A rectangular film of a certain liquid is 5 cms long and 3 cms is breadth. If the amount of work done in increasing its size to `6cmxx5cms" is "3xx10^(-4)J` then the value of surface tension of the liquid is-

Options

A. `0.4J//m^(2)`
B. `0.1J//m^(2)`
C. `3xx10^(-4)J//m^(2)`
D. `5xx10^(-4)J//m^(2)`

Answer

To find the value of surface tension of the liquid from the given data, we can follow these steps:

### Step 1: Understand the Problem
We have a rectangular film of a liquid with initial dimensions of 5 cm (length) and 3 cm (breadth). The film is expanded to dimensions of 6 cm (length) and 5 cm (breadth). The work done in this process is given as \(3 \times 10^{-4} \, \text{J}\).

### Step 2: Calculate the Initial and Final Perimeter
The perimeter of a rectangle is given by the formula:
\[ P = 2 \times (\text{length} + \text{breadth}) \]
- **Initial Perimeter (P1)**:
\[ P_1 = 2 \times (5 \, \text{cm} + 3 \, \text{cm}) = 2 \times 8 \, \text{cm} = 16 \, \text{cm} \]
- **Final Perimeter (P2)**:
\[ P_2 = 2 \times (6 \, \text{cm} + 5 \, \text{cm}) = 2 \times 11 \, \text{cm} = 22 \, \text{cm} \]

### Step 3: Calculate the Change in Perimeter
The change in perimeter (\(\Delta P\)) is:
\[ \Delta P = P_2 - P_1 = 22 \, \text{cm} - 16 \, \text{cm} = 6 \, \text{cm} \]

### Step 4: Convert Units
Convert the change in perimeter from centimeters to meters:
\[ \Delta P = 6 \, \text{cm} = 0.06 \, \text{m} \]

### Step 5: Relate Work Done to Surface Tension
The work done (W) in expanding the surface area of the film is given by:
\[ W = \text{Surface Tension} \times \Delta P \]
Let \(T\) be the surface tension. Therefore:
\[ W = T \times \Delta P \]
Substituting the values we have:
\[ 3 \times 10^{-4} \, \text{J} = T \times 0.06 \, \text{m} \]

### Step 6: Solve for Surface Tension
Rearranging the equation to solve for \(T\):
\[ T = \frac{3 \times 10^{-4} \, \text{J}}{0.06 \, \text{m}} = 5 \times 10^{-3} \, \text{J/m} \]

### Step 7: Final Answer
The value of the surface tension of the liquid is:
\[ T = 5 \times 10^{-3} \, \text{J/m} = 5 \times 10^{-4} \, \text{N/m} \]