Question

The work done in blowing a soap bubble of surface tension $0.06N{m}^{-1}$ from $2cm$ radius to $5cm$ radius is

• A. $3.1mJ$
• B. $1.25mJ$
• C. $2.51mJ$
• D. $4.55mJ$

Answer 1

Answer: (A)

Here, $S=0.06N{m}^{-1},r_1=2cm=2\times 10^{-2}m$ ,
$r_2=5cm=5\times 10^{-2}m$
Since bubble has two surfaces, initial surface area of the bubble
$=2\times 4\pi r_1^2=2\times 4\pi \times {\left(2\times 10^{-2}\right)}^2$
$=32\pi \times 10^{-4}{m}^2$
Final surface area of the bubble
$=2\times 4\pi r_2^2=2\times 4\pi {\left(5\times 10^{-2}\right)}^2$
$=200\pi \times 10^{-4}{m}^2$
Increase in surface area $=200\pi \times 10^{-4}-32\pi \times 10^{-4}$
$=168\pi \times 10^{-4}{m}^2$
$\therefore$ Work done = surface tension $\times$ increase in surface area
$=0.06\times 168\pi \times 10^{-4}$
$=3.12\times 10^{-3}J=3.12mJ$