Question

Air is pushed into a soap bubble of radius 'r' to double its radius. If the surface tension of the soap solution is S, the work done in the process is?

• A. 8 π r² S
• B. 12 π r² S
• C. 16 π r² S
• D. 24 π r² S

Answer 1

The correct option is D

24 π r² S

Initial surface area of bubble =$4\pi r^2$
Surface energy=$2\times S\times 4\pi r^2=8\pi r^{2}S$
{As there are two surfaces one inside and one outside and due to negligible thickness of soap, both can be assumed to have same radius}
Now final surface area=$4\pi (2r{)}^2$
=$16\pi r^2$
Final surface energy = 2S$\left(16\pi r^2\right)$ {same reason}
=$32\pi r^{2}S$
$V_f-V_i=32\pi r^{2}S-8\pi r^{2}S$
=$24\pi r^{2}S$
So work done in the process = $24\pi r^{2}S$