Question

A rectangular film of liquid is expanded from (\(6\) cm \(\times 5\) cm) to (\(8\) cm \(\times 9\) cm). If the work done is \(4\times 10^{-4}\) J, the surface tension of the liquid is

• A. \(0.04\frac{N}{m}\)
• B. \(0.05\frac{N}{m}\)
• C. \(0.06\frac{N}{m}\)
• D. \(0.07\frac{N}{m}\)

Hint:

A film has two surfaces, so the work done equals surface tension times twice the increase in area of one face.

Answer 1

The Correct Option is B

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

Step 2: Area change
Initial area of one face: \(6\times5=30\) cm\(^2\)
Final area of one face: \(8\times9=72\) cm\(^2\)
Increase for both faces: \(2\times42=84\) cm\(^2\) \(=84\times10^{-4}\) m\(^2\)

Step 3: Calculation
\(T=\dfrac{W}{\Delta A}=\dfrac{4\times10^{-4}}{84\times10^{-4}}\approx0.048\) N/m

Step 4: Compare
This is nearly \(0.05\) N/m.

Final Answer:
The surface tension is nearly 0.05 newton per metre. \[ \boxed{\text{(B) }0.05\,\dfrac{\text{N}}{\text{m}}} \]