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

The correct option is B $\frac{\pi }{6}$

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