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.

Step 1: Identify the given values. 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. Step 2: Determine the initial and final velocity vectors. Let's assume the particle starts at position (r, 0) and moves counter-clockwise. The initial velocity vector v_1 will be tangential to the circle, pointing in the positive y-direction. v_1 = v j = 5 j m/s After completing half a revolution, the particle will be at the diametrically opposite position (-r, 0). The final velocity vector 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). v_2 = -v j = -5 j m/s Step 3: Calculate the change in velocity. The change in velocity v is the difference between the final and initial velocity vectors. v = v_2 - v_1 = (-5 j) - (5 j) = -10 j m/s The magnitude of the change in velocity is: | v| = |-10 j| = 10 m/s Step 4: Calculate the time taken for half a revolution. The distance covered in half a revolution is half the circumference of the circle. Distance = (1)/(2) (2 r) = r Substitute the given radius r = 5 m: Distance = (5) = 5 meters The time taken t is the distance divided by the speed: t = DistanceSpeed = 5 m5 m/s = seconds Step 5: Calculate the average acceleration. Average acceleration a_avg is defined as the change in velocity divided by the time taken. a_avg = vt The magnitude of the average acceleration is: |a_avg| = | v|t = 10 m/s s = (10)/() m/s^2 Comparing this result with the given options: (a) शून्य (b) 10/ m s^-2 (c) 10 m s^-2 (d) कोई नहीं The calculated average acceleration matches option (b). The final answer is (b) 10/ m s^-2.