Question

A circular film of a liquid has an area of \(10\ \text{cm}^2\). If the work done in making its radius two times the initial radius is \(8\times10^{-3}\ \text{J}\), the surface tension of the liquid is \[ \left(1+\frac{1}{\alpha}\right)\ \text{N m}^{-1}. \] The value of \(\alpha\) is

• A. \(5\)
• B. \(4\)
• C. \(3\)
• D. \(2\)

Hint:

For a liquid film, there are two surfaces, so the work done in increasing area is \[ W=2T\Delta A. \] Also, for a circular film, \[ A\propto r^2. \] If radius doubles, area becomes four times.

Answer 1

The Correct Option is C

Step 1: Relate work done to surface tension.
Initial radius: \( \pi r_1^2 = 10\text{ cm}^2 \Rightarrow r_1^2 = 10/\pi \). Final radius \( r_2 = 2r_1 \), so \( r_2^2 = 4r_1^2 \). A film has 2 surfaces.
\( W = T\times 2\times\pi(r_2^2 - r_1^2) = T\times 2\times 3\pi r_1^2 = 6\pi T r_1^2 \)

Step 2: Solve for T, then find alpha.
\( 8\times10^{-3} = 6\pi T\times\frac{10\times10^{-4}}{\pi} = 6T\times10^{-3} \Rightarrow T = \frac{8\times10^{-3}}{6\times10^{-3}} = \frac{4}{3}\text{ N/m} \).
Given \( T = \left(1+\frac{1}{\alpha}\right) \Rightarrow \frac{1}{\alpha} = \frac{4}{3}-1 = \frac{1}{3} \Rightarrow \alpha = 3 \).

\[ \boxed{\alpha = 3} \]