Eight drops of mercury of equal radii possessing equal charges combine to from a big drop. Then the capacitance of bigger drop compared to each individual small drop is

Options

A. `8` times
B. `4` times
C. `2` times
D. `32` times

Explanation

To solve the problem of finding the capacitance of a big drop formed by combining eight smaller drops of mercury, we can follow these steps:

### Step 1: Understand the Concept of Capacitance
The capacitance \( C \) of a spherical drop of radius \( r \) is given by the formula:
\[ C = 4 \pi \epsilon_0 r \]
where \( \epsilon_0 \) is the permittivity of free space.

### Step 2: Determine the Volume Relationship
When eight small drops combine to form a larger drop, the volume of the larger drop is equal to the sum of the volumes of the smaller drops. The volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3} \pi r^3 \]
Let the radius of the smaller drops be \( r \) and the radius of the larger drop be \( R \). Therefore, we have:
\[ 8 \left( \frac{4}{3} \pi r^3 \right) = \frac{4}{3} \pi R^3 \]
This simplifies to:
\[ 8 r^3 = R^3 \]

### Step 3: Relate the Radii
From the equation \( 8 r^3 = R^3 \), we can find the relationship between the radii:
\[ R = 2^{1/3} r \]

### Step 4: Calculate the Capacitance of the Larger Drop
Using the capacitance formula for the larger drop, we have:
\[ C' = 4 \pi \epsilon_0 R = 4 \pi \epsilon_0 (2^{1/3} r) \]
This can be rewritten as:
\[ C' = 4 \pi \epsilon_0 r \cdot 2^{1/3} \]

### Step 5: Calculate the Capacitance Ratio
Now, we can find the ratio of the capacitance of the larger drop \( C' \) to the capacitance of the smaller drop \( C \):
\[ \frac{C'}{C} = \frac{4 \pi \epsilon_0 (2^{1/3} r)}{4 \pi \epsilon_0 r} = 2^{1/3} \]

### Step 6: Substitute the Value of N
Since we combined \( N = 8 \) drops, we can also express this as:
\[ C' = N^{1/3} C \]
Substituting \( N = 8 \):
\[ C' = 8^{1/3} C = 2C \]

### Conclusion
Thus, the capacitance of the larger drop compared to each individual small drop is:
\[ \boxed{2C} \]