Define the initial and final velocity vectors.

Step 1: Define the initial and final velocity vectors. Let the particle be at point P initially and point Q finally. The speed of the particle is uniform, v. The velocity vector is always tangential to the circular path. Let v_P be the velocity vector at point P and v_Q be the velocity vector at point Q. The magnitude of both velocity vectors is v, i.e., |v_P| = v and |v_Q| = v. Step 2: Determine the angle between the velocity vectors. The angle between the position vectors OP and OQ is given as . Since the velocity vector at any point on a circular path is perpendicular to the position vector at that point, the angle between the velocity vectors v_P and v_Q will also be . Step 3: Calculate the magnitude of the change in velocity. The change in velocity is v = v_Q - v_P. The magnitude of the change in velocity can be found using the law of cosines for vector subtraction: | v|^2 = |v_Q|^2 + |v_P|^2 - 2|v_Q||v_P| Substitute |v_P| = v and |v_Q| = v: | v|^2 = v^2 + v^2 - 2(v)(v) | v|^2 = 2v^2 - 2v^2 Factor out 2v^2: | v|^2 = 2v^2 (1 - ) Step 4: Use a trigonometric identity to simplify. Recall the trigonometric identity: 1 - = 2 ^2 (()/(2)). Substitute this into the equation for | v|^2: | v|^2 = 2v^2 (2 ^2 (()/(2))) | v|^2 = 4v^2 ^2 (()/(2)) Step 5: Take the square root to find the magnitude. | v| = sqrt(4v^2 ^2 (()/(2))) | v| = 2v | (()/(2))| For typical angles swept in a circular motion (0 2), ()/(2) will be in the range 0 ()/(2) . In this range, (()/(2)) 0. Therefore, we can remove the absolute value sign: | v| = 2v (()/(2)) Comparing this result with the given options: (a) 2v (2) (b) शून्य (c) 2v (()/(2)) (d) 2v (()/(2)) The calculated magnitude of the change in velocity matches option (c). The final answer is (c) 2v (()/(2)).