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(C) The velocity of the particle at point $P$ is $\vec{v}_P$ and at point $Q$ is $\vec{v}_Q$. Both have the same magnitude $v$.The change in velocity

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$2v\sin 20^\circ$

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(C) The velocity of the particle at point $P$ is $\vec{v}_P$ and at point $Q$ is $\vec{v}_Q$. Both have the same magnitude $v$.The change in velocity is given by $\Delta \vec{v} = \vec{v}_Q - \vec{v}_P$.The magnitude of the change in velocity for a particle moving in a circle with uniform speed is given by the formula $|\Delta \vec{v}| = 2v \sin(	heta/2)$, where $	heta$ is the angle subtended at the center.Given $	heta = 40^\circ$, we have:$|\Delta \vec{v}| = 2v \sin(40^\circ / 2) = 2v \sin 20^\circ$.Thus, the correct option is $C$.

$A$ particle is moving on a circular path of radius $r$ with uniform speed $v$. The change in velocity when the particle moves from $P$ to $Q$ is $(\angle POQ = 40^\circ)$.

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