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?

Circular Motion PYQ — JEE Main 2019 · 11 Jan · Shift 1 · Q61 with Solution

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Correct answer: B

Given
- Speed of the particle is constant: $v = 10\,\text{m/s}$
- Angular displacement around the center: $\theta = 60^\circ$
Concept used In uniform circular motion, the velocity changes direction continuously while its magnitude remains constant.

If two velocity vectors have the same magnitude $v$ and the angle between them is $\theta$ , then the magnitude of change in velocity is
$|\Delta \vec v| = 2v\sin\frac{\theta}{2}$
Substitute the values
$|\Delta \vec v| = 2(10)\sin\frac{60^\circ}{2}$ $= 20\sin 30^\circ$
Since,
$\sin 30^\circ = \frac{1}{2}$
therefore,
$|\Delta \vec v| = 20 \times \frac{1}{2} = 10\,\text{m/s}$
Check options
- A: zero $\rightarrow$ incorrect
- B: $10\,\text{m/s}$ $\rightarrow$ correct
- C: $10\sqrt{2}\,\text{m/s}$ $\rightarrow$ incorrect
- D: $10\sqrt{3}\,\text{m/s}$ $\rightarrow$ incorrect
Final Answer
$\boxed{10\,\text{m/s}}$
So, the correct option is B.