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Question

# Water is flowing through a capillary tube at the rate of 20×10−6m3/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:

A
10×106
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B
3.33×106
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C
6.67×106
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D
20×106
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Solution

## The correct option is B 6.67×10−6Qflow=πr4ΔP8ηL=20×10−6m3/s ----------(1)As now 2 pipes are connected in series, the q through both the pipes will be same and let the pressure be P1 at the function,πr4(P−P1)8ηL=πr4(P1)8η2L⇒2P−2P1=P1⇒2P3=P1-------------(2)So, From when both pipe are connected in series=πr4P18η2L=13×20×10−6m3/s........[from(1)]=6.67×10−6m3/s

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