Two water pipes of diameters 2cm and 4cm are connected with the main supply line. the velocity of flow of water in the pipe of $2cm$ diameter is:
A) $4$ times that in the other pipe
B) $\dfrac{1}{4}$ times that in the other pipe
C) $2$ times that in the other pipe
D) $\dfrac{1}{2}$ times that in the other pipe
Answer
Verified
116.4k+ views
Hint: Here, by using the velocity formula, we have to calculate the radius of the pipes and separate the diameter, then proceed to find the flow of water in a particular diameter and then the vector defines the solution and compares the diameter of the other pipe.
Formula used:
According to that velocity of pipes
${A_x}{v_x} = {A_y}{v_y}$
Where,
${v_x}$ and ${v_y}$ are the velocity of water in pipes\[X\]and $Y$respectively.
The radius of $X,\,{r_x} = \dfrac{{dx}}{2}$
The radius of $Y,{r_y} = \dfrac{{dy}}{2}$
$dx,dy$ is differentiation with respect to a particular variable.
Complete step by step solution:
Given by ,
The diameter of the vector pipes
Let, $X = 2cm$ and $Y = 4cm$ respectively
We know that ,
Radius formula
The radius of $X,\,{r_x} = \dfrac{{dx}}{2}$
The radius of $Y,{r_y} = \dfrac{{dy}}{2}$
The value of $dx = 2$ and \[dy = 4\]
Substituting the given value in above equation,
We get,
The radius of $X$ and \[Y\]
${r_x} = 0.01\,m$
${r_y} = 0.02\,m$
According to the equation of continuity:
$\Rightarrow {A_x}{v_x} = {A_y}{v_y}$
$\Rightarrow {A_x} = \pi \left( {{r_x}} \right)$ and ${A_y} = \pi \left( {{r_y}} \right)$
Substituting the given value in above equation,
we get,
$\Rightarrow \pi {\left( {0.01\,m} \right)^2} \times {v_x} = \pi {\left( {0.02\,m} \right)^2}{v_y}$
Rearranging the given equation,
We have,
$\Rightarrow {v_x} = \dfrac{{{{\left( {0.02\,m} \right)}^2}}}{{{{\left( {0.01\,m} \right)}^2}}}{v_y}$
On simplifying,
We get, ${v_x} = 4{v_y}$
Therefore the velocity of flow of water in the pipe of $2cm$ diameter $X$ in four times that of in pipe $Y$
Hence, option A is the correct answer.
Note: If we have to calculate the pipe radius and compare the diameter of the other pipe. A flow rate increase or decrease will result in a corresponding velocity increase or decrease. Velocity is also influenced by pipe size. Reducing pipe size increases the velocity, which increases friction, given a constant flow rate.
Formula used:
According to that velocity of pipes
${A_x}{v_x} = {A_y}{v_y}$
Where,
${v_x}$ and ${v_y}$ are the velocity of water in pipes\[X\]and $Y$respectively.
The radius of $X,\,{r_x} = \dfrac{{dx}}{2}$
The radius of $Y,{r_y} = \dfrac{{dy}}{2}$
$dx,dy$ is differentiation with respect to a particular variable.
Complete step by step solution:
Given by ,
The diameter of the vector pipes
Let, $X = 2cm$ and $Y = 4cm$ respectively
We know that ,
Radius formula
The radius of $X,\,{r_x} = \dfrac{{dx}}{2}$
The radius of $Y,{r_y} = \dfrac{{dy}}{2}$
The value of $dx = 2$ and \[dy = 4\]
Substituting the given value in above equation,
We get,
The radius of $X$ and \[Y\]
${r_x} = 0.01\,m$
${r_y} = 0.02\,m$
According to the equation of continuity:
$\Rightarrow {A_x}{v_x} = {A_y}{v_y}$
$\Rightarrow {A_x} = \pi \left( {{r_x}} \right)$ and ${A_y} = \pi \left( {{r_y}} \right)$
Substituting the given value in above equation,
we get,
$\Rightarrow \pi {\left( {0.01\,m} \right)^2} \times {v_x} = \pi {\left( {0.02\,m} \right)^2}{v_y}$
Rearranging the given equation,
We have,
$\Rightarrow {v_x} = \dfrac{{{{\left( {0.02\,m} \right)}^2}}}{{{{\left( {0.01\,m} \right)}^2}}}{v_y}$
On simplifying,
We get, ${v_x} = 4{v_y}$
Therefore the velocity of flow of water in the pipe of $2cm$ diameter $X$ in four times that of in pipe $Y$
Hence, option A is the correct answer.
Note: If we have to calculate the pipe radius and compare the diameter of the other pipe. A flow rate increase or decrease will result in a corresponding velocity increase or decrease. Velocity is also influenced by pipe size. Reducing pipe size increases the velocity, which increases friction, given a constant flow rate.
Recently Updated Pages
Uniform Acceleration - Definition, Equation, Examples, and FAQs
How to find Oxidation Number - Important Concepts for JEE
How Electromagnetic Waves are Formed - Important Concepts for JEE
Electrical Resistance - Important Concepts and Tips for JEE
Average Atomic Mass - Important Concepts and Tips for JEE
Chemical Equation - Important Concepts and Tips for JEE
Trending doubts
JEE Main 2025: Application Form (Out), Exam Dates (Released), Eligibility & More
JEE Main Login 2045: Step-by-Step Instructions and Details
Class 11 JEE Main Physics Mock Test 2025
JEE Main Chemistry Question Paper with Answer Keys and Solutions
Learn About Angle Of Deviation In Prism: JEE Main Physics 2025
JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics
Other Pages
NCERT Solutions for Class 11 Physics Chapter 7 Gravitation
NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements
NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Fluids
Units and Measurements Class 11 Notes - CBSE Physics Chapter 1
NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line
NCERT Solutions for Class 11 Physics Chapter 8 Mechanical Properties of Solids