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Two capillary tubes AB and BC are joined end to end at B. AB = 12 cm long and of diameter 4 mm. whereas BC is 3 cm long and of diameter 2mm. The composite tube is held horizontally with A connected to a vessel of water giving a constant head of 5 cm and C is open to the air. Find the pressure difference between B and C.
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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"`.

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