JEE PYQ: Laws Of Motion Question 32
Question 32 - 2019 | 11 Jan Shift 2] (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) :
(1) $\frac{\pi}{6}$ [Correct]
(2) $\frac{\pi}{3}$
(3) $\frac{\pi}{8}$
(4) $\frac{\pi}{4}$
Show Answer
Answer: (1)
Solution
$|\vec{\tau}| = rF\sin\theta$. $2.5 = 5 \times 1 \times \sin\theta$. $\sin\theta = 0.5$. $\theta = \frac{\pi}{6}$.
BETA
