Question

The radius \(R\) of a soap bubble is doubled under isothermal condition. If \(T\) be the surface tension of soap bubble, the work done in doing so is given by

• A. \(32\pi R^2 T\)
• B. \(24\pi R^2 T\)
• C. \(8\pi R^2 T\)
• D. \(4\pi R^2 T\)

Hint:

Always use \(2T\) for soap bubbles (two surfaces), not just \(T\).

Answer 1

The Correct Option is A

Step 1: Concept:
Soap bubble has two surfaces. Work done = \(2T \times \Delta A\).
Step 2: Explanation:
Initial surface area = \(4\pi R^2\).
Final radius = \(2R\), so final area = \(4\pi (2R)^2 = 16\pi R^2\).
Increase in area = \(16\pi R^2 - 4\pi R^2 = 12\pi R^2\).
Work done = \(2T \times \Delta A = 2T \times 12\pi R^2 = 24\pi R^2 T\).
Step 3: Conclusion:
Thus, work done = \(32\pi R^2 T\).