If two capillary tubes of radii `r_1 and r_2` and having length `l_1 and l_2` respectively are connected in series across a heaed of pressure p, find the rate of flow of the liqid through the tubes, if `eta` is the coefficient of viscosity of the liquid.

Options

A. `(pi P)/(8 eta) ((l_(1))/(r_(1)^(4)) + (l_(2))/(r_(2)^(4)))^(-1)`
B. `(8pi P)/(eta) ((l_(1))/(r_(1)^(4)) + (l_(2))/(r_(2)^(4)))`
C. `(pi P)/(8eta) ((r_(1)^(4))/(l_(1)) + (r_(2)^(4))/(l_(2)))^(-1)`
D. `(8pi P)/(eta) ((l_(1))/(r_(1)^(4)) + (l_(2))/(r_(2)^(4)))^(-1)`

Answer

A

Explanation

The rate of flow of liquid (V) through a capillary tube is
`V= (pi Pr^(4))/(8 eta l) = P ((pi r^(4))/(8 eta l)) = (P)/(R ) = ("Pressure difference")/("Resistance")`
where, `R= (8 eta l)/(pi r^(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))]) = (pi P)/(8 eta) [(l_(1))/(r_(1)^(4)) + (l_(2))/(r_(2)^(4))]^(-1)`