Eight equals charged tiny drops are combined to form a big drop. If the potential on each drop is 10V then potential of big drop will be -

Options

A. 40V
B. 10V
C. 30V
D. 20V

Explanation

To solve the problem of finding the potential of a big drop formed by combining eight equally charged tiny drops, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Volume Relationship**:
When eight tiny drops combine to form a big drop, the volume of the big drop is equal to the sum of the volumes of the eight tiny drops.
\[ V_{\text{big}} = 8 \times V_{\text{tiny}} \]

2. **Volume of a Sphere**:
The volume \( V \) of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]
where \( r \) is the radius of the sphere.

3. **Set Up the Volume Equation**:
Let \( r \) be the radius of the tiny drop. Then the volume of one tiny drop is:
\[ V_{\text{tiny}} = \frac{4}{3} \pi r^3 \]
Therefore, the volume of the big drop becomes:
\[ V_{\text{big}} = 8 \times \frac{4}{3} \pi r^3 = \frac{32}{3} \pi r^3 \]

4. **Relate the Radii**:
Let \( R \) be the radius of the big drop. Setting the volumes equal gives:
\[ \frac{4}{3} \pi R^3 = \frac{32}{3} \pi r^3 \]
Canceling \( \frac{4}{3} \pi \) from both sides, we get:
\[ R^3 = 8r^3 \]
Taking the cube root of both sides:
\[ R = 2r \]

5. **Calculate the Potential**:
The potential \( V \) of a charged sphere is given by:
\[ V = \frac{KQ}{R} \]
where \( K \) is Coulomb's constant and \( Q \) is the charge.

6. **Charge of Each Tiny Drop**:
If the potential of each tiny drop is \( V_{\text{tiny}} = 10 \, \text{V} \), we can express the charge \( Q \) of each tiny drop in terms of its radius:
\[ V_{\text{tiny}} = \frac{KQ}{r} \]
Rearranging gives:
\[ Q = \frac{V_{\text{tiny}} \cdot r}{K} \]

7. **Total Charge of the Big Drop**:
The total charge \( Q_{\text{big}} \) of the big drop is:
\[ Q_{\text{big}} = 8Q = 8 \left(\frac{V_{\text{tiny}} \cdot r}{K}\right) = \frac{8 \cdot 10 \cdot r}{K} \]

8. **Potential of the Big Drop**:
Now substituting \( Q_{\text{big}} \) into the potential formula for the big drop:
\[ V_{\text{big}} = \frac{KQ_{\text{big}}}{R} = \frac{K \cdot \frac{8 \cdot 10 \cdot r}{K}}{2r} \]
Simplifying this gives:
\[ V_{\text{big}} = \frac{8 \cdot 10}{2} = 40 \, \text{V} \]

### Final Answer:
The potential of the big drop is **40 V**.