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

• A

  $2 \mathrm{v} \sin (2 \theta)$
• B

  zero
• Correct answer
  C

  $2 v \sin \left(\frac{\theta}{2}\right)$
• D

  $2 v \cos \left(\frac{\theta}{2}\right)$

Solution

$\therefore$ Change in magnitude of velocity $\begin{aligned} |\Delta v| &=\sqrt{v^{2}+v^{2}-2 v^{2} \cos \theta} \\ &=2 v \sin \frac{\theta}{2} \end{aligned}$

Hence, the correct answer is option: 'C'.