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

$A$ particle of mass $m$ moves in the $XY$ plane with a velocity $v$ along the straight line $AB.$ If the angular momentum of the particle with respect to origin $O$ is $L_A$ when it is at $A$ and $L_B$ when it is at $B,$ then

Two cars $A$ and $B$ each of mass $10^3 \text{ kg}$ are moving on parallel tracks separated by a distance of $10 \text{ m}$,in the same direction with speeds $72 \text{ km/h}$ and $36 \text{ km/h}$. The magnitude of angular momentum of car $A$ with respect to car $B$ is . . . . . . $\text{J} \cdot \text{s}$.

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

As shown in the figure,a particle of mass $m$ moves with a velocity $v$ along the line $PC$. What is the angular momentum of this particle with respect to point $P$?

$A$ torque of $50 \ Nm$ acts on a body for $8 \ s$,which is initially at rest. The change in its angular momentum is: