A particle is moving in a circular path of radius r under the action of a force F. If at an instant velocity of particle is v, and speed of particle is increasing, then

Options

A. `vecF.vecv=0`
B. `vecF.vecc gt 0`
C. `vecF.vecv lt0`
D. `vecF.vecvge0`

Answer

To solve the problem, we need to analyze the motion of a particle moving in a circular path under the influence of a force \( F \), given that the speed of the particle is increasing.

### Step-by-Step Solution:

1. **Understanding the Motion**:
- The particle is moving in a circular path with radius \( r \).
- The velocity \( \vec{v} \) of the particle is tangential to the circular path at any point.

2. **Identifying the Forces**:
- The force \( \vec{F} \) acting on the particle can be resolved into two components: one that acts towards the center of the circular path (centripetal force) and another that acts tangentially to the path.

3. **Condition of Increasing Speed**:
- If the speed of the particle is increasing, it implies that there is a tangential component of the force \( \vec{F} \) acting in the direction of the velocity \( \vec{v} \).
- This means that the angle \( \theta \) between the force vector \( \vec{F} \) and the velocity vector \( \vec{v} \) is less than 90 degrees.

4. **Dot Product Analysis**:
- The work done by the force \( \vec{F} \) on the particle can be expressed using the dot product:
\[ \vec{F} \cdot \vec{v} = |\vec{F}| |\vec{v}| \cos(\theta) \]
- Since the speed is increasing, the work done must be positive, which means:
\[ \vec{F} \cdot \vec{v} > 0 \]
- This indicates that \( \cos(\theta) > 0 \), confirming that \( \theta < 90^\circ \).

5. **Conclusion**:
- Therefore, the correct conclusion is that the dot product \( \vec{F} \cdot \vec{v} \) is greater than zero, which implies that the force has a component in the direction of the velocity, contributing to the increase in speed.

### Final Answer:
The correct statement is \( \vec{F} \cdot \vec{v} > 0 \).