Statement-I : A point charge Q is rotated in a circle of radius r around a charge q. The work done will be zero.
Statement-II : For this motion the force is along the radius and direction of motion is perpendicular

Options

A. If both Statement-I and Statement-II are true, and Statement-II is the correct explanation of Statement -I.
B. If both Statement-I and Statement-II are true but Statement-II is not the correct explanation of Statement -I.
C. If Statement -I is true but Statement-II is false
D. If Statement-I is false but Statement -II is true

Answer

To solve the given problem, we need to analyze the two statements provided and understand the physics behind them.

### Step-by-Step Solution:

1. **Understanding the System**:
- We have a point charge \( Q \) that is rotating in a circular path of radius \( r \) around another charge \( q \).

2. **Identifying Forces**:
- The force acting on charge \( Q \) due to charge \( q \) is the electrostatic force, which acts along the line connecting the two charges. This force can be calculated using Coulomb's law:
\[ F = k \frac{|Q \cdot q|}{r^2} \]
where \( k \) is Coulomb's constant.

3. **Direction of Motion**:
- As charge \( Q \) moves in a circular path, its instantaneous displacement is tangential to the circle at any point. The direction of motion is thus perpendicular to the radius of the circle at that point.

4. **Work Done Calculation**:
- Work done \( W \) by a force is given by the dot product of force and displacement:
\[ W = \int \vec{F} \cdot d\vec{s} \]
- In this case, since the force \( \vec{F} \) is directed radially (towards charge \( q \)) and the displacement \( d\vec{s} \) is tangential to the circular path, the angle \( \theta \) between \( \vec{F} \) and \( d\vec{s} \) is \( 90^\circ \).

5. **Using the Cosine of the Angle**:
- The work done can be expressed as:
\[ W = F \cdot d\vec{s} \cdot \cos(90^\circ) \]
- Since \( \cos(90^\circ) = 0 \), it follows that:
\[ W = 0 \]

6. **Conclusion**:
- Therefore, the work done in rotating charge \( Q \) around charge \( q \) in a circular path is indeed zero. This confirms Statement-I is true.

7. **Analyzing Statement-II**:
- Statement-II states that the force is along the radius and the direction of motion is perpendicular. This is consistent with our analysis, as the radial force does not do any work on the charge moving in a circular path.

8. **Final Verification**:
- Both statements are true, and Statement-II correctly explains Statement-I. Thus, the conclusion is that the work done is zero due to the perpendicular nature of the force and displacement.

### Final Answer:
Both statements are true, and Statement-II is the correct explanation of Statement-I.

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