
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
225.9k+ 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.
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