A big water drop is formed by the combination of 'n' small water drops of equal radii. The ratio of the surface energy of 'n' drops to the surface energy of big drop is

Options

A. `n^2:1`
B. `n:1`
C. `sqrtn:1`
D. `root3(n):1`

Answer

To solve the problem of finding the ratio of the surface energy of 'n' small water drops to the surface energy of a big drop formed by their combination, we can follow these steps:

### Step 1: Understand Surface Energy
The surface energy (E) of a drop is given by the formula:
\[ E = \text{Surface Tension} \times \text{Surface Area} \]

### Step 2: Calculate Surface Energy of Small Drops
For a small spherical drop of radius \( r \):
- The surface area \( A \) is given by:
\[ A = 4\pi r^2 \]
- Therefore, the surface energy \( E_s \) of one small drop is:
\[ E_s = T \times A = T \times 4\pi r^2 \]

For 'n' small drops, the total surface energy \( E_n \) is:
\[ E_n = n \times E_s = n \times (T \times 4\pi r^2) = 4\pi T n r^2 \]

### Step 3: Calculate Surface Energy of the Big Drop
When 'n' small drops combine to form a big drop of radius \( R \):
- The volume of 'n' small drops is equal to the volume of the big drop. The volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3} \pi r^3 \]
- Therefore, the total volume of 'n' small drops is:
\[ V_n = n \times \frac{4}{3} \pi r^3 \]

The volume of the big drop is:
\[ V_B = \frac{4}{3} \pi R^3 \]

Setting these equal gives:
\[ n \times \frac{4}{3} \pi r^3 = \frac{4}{3} \pi R^3 \]
Cancelling \( \frac{4}{3} \pi \) from both sides:
\[ n r^3 = R^3 \]

### Step 4: Relate the Radii
From the equation \( n r^3 = R^3 \), we can express \( R \) in terms of \( r \):
\[ R = n^{1/3} r \]

### Step 5: Calculate Surface Energy of the Big Drop
The surface energy \( E_B \) of the big drop is:
\[ E_B = T \times \text{Surface Area of big drop} = T \times 4\pi R^2 \]
Substituting \( R \):
\[ E_B = T \times 4\pi (n^{1/3} r)^2 = T \times 4\pi n^{2/3} r^2 \]

### Step 6: Find the Ratio of Surface Energies
Now, we can find the ratio of the surface energy of 'n' small drops to the surface energy of the big drop:
\[ \text{Ratio} = \frac{E_n}{E_B} = \frac{4\pi T n r^2}{4\pi T n^{2/3} r^2} \]
Cancelling \( 4\pi T r^2 \):
\[ \text{Ratio} = \frac{n}{n^{2/3}} = n^{1/3} \]

### Final Answer
The ratio of the surface energy of 'n' small drops to the surface energy of the big drop is:
\[ \text{Ratio} = n^{1/3} \]