Question

The work done in increasing the radius of a soap bubble from R to 2R is W. The work done in further increasing its radius from 2R to 3R will be:

• A. 5/3 W
• B. 4/3 W
• C. 7/3 W
• D. W

Hint:

Key Exam Tip:
Work done on a soap bubble is $2S \Delta A$. Work is proportional to the change in surface area, which scales with the square of the radius.

Answer 1

The Correct Option is A

Work done ($W$) to increase the surface area of a soap bubble is $W = 2S \Delta A$, where $S$ is the surface tension and $\Delta A$ is the change in surface area. Surface area $A = 4\pi r^2$.
Work done from R to 2R: $W = 2S (4\pi(2R)^2 - 4\pi R^2) = 2S (16\pi R^2 - 4\pi R^2) = 2S (12\pi R^2) = 24\pi S R^2$.
Work done from 2R to 3R: $W' = 2S (4\pi(3R)^2 - 4\pi(2R)^2) = 2S (36\pi R^2 - 16\pi R^2) = 2S (20\pi R^2) = 40\pi S R^2$.
To express $W'$ in terms of $W$: $W' = \frac{40\pi S R^2}{24\pi S R^2} \times W = \frac{40}{24} W = \frac{5}{3}W$.
Final Answer: \(\boxed{A}\)