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Two capillary tubes of the same length l and radii r and 2r are fitted to the bottom of a vessel with pressure head p in parallel with each other. What should be the radius of the single tube of the same length l that can replace the two so that the rate of flow is not affected?

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$\sqrt{17} r$

8.5 r

${(17)}^{1 / 4} r$

17 r

answer is A.

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Detailed Solution

Radius R of the single tube is given by

$\frac{1}{R_{E F F}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$

$\frac{{πρR}^{4}}{8 η l} = \frac{πρ r^{4}}{8 η l} + \frac{πρ {(2 r)}^{4}}{8 η l}$

or $R^{4} = r^{4} + 16 r^{4} o r R = {(17)}^{1 / 4} r$ which is choice (1).

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