The magnitude of torque on a particle of mass 1 kg is 2.5 $\sqrt{3}$ Nm 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]

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π/6

π/3

π/4

π/2

answer is B.

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Detailed Solution

Torque about the origin $\bar{τ} = \bar{r} \times \bar{F}$
$\begin{array}{l} τ = rFsin θ \\ 2.5 \sqrt{3} = 5 \times 1 \sin θ \\ \sin θ = \frac{\sqrt{3}}{2} \\ θ = \frac{π}{3} \end{array}$

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