The Poiseuilles Law Formula gives the volumetric flow rate of a viscous, incompressible fluid flowing through a long cylindrical pipe under a pressure difference. Expressed as \ ( Q = \frac{\pi P r^4}{8 \eta l} \), this fundamental relation is covered in Class 11 Physics (NCERT Chapter 10 — Mechanical Properties of Fluids) and appears regularly in JEE Main, JEE Advanced, and NEET examinations. This article covers the complete expression, variable definitions, step-by-step derivation, a full formula sheet, three solved examples at progressive difficulty levels, CBSE exam tips, common mistakes, and JEE/NEET application patterns.
Key Poiseuilles Law Formulas at a Glance
Quick reference for the most important formulas related to Poiseuille’s Law.
- Volumetric flow rate: \( Q = \dfrac{\pi P r^4}{8 \eta l} \)
- Pressure difference form: \( P = \dfrac{8 \eta l Q}{\pi r^4} \)
- Hydraulic resistance: \( R_f = \dfrac{8 \eta l}{\pi r^4} \)
- Velocity profile: \( v(r) = \dfrac{P}{4 \eta l}(R^2 – r^2) \)
- Maximum velocity at centre: \( v_{max} = \dfrac{P R^2}{4 \eta l} \)
- Average velocity: \( v_{avg} = \dfrac{v_{max}}{2} = \dfrac{P R^2}{8 \eta l} \)
- Reynolds number: \( Re = \dfrac{\rho v D}{\eta} \)
What is Poiseuilles Law Formula?
The Poiseuilles Law Formula describes the steady, laminar flow of a Newtonian, viscous, and incompressible fluid through a straight cylindrical pipe of uniform cross-section. It was derived experimentally by the French physician Jean Louis Marie Poiseuille in 1838 and later given a theoretical basis by Gotthilf Hagen. The law states that the volumetric flow rate \ ( Q \) is directly proportional to the fourth power of the pipe radius and the applied pressure difference, and inversely proportional to the dynamic viscosity of the fluid and the length of the pipe.
In the NCERT Class 11 Physics textbook, Poiseuille’s Law appears in Chapter 10 — Mechanical Properties of Fluids. It is one of the most important quantitative results in viscous flow. The formula has wide applications in medical science (blood flow through arteries), engineering (pipeline design), and laboratory fluid dynamics. Understanding the Poiseuilles Law Formula is essential for scoring well in CBSE board examinations and for clearing JEE and NEET entrance tests.
The key condition for applying this formula is that the flow must be laminar, meaning the Reynolds number must be less than 2000. The fluid must also be Newtonian (constant viscosity) and the pipe must be long compared to its radius.
Poiseuilles Law Formula — Expression and Variables
The Poiseuilles Law Formula for volumetric flow rate is:
\[ Q = \frac{\pi P r^4}{8 \eta l} \]
Here, \( Q \) is the volume of fluid flowing per unit time through the pipe. The formula can also be rearranged to find the pressure difference required for a given flow rate:
\[ P = \frac{8 \eta l Q}{\pi r^4} \]
The hydraulic resistance of the pipe, analogous to electrical resistance, is defined as:
\[ R_f = \frac{P}{Q} = \frac{8 \eta l}{\pi r^4} \]
| Symbol | Quantity | SI Unit |
|---|---|---|
| \( Q \) | Volumetric flow rate | m³ s−1 |
| \( P \) | Pressure difference between the two ends of the pipe | Pascal (Pa) or N m−2 |
| \( r \) | Radius of the pipe | metre (m) |
| \( \eta \) | Dynamic (coefficient of) viscosity of the fluid | Pa·s or N s m−2 |
| \( l \) | Length of the pipe | metre (m) |
| \( \pi \) | Mathematical constant (approximately 3.14159) | Dimensionless |
| \( R_f \) | Hydraulic (fluid) resistance | Pa s m−3 |
Derivation of Poiseuilles Law Formula
Consider a long cylindrical pipe of radius \( R \) and length \( l \). A pressure difference \( P \) is maintained across its two ends. We assume the flow is steady, laminar, and the fluid is incompressible and Newtonian.
