Two capillaries of length $L\ and\ 2L$ and of radii $R\ and\ 2R$ respectively are connected in series. The net rate of flow of fluid through them will be (Given, rate of the flow through single capillary, $X = \pi pR^{4}$/8ɳL )

Question

Dr. Abubacker · Accepted Answer

(a) If $l$ and $r$ be the length and radius of the tube and $p\ $the pressure difference, then from Poiseuille’s formula, volume of liquid flowing per second is given by
$Q = \frac{\pi\ pr^{4}}{8ɳl}$
$	herefore Fluid\ resistance = \frac{8ɳl}{\pi\ r^{4}}$
When capillaries are connected in series then equivalent fluid resistance is
$R_{eq} = R_{1} + R_{2}$
$= \frac{8ɳl}{\pi\ pr^{4}} + \frac{8ɳ(2L)}{\pi\ (2R)^{4}} = \frac{9}{8}\left( \frac{8ɳl}{\pi\ pr^{4}} ight)$
$	herefore\ \ \ \ \ Net\ rate\ of\ flow = \frac{p}{R_{eq}} = \frac{p}{\frac{9}{8}\left( \frac{8ɳl}{\pi\ pr^{4}} ight)} = \frac{8}{9}$

Q. Two capillaries of length $L\ and\ 2L$ and of radii $R\ and\ 2R$ respectively are connected in series. The net rate of flow of fluid through them will be (Given, rate of the flow through single capillary, $X = \pi pR^{4}$/8ɳL )

Explanation

Reviewed by Dr. Abubacker

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