A rectangular film of liquid is expanded from $(5\text{ cm} \times 4\text{ cm})$ to $(7\text{ cm} \times 8\text{ cm})$. If the work done is $3 \times 10^{-4}\text{ J}$, the surface tension of the liquid is (nearly)}
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For films, do not forget the factor \(2\), because a film has two free surfaces.
Step 1: Understanding the Concept:
Work done in expanding a liquid film is stored as surface energy. A liquid film has two free surfaces. Step 2: Key Formula or Approach:
Work done $W = 2 \times T \times \Delta A$. Step 3: Detailed Explanation:
Initial area $A_1 = 20\text{ cm}^2 = 20 \times 10^{-4}\text{ m}^2$.
Final area $A_2 = 56\text{ cm}^2 = 56 \times 10^{-4}\text{ m}^2$.
$\Delta A = 36 \times 10^{-4}\text{ m}^2$.
\[ 3 \times 10^{-4} = 2 \times T \times (36 \times 10^{-4}) \]
\[ T = \frac{3}{72} \approx 0.0416\text{ N/m} \] Step 4: Final Answer:
The surface tension is nearly $0.04\text{ N/m}$.