Step 1: Consider a coaxial cylindrical shell of radius \( r \) and thickness \( dr \). The viscous force on this shell is \( F = -\eta \cdot 2\pi r l \cdot \dfrac{dv}{dr} \).
Step 2: The driving force due to pressure is \( F = P \cdot \pi r^2 \).
Step 3: At steady state, the two forces balance. This gives \( \dfrac{dv}{dr} = -\dfrac{Pr}{2\eta l} \).
Step 4: Integrating with the no-slip boundary condition \( v = 0 \) at \( r = R \), we get the parabolic velocity profile: \( v(r) = \dfrac{P}{4\eta l}(R^2 – r^2) \).
Step 5: The flow rate is obtained by integrating over the cross-section: \( Q = \int_0^R v(r) \cdot 2\pi r \, dr = \dfrac{\pi P R^4}{8 \eta l} \).
This completes the derivation of the Poiseuilles Law Formula.
Complete Fluid Mechanics Formula Sheet
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Poiseuille’s Law (Flow Rate) | \( Q = \dfrac{\pi P r^4}{8 \eta l} \) | P = pressure diff., r = radius, η = viscosity, l = length | m³ s−1 | Class 11, Ch 10 |
| Hydraulic Resistance | \( R_f = \dfrac{8 \eta l}{\pi r^4} \) | η = viscosity, l = length, r = radius | Pa s m−3 | Class 11, Ch 10 |
| Stokes’ Law (Drag Force) | \( F = 6\pi \eta r v \) | η = viscosity, r = radius, v = velocity | N | Class 11, Ch 10 |
| Terminal Velocity | \( v_t = \dfrac{2r^2(\rho – \sigma)g}{9\eta} \) | r = radius, ρ = density of sphere, σ = fluid density | m s−1 | Class 11, Ch 10 |
| Continuity Equation | \( A_1 v_1 = A_2 v_2 \) | A = cross-sectional area, v = velocity | m³ s−1 | Class 11, Ch 10 |
| Bernoulli’s Equation | \( P + \frac{1}{2}\rho v^2 + \rho g h = \text{const} \) | P = pressure, ρ = density, v = velocity, h = height | Pa | Class 11, Ch 10 |
| Reynolds Number | \( Re = \dfrac{\rho v D}{\eta} \) | ρ = density, v = velocity, D = diameter, η = viscosity | Dimensionless | Class 11, Ch 10 |
| Velocity of Efflux (Torricelli) | \( v = \sqrt{2gh} \) | g = acceleration due to gravity, h = height of liquid | m s−1 | Class 11, Ch 10 |
| Surface Tension (Excess Pressure in Bubble) | \( \Delta P = \dfrac{4T}{r} \) | T = surface tension, r = radius of bubble | Pa | Class 11, Ch 10 |
| Capillary Rise | \( h = \dfrac{2T\cos\theta}{\rho g r} \) | T = surface tension, θ = contact angle, r = tube radius | m | Class 11, Ch 10 |
Poiseuilles Law Formula — Solved Examples
Example 1 (Class 11 CBSE Level)
Problem: Water flows through a horizontal pipe of radius 2 mm and length 10 cm. The pressure difference between the two ends is 80 Pa. The dynamic viscosity of water is \( 1 \times 10^{-3} \) Pa·s. Find the volumetric flow rate.
Given:
- Radius, \( r = 2 \) mm \( = 2 \times 10^{-3} \) m
- Length, \( l = 10 \) cm \( = 0.10 \) m
- Pressure difference, \( P = 80 \) Pa
- Viscosity, \( \eta = 1 \times 10^{-3} \) Pa·s
Step 1: Write the Poiseuilles Law Formula: \( Q = \dfrac{\pi P r^4}{8 \eta l} \)
Step 2: Calculate \( r^4 \): \( r^4 = (2 \times 10^{-3})^4 = 16 \times 10^{-12} \) m&sup4;
Step 3: Substitute all values:
\[ Q = \frac{3.14159 \times 80 \times 16 \times 10^{-12}}{8 \times 1 \times 10^{-3} \times 0.10} \]
Step 4: Calculate numerator: \( 3.14159 \times 80 \times 16 \times 10^{-12} = 4021.24 \times 10^{-12} \)
Step 5: Calculate denominator: \( 8 \times 10^{-3} \times 0.10 = 8 \times 10^{-4} \)
Step 6: Divide: \( Q = \dfrac{4021.24 \times 10^{-12}}{8 \times 10^{-4}} = 5.03 \times 10^{-6} \) m³ s−1
Answer
Volumetric flow rate \( Q \approx 5.03 \times 10^{-6} \) m³ s−1 (approximately 5.03 mL/s).
