Q. Two capillaries of radii $r_{1}$ and$r_{2}$, length $l_{1}$ and $l_{2}$ respectively are in series. A liquid of viscosity $\eta$ is flowing through the combination under a pressure difference $p$. What is the rate of volume flow of liquid?
A) $\frac{\pi\ p}{8\ \eta}\ \left( \frac{l_{4}}{r_{1}^{4}} + \frac{l_{4}}{r_{2}^{4}} \right)^{- 1}$
B) $\frac{8\pi\ p}{\eta}\ \left( \frac{l_{1}}{r_{1}^{4}} + \frac{l_{2}}{r_{2}^{4}} \right)$
C) $\frac{\pi\ p}{8\ \eta}\ \left( \frac{r_{1}^{4}}{l_{1}} + \frac{r_{2}^{4}}{l_{2}} \right)^{- 1}$
D) $\frac{\pi\ p}{8\ \eta}\ \left( \frac{l_{1}}{r_{1}^{4}} + \frac{l_{2}}{r_{2}^{4}} \right)^{- 1}$
Explanation
(d) The rate of flow of liquid $(V)$ through capillary tube is
$V = \frac{\pi\ p\ r^{4}}{8\eta l} = p\left( \frac{\pi\ r^{4}}{8\eta l} \right) = \frac{p}{R} = \frac{pressure\ difference}{resistance}$
Where, $R = \frac{8\eta hl}{\pi\ r^{4}}$
When two tubes are in series
Total resistance $R = R_{1} + R_{2}$
Rate of flow of liquid, $V^{'} = \frac{p}{R_{1} + R_{2}}$
$= \frac{p}{\frac{8\eta}{\pi}\ \left\lbrack \frac{l_{1}}{r_{1}^{4}} + \frac{l_{2}}{r_{2}^{4}} \right\rbrack} = \frac{\pi p}{8\eta}\ \left\lbrack \frac{l_{1}}{r_{1}^{4}} + \frac{l_{2}}{r_{2}^{4}} \right\rbrack^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.