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 step by step, we need to analyze the situation involving a large hollow metal sphere and small mercury drops.

### Step 1: Understand the Problem
We have a large hollow metal sphere of radius \( R \) with a small opening at the top. Small mercury drops, each with a radius \( r \), are introduced into the sphere. After \( N \) drops fall into the sphere, the potential of the sphere becomes \( V' \).

### Step 2: Determine the Charge of One Drop
Let's denote the charge on one small drop of mercury as \( q \). The potential \( V \) of a single drop at its surface can be expressed using the formula:
\[ V = \frac{kq}{r} \]
where \( k \) is Coulomb's constant.

### Step 3: Total Charge from N Drops
If \( N \) drops fall into the sphere, the total charge \( Q_T \) contributed by these \( N \) drops is:
\[ Q_T = N \cdot q \]

### Step 4: Calculate the Potential of the Sphere
The potential \( V' \) of the hollow sphere after \( N \) drops have been added can be calculated using the total charge:
\[ V' = \frac{kQ_T}{R} \]
Substituting \( Q_T \) from the previous step:
\[ V' = \frac{k(N \cdot q)}{R} \]

### Step 5: Relate the Two Potentials
From the earlier steps, we have two expressions for potential:
1. Potential of one drop: \( V = \frac{kq}{r} \)
2. Potential of the sphere after \( N \) drops: \( V' = \frac{k(N \cdot q)}{R} \)

Since we are given that \( V = V' \), we can set these two equations equal to each other:
\[ \frac{kq}{r} = \frac{k(N \cdot q)}{R} \]

### Step 6: Simplify the Equation
We can cancel \( k \) and \( q \) (assuming \( q \neq 0 \)):
\[ \frac{1}{r} = \frac{N}{R} \]

### Step 7: Solve for N
Rearranging the equation gives:
\[ N = \frac{R}{r} \]

### Conclusion
Thus, the relationship between the number of drops \( N \), the radius of the sphere \( R \), and the radius of the drops \( r \) is:
\[ N = \frac{R}{r} \]