Question

The amount of work done in blowing a soap bubble such that its diameter increases from 'd' to 'D' is ($T$ = surface tension of solution) ______.

• A. $\pi (D^2 - d^2) T$
• B. $2\pi (D^2 - d^2) T$
• C. $4\pi (D^2 - d^2) T$
• D. $8\pi (D^2 - d^2) T$

Hint:

Always read carefully: is it a liquid drop or a soap bubble? A drop has 1 surface (area = $4\pi r^2$). A bubble has 2 surfaces (area = $8\pi r^2$). Failing to multiply by 2 is the most common error in these problems!

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Work done in increasing the surface area of a liquid film is $W = T \times \Delta A$. A soap bubble has two free surfaces (inner and outer).

Step 2: Formula Application:
Surface area of a sphere is $4\pi r^2$ or $\pi (\text{diameter})^2$. Total surface area $A = 2 \times \pi (\text{diameter})^2$.

Step 3: Explanation:
Initial area $A_1 = 2\pi d^2$. Final area $A_2 = 2\pi D^2$. Change in area $\Delta A = 2\pi (D^2 - d^2)$. Work done $W = T \times \Delta A = 2\pi (D^2 - d^2) T$.

Step 4: Final Answer:
The work done is $2\pi (D^2 - d^2) T$.