The potential of a large liquid drop when eight liquid drops are combined is 20 V. Then, the potential of each single drop was

Options

A. 10 V
B. 7.5 V
C. 5 V
D. 2.5 V

Answer

To solve the problem, we need to understand the relationship between the potential of a charged drop and its radius. The potential \( V \) of a charged sphere (or drop) is given by the formula:

\[ V = \frac{k \cdot Q}{R} \]

where:
- \( V \) is the potential,
- \( k \) is Coulomb's constant,
- \( Q \) is the charge, and
- \( R \) is the radius of the sphere.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We know that when 8 small drops combine to form one large drop, the potential of the large drop is given as 20 V. We need to find the potential of each small drop.

2. **Volume Conservation**:
The volume of the large drop is equal to the combined volume of the 8 small drops. The volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3} \pi r^3 \]
Therefore, for 8 small drops:
\[ 8 \left(\frac{4}{3} \pi r^3\right) = \frac{4}{3} \pi R^3 \]
where \( r \) is the radius of the small drop and \( R \) is the radius of the large drop.

3. **Equating Volumes**:
Simplifying the equation:
\[ 8 \cdot r^3 = R^3 \]
This implies:
\[ R = 2r \]
(since \( R^3 = 8r^3 \) leads to \( R = 2r \)).

4. **Relating Potential of Small and Large Drops**:
The potential of the large drop is given as 20 V. Using the potential formula:
\[ V_{large} = \frac{k \cdot Q_{large}}{R} \]
and for the small drop:
\[ V_{small} = \frac{k \cdot Q_{small}}{r} \]

5. **Charge Conservation**:
The total charge is conserved when the drops combine:
\[ Q_{large} = Q_{small} \times 8 \]
Therefore:
\[ V_{large} = \frac{k \cdot (8Q_{small})}{R} = \frac{k \cdot (8Q_{small})}{2r} = \frac{4k \cdot Q_{small}}{r} \]

6. **Finding the Potential of the Small Drop**:
From the above, we can relate the potentials:
\[ 20 V = 4 \cdot V_{small} \]
Thus:
\[ V_{small} = \frac{20 V}{4} = 5 V \]

### Final Answer:
The potential of each single drop is **5 V**.