The potential energy of a particle of mass $m$ at a distance $r$ from a fixed point $O$ is given by $\mathrm{V}(\mathrm{r})=\mathrm{kr}^2 / 2$, where $\mathrm{k}$ is a positive constant of appropriate dimensions. This particle is moving in a circular orbit of radius $\mathrm{R}$ about the point $\mathrm{O}$. If $\mathrm{v}$ is the speed of the particle and $\mathrm{L}$ is the magnitude of its angular momentum about $\mathrm{O}$, which of the following statements is (are) true?

$(A)$ $v=\sqrt{\frac{k}{2 m}} R$

$(B)$ $v=\sqrt{\frac{k}{m}} R$

$(C)$ $\mathrm{L}=\sqrt{\mathrm{mk}} \mathrm{R}^2$

$(D)$ $\mathrm{L}=\sqrt{\frac{\mathrm{mk}}{2}} \mathrm{R}^2$

The position vector of $1\,kg$ object is $\overrightarrow{ r }=(3 \hat{ i }-\hat{ j })\,m$ and its velocity $\overrightarrow{ v }=(3 \hat{ j }+ k )\,ms ^{-1}$. The magnitude of its angular momentum is $\sqrt{ x } Nm$ where $x$ is

$A$ uniform rod is fixed to a rotating turntable so that its lower end is on the axis of the turntable and it makes an angle of $20^o$ to the vertical. (The rod is thus rotating with uniform angular velocity about a vertical axis passing through one end.) If the turntable is rotating clockwise as seen from above. What is the direction of the rod's angular momentum vector (calculated about its lower end)?

If the resultant of all the external forces acting on a system of particles is zero, then from an inertial frame, one can surely say that

A particle is moving along a straight line with increasing speed. Its angular momentum about a fixed point on this line