JEE MAIN - Physics (2019 - 11th January Evening Slot - No. 9)
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 5m, the angle between the force and the position vector is (in radians) :
$${\pi \over 8}$$
$${\pi \over 6}$$
$${\pi \over 4}$$
$${\pi \over 3}$$
Explanation
2.5 = 1 $$ \times $$ 5 sin $$\theta $$
sin$$\theta $$ = 0.5 = $${1 \over 2}$$
$$\theta $$ = $${\pi \over 6}$$
sin$$\theta $$ = 0.5 = $${1 \over 2}$$
$$\theta $$ = $${\pi \over 6}$$
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