Question

The work done in blowing a soap bubble of volume \( V \) is \( W \). The work done in blowing the bubble of volume \( 2V \) from the same soap solution is?

• A. \( \frac{W}{2} \)
• B. \( \sqrt{2} W \)
• C. \( (2)^{\frac{1}{3}} W \)
• D. \( (4)^{\frac{1}{3}} W \) \

Hint:

For problems involving bubbles and expansion, always relate work done to the surface area of the bubble and use the relationship between volume and radius to determine the proportionality.

Answer 1

The Correct Option is D

The work done in blowing a soap bubble is proportional to the surface area of the bubble. The volume \( V \) of a spherical bubble is related to its radius \( r \) as: \[ V = \frac{4}{3} \pi r^3 \] Step 1: Expressing work in terms of radius
Since the work done \( W \) is proportional to the surface area, and surface area \( A \) of a sphere is: \[ A = 4\pi r^2 \] We get: \[ W \propto r^2 \] Step 2: Finding new work for volume \( 2V \)
If the volume becomes \( 2V \), then: \[ 2V = \frac{4}{3} \pi r'^3 \] Dividing by the original volume equation: \[ \frac{r'^3}{r^3} = 2 \Rightarrow r' = (2)^{\frac{1}{3}} r \] Step 3: Compute the ratio of work done
Since \( W \propto r^2 \), we have: \[ W' = W \left( \frac{r'^2}{r^2} \right) = W \left( \frac{(2)^{\frac{2}{3}} r^2}{r^2} \right) = W (2)^{\frac{2}{3}} \] Step 4: Express in terms of \( 4^{\frac{1}{3}} \)
\[ (2)^{\frac{2}{3}} = (4)^{\frac{1}{3}} \] Thus, \[ W' = (4)^{\frac{1}{3}} W \] Hence, the work done in blowing a bubble of volume \( 2V \) is \( (4)^{\frac{1}{3}} W \). \bigskip