JEE Mains · Physics · STD 11 - System of particles and rotational motion
- JEE Mains 2019
- Medium
- MCQ
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)
Pick an option to check your answer.
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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