A large hollow metal sphere of radius R has a small opening at the top. Small drops of mercury, each of radius r and charged to a potential of the sphere becomes V' after N drops fall into it.Then

Options

A. `V'="V for any value of N"`
B. `V'=V" for "N=((R)/(r))^(2//3)`
C. `V'=V" for "N=((R)/(r))^(1//3)`
D. `V'=V" for " N=(R)/(r)`

Answer

To solve the problem, we need to analyze the situation step by step, focusing on the relationship between the potential of the hollow metal sphere and the potential after N drops of mercury have been added.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a large hollow metal sphere with radius \( R \).
- There is a small opening at the top of the sphere.
- Small drops of mercury, each with radius \( r \), are dropped into the sphere.

2. **Charge of a Single Drop**:
- Each drop of mercury has a charge \( Q \).
- The potential \( V \) of a single drop is given by the formula:
\[ V = \frac{kQ}{r} \]
- Rearranging this gives us the charge on a single drop:
\[ Q = \frac{Vr}{k} \]

3. **Total Charge from N Drops**:
- If \( N \) drops are added, the total charge \( Q' \) on the sphere becomes:
\[ Q' = NQ = N \left(\frac{Vr}{k}\right) = \frac{N Vr}{k} \]

4. **Potential of the Sphere After Adding N Drops**:
- The potential \( V' \) of the sphere after adding \( N \) drops can be expressed as:
\[ V' = \frac{kQ'}{R} = \frac{k \left(\frac{N Vr}{k}\right)}{R} = \frac{N Vr}{R} \]

5. **Comparing Potentials**:
- We now have expressions for \( V \) and \( V' \):
- Original potential: \( V \)
- Potential after adding drops: \( V' = \frac{N Vr}{R} \)
- To find the relationship between \( V' \) and \( V \), we can set:
\[ V' = \frac{N Vr}{R} \]

6. **Finding Condition for \( V' = V \)**:
- For \( V' \) to equal \( V \):
\[ \frac{N Vr}{R} = V \]
- Dividing both sides by \( V \) (assuming \( V \neq 0 \)):
\[ \frac{N r}{R} = 1 \implies N = \frac{R}{r} \]

### Conclusion:
The relationship between \( V' \) and \( V \) is such that \( V' = V \) when \( N = \frac{R}{r} \).