27 drops of same size are charged `220 V` each. They coalesce to form a bigger drop. Calculate the potential of bigger drop.

Explanation

To solve the problem of finding the potential of a bigger drop formed by the coalescence of 27 smaller drops, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have 27 drops, each with a potential of \( V = 220 \, \text{V} \).
- When these drops coalesce, they form a single larger drop.

2. **Volume Conservation**:
- The volume of the smaller drops combined must equal the volume of the larger drop.
- The volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3} \pi r^3 \]
- For 27 smaller drops, the total volume is:
\[ V_{\text{total}} = 27 \times \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (27r^3) \]
- The volume of the larger drop is:
\[ V_{\text{big}} = \frac{4}{3} \pi R^3 \]
- Setting these equal gives:
\[ \frac{4}{3} \pi R^3 = \frac{4}{3} \pi (27r^3) \]
- Simplifying, we find:
\[ R^3 = 27r^3 \implies R = 3r \]

3. **Charge Calculation**:
- Let \( q \) be the charge on each small drop.
- The total charge \( Q \) on the larger drop is:
\[ Q = 27q \]

4. **Capacitance Calculation**:
- The capacitance \( C \) of a sphere is given by:
\[ C = 4 \pi \epsilon_0 R \]
- For the larger drop:
\[ C_{\text{big}} = 4 \pi \epsilon_0 (3r) = 12 \pi \epsilon_0 r \]

5. **Potential of the Larger Drop**:
- The potential \( V' \) of the larger drop is given by:
\[ V' = \frac{Q}{C_{\text{big}}} \]
- Substituting the values:
\[ V' = \frac{27q}{12 \pi \epsilon_0 r} \]

6. **Relating Charge and Potential of Smaller Drops**:
- The potential of a small drop is:
\[ V = \frac{q}{4 \pi \epsilon_0 r} \]
- Rearranging gives:
\[ q = 4 \pi \epsilon_0 r V \]
- Substituting this into the potential of the larger drop:
\[ V' = \frac{27(4 \pi \epsilon_0 r V)}{12 \pi \epsilon_0 (3r)} \]
- Simplifying:
\[ V' = \frac{27 \cdot 4 V}{12 \cdot 3} = \frac{27 \cdot 4 V}{36} = \frac{3}{4} \cdot 27 V = 9V \]

7. **Final Calculation**:
- Given \( V = 220 \, \text{V} \):
\[ V' = 9 \times 220 = 1980 \, \text{V} \]

### Conclusion:
The potential of the bigger drop is \( \boxed{1980 \, \text{V}} \).