2t^2\hat{j}$. The angular momentum of the particle at $t = 2\, s$ about the origin in $kg\, m^2\, s^{-1}$ is:

$A$ body of mass $2 \ kg$ is moving in a circular path of radius $2 \ m$ with uniform speed. If the centripetal force acting on it is $100 \ N$,then its angular momentum is ....... $J \cdot s$.

$A$ particle of mass $m$ moves along a straight line $AB$ in the $XY$ plane. Let $L_A$ and $L_B$ be the angular momentum of the particle about the origin $O$ when it is at points $A$ and $B$ respectively. Then:

The moment of inertia of a uniform rod of mass $M$ and length $L$ about an axis through its centre and perpendicular to its length is given by $\frac{ML^2}{12}$. Now,consider one such rod pivoted at its centre,free to rotate in a vertical plane. The rod is at rest in the vertical position. $A$ bullet of mass $M$ moving horizontally at a speed $v$ strikes and embeds itself in one end of the rod. The angular velocity of the rod just after the collision will be