Three capillary tubes of the same radius r but of lengths `l_(1),l_(2)andl_(3)` are fitted horizontally to the bottom of a tall vessel containing a liquid at constant head and flowing through these tubes. Calculate the length of a single outflow tube of the same radius r which can replace the three capillaries.

Explanation

Let P be the liquid pressure at the bottom of the vessel. Then the rates of flow through the three tubes will be
`Q_(1)=(piPr^4)/(8etal_1),Q_(2)=(piPr^4)/(8etal_2),Q_(3)=(piPr^4)/(8etal_3)`
Let l be the length of the single tube of radius r which can replace the three tubes. The rate of flow through it will be
`Q=(piPr^4)/(8etal)`
but `Q=Q_(1)+Q_(2)+Q_(3)`
`therefore" "(piPr^4)/(8etal)=(piPr^4)/(8etal_1)+(piPr^4)/(8etal_2)+(piPr^4)/(8etal_3)`
(or) `(1)/(l)=(1)/(l_1)+(1)/(l_2)+(1)/(l_3)=(l_(2)l_(3)+l_(1)l_(3)+l_(1)l_(2))/(l_(1)l_(2)l_(3))`
(or) `l=(l_(1)l_(2)l_(3))/(l_(2)l_(3)+l_(1)l_(3)+l_(1)l_(2))`