A water pipe 4cm in diameter has a constriction of diameter 2cm. Find the rate of discharge of water through the pipe if the velocity of flow of water in the main pipe is $6cm/s$ . Find also the velocity of flow at the constriction. ($75.36 \times {10^{ - 6}}{m^3}/s;24cm/s$ )

Hint: Calculate the cross-sectional area of the pipe. We are already given the velocity of water through the pipe. Multiply the values to get the rate of flow. Then, calculate the velocity of flow at the constriction from the equation of continuity.

Formula used: If the velocity of flow is $v$, and the area of cross-section of the tube is $a$ , then, the rate of flow of liquid through the tube is $va$.
If two sections of a tube have areas ${a_1}$ and ${a_2}$ having velocity ${v_1}$ and ${v_2}$ respectively then, from the equation of continuity we get, ${v_1}{a_1} = {v_2}{a_2}$ .

Complete step by step solution: The rate of flow of a liquid through a tube means the volume of liquid flowing through any cross-section of the tube per second.
So, if the velocity of flow is $v$ and the area of cross-section of the tube is $a$ , then, the rate of flow of liquid through the tube is $va$.
Now, let us focus on the point that is the equation of continuity here. For a streamline flow of any liquid through a tube, be it liquid or gas, the mass of the fluid flowing per second through any cross-section of the tube remains constant.
So, let us have two sections of a tube have areas ${a_1}$ and ${a_2}$ having velocity ${v_1}$ and ${v_2}$ respectively.
Then, from the equation of continuity, ${v_1}{a_1} = {v_2}{a_2}$.
It is given that the velocity of the flow of water in the main pipe is $6cm/s$.
So, ${V_P} = 6cm/s$
Also, the diameter of the pipe is $4cm$.
So, the radius of the pipe is $\dfrac{4}{2}cm = 2cm$
So, the cross-sectional area of the pipe is, $\pi {\left( 2 \right)^2} = 3.14 \times {(2)^2} = 12.56c{m^2}$
So, the rate of discharge of water through the pipe is $\left( {12.56 \times 6} \right) = 75.36c{m^3}/s$
Now, as we know that,
$1cm = {10^{ - 2}}m$
$ \Rightarrow 1c{m^3} = {10^{ - 6}}{m^3}$
Now we will convert the value of rate of discharge of water through pipe in meter,
$\therefore 75.36c{m^3} = 75.36 \times {10^{ - 6}}{m^3}$
$\therefore $ The rate of discharge of water through the pipe is $ = 75.36 \times {10^{ - 6}}{m^3}/s$
Now, let say that the velocity of flow of water in the main pipe be ${V_P}$ and the cross-sectional area of the pipe be ${A_P}$
Also, let, the velocity of flow at the constriction be ${V_C}$ and cross-sectional area of the constriction is ${A_C}$ .
Now, it is given that ${V_P} = 6cm/s$ and ${A_P} = \pi {\left( {\dfrac{4}{2}} \right)^2} = \pi {\left( 2 \right)^2}c{m^2}$
And, ${A_C} = \pi {\left( {\dfrac{2}{2}} \right)^2} = \pi c{m^2}$
So, from the equation of continuity,
${A_P}{V_P} = {A_C}{V_C}$
or, ${V_C} = \dfrac{{{A_P}{V_P}}}{{{A_C}}}$
Putting all the values in the equation, we get,
${V_C} = \dfrac{{\pi {{\left( 2 \right)}^2} \times 6}}{\pi } = \left( {4 \times 6} \right) = 24cm/s$

So, the velocity of flow at the constriction is $24cm/s$ .

Note: From the equation of continuity, we know that ${v_1}{a_1} = {v_2}{a_2}$ . So, we can say that $va = $ constant.
So, $v \propto \dfrac{1}{a}$ , which means that the velocity of the flow of any liquid through any cross-section of a tube is inversely proportional to its cross-sectional area. Also, the equation of continuity expresses the law of conservation of mass.