JEE Main · 2019 · 11 Jan 2019 · Shift 2 · Rotational Mechanics

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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 IN, 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) $\frac{π}{6}$ ✓ Correct
(B) $\frac{π}{3}$
(C) $\frac{π}{8}$
(D) $\frac{π}{4}$

Answer: (A)

Solution

Torque is

τ = r F sin θ

, with

θ

the angle between the position vector and the force. Here

τ = 2.5

\cdot

r = 5

m and

F = 1

N, so

sin θ = \frac{τ}{r F} = \frac{2.5}{5 \times 1} = \frac{1}{2}

, giving

θ = \frac{π}{6}

. The mass has no bearing on this relation.

JEE Main · 2019 · 11 Jan 2019 · Shift 2 · Single correct option

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