Two capillary tubes of same radius `r` but of lengths `l_(1)` and `l_(2)` are fitted in parallel to the bottom of a vessel. The pressure to the bottom of a vessel. The pressure head is P. What should be the length of a singl tube of same radius that can replace the two tubes so that the rate of flow is same as before?

Options

A. `l_(1)+l_(2)`
B. `(1)/(l_(1))+(1)/(l_(2))`
C. `(l_(1)l_(2))/(l_(1)+l_(2))`
D. `(1)/(l_(1)+l_(2))`

Answer

C

Explanation

Fluid resistance `R=(8etaL)/(pir^(4))`
in parallel `(1)/(R)=(1)/(R_(1))+(1)/(R_(2))`
or `(pir^(4))/(8etal_(eq))=(pir^(4))/(8etal_(1))+(pir^(4))/(8etal_(2))` or `l_(eq)=(l_(1)l_(2))/(l_(1)+l_(2))`