Two small drops of mercury, each of radius `R`, coalesce to form a single large drop. The ratio of the total surface energies before and after the change is

Explanation

To solve the problem of finding the ratio of the total surface energies before and after two small drops of mercury coalesce into a single larger drop, 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 drops, the total volume \( V_{\text{total}} \) of the two small drops is:
\[ V_{\text{total}} = 2 \times \frac{4}{3} \pi R^3 = \frac{8}{3} \pi R^3 \]

### Step 2: Calculate the Radius of the Large Drop
When the two small drops coalesce, they form a single larger drop. Let the radius of the larger drop be \( R' \). The volume of the larger drop is also given by the volume formula:
\[ V' = \frac{4}{3} \pi (R')^3 \]
Setting the total volume of the two small drops equal to the volume of the larger drop gives:
\[ \frac{8}{3} \pi R^3 = \frac{4}{3} \pi (R')^3 \]
We can cancel \( \frac{4}{3} \pi \) from both sides:
\[ 2 R^3 = (R')^3 \]
Taking the cube root of both sides, we find:
\[ R' = R \cdot 2^{1/3} \]

### Step 3: Calculate the Surface Energies
The surface energy \( S \) of a drop is given by:
\[ S = \text{Surface Area} \times \text{Surface Tension} \]
The surface area \( A \) of a sphere is given by:
\[ A = 4 \pi R^2 \]
Thus, the surface energy of one small drop is:
\[ S_{\text{small}} = 4 \pi R^2 \cdot T \]
For two small drops, the total surface energy \( S_{\text{initial}} \) is:
\[ S_{\text{initial}} = 2 \times S_{\text{small}} = 2 \times (4 \pi R^2 \cdot T) = 8 \pi R^2 T \]

For the larger drop with radius \( R' \):
\[ S_{\text{large}} = 4 \pi (R')^2 \cdot T = 4 \pi (R \cdot 2^{1/3})^2 \cdot T = 4 \pi R^2 \cdot 2^{2/3} \cdot T \]

### Step 4: Calculate the Ratio of Surface Energies
Now we can find the ratio of the total surface energies before and after the coalescence:
\[ \text{Ratio} = \frac{S_{\text{initial}}}{S_{\text{large}}} = \frac{8 \pi R^2 T}{4 \pi R^2 \cdot 2^{2/3} T} \]
Canceling out common terms:
\[ \text{Ratio} = \frac{8}{4 \cdot 2^{2/3}} = \frac{2}{2^{2/3}} = 2^{1 - 2/3} = 2^{1/3} \]

### Final Answer
Thus, the ratio of the total surface energies before and after the change is:
\[ \text{Ratio} = 2^{1/3} \]