JEE Mains · Physics · STD 11 - Vectors
- JEE Mains 2019
- Medium
- Numerical
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 .......... \(m/s\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(10\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{l}
\Delta \vec v\, = 2v\,\sin \left( {\frac{\theta }{2}} \right)\\
\,\,\,\,\,\,\, = 2 \times 10 \times \sin \left( {{{30}^ \circ }} \right)\\
\,\,\,\,\,\,\, = 10\,m/s
\end{array}\)
\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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