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 of mass $1 \ kg$ is moving with a linear velocity of $2 \ ms^{-1}$ parallel to the positive $X$-axis. During this motion,its minimum distance from the origin is $12 \ cm$. The angular momentum of this particle about the origin is ....... $Js$.

The units of angular momentum are

$A$ 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:

$A$ 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

$A$ 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: