A membrane is formed on a square wire frame of 10 cm side when taken out from a liquid of surface tension `20 xx 10^(-3)Nm^(-1)` . Calculate the force acting on the surface of frame.

Answer

To solve the problem, we need to calculate the force acting on the surface of a square wire frame when it is taken out from a liquid, given the surface tension of the liquid.

### Step-by-Step Solution:

1. **Identify the dimensions of the square frame**:
The side length of the square frame is given as 10 cm. We need to convert this to meters for our calculations.
\[ \text{Side length (L)} = 10 \, \text{cm} = 0.1 \, \text{m} \]

2. **Determine the total length of the surface formed**:
When the square frame is taken out of the liquid, two surfaces are formed on the frame (one on the front and one on the back). Each side of the square contributes to the total length of the surface.
- The total length of the wire frame is:
\[ \text{Total length} = 4L \quad (\text{since there are 4 sides}) \]
- Since there are two surfaces (front and back), the total length of the surface is:
\[ \text{Total length of surface} = 2 \times 4L = 8L \]

3. **Substitute the value of L**:
Now, substituting the value of L:
\[ \text{Total length of surface} = 8 \times 0.1 \, \text{m} = 0.8 \, \text{m} \]

4. **Use the surface tension to calculate the force**:
The force acting on the surface of the frame can be calculated using the formula:
\[ \text{Force} (F) = \text{Surface Tension} (\sigma) \times \text{Total Length of Surface} \]
Given that the surface tension \(\sigma\) is \(20 \times 10^{-3} \, \text{N/m}\):
\[ F = 20 \times 10^{-3} \, \text{N/m} \times 0.8 \, \text{m} \]

5. **Calculate the force**:
\[ F = 20 \times 10^{-3} \times 0.8 = 16 \times 10^{-3} \, \text{N} \]

6. **Final Answer**:
The force acting on the surface of the frame is:
\[ F = 16 \, \text{mN} \, \text{(milliNewtons)} \]

### Summary:
The force acting on the surface of the square wire frame when taken out from the liquid is \(16 \, \text{mN}\).