A mercury drop has potential 'V' on its surface. 1000 such drops combine to form a new drop. Find the potential on the surface of the new drop.

Options

A. V
B. 10 V
C. 100 V
D. 1000 V

Answer

To solve the problem, we need to find the potential on the surface of a new drop formed by combining 1000 smaller mercury drops, each having a potential \( V \).

### Step-by-Step Solution:

1. **Understanding the Potential of a Single Drop:**
The potential \( V \) on the surface of a small mercury drop can be expressed using the formula:
\[ V = \frac{kQ}{R} \]
where \( k \) is Coulomb's constant, \( Q \) is the charge on the drop, and \( R \) is the radius of the small drop.

2. **Volume Conservation:**
When 1000 smaller drops combine to form a larger drop, the volume of the larger drop will be equal to the total volume of the 1000 smaller drops. The volume \( V \) of a single drop is given by:
\[ V = \frac{4}{3} \pi r^3 \]
Therefore, the total volume of 1000 smaller drops is:
\[ V_{\text{total}} = 1000 \times \frac{4}{3} \pi r^3 \]
The volume of the larger drop is:
\[ V_{\text{large}} = \frac{4}{3} \pi R^3 \]
Setting these equal gives:
\[ 1000 \times \frac{4}{3} \pi r^3 = \frac{4}{3} \pi R^3 \]
Simplifying, we find:
\[ 1000 r^3 = R^3 \]

3. **Finding the Radius of the Larger Drop:**
Taking the cube root of both sides, we have:
\[ R = 10r \]
This means the radius of the larger drop is 10 times the radius of a smaller drop.

4. **Charge of the Larger Drop:**
The charge \( Q \) on a drop is proportional to its volume (and hence its radius). Since the charge of a single drop is \( Q \), the total charge of the larger drop formed from 1000 smaller drops is:
\[ Q_{\text{large}} = 1000Q \]

5. **Calculating the Potential of the Larger Drop:**
Now we can find the potential \( V_{\text{large}} \) of the larger drop using the formula:
\[ V_{\text{large}} = \frac{kQ_{\text{large}}}{R} \]
Substituting \( Q_{\text{large}} = 1000Q \) and \( R = 10r \):
\[ V_{\text{large}} = \frac{k(1000Q)}{10r} \]
We can express \( kQ/r \) in terms of the potential \( V \) of a smaller drop:
\[ V_{\text{large}} = \frac{1000}{10} \cdot \frac{kQ}{r} = 100V \]

6. **Final Result:**
Thus, the potential on the surface of the new drop is:
\[ V_{\text{large}} = 100V \]

### Conclusion:
The potential on the surface of the new drop formed by combining 1000 smaller drops is \( 100V \).