Q 11-03-111JEE MainJEE Main 2022 (25 Jun, Shift 2)Easy
For a particle in uniform circular motion, the acceleration $\vec a$ at any point $P(R, \theta)$ on the circular path of radius $R$ is (when $\theta$ is measured from the positive $x$-axis and $v$ is uniform speed):
Answer: (C) $-\dfrac{v^2}{R}\cos\theta\,\hat i - \dfrac{v^2}{R}\sin\theta\,\hat j$
The acceleration has magnitude $\dfrac{v^2}{R}$ and points from $P$ towards the centre, i.e. along $-(\cos\theta\,\hat i + \sin\theta\,\hat j)$:
$$\vec a = -\frac{v^2}{R}\cos\theta\,\hat i - \frac{v^2}{R}\sin\theta\,\hat j$$
Solution by Sreeraj P, M.Sc Physics