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Two capillary tubes A and B of equal radii r_a = r_b= r and equal lengths l_a = l_b = l are held horizontally. When the same pressure difference P is maintained across each tube, the rate of flow of water in each is Q. If the tubes are connected in series and the same pressure difference P is maintained across the combination, the rate of flow through the combination will be

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$\frac{Q}{\sqrt{2}}$

$2 Q$

$Q$

$\frac{Q}{2}$

answer is A.

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Detailed Solution

$v o l u m e r a t e = Q = \frac{{πPr}^{4}}{8 ηl} & V \propto \frac{1}{l}$

A series combination of two identical capillary tubes, each of length l and radius r, is equivalent to a tube of length 2l and radius r.

Since Q is inversely proportional to l, if l is doubled, Q becomes half. Hence the correct choice is (a).

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