Q 11-03-162JEE MainJEE Main 2019 (11 Jan, Shift 1)Easy
A particle is moving along a circular path with a constant speed of $10\ \text{m s}^{-1}$. What is the magnitude of the change in velocity of the particle, when it moves through an angle of $60^\circ$ around the centre of the circle?
Answer: (D) $10$ m/s
The velocity turns through the same angle as the radius, $60^\circ$, while its magnitude stays $v = 10$ m/s. The two velocity vectors and their difference form an isosceles triangle:
$$|\Delta\vec{v}| = 2v\sin\frac{60^\circ}{2} = 2\times10\times\frac12 = 10\ \text{m/s}$$
Solution by Sreeraj P, M.Sc Physics