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) :

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) (π /8) (B) (π /6) (C) (π /4) (D) (π /3)

Tardigrade Admin · Accepted Answer

2.5 = 1 × 5 sin θ sin θ = 0.5 = (1/2) θ = (π/6)

Q. The magnitude of torque on a particle of mass $1 k g$ is $2.5 N m$ 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) :

$\frac{π}{8}$

$\frac{π}{6}$

$\frac{π}{4}$

$\frac{π}{3}$

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

Q. The magnitude of torque on a particle of mass 1kg is 2.5Nm about the origin. If the force acting on it is 1N, and the distance of the particle from the origin is 5m, the angle between the force and the position vector is (in radians) :

8π​

6π​

4π​

3π​