A particle of mass 2 kg moves in a circular path of radius 5 m with uniform speed 5 m/s. Calculate the magnitude of the average acceleration when it completes half a revolution.

The radius of the circular path is $r = 5$ meters.
The uniform speed of the particle is $v = 5$ meters/second.
We need to find the average acceleration when the particle completes half a revolution.

Let's assume the particle starts at position $(r, 0)$ and moves counter-clockwise.
The initial velocity vector $\vec{v_1}$ will be tangential to the circle, pointing in the positive y-direction.

After completing half a revolution, the particle will be at the diametrically opposite position $(-r, 0)$.
The final velocity vector $\vec{v_2}$ will also be tangential to the circle, but pointing in the negative y-direction (since the speed is uniform, the magnitude remains the same).

The change in velocity $\Delta \vec{v}$ is the difference between the final and initial velocity vectors.

The magnitude of the change in velocity is:

The distance covered in half a revolution is half the circumference of the circle.

Substitute the given radius $r = 5$ m:

The time taken $t$ is the distance divided by the speed:

Average acceleration $\vec{a_{avg}}$ is defined as the change in velocity divided by the time taken.

The magnitude of the average acceleration is:

Comparing this result with the given options:

(a) शून्य

(b) $10/\pi \text{ m s}^{-2}$

(c) $10 \text{ m s}^{-2}$

(d) कोई नहीं

The calculated average acceleration matches option (b).

The final answer is $\boxed{\text{(b) } 10/\pi \text{ m s}^{-2}}$.