At $t=0$, a spherical soap bubble with surface tension T and radius R is formed at one end of a cylindrical pipe of length L and cross-section radius of ${ r }_{ o }$. The other end of the pipe is kept open to the atmosphere as shown.

The air that is inside the soap bubble has density $\rho$, and coefficient of viscosity $\eta$.

Find the time taken for the radius of the sphere to be **halved**. The time can be expressed as :

$\displaystyle{t\quad =\cfrac { a }{ b } (\cfrac { \eta L{ \rho }^{ c }{ R }^{ d } }{ T{ { (r }_{ o }) }^{ e } } )}.$

Compute the Value of $a+b+c+d+e$.

Here $a,b,c,d,e$ are non-negative integers.

**Assumptions**

$\bullet$ $\displaystyle{R\quad >>\quad { r }_{ o }}$.

$\bullet$ Outside The Soap Bubble air is present in atmosphere with constant atmospheric Pressure (1 atm)

$\bullet$ Surface Tension is Enough So that It always Maintains the spherical Shape of the Soap bubble!

$\bullet$ Length of Pipe is not So long (i.e Initial Volume Capacity of sphere is Sufficiently greater Than that of Pipe )

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