JEE Mains · Physics · STD 11 - 3.1 vectors
A particle is moving along a circular path with a constant speed of \(10\,ms^{-1}.\) What is the magnitude of the change in velocity of the particle, when it moves through an angle of \(60^{o}\) around the centre of the circle .......... \(m/s\)
- A \(10\sqrt 3 \)
- B \(0\)
- C \(10\sqrt 2 \)
- D \(10\)
Answer & Solution
Correct Answer
(D) \(10\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{l} \Delta \vec v\, = 2v\,\sin \left( {\frac{\theta }{2}} \right)\\ \,\,\,\,\,\,\, = 2 \times 10 \times \sin \left( {{{30}^ \circ }} \right)\\ \,\,\,\,\,\,\, = 10\,m/s \end{array}\)
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