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We have two (narrow) capillary tubes T a...

We have two (narrow) capillary tubes T and T . Their lengths are l and l and radii of cross-section are r and r respectively. The rate of flow of water under a pressure difference P through tube T is `8 cm 3// sec`. If `l = 2l` and `r = r` what will be the rate of flow when the two tubes are connected in series and pressure difference across the combinatin is same as before `(= P)`

A

`4cm^(3)//sec`

B

`(16//3)cm^(3)//sec`

C

`(8//17)cm^(3)//sec`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
B

`V=(piPr^(4))/(8etal)=(8cm^(3))/(sec)`
For composite tube
`V_(1)=(Ppir^(4))/(8eta(l+(l)/(2)))=(2)/(3)(piPr^(4))/(8etal)=(2)/(3)xx8=(16)/(3)(cm^(3))/(sec)" "[thereforel_(1)=l=2l_(2)" or "l _(2)=(1)/(2)]`
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