Twenty seven drops of mercury are charged simultaneously to the same potential of 10V. What will be the potential if all the charge drops are made to combine to form one large drop? Assume the drops to be spherical.

Options

A. 90
B. 40
C. 160
D. 10

Answer

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

### Step-by-Step Solution:

1. **Understanding the Volume Conservation**:
- The volume of the smaller drops combined must equal the volume of the larger drop.
- The volume \( V \) of a single spherical drop is given by:
\[ V = \frac{4}{3} \pi r^3 \]
- Therefore, the total volume of 27 smaller drops is:
\[ V_{\text{total}} = 27 \times \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (27 r^3) \]

2. **Volume of the Larger Drop**:
- Let the radius of the larger drop be \( R \). The volume of the larger drop is:
\[ V_{\text{large}} = \frac{4}{3} \pi R^3 \]

3. **Setting Volumes Equal**:
- By the conservation of volume:
\[ \frac{4}{3} \pi (27 r^3) = \frac{4}{3} \pi R^3 \]
- Canceling out \( \frac{4}{3} \pi \) from both sides gives:
\[ 27 r^3 = R^3 \]

4. **Finding the Radius of the Larger Drop**:
- Taking the cube root of both sides:
\[ R = 3r \]

5. **Understanding Charge Conservation**:
- The charge \( q \) on a single drop can be expressed in terms of its potential \( V \) and radius \( r \):
\[ q = V \cdot k \cdot r \]
- For 27 drops, the total charge \( Q \) is:
\[ Q = 27q = 27 \cdot (10 \cdot k \cdot r) = 270k \cdot r \]

6. **Finding the Potential of the Larger Drop**:
- The potential \( V' \) of the larger drop is given by:
\[ V' = \frac{Q}{kR} \]
- Substituting \( Q \) and \( R \):
\[ V' = \frac{270k \cdot r}{k \cdot (3r)} = \frac{270}{3} = 90 \text{ volts} \]

### Final Answer:
The potential of the larger drop is **90 volts**.