A particle is moving on a circular path with constant speed v. it moves between two points A and B, which subtends and angle `60^(@)` at the centre of circle, The magnitude of change in tis velocity and change and magnitude of its speed during motion from A to B are respectively

Options

A. Zero,zero
B. v,0
C. 0,v
D. 2v, v

Answer

To solve the problem, we need to find the change in velocity and the change in speed of a particle moving on a circular path with constant speed as it moves from point A to point B, which subtends an angle of 60 degrees at the center of the circle.

### Step-by-Step Solution:

1. **Understanding the Motion**:
- The particle is moving in a circular path with constant speed \( v \).
- The angle subtended at the center of the circle by points A and B is \( 60^\circ \).

2. **Change in Speed**:
- Since the particle is moving with constant speed, the magnitude of its speed does not change.
- Therefore, the change in speed \( \Delta v \) is:
\[ \Delta v = 0 \]

3. **Finding the Velocity Vectors**:
- Let’s denote the position of point A as \( A \) and the position of point B as \( B \).
- The velocity vector at point A, \( \vec{V_A} \), can be represented as:
\[ \vec{V_A} = v \hat{i} \]
- At point B, the direction of the velocity vector changes due to circular motion. The angle between the radius to point A and the radius to point B is \( 60^\circ \).

4. **Calculating the Components of Velocity at Point B**:
- The velocity vector at point B, \( \vec{V_B} \), can be broken down into components:
- The horizontal component (x-direction) is:
\[ V_{B_x} = v \cos(60^\circ) = v \cdot \frac{1}{2} = \frac{v}{2} \]
- The vertical component (y-direction) is:
\[ V_{B_y} = v \sin(60^\circ) = v \cdot \frac{\sqrt{3}}{2} \]
- Since point B is below the x-axis, the vertical component will be negative:
\[ \vec{V_B} = \frac{v}{2} \hat{i} - v \cdot \frac{\sqrt{3}}{2} \hat{j} \]

5. **Finding the Change in Velocity**:
- The change in velocity \( \Delta \vec{V} \) from A to B is given by:
\[ \Delta \vec{V} = \vec{V_B} - \vec{V_A} \]
- Substituting the values:
\[ \Delta \vec{V} = \left( \frac{v}{2} \hat{i} - v \cdot \frac{\sqrt{3}}{2} \hat{j} \right) - (v \hat{i}) \]
\[ = \left( \frac{v}{2} - v \right) \hat{i} - v \cdot \frac{\sqrt{3}}{2} \hat{j} \]
\[ = -\frac{v}{2} \hat{i} - v \cdot \frac{\sqrt{3}}{2} \hat{j} \]

6. **Calculating the Magnitude of Change in Velocity**:
- The magnitude of \( \Delta \vec{V} \) is:
\[ |\Delta \vec{V}| = \sqrt{\left(-\frac{v}{2}\right)^2 + \left(-v \cdot \frac{\sqrt{3}}{2}\right)^2} \]
\[ = \sqrt{\frac{v^2}{4} + \frac{3v^2}{4}} = \sqrt{v^2} = v \]

### Final Results:
- The change in speed is \( 0 \).
- The magnitude of change in velocity is \( v \).

### Summary:
- Change in speed: \( 0 \)
- Change in velocity: \( v \)