There are `27` drops of a conducting fluid. Each drop has radius `r`, and each of them is charged to the same potential `V_1`. They are then combined to from a bigger drop. The potential of the bigger drop is `V_2`. Find the ratio `V_2//V_1`. Ignore the change in density of the fluid on combining the drops.

Explanation

To solve the problem, we need to find the ratio of the potential of the bigger drop \( V_2 \) to the potential of one smaller drop \( V_1 \).

### Step-by-step Solution:

1. **Understanding the Volume of Drops**:
Each smaller drop has a radius \( r \). The volume \( V \) of one drop is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]
Therefore, the total volume of 27 smaller drops is:
\[ V_{\text{total}} = 27 \times \frac{4}{3} \pi r^3 = 36 \pi r^3 \]

2. **Volume of the Bigger Drop**:
Let the radius of the bigger drop be \( R \). The volume of the bigger drop can also be expressed as:
\[ V_{\text{big}} = \frac{4}{3} \pi R^3 \]
Since the total volume of the smaller drops equals the volume of the bigger drop, we can set these equal:
\[ \frac{4}{3} \pi R^3 = 36 \pi r^3 \]

3. **Solving for \( R \)**:
Canceling \( \frac{4}{3} \pi \) from both sides gives:
\[ R^3 = 27 r^3 \]
Taking the cube root of both sides, we find:
\[ R = 3r \]

4. **Potential of Smaller Drops**:
The potential \( V_1 \) of a smaller drop is given by the formula:
\[ V_1 = \frac{kQ}{r} \]
where \( k \) is Coulomb's constant and \( Q \) is the charge on one drop.

5. **Total Charge on the Bigger Drop**:
The total charge \( Q_{\text{total}} \) on the 27 smaller drops is:
\[ Q_{\text{total}} = 27Q \]

6. **Potential of the Bigger Drop**:
The potential \( V_2 \) of the bigger drop is given by:
\[ V_2 = \frac{kQ_{\text{total}}}{R} = \frac{k(27Q)}{3r} = \frac{9kQ}{r} \]

7. **Finding the Ratio \( \frac{V_2}{V_1} \)**:
Now we can find the ratio of the potentials:
\[ \frac{V_2}{V_1} = \frac{\frac{9kQ}{r}}{\frac{kQ}{r}} = 9 \]

### Final Answer:
The ratio \( \frac{V_2}{V_1} = 9 \).