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 determine the change in velocity and the change in speed of a particle moving along a circular path from point A to point B, where the angle subtended at the center of the circle is \(60^\circ\).

### 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 between points A and B is \(60^\circ\).

2. **Identifying Velocity at Point A**:
At point A, let's assume the particle is moving in the positive x-direction. Therefore, the velocity vector at point A can be represented as:
\[ \vec{v_A} = v \hat{i} \]
where \(\hat{i}\) is the unit vector in the x-direction.

3. **Finding Velocity at Point B**:
Since the particle moves along a circular path, we need to find the velocity vector at point B. The angle from point A to point B is \(60^\circ\). The velocity vector at point B can be resolved into its components:
- The x-component: \(v \cos(60^\circ) = v \cdot \frac{1}{2} = \frac{v}{2}\)
- The y-component: \(v \sin(60^\circ) = v \cdot \frac{\sqrt{3}}{2}\)

Since point B is below the x-axis, the y-component will be negative:
\[ \vec{v_B} = \frac{v}{2} \hat{i} - v \frac{\sqrt{3}}{2} \hat{j} \]

4. **Calculating Change in Velocity**:
The change in velocity \(\Delta \vec{v}\) is given by:
\[ \Delta \vec{v} = \vec{v_B} - \vec{v_A} \]
Substituting the values:
\[ \Delta \vec{v} = \left(\frac{v}{2} \hat{i} - v \frac{\sqrt{3}}{2} \hat{j}\right) - (v \hat{i}) \]
Simplifying this:
\[ \Delta \vec{v} = \left(\frac{v}{2} - v\right) \hat{i} - v \frac{\sqrt{3}}{2} \hat{j} = -\frac{v}{2} \hat{i} - v \frac{\sqrt{3}}{2} \hat{j} \]

5. **Magnitude of Change in Velocity**:
To find the magnitude of the change in velocity:
\[ |\Delta \vec{v}| = \sqrt{\left(-\frac{v}{2}\right)^2 + \left(-v \frac{\sqrt{3}}{2}\right)^2} \]
Calculating this:
\[ |\Delta \vec{v}| = \sqrt{\frac{v^2}{4} + \frac{3v^2}{4}} = \sqrt{\frac{4v^2}{4}} = \sqrt{v^2} = v \]

6. **Change in Speed**:
Since the speed of the particle is constant throughout its motion, the change in speed is:
\[ \Delta \text{speed} = 0 \]

### Final Answer:
The magnitude of change in velocity is \(v\) and the change in speed is \(0\).