Question

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

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

Answer 1

Answer: (A)

3.1 mJ

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
$=2\times 4\pi r_1^2=2\times 4\pi \times (2\times {10}^{-2}{)}^2$
$=32\pi \times {10}^{-4}{m}^2$

Final surface area of the bubble
$=2\times 4\pi r_2^2=2\times 4\pi \times (5\times {10}^{-2}{)}^2$
$=200\pi \times {10}^{-4}{m}^2$

Increase is surface area = $200\pi \times {10}^{-4}-32\pi \times {10}^{-4}$
$=168\pi \times {10}^{-4}{m}^2$

Work done = surface tension $\times$ increase is surface area

=$0.06\times 168\pi \times {10}^{-4}$
=$3.12\times {10}^{-3}J=3.12mJ$