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The rate of steady volume flow of water ...

The rate of steady volume flow of water through a capillary tube of length ' l ' and radius ' r ' under a pressure difference of P is V . This tube is connected with another tube of the same length but half the radius in series. Then the rate of steady volume flow through them is (The pressure difference across the combination is P )

Text Solution

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The rate of flow of liquid =`P/R`
where R is fluid resistance and R =`(8 eta l)/(pir^(4))`
For. second tube
R. = `(8etal)/(pi(r//2)^(4))` = 16R
Now for series combination
`R_(eq) = R + R. = 17R`
New value of rate of flow = `p/R_(eq) = P/(17R) = V/17`
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