A liquid drop of radius R is broken into 1000 drops each of radius r. If T is surface tension, change in surface energy is

Options

A. `4piR^2T`
B. `72piR^2T``
C. `16piR^2T`
D. `36piR^2T`

Explanation

To solve the problem of finding the change in surface energy when a liquid drop of radius \( R \) is broken into 1000 smaller drops of radius \( r \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Initial and Final Conditions**:
- We have one large drop of radius \( R \).
- This drop is broken into 1000 smaller drops, each of radius \( r \).

2. **Calculate the Volume of the Large Drop**:
- The volume \( V \) of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi R^3 \]
- Therefore, the volume of the large drop is:
\[ V_{\text{initial}} = \frac{4}{3} \pi R^3 \]

3. **Calculate the Volume of the Smaller Drops**:
- The volume of one small drop of radius \( r \) is:
\[ V_{\text{small}} = \frac{4}{3} \pi r^3 \]
- Since there are 1000 such drops, the total volume of the smaller drops is:
\[ V_{\text{final}} = 1000 \times \frac{4}{3} \pi r^3 = \frac{4000}{3} \pi r^3 \]

4. **Set the Initial and Final Volumes Equal**:
- By conservation of volume, we have:
\[ \frac{4}{3} \pi R^3 = \frac{4000}{3} \pi r^3 \]
- Canceling \( \frac{4}{3} \pi \) from both sides gives:
\[ R^3 = 1000 r^3 \]

5. **Solve for the Radius of the Smaller Drops**:
- Taking the cube root of both sides:
\[ R = 10 r \]

6. **Calculate the Surface Energy of the Large Drop**:
- The surface area \( A \) of the large drop is:
\[ A_{\text{initial}} = 4 \pi R^2 \]
- The surface energy \( E_{\text{initial}} \) is given by:
\[ E_{\text{initial}} = T \times A_{\text{initial}} = T \times 4 \pi R^2 \]

7. **Calculate the Surface Energy of the Small Drops**:
- The surface area of one small drop is:
\[ A_{\text{small}} = 4 \pi r^2 \]
- Therefore, the total surface area of the 1000 small drops is:
\[ A_{\text{final}} = 1000 \times 4 \pi r^2 = 4000 \pi r^2 \]
- The surface energy \( E_{\text{final}} \) is:
\[ E_{\text{final}} = T \times A_{\text{final}} = T \times 4000 \pi r^2 \]

8. **Calculate the Change in Surface Energy**:
- The change in surface energy \( \Delta E \) is given by:
\[ \Delta E = E_{\text{final}} - E_{\text{initial}} \]
- Substituting the expressions we found:
\[ \Delta E = (4000 \pi r^2 T) - (4 \pi R^2 T) \]
- Substitute \( R = 10r \):
\[ \Delta E = 4000 \pi r^2 T - 4 \pi (10r)^2 T \]
- Simplifying gives:
\[ \Delta E = 4000 \pi r^2 T - 400 \pi r^2 T = 3600 \pi r^2 T \]

### Final Answer:
The change in surface energy is:
\[ \Delta E = 3600 \pi r^2 T \]