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Two capillary tubes of the same length l and radii r and 2 r 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?

Option: 1

\mathrm{(17)^{1 / 4} r}


Option: 2

\mathrm{17 r}


Option: 3

\mathrm{8.5 r}


Option: 4

\mathrm{\sqrt{17} r}


Answers (1)

best_answer

Radius R of the single tube is given by

                     \mathrm{ \frac{\pi p R^4}{8 \eta l}=\frac{\pi p r^4}{8 \eta l}+\frac{\pi p(2 r)^4}{8 \eta l} }

or        \mathrm{R^4=r^4+16 r^4\, \, or\, \, R=(17)^{1 / 4} r}

                which is choice (a).

Posted by

Ritika Harsh

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