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

Question

Dr. Abubacker · Accepted Answer

(a) $Fluid\ resistance\ is\ given\ by\ \ \ \ \ R = \frac{8ɳL}{{\pi r}^{4}}$
When two capillary tubes of same size are joined in parallel, then equivalent fluid resistance is
$R_{eq} = R_{1} + R_{2} = \frac{8ɳL}{{\pi R}^{4}} + \frac{8ɳ 	imes 2L}{\pi(2R)^{4}}$
Equivalent resistance becomes$\ \ \ \frac{9}{8}$ times so,$\ $
$rate\ of\ flow\ will\ be\frac{8}{9}X.$

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

Explanation

Reviewed by Dr. Abubacker

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