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:

Work done equals surface tension times the increase in surface area, and the film has two free surfaces.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Work done \(W = T \times \Delta A\). A liquid film has two surfaces, so the total area increase is twice the change in the rectangle area.

Step 2: Change in Area
Initial area \(= 5 \times 4 = 20\ \text{cm}^2\). Final area \(= 7 \times 8 = 56\ \text{cm}^2\). Change \(= 36\ \text{cm}^2 = 36 \times 10^{-4}\ \text{m}^2\). Two surfaces give \(\Delta A = 72 \times 10^{-4}\ \text{m}^2\).

Step 3: Calculation
\(T = \dfrac{W}{\Delta A} = \dfrac{3 \times 10^{-4}}{72 \times 10^{-4}} \approx 0.042\) N/m.

Step 4: Compare the options
This is nearly 0.04 N/m, the second option.

Final Answer:
The surface tension is nearly 0.04 N/m. \[ \boxed{\text{(B) }0.04\ \text{N/m}} \]