There are `27` drops of a conducting fluid. Each has a radius r and they are charged to a potential `V_(0)`. They are then combined to form a bigger drop. Find its potential.

Explanation

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

### Step-by-Step Solution:

1. **Understand the 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 the formula:
\[ V = \frac{4}{3} \pi r^3 \]
For 27 smaller drops, the total volume is:
\[ V_{total} = 27 \times \frac{4}{3} \pi r^3 \]
For the larger drop with radius \( R \), the volume is:
\[ V_{big} = \frac{4}{3} \pi R^3 \]

2. **Set the Volumes Equal**:
Set the total volume of the smaller drops equal to the volume of the larger drop:
\[ 27 \times \frac{4}{3} \pi r^3 = \frac{4}{3} \pi R^3 \]
We can cancel \( \frac{4}{3} \pi \) from both sides:
\[ 27 r^3 = R^3 \]

3. **Solve for the Radius of the Bigger Drop**:
Taking the cube root of both sides gives:
\[ R = \sqrt[3]{27} r = 3r \]

4. **Relate Charge and Potential**:
The potential \( V \) of a charged sphere is given by:
\[ V = \frac{kQ}{R} \]
where \( k \) is Coulomb's constant, \( Q \) is the charge, and \( R \) is the radius of the sphere.

5. **Charge of the Smaller Drops**:
Each smaller drop has a potential \( V_0 \):
\[ V_0 = \frac{kQ}{r} \]
Rearranging this gives:
\[ Q = \frac{V_0 r}{k} \]

6. **Total Charge of the Bigger Drop**:
Since there are 27 drops, the total charge \( Q_{big} \) of the larger drop is:
\[ Q_{big} = 27Q = 27 \left(\frac{V_0 r}{k}\right) = \frac{27 V_0 r}{k} \]

7. **Potential of the Bigger Drop**:
Substituting \( Q_{big} \) and \( R = 3r \) into the potential formula for the larger drop:
\[ V_{big} = \frac{kQ_{big}}{R} = \frac{k \left(\frac{27 V_0 r}{k}\right)}{3r} \]
Simplifying this gives:
\[ V_{big} = \frac{27 V_0}{3} = 9 V_0 \]

### Final Answer:
The potential of the bigger drop is \( 9 V_0 \).