8000 identical water drops combine together to form a big drop. Then the ratio of the initial surface enrgy of all the initial surface energy of all the drops together is

Options

A. `1:10`
B. `1:15`
C. `1:20`
D. `1:25`

Explanation

To solve the problem of finding the ratio of the initial surface energy of all the small drops to the final surface energy of the big drop formed by combining 8000 identical water drops, we can follow these steps:

### Step 1: Understand the Surface Energy Formula
The surface energy (E) of a drop can be expressed as:
\[ E = \sigma \times A \]
where \( \sigma \) is the surface tension and \( A \) is the surface area of the drop.

### Step 2: Calculate Initial Surface Energy
For small identical drops with radius \( r \):
- The surface area \( A \) of one small drop is:
\[ A = 4\pi r^2 \]
- Therefore, the total surface area of \( n \) drops (where \( n = 8000 \)) is:
\[ A_{\text{initial}} = n \times 4\pi r^2 = 8000 \times 4\pi r^2 \]
- The initial surface energy \( E_{\text{initial}} \) is:
\[ E_{\text{initial}} = \sigma \times A_{\text{initial}} = \sigma \times (8000 \times 4\pi r^2) \]

### Step 3: Calculate Final Surface Energy
When these drops combine to form a big drop with radius \( R \):
- The surface area of the big drop is:
\[ A_{\text{final}} = 4\pi R^2 \]
- Therefore, the final surface energy \( E_{\text{final}} \) is:
\[ E_{\text{final}} = \sigma \times A_{\text{final}} = \sigma \times (4\pi R^2) \]

### Step 4: Relate the Radii Using Volume Conservation
The volume of the small drops combined must equal the volume of the big drop:
\[ n \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3 \]
This simplifies to:
\[ n r^3 = R^3 \]
From this, we can express \( R \) in terms of \( r \):
\[ R = n^{1/3} r \]

### Step 5: Substitute \( R \) into Final Surface Energy
Now substitute \( R \) into the expression for \( E_{\text{final}} \):
\[ E_{\text{final}} = \sigma \times (4\pi (n^{1/3} r)^2) = \sigma \times (4\pi n^{2/3} r^2) \]

### Step 6: Find the Ratio of Final to Initial Surface Energy
Now we can find the ratio:
\[ \frac{E_{\text{final}}}{E_{\text{initial}}} = \frac{\sigma \times (4\pi n^{2/3} r^2)}{\sigma \times (8000 \times 4\pi r^2)} = \frac{n^{2/3}}{8000} \]

### Step 7: Substitute \( n = 8000 \)
Substituting \( n = 8000 \):
\[ \frac{E_{\text{final}}}{E_{\text{initial}}} = \frac{8000^{2/3}}{8000} = \frac{8000^{2/3}}{8000^{1}} = \frac{1}{8000^{1/3}} = \frac{1}{20} \]

### Final Answer
Thus, the ratio of the initial surface energy of all the small drops to the final surface energy of the big drop is:
\[ \frac{E_{\text{initial}}}{E_{\text{final}}} = 20 \]