Water flows through a capillary tube of radius r and length l at a rate of 40 mL per second, when connected to a pressure difference of h cm of water. Another tube of the same length but radius `r//2` is conncected in series with this tube and the combination is connected to the same pressure head. Calculate the pressure difference across each tube and the rate of flow of water through the combination.

Explanation

Volume of the liquid flowing per second through the tube of length l, radius r, under a pressure difference `P=(h rho g)` at the two ends of tube is
`V = (pi P r^(4))/(8 eta l) = (pi rho gh r^(4))/(8etal) = 40` …(i)
When tubes are connected in series the volume of liquid flowing per second `(V')` , through each tube is same, i.e.,
`V'= (pi P_(1)r^(4))/(8eta l) =(pi P_(2)(r//2)^(4))/(8 eta l)` ...(ii)
where, `P_(1)+P_(2) = P = h rho g` ...(iii)
Putting this value in (iii) we have
`P_(1)+16 P_(1) = rho gh" or "P_(1)=(rho gh)/(17)`
and `P_(2) = 16/17 rho gh`
From (ii), `V'=(pi(rho gh//17)r^(4))/(8 eta l)`
`1/17 xx (pi rho gh r^(4))/(8 eta l) = 1/17 xx 40`
`=2.353 mL//s`.