JEE MAIN - Physics (2019 - 11th January Morning Slot - No. 20)
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 60o around the centre of the circle?
zero
10 m/s
$$10\sqrt 2 m/s$$
$$10\sqrt 3 m/s$$
Explanation
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$$\left| {\Delta \overrightarrow v } \right| = \sqrt {v_1^2 + v_2^2 + 2{v_1}{v_2}\cos \left( {\pi - \theta } \right)} $$
$$ = 2v\sin {\theta \over 2}$$ since $$\left[ {\left| {\overline v {}_1} \right| = \left| {{{\overline v }_2}} \right|} \right]$$
$$ = \left( {2 \times 10} \right) \times \sin \left( {{{30}^o}} \right)$$
= 10 m/s
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