27 similar drops of mercury are maintained at 10V each. All these spherical drops combine into a single big drop. The potential energy of the bigger drop is ............ times that of a smaller drop.

Answer

To solve the problem, we need to find the potential energy of the bigger drop formed by combining 27 smaller mercury drops, each maintained at a potential of 10V.

### Step-by-Step Solution:

1. **Identify the Charge on Smaller Drops**:
Each smaller drop has a charge \( Q \) and is maintained at a voltage \( V = 10V \).

2. **Calculate the Potential Energy of a Smaller Drop**:
The potential energy \( U_S \) of a single smaller drop can be calculated using the formula for the potential energy of a charged sphere:
\[ U_S = \frac{K Q^2}{2r} \]
where \( K \) is Coulomb's constant and \( r \) is the radius of the smaller drop.

3. **Combine the Smaller Drops**:
When 27 smaller drops combine, the total charge \( Q_T \) on the larger drop becomes:
\[ Q_T = 27Q \]

4. **Volume Conservation**:
The volume of the smaller drops combined equals the volume of the larger drop. The volume of a sphere is given by:
\[ V = \frac{4}{3} \pi r^3 \]
Therefore, for 27 smaller drops:
\[ 27 \left(\frac{4}{3} \pi r^3\right) = \frac{4}{3} \pi R^3 \]
where \( R \) is the radius of the larger drop. Simplifying this gives:
\[ 27r^3 = R^3 \implies R = 3r \]

5. **Calculate the Potential Energy of the Bigger Drop**:
The potential energy \( U_B \) of the larger drop can be expressed as:
\[ U_B = \frac{K (27Q)^2}{2R} \]
Substituting \( R = 3r \):
\[ U_B = \frac{K (27Q)^2}{2(3r)} = \frac{K \cdot 729Q^2}{6r} = \frac{243KQ^2}{2r} \]

6. **Find the Ratio of Potential Energies**:
Now, we can find the ratio of the potential energy of the bigger drop to that of a smaller drop:
\[ \frac{U_B}{U_S} = \frac{\frac{243KQ^2}{2r}}{\frac{KQ^2}{2r}} = 243 \]

### Final Answer:
The potential energy of the bigger drop is **243 times** that of a smaller drop.