Solveeit Logo
Login

Question

Question: Water is flowing through a capillary tube at the rate of \(20 \times 10^{−6} m^3/s\). Another tube o...

Water is flowing through a capillary tube at the rate of 20×10−6m3/s20 \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 m3/s is
A. 10×10−610 \times {10^{ - 6}}
B. 3.33×10−63.33 \times {10^{ - 6}}
C. 6.67×10−66.67 \times {10^{ - 6}}
D. 20×10−620 \times {10^{ - 6}}

Explanation

Solution

From Hagen-Poiseuille equation, when two or more pipes are connected in series then the flow rate of water is inversely proportional to their length.

Complete step by step solution:
Given,
The flow rate in the first pipe is
Q1=20×10−6{Q_1} = 20 \times {10^{ - 6}} m3/s
From Hagen-Poiseuille equation,
We know that the flow rate
Flow rate, Q=πPr48ηlQ = \dfrac{{\pi P{r^4}}}{{8\eta l}} ​. . . . . . .(1)
where, l = length of capillary tube
P = pressure
r = radius of capillary tube
From equation (1),
We can say flow rate is inversely proportional to the length of pipe

Q∝1lQ \propto \dfrac{1}{l}
if we take ratio of flow rate of two different pipe we get
Q2Q1=l1l2\dfrac{{{Q_{2}}}}{{{Q_1}}} = \dfrac{{{l_1}}}{{{l_2}}}
Multiplying Q1Q_1 on right side we get
Q2=Q1l1l2{Q_{2}} = {Q_1}\dfrac{{{l_1}}}{{{l_2}}}
On putting the given values in the above equation

We get the flow rate of second pipe as
Q2=20×10−6×12{Q_2} = 20 \times {10^{ - 6}} \times \dfrac{1}{2} m3/s
Or,
Q2=10×10−6{Q_2} = 10 \times {10^{ - 6}} m3/s
The flow rate in the second pipe is 10×10−610 \times {10^{ - 6}} m3/s

Hence option A is the correct answer.

Additional information: The flow rate is the same when pipes are connected in series and flow rate gets changed when pipes are connected in parallel.
And the flow rate is directly proportional to the radius to the four of that pipe.

Note: If length of pipe doubled flow rate get decreased by 50%, and if length is half then the flow rate is increased by 50%.