Question

The amount of work done in blowing a soap bubble such that its diameter increases from $d_1$ to $d_2$ is (T = surface tension of soap solution)

• A. $4\pi (d_2^2 - d_1^2) T$
• B. $8\pi (d_2^2 - d_1^2) T$
• C. $\pi (d_2^2 - d_1^2) T$
• D. $2\pi (d_2^2 - d_1^2) T$

Hint:

Always multiply by 2 for "bubbles" and 1 for "drops" due to the number of free surfaces.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
Work done in increasing the surface area of a liquid film is the product of surface tension and the increase in surface area. A soap bubble has two free surfaces (inner and outer).
Step 2: Key Formula or Approach:
\[ W = T \times \Delta A_{total} \]
For a soap bubble: $\Delta A_{total} = 2 \times (4\pi R_2^2 - 4\pi R_1^2)$
Step 3: Detailed Explanation:
Given diameters $d_1$ and $d_2$, the radii are $R_1 = d_1/2$ and $R_2 = d_2/2$.
The total surface area of a soap bubble is $2 \times (4\pi R^2)$.
Initial Area $A_1 = 2 \times 4\pi (d_1/2)^2 = 2\pi d_1^2$
Final Area $A_2 = 2 \times 4\pi (d_2/2)^2 = 2\pi d_2^2$
Increase in area $\Delta A = A_2 - A_1 = 2\pi (d_2^2 - d_1^2)$
Work done $W = T \cdot \Delta A = T \cdot 2\pi (d_2^2 - d_1^2)$
Step 4: Final Answer:
The work done is $2\pi (d_2^2 - d_1^2) T$.