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 is 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

What is the physical quantity of the time rate of the angular momentum ?

$A$ paritcle falls freely near the surface of the earth. Consider $a$ fixed point $O$ (not vertically below the particle) on the ground.

In an orbital motion, the angular momentum vector is

A particle of mass $M=0.2 kg$ is initially at rest in the $x y$-plane at a point $( x =-l, y =-h)$, where $l=10 m$ and $h=1 m$. The particle is accelerated at time $t =0$ with a constant acceleration $a =10 m / s ^2$ along the positive $x$-direction. Its angular momentum and torque with respect to the origin, in SI units, are represented by $\vec{L}$ and $\vec{\tau}$, respectively. $\hat{i}, \hat{j}$ and $\hat{k}$ are unit vectors along the positive $x , y$ and $z$-directions, respectively. If $\hat{k}=\hat{i} \times \hat{j}$ then which of the following statement($s$) is(are) correct?

$(A)$ The particle arrives at the point $(x=l, y=-h)$ at time $t =2 s$.

$(B)$ $\vec{\tau}=2 \hat{ k }$ when the particle passes through the point $(x=l, y=-h)$

$(C)$ $\overrightarrow{ L }=4 \hat{ k }$ when the particle passes through the point $(x=l, y=-h)$

$(D)$ $\vec{\tau}=\hat{ k }$ when the particle passes through the point $(x=0, y=-h)$