Three horizontal capillary tubes of same radii and lengths $L_{1}, L_{2} a n d L_{3}$ are fitted side by side a little above the bottom, to the wall of a tank that is filled with water. The length of a single capillary tube of same radius that can replace the three tubes such that the rate of flow of water through the single tube equals the combined rate of flow through the three tubes is :

\frac{L_{1} L_{2} L_{3}}{L_{1} + L_{2} + L_{3}}

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\frac{L_{1} L_{2} L_{3}}{L_{1} L_{2} + L_{2} L_{3} + L_{3} L_{1}}

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\frac{L_{1} + L_{2} + L_{3}}{L_{1} L_{2} L_{3}}

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\frac{L_{1} L_{2} + L_{2} L_{3} + L_{3} L_{1}}{L_{1} L_{2} L_{3}}

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Solution

The correct option is B $\frac{L_{1} L_{2} L_{3}}{L_{1} L_{2} + L_{2} L_{3} + L_{3} L_{1}}$
$(Q_{1})$ Rate of flow through 1 $= \frac{π r^{4} p}{8 η L_{1}}$
$(Q_{2})$ Rate of flow through 2 $= \frac{π r^{4} p}{8 η L_{2}}$
$(Q_{3})$ Rate of flow through 3 $= \frac{π r^{4} p}{8 η L_{3}}$
So,
$Q (c o m b i n e d) = \frac{π r^{4} p}{8 η L} = Q_{1} + Q_{2} + Q_{3}$
$= \frac{π r^{4} p}{8 η L} = \frac{π r^{4} p}{8 η L} (\frac{1}{L_{1}} + \frac{1}{L_{2}} + \frac{1}{L_{3}})$
$\frac{1}{L} = \frac{L_{2} L_{3} + L_{1} L_{2} + L_{1} L_{3}}{L_{1} L_{2} L_{3}}$
$L = \frac{L_{1} L_{2} L_{3}}{L_{1} L_{3} + L_{1} L_{2} + L_{2} L_{3}}$

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