JEE Main 2019 — Physics solution

JEE Main2019Physics

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
$\frac{π}{8}$
B
$\frac{π}{6}$
C
$\frac{π}{3}$
D
$\frac{π}{4}$

Solution

$2.5 = 1 \times 5 sin θ sin θ = 0.5 = \frac{1}{2} θ = \frac{π}{6}$

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