Question

Two capillary 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 to the flow through single capillary, $X=\frac{\pi P{R}^4}{8\eta L}$)

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

Answer 1

The correct option is A $\frac{8}{9}X$

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
$R_s=R_1+R_2=\frac{8\eta L}{\pi R^4}+\frac{8\eta \times 2L}{\pi (2R{)}^4}=\left(\frac{8\eta L}{\pi R^4}\right)\times \frac{9}{8}$
Rate
of flow = $\frac{P}{R_s}=\frac{\pi P{R}^4}{8\eta L}\times \frac{8}{9}=\frac{8}{9}X[asX=\frac{\pi P{R}^4}{8\eta L}]$