The viscous resistance of a tube to liquid flow is R. Its resistance for a narrow tube of same length and $\dfrac{1}{3}$ times radius is:
$\begin{align}
& \left( A \right)\dfrac{R}{3} \\
& \left( B \right)3R \\
& \left( C \right)27R \\
& \left( D \right)81R \\
\end{align}$
Answer
609.9k+ views
Hint: Use Poiseuille's equation. By Newton’s law of viscous flow, the viscous force is directly proportional to the surface area and change in its velocity. The flow of liquid is related to the factors like viscosity of the fluid , the pressure, length and the diameter of the tube. Hence by substituting the values in poiseuille's we will get the solution.
Formula used:
According to Poiseulli’s equation,
$Q=\dfrac{\pi (\vartriangle \operatorname{P}){{r}^{4}}}{8\eta L}$
where, Q is the rate of flow
P is the pressure
r is the inner radius of the tube
$\eta _{{}}^{{}}$ is the viscosity of the liquid
L is the length of the tube
Complete step by step solution:
According to Poiseulli’s equation,
$Q=\dfrac{\pi (\vartriangle \operatorname{P}){{r}^{4}}}{8\eta L}$
According to Ohm’s law,
V=IR …………..(1)
Which is similar to the poiseuille's equation.
That is, here the pressure difference is similar to potential difference, current is equal to the rate of flow of liquid and the constants are equal to resistance.
Thus rearranging the equation for pressure difference we get,
$\vartriangle P=Q\left[ \dfrac{8\eta L}{\pi {{r}^{4}}} \right]$ …………….(2)
Comparing equation (1) and (2),
Resistance,$R=\dfrac{8\eta L}{\pi {{r}^{4}}}$
Here, resistance is directly proportional to its length and inversely proportional to its radius.
New Resistance for the narrow tube,
$R'=\dfrac{8\eta L}{\pi {{\left( \dfrac{r}{3} \right)}^{4}}}$
$\begin{align}
& R'=\left( \dfrac{8\eta L}{\pi {{r}^{4}}} \right)\times {{3}^{4}} \\
& \\
\end{align}$
$\therefore R'=81R$
Thus option (D) is correct.
Note:
For deriving the formula for volume of a liquid flowing through a tube Poiseulli adopted some assumptions. That is, the flow of liquid in the tube is steady, streamline and parallel to the axis of the tube. There is no radial flow. The pressure over any cross section at right angles to the axis of the tube is constant. The velocity of the liquid layer in contact with the side of the tube is zero and increases in a regular manner as the axis of the tube is approached. The limitations of poiseuille's formula was it holds only for streamline flow of liquid. If the layers of the liquid in contact with the walls of the tube are not at rest and thus having relative slipping. This formula is true for fluid velocities, involving negligible amounts of kinetic energy.
Formula used:
According to Poiseulli’s equation,
$Q=\dfrac{\pi (\vartriangle \operatorname{P}){{r}^{4}}}{8\eta L}$
where, Q is the rate of flow
P is the pressure
r is the inner radius of the tube
$\eta _{{}}^{{}}$ is the viscosity of the liquid
L is the length of the tube
Complete step by step solution:
According to Poiseulli’s equation,
$Q=\dfrac{\pi (\vartriangle \operatorname{P}){{r}^{4}}}{8\eta L}$
According to Ohm’s law,
V=IR …………..(1)
Which is similar to the poiseuille's equation.
That is, here the pressure difference is similar to potential difference, current is equal to the rate of flow of liquid and the constants are equal to resistance.
Thus rearranging the equation for pressure difference we get,
$\vartriangle P=Q\left[ \dfrac{8\eta L}{\pi {{r}^{4}}} \right]$ …………….(2)
Comparing equation (1) and (2),
Resistance,$R=\dfrac{8\eta L}{\pi {{r}^{4}}}$
Here, resistance is directly proportional to its length and inversely proportional to its radius.
New Resistance for the narrow tube,
$R'=\dfrac{8\eta L}{\pi {{\left( \dfrac{r}{3} \right)}^{4}}}$
$\begin{align}
& R'=\left( \dfrac{8\eta L}{\pi {{r}^{4}}} \right)\times {{3}^{4}} \\
& \\
\end{align}$
$\therefore R'=81R$
Thus option (D) is correct.
Note:
For deriving the formula for volume of a liquid flowing through a tube Poiseulli adopted some assumptions. That is, the flow of liquid in the tube is steady, streamline and parallel to the axis of the tube. There is no radial flow. The pressure over any cross section at right angles to the axis of the tube is constant. The velocity of the liquid layer in contact with the side of the tube is zero and increases in a regular manner as the axis of the tube is approached. The limitations of poiseuille's formula was it holds only for streamline flow of liquid. If the layers of the liquid in contact with the walls of the tube are not at rest and thus having relative slipping. This formula is true for fluid velocities, involving negligible amounts of kinetic energy.
Recently Updated Pages
Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Trending doubts
Find the value of the expression given below sin 30circ class 11 maths CBSE

Difference between physical and chemical change class 11 chemistry CBSE

How do I convert ms to kmh Give an example class 11 physics CBSE

What is the atomic mass of Sodium a 22 gmol b 23 gmol class 11 chemistry CBSE

In which country Lascaux cave is located A India B class 11 social science CBSE

1 ton equals to A 100 kg B 1000 kg C 10 kg D 10000 class 11 physics CBSE

