
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
179.1k+ 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

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

What is Hybridisation in Chemistry?

Other Pages
NCERT Solutions For Class 11 Physics Chapter 2 Motion In A Straight Line - 2025-26

NCERT Solutions For Class 11 Physics Chapter 1 Units and Measurements - 2025-26

NCERT Solutions For Class 11 Physics Chapter 3 Motion In A Plane - 2025-26

JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Units And Measurements Class 11 Physics Chapter 1 CBSE Notes - 2025-26

Motion in a Straight Line Class 11 Physics Chapter 2 CBSE Notes - 2025-26
