A liquid drop having surface energy E is spread into 512 droplets of same size. The final surface energy of the droplets is

Options

A. 2E
B. 4E
C. 8E
D. 12E

Answer

To solve the problem, we need to find the final surface energy of the droplets when a liquid drop with surface energy \( E \) is divided into 512 smaller droplets of the same size.

### Step-by-Step Solution:

1. **Understanding the Volume Conservation**:
- The volume of the original liquid drop must equal the total volume of the 512 smaller droplets.
- Let the radius of the original drop be \( a \) and the radius of each smaller drop be \( b \).
- The volume of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]
- Therefore, the volume of the original drop is:
\[ V_{\text{original}} = \frac{4}{3} \pi a^3 \]
- The total volume of the 512 smaller droplets is:
\[ V_{\text{small}} = 512 \times \frac{4}{3} \pi b^3 \]
- Setting these equal gives:
\[ \frac{4}{3} \pi a^3 = 512 \times \frac{4}{3} \pi b^3 \]
- Canceling \( \frac{4}{3} \pi \) from both sides, we have:
\[ a^3 = 512 b^3 \]

2. **Finding the Relationship Between Radii**:
- From \( a^3 = 512 b^3 \), we can take the cube root:
\[ a = 8b \]

3. **Calculating Initial Surface Energy**:
- The surface energy \( E \) of the original drop is given by:
\[ E = \text{Surface Tension} \times \text{Surface Area} \]
- The surface area of the original drop is:
\[ A_{\text{original}} = 4 \pi a^2 \]
- Thus, the initial surface energy is:
\[ E = T \times 4 \pi a^2 \]

4. **Calculating Final Surface Energy**:
- The surface area of each smaller droplet is:
\[ A_{\text{small}} = 4 \pi b^2 \]
- The total surface area for 512 droplets is:
\[ A_{\text{total}} = 512 \times 4 \pi b^2 \]
- Therefore, the final surface energy \( E' \) is:
\[ E' = T \times A_{\text{total}} = T \times (512 \times 4 \pi b^2) \]

5. **Relating Initial and Final Surface Energies**:
- We can now relate \( E \) and \( E' \):
\[ E = T \times 4 \pi (8b)^2 = T \times 4 \pi \times 64b^2 = 256 T \pi b^2 \]
- And:
\[ E' = T \times (512 \times 4 \pi b^2) = 2048 T \pi b^2 \]

6. **Finding the Ratio**:
- Now we can find the ratio of \( E' \) to \( E \):
\[ \frac{E'}{E} = \frac{2048 T \pi b^2}{256 T \pi b^2} = \frac{2048}{256} = 8 \]
- Thus, we have:
\[ E' = 8E \]

### Final Answer:
The final surface energy of the droplets is \( 8E \).