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) :

• A. $\frac{\pi }{8}$
• B. $\frac{\pi }{6}$
• C. $\frac{\pi }{4}$
• D. $\frac{\pi }{3}$

Answer 1

Answer: (B)

$\frac{\pi }{6}$

Answer 2

$2.5 = 1 \times 5 \sin \theta$ $\sin \theta = 0.5 = \frac{1}{2}$ $\theta = \frac{\pi}{6}$