Question

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

• A. $60\pi r^{2}S$
• B. $42\pi r^{2}S$
• C. $32\pi r^{2}S$
• D. $\pi r^{2}S$

Answer 1

The correct option is C $32\pi r^{2}S$

Initial area, $A_1=4\pi r^2$
Final area, $A_2=4\pi [3r{]}^2=36\pi r^2$

Increase in surface area = $A_2-A_1=36\pi r^2-4\pi r^2=32\pi r^2$

No. of surface of soap bubble = 2

Therefore total increase in surface area = $2\times 32\pi r^2=64\pi r^2$

Work done = change in surface energy
$\Rightarrow$ Work done $=W=S\left[△A\right]=S\times 64\pi r^2=64\pi r^{2}S$.