Two capillary tubes AB and BC are joined end to end at B, AB is 16cm long and of diameter 4mm whereas BC is 4cm long and of diameter 2mm. The composite tube is held horizontally with A connected to a vessel of water giving a constant head of 3cm and C is open to the air. Calculate the pressure difference between B and C.

Explanation

Let `P_1, P_2, P_3` are pressures across points X, Y, Z
As Z is open to air-
`P_3=P_0` (atmospheric pressure)
`P_(1)=P_(0)+pg(h)`
`P_(2)=P," let l"_(1)=15cm ,l_(2)=5cm`
Rate of flow of water through XY `Q_(1)=(pi r_(1)^(4)))/(8 eta) ((P_(1)-P)/(l_(1))`
`=(pi (2r)^(4))/(8 eta) . ((P_(0)+pgh-P))/(l_(1))`
`Q_(2)=(pir^(4))/(8 eta) (P-P_(0))/(l_(2))`
As `Q_(1)=Q_(2)`
`rArr 16 ((P_(0)+pgh-P))/(15)=(P-P_(0))/(5)`
`16pgh -16 (P-P_(0))=3(P-P_(0))`
`19(P-P_(0))=16pgh`
`19(P-P_(0))=16pgh`
`P-P_(0)=(16h)/(19) pg .........(i)`
Put h=2cm in (i)
`P-P_(0)=(32)/(19) pg`
Thus the pressure difference between Y and Z is `(32)/(19) cm` of water column.