Q 11-06-225JEE MainJEE Main 2019 (11 Jan, Shift 2)Easy
The magnitude of torque on a particle of mass $1$ kg is $2.5$ N m about the origin. If the force acting on it is $1$ N, and the distance of the particle from the origin is $5$ m, the angle between the force and the position vector is (in radians):
Answer: (A) $\dfrac\pi6$
$$\tau = rF\sin\theta \Rightarrow 2.5 = 5\times1\times\sin\theta \Rightarrow \sin\theta = \frac12 \Rightarrow \theta = \frac\pi6$$
Solution by Sreeraj P, M.Sc Physics