Example 2 (Class 11-12 Multi-Step Level)
Problem: Two pipes A and B are connected in series. Pipe A has radius \( r \) and length \( l \). Pipe B has radius \( 2r \) and length \( 2l \). The same fluid of viscosity \( \eta \) flows through both. If the total pressure drop across the combination is \( P_0 \), find the flow rate through the system and the pressure drop across each pipe.
Given:
- Pipe A: radius \( r \), length \( l \)
- Pipe B: radius \( 2r \), length \( 2l \)
- Total pressure drop: \( P_0 \)
- Viscosity: \( \eta \)
Step 1: Find hydraulic resistance of each pipe using \( R_f = \dfrac{8\eta l}{\pi r^4} \).
Step 2: Resistance of Pipe A: \( R_A = \dfrac{8\eta l}{\pi r^4} \)
Step 3: Resistance of Pipe B: \( R_B = \dfrac{8\eta (2l)}{\pi (2r)^4} = \dfrac{16\eta l}{16\pi r^4} = \dfrac{\eta l}{\pi r^4} \)
Step 4: Note that \( R_B = \dfrac{R_A}{8} \). For pipes in series, total resistance is \( R_{total} = R_A + R_B = R_A + \dfrac{R_A}{8} = \dfrac{9R_A}{8} \).
Step 5: Flow rate (same through both pipes in series): \( Q = \dfrac{P_0}{R_{total}} = \dfrac{P_0}{\frac{9R_A}{8}} = \dfrac{8P_0}{9R_A} = \dfrac{8P_0 \pi r^4}{9 \times 8\eta l} = \dfrac{\pi P_0 r^4}{9\eta l} \)
Step 6: Pressure drop across A: \( P_A = Q \cdot R_A = \dfrac{\pi P_0 r^4}{9\eta l} \times \dfrac{8\eta l}{\pi r^4} = \dfrac{8P_0}{9} \)
Step 7: Pressure drop across B: \( P_B = P_0 – P_A = P_0 – \dfrac{8P_0}{9} = \dfrac{P_0}{9} \)
Answer
Flow rate: \( Q = \dfrac{\pi P_0 r^4}{9\eta l} \). Pressure drop across A: \( \dfrac{8P_0}{9} \). Pressure drop across B: \( \dfrac{P_0}{9} \). The narrower pipe (A) carries most of the pressure drop, demonstrating the dominant role of radius in the Poiseuilles Law Formula.
Example 3 (JEE/NEET Application Level)
Problem: Blood flows through a capillary of length 1 mm and radius 4 μm. The viscosity of blood is \( 2.5 \times 10^{-3} \) Pa·s. If the pressure difference across the capillary is 2660 Pa (equivalent to 20 mmHg), find: (a) the volumetric flow rate, (b) the maximum velocity of blood in the capillary, and (c) the average velocity.
