Question

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 a cylindrical column. The radius of the bubble formed $R>>b$. If the air is incident normally on the surface of the bubble and air stops after striking the surface, then the radius $R$ of the bubble when the soap bubble separates from the ring is:
(Surface tension of liquid is $S$ and density of air is $\rho$)

• A. $\frac{S}{\rho v^2}$
• B. $\frac{4S}{\rho v^2}$
• C. $\frac{Sb}{\rho v}$
• D. $\frac{4Sb}{\rho v^2}$

Answer 1

The correct option is B $\frac{4S}{\rho v^2}$

Force exerted by flowing fluid,
$F=\frac{\text{momentum}\left(P\right)}{\text{time}\left(t\right)}=\frac{\text{mass}\times \text{velocity}}{\text{time}\left(t\right)}=\frac{\text{density}\times \text{volume}\times \text{velocity}}{\text{time}\left(t\right)}=\frac{\rho \times Ah\times v}{t}=\rho A{v}^2$


Bubble will separate cylindrical frame when,
Force due to air molecules = Force due to surface tension $\left(S\right)$
$\Rightarrow \rho A{v}^2=\left(2\pi b\right)2S\sin \theta$
$\Rightarrow \rho A{v}^2=4\pi bS\frac{b}{R}$ $(\because sin\theta =\frac{b}{R})$
$\Rightarrow \rho \pi b^2{v}^2=4\pi S\frac{b^2}{R}$ $(\because A=\pi b^2)$
$\Rightarrow R=\frac{4S}{\rho v^2}$
Hence, option $\left(b\right)$ is the correct answer.