8000 identical water drops are combined to form a big drop then the ratio of the final surface energy to the initial surface energy of all the drops together is

Options

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

Answer

To solve the problem of finding the ratio of the final surface energy to the initial surface energy when 8000 identical water drops combine to form a larger drop, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have 8000 identical small water drops, and we need to find the ratio of the surface energy of the final big drop to the total surface energy of the initial small drops.

2. **Defining Variables**:
Let the radius of each small drop be \( r \) and the radius of the big drop be \( R \).

3. **Volume Conservation**:
The volume of the big drop must equal the total volume of the 8000 small drops. The volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3} \pi r^3 \]
Therefore, the total volume of the 8000 small drops is:
\[ V_{\text{initial}} = 8000 \times \frac{4}{3} \pi r^3 \]
The volume of the big drop is:
\[ V_{\text{final}} = \frac{4}{3} \pi R^3 \]
Setting these equal gives:
\[ \frac{4}{3} \pi R^3 = 8000 \times \frac{4}{3} \pi r^3 \]
Cancelling \( \frac{4}{3} \pi \) from both sides:
\[ R^3 = 8000 r^3 \]

4. **Finding the Radius Relationship**:
Taking the cube root of both sides:
\[ R = 8000^{1/3} r = 20r \]

5. **Calculating Surface Energy**:
The surface energy \( E \) is given by:
\[ E = \text{Surface Tension} \times \text{Surface Area} \]
The surface area \( A \) of a sphere is given by:
\[ A = 4 \pi r^2 \]
Thus, the initial surface energy \( E_i \) for 8000 small drops is:
\[ E_i = 8000 \times \sigma \times 4 \pi r^2 = 32000 \pi \sigma r^2 \]
The final surface energy \( E_f \) for the big drop is:
\[ E_f = \sigma \times 4 \pi R^2 = 4 \pi \sigma (20r)^2 = 1600 \pi \sigma r^2 \]

6. **Finding the Ratio**:
Now, we can find the ratio of the final surface energy to the initial surface energy:
\[ \frac{E_f}{E_i} = \frac{1600 \pi \sigma r^2}{32000 \pi \sigma r^2} \]
Simplifying this gives:
\[ \frac{E_f}{E_i} = \frac{1600}{32000} = \frac{1}{20} \]

### Final Answer:
The ratio of the final surface energy to the initial surface energy is:
\[ \frac{E_f}{E_i} = \frac{1}{20} \]