Angular momentum is

$A$ thin uniform rod of length $L$ and mass $m$ is kept on a frictionless horizontal table with a massless string of length $L$ fixed to one end (top view is shown in the figure). The other end of the string is pivoted to a point $O$. If a horizontal impulse $P$ is imparted to the rod at a distance $x = L/n$ from the mid-point of the rod (see figure), then the rod and string revolve together around the point $O$, with the rod remaining aligned with the string. In such a case, the value of $n$ 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 ....... $J \cdot s$.

Two rigid bodies $A$ and $B$ rotate with rotational kinetic energies $E_A$ and $E_B$ respectively. The moments of inertia of $A$ and $B$ about the axis of rotation are $I_A$ and $I_B$ respectively. If $I_A = I_B/4$ and $E_A = 100 E_B$,the ratio of angular momentum $(L_A)$ of $A$ to the angular momentum $(L_B)$ of $B$ 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:

$A$ small mass $m$ is attached to a massless string whose other end is fixed at $P$ as shown in the figure. The mass is undergoing circular motion in the $x-y$ plane with centre at $O$ and constant angular speed $\omega$. If the angular momentum of the system,calculated about $O$ and $P$ are denoted by $\vec{L}_O$ and $\vec{L}_P$ respectively,then