Question 13 Marks
A soap bubble of radius 4 cm and surface tension 30 dyne $cm ^{-1}$ is blown at the end of a tube of length 10 cm and internal radius 0.20 cm . If the viscosity of air is $1.89 \times 10^{-4}$ poise, find the time taken by the bubble to be reduced to a radius of 2 cm .
Answer
View full question & answer→Let R be the radius of the bubble at any instant. Its volume is
$
V=\frac{4}{3} \pi R^3
$
$\therefore$ Rate of flow of air
$
=\frac{d V}{d t}=\frac{4}{3} \pi \times 3 R^2 \frac{d R}{d t}=4 \pi R^2 \frac{d R}{d t}
$
But $\frac{d V}{d t}=\frac{\pi p r^4}{8 \eta l}$ and for a soap bubble, $p =\frac{4 \sigma}{R}$
$
\therefore \frac{d V}{d t}=\frac{\pi r^4}{8 \eta l} \cdot \frac{4 \sigma}{R}=4 \pi R^2 \frac{d R}{d t} \text { or } dt=\frac{8 \ln }{\sigma r^4} R^3 dR
$
Time taken by the bubble when its radius changes from $R_1$ to $R_2$ is
$
t=\int d t=\frac{8 \ln }{\sigma r^4} \int_{R_2}^{R_1} R^3 d R=\frac{8 \ln }{\sigma r^4}\left(\frac{R_1^4-R_2^4}{4}\right)
$
But $R _1=4 cm, R _2=2 cm, \sigma=30$ dyne $cm ^{-1}, l =10 cm, r =0.2 cm, \eta=1.85 \times 10^{-4}$ poise
$
\therefore t=\frac{8 \times 10 \times 1.85 \times 10^{-4}}{30 \times(0.2)^4} \times\left(\frac{4^4-2^2}{4}\right)=296 s
$
$
V=\frac{4}{3} \pi R^3
$
$\therefore$ Rate of flow of air
$
=\frac{d V}{d t}=\frac{4}{3} \pi \times 3 R^2 \frac{d R}{d t}=4 \pi R^2 \frac{d R}{d t}
$
But $\frac{d V}{d t}=\frac{\pi p r^4}{8 \eta l}$ and for a soap bubble, $p =\frac{4 \sigma}{R}$
$
\therefore \frac{d V}{d t}=\frac{\pi r^4}{8 \eta l} \cdot \frac{4 \sigma}{R}=4 \pi R^2 \frac{d R}{d t} \text { or } dt=\frac{8 \ln }{\sigma r^4} R^3 dR
$
Time taken by the bubble when its radius changes from $R_1$ to $R_2$ is
$
t=\int d t=\frac{8 \ln }{\sigma r^4} \int_{R_2}^{R_1} R^3 d R=\frac{8 \ln }{\sigma r^4}\left(\frac{R_1^4-R_2^4}{4}\right)
$
But $R _1=4 cm, R _2=2 cm, \sigma=30$ dyne $cm ^{-1}, l =10 cm, r =0.2 cm, \eta=1.85 \times 10^{-4}$ poise
$
\therefore t=\frac{8 \times 10 \times 1.85 \times 10^{-4}}{30 \times(0.2)^4} \times\left(\frac{4^4-2^2}{4}\right)=296 s
$
