Answer
64.8k+ views
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.
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.
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What is the common property of the oxides CONO and class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What happens when dilute hydrochloric acid is added class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
If four points A63B 35C4 2 and Dx3x are given in such class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
The area of square inscribed in a circle of diameter class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Other Pages
Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
In the ground state an element has 13 electrons in class 11 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Electric field due to uniformly charged sphere class 12 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
A boat takes 2 hours to go 8 km and come back to a class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
According to classical free electron theory A There class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)