Question

Water is flowing through a capillary tube at the rate of $20\times {10}^{-6}{m}^3/s$. Another tube of same radius and double the length is connected in series to the first tube. Now the rate of flow of water in $m^3{s}^{-1}$ 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)

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

$Q_{flow}=\frac{\pi r^4\Delta P}{8\eta L}=20\times {10}^{-6}{m}^3/s$ ----------(1)
As now 2 pipes are connected in series, the q through both the pipes will be same and let the pressure be $P_1$ at the function,
$\frac{\pi r^4(P-P_1)}{8\eta L}=\frac{\pi r^4\left(P_1\right)}{8\eta 2L}$

$\Rightarrow 2P-2P_1=P_1\Rightarrow \frac{2P}{3}=P_1$-------------(2)
So, From when both pipe are connected in series

$=\frac{\pi r^4{P}_1}{8\eta 2L}=\frac{1}{3}\times 20\times {10}^{-6}{m}^3/s$........[from(1)]
$=6.67\times {10}^{-6}{m}^3/s$