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 h cm be the head of pressure at B over C. Then pressure difference between A and B,
`p_(1) = (5-h)" cm of "H_(2)O`
For tube AB : `L_(1) = 12 cm`,
`r_(1) = 2mm -=0.2 cm`,
`p_(1) = (5-h)" cm of water"`
`= (5-h) xx 1xx 980 dyn e//cm^(2)`
Rate of flow of water through AB is
`Q_(1) =(pi p_(1)r_(1)^(4))/(8 eta l_(1)) = (pi(5-h) xx 1 xx 980 xx (0.2)^(4))/(8 eta xx 12) cm^(3) s^(-1)`
For tube BC: `l_(2) = 3 cm` ,
`r_(2) 1 mm = 0.1 cm`,
`p_(2)" h cm of water "= h xx 1 xx 980 dyn e//cm^(2)`
Rate of flow of water through BC is
`Q_(2) =(pi p_(2)r_(2)^(4))/(8 eta l_(2)) = (pi xx h xx 1 xx 980 xx (0.1)^(4))/(8 eta xx 3) cm^(3) s^(-1)`
As flow through the two tubes in series is steady, so `Q_(1) = Q_(2)`
or `(pi(5-h) xx 1 xx 980 xx (0.2)^(4))/(8 eta xx 12) = (pi h xx 1 xx 980 xx (0.1)^(4))/(8 eta xx 3)`
`3(5-h) xx 16 = 12 h`
On solving, `h = 4" cm of water column"`.