
The rate of flow of liquid in a tube of radius r, length whose ends are maintained at a pressure difference p is \[V = \dfrac{{\pi Qp{r^4}}}{{\eta l}}\] , where \[\eta \] is coefficient of the viscosity and Q.
A. 8
B. \[\dfrac{1}{8} \\ \]
C. \[\dfrac{1}{6} \\ \]
D. \[\dfrac{1}{{16}}\]
Answer
163.8k+ views
Hint: When the viscosity of the liquid is not considered then the total energy of the flow is constant at every point on the line of the flow. But, when the viscosity is considered, then the energy is dissipated in the form of frictional work, the same as the energy dissipated in a resistor carrying electric current through itself when electric potential is applied across the resistor.
Complete step by step solution:
The radius of the tube in which the liquid is flowing is given as r. The length of the tube is given as $l$. The pressure difference between the ends of the tube is constant and it is equal to $p$. The rate of flow is given as,
\[V = \dfrac{{\pi Qp{r^2}}}{{\eta l}}\]
We need to determine Q. As in the expression for the flow rate there is the term coefficient of viscosity, so the fluid’s drag is considered while finding the expression for the flow rate of the liquid in the tube.
Hagen-Poiseuille equation gives the relation between the fluid resistance to the flow through the tube, the flow rate, the length of the flow and the pressure between the ends of the tube which is analogous to the flow of electric current in the resistance when electric potential is applied across the ends of the resistor as,
\[V = \dfrac{{\pi p{r^4}}}{{8\eta l}}\]
When we compare both the equations, we find the value of Q as \[\dfrac{1}{8}\].
Therefore, the correct option is B.
Note: When the tube is frictionless then we simply use Bernoulli’s equation to determine the flow rate as total energy of the flow is constant when the surface of the tube is frictionless. The speed of the flow at both the ends is related to the continuity equation. But when it is given that we need to consider the friction of the tube or the viscosity of the fluid then we use the Hagen-Poiseuille equation.
Complete step by step solution:
The radius of the tube in which the liquid is flowing is given as r. The length of the tube is given as $l$. The pressure difference between the ends of the tube is constant and it is equal to $p$. The rate of flow is given as,
\[V = \dfrac{{\pi Qp{r^2}}}{{\eta l}}\]
We need to determine Q. As in the expression for the flow rate there is the term coefficient of viscosity, so the fluid’s drag is considered while finding the expression for the flow rate of the liquid in the tube.
Hagen-Poiseuille equation gives the relation between the fluid resistance to the flow through the tube, the flow rate, the length of the flow and the pressure between the ends of the tube which is analogous to the flow of electric current in the resistance when electric potential is applied across the ends of the resistor as,
\[V = \dfrac{{\pi p{r^4}}}{{8\eta l}}\]
When we compare both the equations, we find the value of Q as \[\dfrac{1}{8}\].
Therefore, the correct option is B.
Note: When the tube is frictionless then we simply use Bernoulli’s equation to determine the flow rate as total energy of the flow is constant when the surface of the tube is frictionless. The speed of the flow at both the ends is related to the continuity equation. But when it is given that we need to consider the friction of the tube or the viscosity of the fluid then we use the Hagen-Poiseuille equation.
Recently Updated Pages
Uniform Acceleration - Definition, Equation, Examples, and FAQs

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

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

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

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2025: Derivation of Equation of Trajectory in Physics

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

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

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

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line

Motion in a Straight Line Class 11 Notes: CBSE Physics Chapter 2

Important Questions for CBSE Class 11 Physics Chapter 1 - Units and Measurement
