A particle is moving along a circular path of radius 2 m and with uniform speed of 5 ms⁻¹ . What will be the change in velocity when the particle completes half of the revolution

The correct option is 10 ms⁻¹.

Step-by-step Explanation:
Let a particle be moving along a circular path of radius $r=2m$ with a uniform speed of $v=5ms{⁻}^1$.

Velocity is a vector quantity that has both magnitude (speed) and direction. Let the initial velocity vector of the particle be:
${\mathbf{v}}_{\mathbf{i}}=v\hat{\mathbf{i}}$
where $\hat{\mathbf{i}}$ represents the unit vector in the direction of the initial motion.

When the particle completes half of a revolution, it reaches the diametrically opposite point of the circular path. At this point, the direction of its motion is exactly reversed, while its speed remains the same.
Therefore, the final velocity vector is:
${\mathbf{v}}_{\mathbf{f}}=-v\hat{\mathbf{i}}$

The change in velocity ($\Delta \mathbf{v}$) is the difference between the final velocity and the initial velocity:
$\Delta \mathbf{v}={\mathbf{v}}_{\mathbf{f}}-{\mathbf{v}}_{\mathbf{i}}$
Substituting the values of ${\mathbf{v}}_{\mathbf{i}}$ and ${\mathbf{v}}_{\mathbf{f}}$:
$\Delta \mathbf{v}=-v\hat{\mathbf{i}}-v\hat{\mathbf{i}}=-2v\hat{\mathbf{i}}$

The magnitude of the change in velocity is:
$|\Delta \mathbf{v}|=2v$
Given that $v=5ms{⁻}^1$, we substitute this value:
$|\Delta \mathbf{v}|=2\times 5ms{⁻}^1=10ms{⁻}^1$

Thus, the magnitude of the change in velocity is 10 ms⁻¹.