Question

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)}

• A. $0.4\text{ N/m}$
• B. $0.04\text{ N/m}$
• C. $0.4\text{ dyne /cm}$
• D. $4.0\text{ N/m}$

Hint:

For films, do not forget the factor \(2\), because a film has two free surfaces.

Answer 1

The Correct Option is A

Concept:
For a soap film or liquid film with two surfaces: \[ W=2T\Delta A \] ip

Step 1: Find initial and final areas.
Initial area: \[ A_1=5\times 4=20\text{ cm}^2 \] Final area: \[ A_2=7\times 8=56\text{ cm}^2 \] So change in area: \[ \Delta A=56-20=36\text{ cm}^2 \] Convert to SI: \[ 36\text{ cm}^2=36\times10^{-4}\text{ m}^2=3.6\times10^{-3}\text{ m}^2 \] ip

Step 2: Use work done formula.
\[ W=2T\Delta A \] \[ 3\times10^{-4}=2T(3.6\times10^{-3}) \] ip

Step 3: Solve for surface tension.
\[ T=\frac{3\times10^{-4}}{7.2\times10^{-3}} \] \[ T\approx 4.17\times10^{-2}\text{ N/m}\approx 0.04\text{ N/m} \] So the direct calculation gives: \[ 0.04\text{ N/m} \] which matches option (B), not option (A). ip The direct calculation gives:
\[ \boxed{0.04\text{ N/m}} \]