A soap film in formed on a frame of area `4xx10^(-3)m^(2)`. If the area of the film in reduced to half, then the change in the potential energy of the film is (surface tension of soap solution`=40xx10^(-3)N//m)`

Options

A. `32xx10^(-5)` J
B. `16xx10^(-5)` J
C. `8xx10^(-5)` J
D. `16xx10^(-5)` J

Answer

To solve the problem step by step, we will follow these steps:

### Step 1: Understand the problem
We are given the initial area of a soap film and the surface tension of the soap solution. We need to find the change in potential energy when the area of the soap film is reduced to half.

### Step 2: Identify the given values
- Initial area of the soap film, \( A = 4 \times 10^{-3} \, m^2 \)
- Final area of the soap film, \( A_f = \frac{A}{2} = \frac{4 \times 10^{-3}}{2} = 2 \times 10^{-3} \, m^2 \)
- Surface tension of the soap solution, \( T = 40 \times 10^{-3} \, N/m \)

### Step 3: Calculate the change in area
The change in area, \( \Delta S \), can be calculated as:
\[ \Delta S = A_f - A = 2 \times 10^{-3} - 4 \times 10^{-3} = -2 \times 10^{-3} \, m^2 \]
Since we are interested in the magnitude of the change in area, we take the absolute value:
\[ |\Delta S| = 2 \times 10^{-3} \, m^2 \]

### Step 4: Calculate the change in potential energy
The change in potential energy \( \Delta U \) of the soap film can be calculated using the formula:
\[ \Delta U = T \cdot \Delta S_{total} \]
Since the soap film has two surfaces, the total change in area is:
\[ \Delta S_{total} = 2 \cdot |\Delta S| = 2 \cdot (2 \times 10^{-3}) = 4 \times 10^{-3} \, m^2 \]
Now substituting the values into the potential energy formula:
\[ \Delta U = T \cdot \Delta S_{total} = (40 \times 10^{-3}) \cdot (4 \times 10^{-3}) \]
Calculating this gives:
\[ \Delta U = 160 \times 10^{-6} \, J = 16 \times 10^{-5} \, J \]

### Step 5: Final answer
The change in potential energy of the soap film when the area is reduced to half is:
\[ \Delta U = 16 \times 10^{-5} \, J \]