Fluid Mechanics Viscosity Question 325
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 PR^{4}}{8\eta L} $ )
Options:
A) $ \frac{8}{9}X $
B) $ \frac{9}{8}X $
C) $ \frac{5}{7}X $
D) $ \frac{7}{5}X $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] Fluid resistance is given by $ R=\frac{8\eta L}{\pi r^{4}} $ When two capillary tubes of same size are joined in series, then equivalent fluid resistance is $ R_{S}=\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 PR^{4}}{8\eta L}\times \frac{8}{9}=\frac{8}{9}X\left[ as,X=\frac{\pi PR^{4}}{8\eta L} \right] $
BETA
