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. 2 times
B. 8 times
C. 4 times
D. 16 times

Answer

To solve the problem, we need to compare the capacitance of a larger drop formed by combining eight smaller drops of mercury, each with equal radius and charge.

### Step-by-Step Solution:

1. **Understanding the Problem:**
- We have 8 small drops of mercury, each with a radius \( r \) and charge \( Q \).
- These drops combine to form a larger drop with radius \( R \).
- We need to find the relationship between the capacitance of the larger drop \( C_1 \) and the capacitance of one small drop \( C \).

2. **Volume Conservation:**
- The volume of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]
- The total volume of the 8 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:
\[ V_{\text{large}} = \frac{4}{3} \pi R^3 \]
- Setting these volumes equal gives:
\[ \frac{4}{3} \pi R^3 = \frac{32}{3} \pi r^3 \]
- Canceling \( \frac{4}{3} \pi \) from both sides, we have:
\[ R^3 = 8r^3 \]
- Taking the cube root of both sides:
\[ R = 2r \]

3. **Capacitance Calculation:**
- The capacitance \( C \) of a sphere is given by:
\[ C = 4 \pi \epsilon_0 r \]
- For the small drop, the capacitance \( C \) is:
\[ C = 4 \pi \epsilon_0 r \]
- For the larger drop, the capacitance \( C_1 \) is:
\[ C_1 = 4 \pi \epsilon_0 R = 4 \pi \epsilon_0 (2r) = 8 \pi \epsilon_0 r \]

4. **Finding the Ratio of Capacitances:**
- Now, we can find the ratio of the capacitance of the larger drop to the smaller drop:
\[ \frac{C_1}{C} = \frac{8 \pi \epsilon_0 r}{4 \pi \epsilon_0 r} = \frac{8}{4} = 2 \]

5. **Conclusion:**
- Therefore, the capacitance of the larger drop compared to each individual small drop is:
\[ C_1 = 2C \]

### Final Answer:
The capacitance of the bigger drop compared to each individual small drop is \( 2 \).