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(B) The angular velocity $\omega$ of a particle moving with velocity $v$ at a distance $r$ from a point is given by $\omega = \frac{v_{\perp}}{r}$,whe

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(B) The angular velocity $\omega$ of a particle moving with velocity $v$ at a distance $r$ from a point is given by $\omega = \frac{v_{\perp}}{r}$,where $v_{\perp}$ is the component of velocity perpendicular to the position vector.At point $B$,the velocity $v$ is perpendicular to the diameter $AB$.For point $C$ (the centre): The distance $r_C = a$. The velocity $v$ is perpendicular to $CB$. Thus,$\omega_C = \frac{v}{a}$.For point $A$: The distance $r_A = AB = 2a$. The velocity $v$ is perpendicular to $AB$. Thus,$\omega_A = \frac{v}{2a}$.The ratio of angular velocity about $A$ to that about $C$ is:$\frac{\omega_A}{\omega_C} = \frac{v / 2a}{v / a} = \frac{a}{2a} = \frac{1}{2}$.

$A$ particle $P$ is moving in a circle of radius $a$ with a uniform speed $v$. $C$ is the centre of the circle and $AB$ is a diameter. When passing through $B$,the angular velocity of $P$ about $A$ and $C$ are in the ratio:

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