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):

We recall the definition of torque produced by a force about the origin. In vector form, the torque is given by $$\vec \tau = \vec r \times \vec F$$, where $$\vec r$$ is the position vector (whose magnitude is the distance of the particle from the origin) and $$\vec F$$ is the force vector.

For magnitudes alone, the cross-product gives

$$|\vec \tau| = |\vec r|\,|\vec F|\,\sin\theta,$$

where $$\theta$$ is the angle between $$\vec r$$ and $$\vec F$$. This is the relationship we shall use.

Now we substitute the given numerical values. The magnitude of torque is $$|\vec \tau| = 2.5\ \text{N\,m}$$, the magnitude of the force is $$|\vec F| = 1\ \text{N}$$, and the distance from the origin is $$|\vec r| = 5\ \text{m}$$.

So we write

$$2.5 = 5 \times 1 \times \sin\theta.$$

Dividing both sides by 5, we obtain

$$\sin\theta = \frac{2.5}{5}.$$

The fraction simplifies to

$$\sin\theta = 0.5 = \frac{1}{2}.$$

We now look for an angle whose sine is $$\tfrac12$$. The principal angle in the first quadrant with this sine value is

$$\theta = \frac{\pi}{6}\ \text{radians}.$$

(Another mathematical solution is $$\theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$$, but that corresponds to an obtuse angle of 150°, whereas the usual convention for the angle between two vectors in such problems is the acute angle. Hence we select $$\theta = \frac{\pi}{6}$$.)

Hence, the correct answer is Option A.