The magnitude of torque on a particle of...

The magnitude of torque on a particle of mass 1 kg is 2.5 Nm about the orifgin.If the force acting 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

`pi/4`

B

`pi/6`

C

`pi/3`

D

`pi/8`

Verified by Experts

The correct Answer is:

B

`tau = FR sin theta, sin theta =(2.5)/(1 xx 5) therefore theta = 30^(@)`.

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