Question

Work done in increasing the size of a soap bubble from a radius of $ 3 \,cm $ to $ 5 \,cm $ is nearly (surface tension of soap solution $ = 0.03\, Nm^{-1} $ ).

• A. $ 4\pi \,mJ $
• B. $ 0.4 \pi \,mJ $
• C. $ 0.2 \pi \,mJ $
• D. $ 2 \pi \,mJ $

Answer 1

Answer: (B)

Given, $R_1 =3\, cm = 3 \times 10^{-2}\, m$
$R_2 = 5 \,cm = 5 \times 10^{-2}\, m$ and $T = 0.03\, N/m$
Original surface area $= 2 \times 4 \,\pi R_1$
For second bubble $= 2 \times 4 \,\pi R_2$
Work done $=$ Surface tension $\times$ extension in area
$= T \times \Delta A$
$ = 0.03 \times 2[4\,\pi R_2^2 - 4 \,\pi R_1^2]$
$= 0.03 \times 8 \pi ((5)^2 - (3)^2] \times 10^{-4}$
$= 0.03 \times 8 \pi \times 16 \times 10^{-4}$
$= 0.384 \,\pi \times 10^{-3}\, J$
$ \simeq 0.4\,\pi \,mJ$