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 Question:
We must calculate the physical work required to expand a soap bubble from an initial diameter $d$ to a final diameter $D$.

Step 2: Key Formula or Approach:
The work done in expanding a liquid surface is given by the product of Surface Tension ($T$) and the change in surface area ($\Delta A$):
$$W = T \times \Delta A$$
Crucially, a soap bubble in air has two free liquid surfaces (an inner wall and an outer wall). Therefore, the effective surface area is twice the geometric spherical area.
$\text{Effective Area} = 2 \times 4\pi r^2 = 8\pi r^2$.
We must also remember to convert radii ($r$) to diameters ($d$): $r = d/2$.

Step 3: Detailed Explanation:
Let's express the effective surface area entirely in terms of diameter:
$$\text{Area} = 8\pi \left(\frac{d}{2}\right)^2 = 8\pi \left(\frac{d^2}{4}\right) = 2\pi d^2$$
Initial effective surface area: $A_1 = 2\pi d^2$
Final effective surface area: $A_2 = 2\pi D^2$
Calculate the change in area ($\Delta A$):
$$\Delta A = A_2 - A_1 = 2\pi D^2 - 2\pi d^2 = 2\pi(D^2 - d^2)$$
Calculate the work done:
$$W = T \times \Delta A = T \times 2\pi(D^2 - d^2)$$
$$W = 2\pi (D^2 - d^2) T$$

Step 4: Final Answer:
The work done matches option (b).