Assume that 64 water droplets combine to form a large drop. Determine the ratio of the total surface energy of 64 droplets to that of large drop. Given, surface tension of water `= 0.072 Nm^(-1)`.

Explanation

To determine the ratio of the total surface energy of 64 water droplets to that of a large drop formed by these droplets, we can follow these steps:

### Step 1: Understand Surface Energy
Surface energy (E) is given by the formula:
\[ E = \text{Surface Area} \times \text{Surface Tension} \]
where the surface tension of water is given as \( \sigma = 0.072 \, \text{N/m} \).

### Step 2: Calculate the Volume of the Droplets
Let the radius of a small droplet be \( r \). The volume \( V \) of one small droplet is:
\[ V = \frac{4}{3} \pi r^3 \]
For 64 droplets, the total volume \( V_{total} \) is:
\[ V_{total} = 64 \times \frac{4}{3} \pi r^3 = \frac{256}{3} \pi r^3 \]

### Step 3: Calculate the Volume of the Large Drop
Let the radius of the large drop be \( R \). The volume of the large drop is:
\[ V_{large} = \frac{4}{3} \pi R^3 \]
Since the volume remains constant when the droplets combine, we have:
\[ \frac{256}{3} \pi r^3 = \frac{4}{3} \pi R^3 \]
Cancelling \( \frac{4}{3} \pi \) from both sides gives:
\[ 256 r^3 = 4 R^3 \]
From this, we can simplify to find \( R \):
\[ R^3 = 64 r^3 \]
Taking the cube root:
\[ R = 4r \]

### Step 4: Calculate Surface Areas
The surface area \( A \) of one small droplet is:
\[ A_{small} = 4 \pi r^2 \]
For 64 droplets, the total surface area \( A_{total} \) is:
\[ A_{total} = 64 \times 4 \pi r^2 = 256 \pi r^2 \]

The surface area of the large drop is:
\[ A_{large} = 4 \pi R^2 = 4 \pi (4r)^2 = 4 \pi \times 16 r^2 = 64 \pi r^2 \]

### Step 5: Calculate Surface Energies
Now we can calculate the surface energies:
- Total surface energy of 64 droplets:
\[ E_{small} = A_{total} \times \sigma = 256 \pi r^2 \times 0.072 \]
- Surface energy of the large drop:
\[ E_{large} = A_{large} \times \sigma = 64 \pi r^2 \times 0.072 \]

### Step 6: Find the Ratio of Surface Energies
Now we can find the ratio of the total surface energy of the 64 droplets to that of the large drop:
\[ \text{Ratio} = \frac{E_{small}}{E_{large}} = \frac{256 \pi r^2 \times 0.072}{64 \pi r^2 \times 0.072} \]
The \( \pi r^2 \) and \( 0.072 \) cancel out:
\[ \text{Ratio} = \frac{256}{64} = 4 \]

### Conclusion
Thus, the ratio of the total surface energy of 64 droplets to that of the large drop is:
\[ \text{Ratio} = 4 : 1 \]