Question

Two capillary of lengths 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.$X=\frac{\pi pR^4}{8\eta L})$

• A. $\frac{8}{9}X$
• B. $\frac{9}{8}X$
• C. $\frac{5}{7}X$
• D. $\frac{7}{5}X$

Answer 1

Answer: (A)

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
$R_{eq}=R_1 + R_2=\frac{8\eta L}{\pi R^4}+\frac{8\eta\times2L}{\pi(2R)^4}$
$=\left[\frac{8\eta L}{\pi R^4}\times\frac{9}{8}\right]$
Equivalent resistance becomes $\frac{9}{8}$ times so, rate of flow will be $\frac{8}{9}X$