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Two capillary of lengths  l  and 2l  and of radius R and 2R are connected in series. The net rate of flow of fluid through them will the  x = πPR4/8ηl

 

Option: 1

\frac{8}{9}X


Option: 2

\frac{9}{8}X


Option: 3

\frac{5}{7}X


Option: 4

\frac{7}{5}X


Answers (1)

best_answer

fluid resistance is given by   R=\frac{8 \eta l}{\pi r^{4}}

when two capillary tubes of same size are joined in parallel then equivalent fluid resistance is

\begin{aligned} R_e & =R_1+R_2 \\ & =\frac{8 \eta l}{\pi r^4}+\frac{8 \eta \times 2 l}{\pi(2 R)^4}=\left(\frac{8 \eta l}{\pi r^4}\right) \times \frac{9}{8} \end{aligned}

Equivalent resistance become 9/8 times, so rate of flow will be (8/9)x

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Anam Khan

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