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Q.

Water is flowing through a capillary tube at the rate of 20 x 10-6 m3 /s. Another tube of same radius and double the length is connected in series to the first tube. Now the rate of flow of water in m3s-1 is

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a

10×10-6

b

3.33×10-6

c

20×10-6

d

6.67×10-6

answer is C.

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Detailed Solution

Q=πPr48ηl; Q∞1l

after joining length of the tube becomes 3l so volume rate of flow decreases by three times it will become 20/3=6.67×10-6 m-3s-1

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