Two capillaries of radii `r_(1)andr_(2)` and lengths `l_(1)andl_(2)` respectively are in series. A liquid of viscosity `eta` is flowing through the combination under a pressure difference P. What is the rate of volume flow of liquid?

Options

A. `(8piP)/(eta)((l_1)/(r_(1)^(4))+(l_2)/(r_(2)^(4)))`
B. `(piP)/(8eta)((r_(1)^(4))/(l_1)+(r_(2)^(4))/(l_2))^(-1)`
C. `(8piP)/(eta)((l_1)/(r_(1)^(4))+(l_2)/(r_(2)^(4)))^(-1)`
D. `(piP)/(8eta)((l_1)/(r_(1)^(4))+(l_2)/(r_(2)^(4)))^(-1)`

Answer

D

Explanation

The rate of flow of liquid (v) through a capillary tube is,
`v=(piPr^4)/(8etal)=P((pir^4)/(8etal))`
`=(P)/(R)=("pressure difference")/("resistance")`
Where, `R=(8etal)/(pir^4)`
When two tubes are in series,
Total resistance `=R_(1)+R_(2)`
Rate of flow of liquid,
`v'=(P)/(R_(1)+R_(2))=(P)/((8eta)/(pi)[(l_1)/(r_(1)^(4))+(l_2)/(r_(2)^(4))])`
`=(piP)/(8eta)[(l_1)/(r_(1)^(4))+(l_2)/(r_(2)^(4))]^(-1)`