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Q.

We have two (narrow) capillary tubes T1, and T2 . Their lengths are l1 and l2 and radii of cross-sections arc r1, and r2 respectively. The rate of flow of water under a pressure difference P through tube T1, is 8 cm3/s. If l1 =2 l2 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 (P) as before?

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a

4cm3/s

b

(16/3)cm3/s

c

(8/17)cm3/m

d

none of the above

answer is B.

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Detailed Solution

Given that, V=πPr48ηl=8cm3/ s where l1=l For composite tube V1=πPr48η(l+l/2) ∵l1=l=2l2  or  l2=l/2 ∴ V1=23×πPr48ηl =23×8cm3/s=163cm3/s
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