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)

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Answer & Solution

Correct Answer

(A) \(\frac{\pi }{6}\)

Step-by-step Solution

Detailed explanation

\(2.5 = 1 \times 5\,\sin \,\theta \)

\(\,\sin \,\theta = 0.5 = \frac{1}{2}\)

\(\,\theta = \frac{\pi }{6}\)

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