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# Water is flowing through a cylindrical pipe of cross sectional area $0.09\pi \,{m^{ - 2}}$ at a speed of $1.0\,m{s^{ - 1}}$. If the diameter of the pipe is halved, then find the speed of flow of water through it.

Last updated date: 15th Aug 2024
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Hint: Use the continuity equation given below and substitute the condition of the cylindrical pipe, area and the velocity before and the after changing the diameter of the pipe. Simplifying the substitution provides the value of the velocity of the pipe.

Formula used:
The continuity equation is given by
${A_1}{v_1} = {A_2}{v_2}$
Where ${A_1}$ is the cross sectional area of the first cylinder considered, ${v_1}$ is the velocity of the water through pipe, ${A_2}$ is the cross sectional area of the pipe after changing the diameter of the pipe and ${v_2}$ is the velocity of the pipe after changing the diameter.

Complete step by step solution:
It is given that the
Cross sectional area of the cylindrical pipe, $A = 0.09\pi \,{m^2}$
The speed of the water which flows through the cylindrical pipe, $v = 1.0\,m{s^{ - 1}}$
Let us consider the continuity equation,
${A_1}{v_1} = {A_2}{v_2}$
By substituting the known values in the above equation,
The area of the second condition after the diameter gets halved, $A = \dfrac{A}{{{2^2}}}$
$\Rightarrow$ ${A_2} = \dfrac{{0.09}}{4}$
$\Rightarrow$ $0.09\pi \left( 1 \right) = \dfrac{1}{4} \times 0.09\pi \times {V_2}$
By simplifying the above equation, we get
$\Rightarrow$ ${v_2} = 4\,m{s^{ - 1}}$

Hence the velocity of the water flowing through the cylindrical pipe, after making the half of the diameter it changes to $4\,m{s^{ - 1}}$.

Note: The continuity equation holds for the conservation of the mass in the steady state one dimensional flow. The mass of the system before is equal to the mass of the system after the process. This equation is only possible for the flow of fluid.