JEE Main 2019 Physics question
The magnitude of torque on a particle of mass 1 kg is 2.5 Nm 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: (B)
Solution
5 × 1 × sin θ = 2.5 ⇒ θ = π/6.