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Q.

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 =r2, what will be the rate of flow when the two tubes are connected in series and pressure difference across the combination is same as before (= P)

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a

4 cm3/sec

b

(16/3) cm3/sec

c

(8/17) cm3/sec

d

None of these

answer is B.

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Detailed Solution

V=πPr48ηl=8cm3secFor composite tube V1=Pπr48ηl+l2=23πPr48ηl=23×8=163cm3sec∵l1=l=2l2 or l2=l2
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