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

Answer

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

### Step 1: Understand Volume Conservation
When the eight small drops combine to form a larger drop, the total volume of the drops remains constant. The volume \( V \) of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]
where \( r \) is the radius of the sphere.

### Step 2: Set Up the Volume Equation
Let the radius of each small drop be \( r \) and the radius of the larger drop be \( R \). The total volume of the eight small drops is:
\[ V_{\text{small}} = 8 \times \frac{4}{3} \pi r^3 = \frac{32}{3} \pi r^3 \]
The volume of the larger drop is:
\[ V_{\text{large}} = \frac{4}{3} \pi R^3 \]
Since the volumes are equal, we can set them equal to each other:
\[ \frac{32}{3} \pi r^3 = \frac{4}{3} \pi R^3 \]

### Step 3: Simplify the Equation
We can cancel out \( \frac{4}{3} \pi \) from both sides:
\[ 32 r^3 = R^3 \]

### Step 4: Solve for the Radius of the Larger Drop
Taking the cube root of both sides, we find:
\[ R = \sqrt[3]{32} r = 2 r \]
This means the radius of the larger drop is twice the radius of each small drop.

### Step 5: Find the Capacitance
The capacitance \( C \) of a sphere is given by the formula:
\[ C = 4 \pi \epsilon_0 r \]
where \( \epsilon_0 \) is the permittivity of free space.

### Step 6: Calculate the Capacitance of the Small and Large Drops
- Capacitance of each small drop:
\[ C_{\text{small}} = 4 \pi \epsilon_0 r \]
- Capacitance of the larger drop:
\[ C_{\text{large}} = 4 \pi \epsilon_0 R = 4 \pi \epsilon_0 (2r) = 8 \pi \epsilon_0 r \]

### Step 7: Compare the Capacitances
Now, we can find the ratio of the capacitance of the larger drop to that of the smaller drop:
\[ \frac{C_{\text{large}}}{C_{\text{small}}} = \frac{8 \pi \epsilon_0 r}{4 \pi \epsilon_0 r} = 2 \]

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