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A particle is moving with a uniform speed $v$ in a circular path of radius $r$ with the centre at $O$. When the particle moves from a point $P$ to $Q$ on the circle such that $\angle P O Q=\theta$, then the magnitude of the change in velocity is
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The correct answer is:
$2 v \sin \left(\frac{\theta}{2}\right)$
$\therefore$ Change in magnitude of velocity
$\begin{aligned}
|\Delta v| &=\sqrt{v^{2}+v^{2}-2 v^{2} \cos \theta} \\
&=2 v \sin \frac{\theta}{2}
\end{aligned}$
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