Two capillary tubes of length L and 2L and of | vedclass

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Question

(A) The fluid resistance $R_f$ of a capillary tube is given by $R_f = \frac{8 \eta l}{\pi r^4}$.For the first capillary,$R_1 = \frac{8 \eta L}{\pi R^4

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$\frac{8}{9}X$

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(A) The fluid resistance $R_f$ of a capillary tube is given by $R_f = \frac{8 \eta l}{\pi r^4}$.For the first capillary,$R_1 = \frac{8 \eta L}{\pi R^4}$.For the second capillary,$R_2 = \frac{8 \eta (2L)}{\pi (2R)^4} = \frac{16 \eta L}{\pi (16 R^4)} = \frac{\eta L}{\pi R^4} = \frac{1}{8} R_1$.When connected in series,the equivalent resistance is $R_{eq} = R_1 + R_2 = R_1 + \frac{1}{8} R_1 = \frac{9}{8} R_1$.The rate of flow $Q$ is given by $Q = \frac{P}{R_{eq}}$.Since $X = \frac{P}{R_1}$,the new rate of flow $Q = \frac{P}{\frac{9}{8} R_1} = \frac{8}{9} \left( \frac{P}{R_1} \right) = \frac{8}{9} X$.

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

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