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Question: Answered & Verified by Expert
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
PhysicsMathematics in PhysicsJEE Main
Options:
  • A $2 \mathrm{v} \sin (2 \theta)$
  • B zero
  • C $2 v \sin \left(\frac{\theta}{2}\right)$
  • D $2 v \cos \left(\frac{\theta}{2}\right)$
Solution:
2999 Upvotes Verified Answer
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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