Two small drops of mercury each of radius r form a single large drop. The ratio of surface energy before and after this change is

Options

A. `2 : 2^(2//3)`
B. `2^(2//3) : 1`
C. `2 : 1`
D. `1 :2`

Explanation

To find the ratio of surface energy before and after the merging of two small mercury drops into a larger one, we can follow these steps:

### Step 1: Calculate the volume of the small drops
The volume \( V \) of a single small drop of mercury with radius \( r \) is given by the formula for the volume of a sphere:
\[ V = \frac{4}{3} \pi r^3 \]
Since there are two small drops, the total volume \( V_{small} \) is:
\[ V_{small} = 2 \times \frac{4}{3} \pi r^3 = \frac{8}{3} \pi r^3 \]

### Step 2: Calculate the radius of the large drop
When the two small drops merge into a single larger drop, the volume of the larger drop \( V_{large} \) is equal to the total volume of the two small drops:
\[ V_{large} = \frac{4}{3} \pi R^3 \]
Setting the volumes equal gives:
\[ \frac{8}{3} \pi r^3 = \frac{4}{3} \pi R^3 \]
We can simplify this by canceling \( \frac{4}{3} \pi \) from both sides:
\[ 2r^3 = R^3 \]
Taking the cube root of both sides, we find:
\[ R = 2^{1/3} r \]

### Step 3: Calculate the surface areas of the drops
The surface area \( A \) of a sphere is given by:
\[ A = 4 \pi r^2 \]
For the two small drops, the total surface area \( A_{small} \) is:
\[ A_{small} = 2 \times 4 \pi r^2 = 8 \pi r^2 \]
For the larger drop, the surface area \( A_{large} \) is:
\[ A_{large} = 4 \pi R^2 = 4 \pi (2^{1/3} r)^2 = 4 \pi \cdot 2^{2/3} r^2 = 4 \cdot 2^{2/3} \pi r^2 \]

### Step 4: Calculate the surface energy
The surface energy \( U \) is directly proportional to the surface area. Thus, we can express the surface energies:
- Surface energy before merging (for two small drops):
\[ U_{small} \propto A_{small} = 8 \pi r^2 \]
- Surface energy after merging (for the large drop):
\[ U_{large} \propto A_{large} = 4 \cdot 2^{2/3} \pi r^2 \]

### Step 5: Calculate the ratio of surface energies
The ratio of the surface energy before and after merging is:
\[ \frac{U_{small}}{U_{large}} = \frac{8 \pi r^2}{4 \cdot 2^{2/3} \pi r^2} = \frac{8}{4 \cdot 2^{2/3}} = \frac{2}{2^{2/3}} = 2^{1 - 2/3} = 2^{1/3} \]

### Final Answer
The ratio of surface energy before and after the change is:
\[ \frac{U_{small}}{U_{large}} = 2^{1/3} \]