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Question

We have two (narrow) capillary tubes T1 and T2. their lengths are l1 and l2 and radii of cross section are r1 and r2 respectively. The rate of flow of water under a pressure difference P through tube T1 is 8cm3/sec. If l1=2l2 and r1=2r2. What will be the rate of flow when the two tuves are connected in series and pressure difference across the combinatin is same as before [=P]

A
4cm3/sec
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B
(16/3)cm3/sec
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C
(8/17)cm3/sec
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D
None of these
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Solution

The correct option is C (16/3)cm3/sec
The flow rate of water under a pressure difference is given by
V=πr4P8nl=8cm3sec
For composite tube
V=πr4P8n(3l/2)
=23×πr4P8nl
=2×83
=168
Since [l1=l=2l2]

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