Soap Bubble gets smaller and smaller !

Let at time t=t let radius of bubble is 'r'

Volume Flow out through tube is : (Use Poiseuille Law for Volume Rate flow through pipes)

$\displaystyle{\cfrac { dV }{ dt } =\cfrac { \Delta P(\pi { { r }_{ o } }^{ 4 }) }{ 8\eta L } \quad (1)}$ .

Also Volume that flow out through Spherical soap bubble decrees , and equal to volume rate flow through Pipe (Using continuity theorem)

$\displaystyle{\because \Delta P=\cfrac { 4T }{ r } \quad (2)\\ \cfrac { dV }{ dt } =4\pi r^{ 2 }(\cfrac { -dr }{ dt } )\quad (3)\\ }$ .

Using Continuity Theorem : $\displaystyle{-4\pi r^{ 2 }\cfrac { dr }{ dt } =\cfrac { 4T(\pi { { r }_{ o } }^{ 4 }) }{ 8\eta Lr } \\ \\ -\int _{ R }^{ R/2 }{ r^{ 3 }dr } =\cfrac { T(\pi { { r }_{ o } }^{ 4 }) }{ 8\eta L } \int _{ 0 }^{ T }{ dt } }$ .

we get : $\boxed { { T\quad =\cfrac { 15 }{ 8 } (\cfrac { \eta L{ R }^{ 4 } }{ T{ { (r }_{ o }) }^{ 4 } } ) } }$ .

This is an interesting problem. I will give some hints:

1. Use Poiseuille law on the cylindrical pipe.

2. Try to find the excess pressure on the sphere in terms of surface tension and radius of sphere.

3. Express the volumetric flow rate in the Poiseuille law in terms of the change in radius with respect to time.

4.Substitute the value for pressure loss , integrate the equation and you have the answer.

The important point is to notice that the rate does not depend on the density whatsoever and then apply Poiseuille's volume flux equation with a little calculus.