Area of liquid film is `6xx10cm^(2)` and surface tension is T = 20 dyne/cm, what is the work done to change area up to `12xx10cm^(2)`

Options

A. 120 joule
B. 120 erg
C. 1200 joule
D. 2400 erg

Answer

To solve the problem of calculating the work done to change the area of a liquid film, we will follow these steps:

### Step 1: Understand the Given Information
- Initial Area (A1) = \(6 \times 10 \, \text{cm}^2\)
- Final Area (A2) = \(12 \times 10 \, \text{cm}^2\)
- Surface Tension (T) = \(20 \, \text{dyne/cm}\)

### Step 2: Calculate the Initial Surface Energy
The surface energy (U) of a liquid film is given by the formula:
\[ U = T \times A \times n \]
where \(n\) is the number of free surfaces. For a liquid film, there are two free surfaces (top and bottom).

- Initial Surface Energy (U1):
\[ U_1 = T \times A_1 \times 2 = 20 \, \text{dyne/cm} \times (6 \times 10) \, \text{cm}^2 \times 2 \]
\[ U_1 = 20 \times 60 \, \text{dyne cm} = 1200 \, \text{dyne cm} \]

### Step 3: Calculate the Final Surface Energy
- Final Surface Energy (U2):
\[ U_2 = T \times A_2 \times 2 = 20 \, \text{dyne/cm} \times (12 \times 10) \, \text{cm}^2 \times 2 \]
\[ U_2 = 20 \times 120 \, \text{dyne cm} = 2400 \, \text{dyne cm} \]

### Step 4: Calculate the Change in Surface Energy
The work done (W) to change the area is equal to the change in surface energy:
\[ W = U_2 - U_1 \]
\[ W = 2400 \, \text{dyne cm} - 1200 \, \text{dyne cm} = 1200 \, \text{dyne cm} \]

### Step 5: Convert the Units (if necessary)
Since the work done is typically expressed in erg, we note that \(1 \, \text{dyne cm} = 1 \, \text{erg}\). Thus:
\[ W = 1200 \, \text{erg} \]

### Final Answer
The work done to change the area of the liquid film from \(6 \times 10 \, \text{cm}^2\) to \(12 \times 10 \, \text{cm}^2\) is **1200 erg**.

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