A particle is moving on a circular path with constant speed v. The magnitude of the change in its velocity after it has described an angle of `90^(@)` is :

Options

A. v
B. `sqrt(2) v`
C. 0
D. `sqrt(3)` v

Explanation

To solve the problem of finding the magnitude of the change in velocity of a particle moving in a circular path after it has described an angle of 90 degrees, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Initial and Final Velocity**:
- The particle is moving with a constant speed \( v \) in a circular path.
- At the initial position, let's assume the particle is at point A, moving along the positive x-axis. Thus, the initial velocity vector \( \vec{V_i} \) can be represented as:
\[ \vec{V_i} = v \hat{i} \]
- After moving through an angle of 90 degrees, the particle will be at point B, moving along the positive y-axis. Therefore, the final velocity vector \( \vec{V_f} \) can be represented as:
\[ \vec{V_f} = v \hat{j} \]

2. **Calculating Change in Velocity**:
- The change in velocity \( \Delta \vec{V} \) is given by:
\[ \Delta \vec{V} = \vec{V_f} - \vec{V_i} \]
- Substituting the values of \( \vec{V_f} \) and \( \vec{V_i} \):
\[ \Delta \vec{V} = v \hat{j} - v \hat{i} \]
- This can be rewritten as:
\[ \Delta \vec{V} = -v \hat{i} + v \hat{j} \]

3. **Finding the Magnitude of Change in Velocity**:
- The magnitude of the change in velocity \( |\Delta \vec{V}| \) can be calculated using the Pythagorean theorem:
\[ |\Delta \vec{V}| = \sqrt{(-v)^2 + (v)^2} \]
- Simplifying this:
\[ |\Delta \vec{V}| = \sqrt{v^2 + v^2} = \sqrt{2v^2} = v\sqrt{2} \]

4. **Conclusion**:
- Therefore, the magnitude of the change in velocity after the particle has described an angle of 90 degrees is:
\[ |\Delta \vec{V}| = \sqrt{2}v \]

### Final Answer:
The magnitude of the change in velocity is \( \sqrt{2}v \).

---