Question

Water is flowing through a capillary tube at the rate of $20\times {10}^{-6}{m}^3/s.$. Another tube of the same radius and double the length is connected in series to the first tube. Now the rate of flow of water in $m^3/s$ is

• A. $10\times {10}^{-6}$
• B. $3.33\times {10}^{-6}$
• C. $6.67\times {10}^{-6}$
• D. $20\times {10}^{-6}$

Answer 1

Answer: (C)

$6.67\times {10}^{-6}$

Given,

$Q_1=20\times {10}^{-6}{m}^3/s$

From Hagen-Poiseuille equation,

Flow rate, $Q=\frac{\pi P{r}^4}{8\eta l}$. . . . . . .(1)

where, $l=$ length of capillary tube

$P=$ pressure

$r=$ radius of capillary tube

From equation (1),

$Q\propto \frac{1}{l}$

$\frac{Q_2}{Q_1}=\frac{l_1}{l_2}$

$Q_2=Q_1\frac{l}{3l}$

$Q_2=\frac{Q_1}{3}=\frac{20\times {10}^{-6}}{3}=6.67\times {10}^{-6}{m}^3/s$

The correct option is C.