Q. A soap film is made by dipping a circular frame of radius $b$ in soap solution. A bubble is formed by blowing air with speed $v$ in the form of cylinder. The radius of the bubble formed $R \gg b$ so that the air is incident normally on the surface of bubble. Air stops after striking surface of soap bubble. Density of air is $\rho$. The radius $R$ of the bubble when the soap bubble separates from the ring is (surface tension of liquid is $S$)
A) $\frac{S}{\rho{\ v}^{2}}$
B) $\frac{4\ S}{\rho{\ v}^{2}}$
C) $\frac{S\ b}{\rho\ v}$
D) $\frac{4\ Sb}{\rho{\ v}^{2}}$
Explanation
(b) In figure total force on the ring due to surface tension of soap film $= (2\pi b) \times 2S\sin\theta$
Mass of air entering per second the bubble
$=$ volume $\times$ density $= (A\ v)\rho = \pi\ b^{2} \times v\rho$
Momentum of air energy per sec
$= \pi b^{2}v\rho \times v = \pi^{2}b^{2}v^{2}\rho$
The soap bubble will separate from the ring, when force of surface tension of ring is equal to the force
Or $2\pi b \times 2S \times \frac{b}{R} = \pi b^{2}v^{2}\rho$ or $R = \frac{4\ S}{\rho{\ v}^{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.