Soap Bubble gets smaller and smaller !

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.

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

This is Part of my Set Deepanshu's Mechanics Blast

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