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 $\vec{v_P}$ be the velocity vector at point $P$ and $\vec{v_Q}$ be the velocity vector at point $Q$.
The magnitude of both velocity vectors is $v$, i.e., $|\vec{v_P}| = v$ and $|\vec{v_Q}| = v$.

The angle between the position vectors $\vec{OP}$ and $\vec{OQ}$ is given as $\theta$.
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 $\vec{v_P}$ and $\vec{v_Q}$ will also be $\theta$.

The change in velocity is $\Delta \vec{v} = \vec{v_Q} - \vec{v_P}$.
The magnitude of the change in velocity can be found using the law of cosines for vector subtraction:

Substitute $|\vec{v_P}| = v$ and $|\vec{v_Q}| = v$:

Factor out $2v^2$:

Recall the trigonometric identity: $1 - \cos \theta = 2 \sin^2 \left(\frac{\theta}{2}\right)$.
Substitute this into the equation for $|\Delta \vec{v}|^2$:

For typical angles $\theta$ swept in a circular motion ($0 \le \theta \le 2\pi$), $\frac{\theta}{2}$ will be in the range $0 \le \frac{\theta}{2} \le \pi$. In this range, $\sin\left(\frac{\theta}{2}\right) \ge 0$.
Therefore, we can remove the absolute value sign:

Comparing this result with the given options:

(a) $2v \sin (2\theta)$

(b) शून्य

(c) $2v \sin \left(\frac{\theta}{2}\right)$

(d) $2v \cos \left(\frac{\theta}{2}\right)$

The calculated magnitude of the change in velocity matches option (c).

The final answer is $\boxed{\text{(c) } 2v \sin \left(\frac{\theta}{2}\right)}$.