WBJEE · Physics · Mathematics in Physics
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
- C \(2 v \sin \left(\frac{\theta}{2}\right)\)
- D \(2 v \cos \left(\frac{\theta}{2}\right)\)
Answer & Solution
Correct Answer
(C) \(2 v \sin \left(\frac{\theta}{2}\right)\)
Step-by-step Solution
Detailed explanation
\(\therefore\) 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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