When the positively charged hanging pendulum bob is made fixed, the work done in slowly shifting a unit positive charge from infinity to P is V. If the pendulum is free to move, the corresponding work done id `V'`. Then.

Options

A. `V = V'`
B. `V gt V'`
C. `V lt V'`
D. `V le V'`

Explanation

To solve the problem, we need to establish a relationship between the work done in bringing a unit positive charge from infinity to a point P when a positively charged pendulum bob is fixed (denoted as \( V \)) and when the pendulum bob is free to move (denoted as \( V' \)).

### Step-by-Step Solution:

1. **Understanding the Fixed Pendulum Case:**
- When the pendulum bob is fixed, it has a positive charge \( +Q \).
- The work done \( V \) in bringing a unit positive charge \( +1 \) from infinity to point \( P \) is given by the formula for electric potential energy:
\[ V = \frac{k \cdot Q \cdot 1}{r} \]
- Here, \( k \) is Coulomb's constant, \( Q \) is the charge of the pendulum bob, and \( r \) is the distance from the charge to point \( P \).

2. **Understanding the Free Pendulum Case:**
- When the pendulum bob is free to move, it will experience a repulsive force due to the positive charge \( +Q \).
- This repulsion will cause the pendulum to move away from point \( P \), increasing the distance from \( r \) to \( r' \) (where \( r' > r \)).
- The work done \( V' \) in this case is:
\[ V' = \frac{k \cdot Q \cdot 1}{r'} \]

3. **Comparing the Two Cases:**
- Since \( r' > r \), it follows that:
\[ \frac{1}{r'} < \frac{1}{r} \]
- Therefore, we can conclude:
\[ V' < V \]
- This implies that the work done \( V' \) when the pendulum is free to move is less than the work done \( V \) when it is fixed.

4. **Final Relationship:**
- The final relationship we establish is:
\[ V > V' \]

### Conclusion:
The work done in bringing a unit positive charge from infinity to point \( P \) when the pendulum bob is fixed is greater than the work done when the pendulum bob is free to move.