Given:
- Length, \( l = 1 \) mm \( = 1 \times 10^{-3} \) m
- Radius, \( R = 4 \) μm \( = 4 \times 10^{-6} \) m
- Viscosity, \( \eta = 2.5 \times 10^{-3} \) Pa·s
- Pressure difference, \( P = 2660 \) Pa
Step 1 (Flow Rate): Apply the Poiseuilles Law Formula:
\[ Q = \frac{\pi P R^4}{8 \eta l} \]
Step 2: Calculate \( R^4 = (4 \times 10^{-6})^4 = 256 \times 10^{-24} \) m&sup4;
Step 3: Numerator: \( \pi \times 2660 \times 256 \times 10^{-24} = 3.14159 \times 681,000 \times 10^{-24} \approx 2.139 \times 10^{-18} \)
Step 4: Denominator: \( 8 \times 2.5 \times 10^{-3} \times 10^{-3} = 2 \times 10^{-5} \)
Step 5: \( Q = \dfrac{2.139 \times 10^{-18}}{2 \times 10^{-5}} \approx 1.07 \times 10^{-13} \) m³ s−1
Step 6 (Maximum Velocity): \( v_{max} = \dfrac{PR^2}{4\eta l} = \dfrac{2660 \times (4 \times 10^{-6})^2}{4 \times 2.5 \times 10^{-3} \times 10^{-3}} \)
Step 7: \( v_{max} = \dfrac{2660 \times 16 \times 10^{-12}}{10^{-5}} = \dfrac{4.256 \times 10^{-8}}{10^{-5}} = 4.256 \times 10^{-3} \) m s−1 \( \approx 4.26 \) mm s−1
Step 8 (Average Velocity): \( v_{avg} = \dfrac{v_{max}}{2} \approx 2.13 \) mm s−1
Answer
(a) Flow rate \( Q \approx 1.07 \times 10^{-13} \) m³ s−1. (b) Maximum velocity \( v_{max} \approx 4.26 \) mm s−1. (c) Average velocity \( v_{avg} \approx 2.13 \) mm s−1. This example models real capillary blood flow, a common NEET biology-physics crossover topic.
CBSE Exam Tips 2025-26
- Memorise the power of radius: The most tested aspect of the Poiseuilles Law Formula is that flow rate varies as the fourth power of radius. If radius doubles, flow rate increases 16 times. This fact appears in both MCQ and short-answer questions every year.
- Know the analogy with Ohm’s Law: CBSE Class 11 exam questions often ask students to compare \( Q = P / R_f \) with \( I = V/R \). We recommend practising this analogy thoroughly. Hydraulic resistance \( R_f = 8\eta l / \pi r^4 \) is analogous to electrical resistance.
- Unit conversion is critical: Always convert radius to metres and length to metres before substituting. A common error is using mm or cm directly, which leads to wrong answers.
- State the conditions: In 3-mark and 5-mark descriptive answers, always mention the conditions — laminar flow, Newtonian fluid, incompressible fluid, and steady flow. This earns full marks in 2025-26 board exams.
- Practice the derivation: The derivation of the Poiseuilles Law Formula is a frequently asked 5-mark question in CBSE Class 11 Physics. Our experts suggest practising it at least three times before the exam.
- Series and parallel pipes: Questions on pipes in series (resistances add) and pipes in parallel (reciprocal rule) are common in the 2025-26 CBSE pattern. Treat them exactly like series/parallel resistors.
Common Mistakes to Avoid with Poiseuilles Law Formula
- Mistake 1 — Using diameter instead of radius: The formula uses radius \( r \), not diameter \( d \). If you are given the diameter, divide by 2 first. Forgetting this step gives a flow rate that is 16 times too large.
- Mistake 2 — Ignoring the \( \pi / 8 \) factor: Many students write \( Q = Pr^4 / \eta l \) and forget the \( \pi / 8 \) prefactor. Always write the complete expression \( Q = \pi P r^4 / (8 \eta l) \).
- Mistake 3 — Applying the formula to turbulent flow: Poiseuille’s Law is valid only for laminar flow (Reynolds number \( Re < 2000 \)). Applying it to turbulent flow gives incorrect results. Always check the Reynolds number if flow conditions are given.
- Mistake 4 — Confusing viscosity symbols: Dynamic viscosity is denoted \( \eta \) (eta) in the NCERT textbook. Some students confuse it with kinematic viscosity \( \nu = \eta / \rho \). Use \( \eta \) in the Poiseuilles Law Formula, not \( \nu \).
- Mistake 5 — Wrong series/parallel rule: When two pipes are in series, the same flow rate passes through both and pressure drops add. When in parallel, the same pressure difference applies and flow rates add. Mixing these rules is a very common error in numerical problems.
JEE/NEET Application of Poiseuilles Law Formula
In our experience, JEE aspirants encounter the Poiseuilles Law Formula in two main contexts: fluid mechanics problems in JEE Main Paper 1 and biofluid questions in NEET Biology-Physics crossovers. Understanding the formula deeply — not just memorising it — is the key to solving these problems quickly.
Pattern 1 — Ratio Problems (JEE Main)
JEE Main frequently asks questions of the type: “If the radius of a pipe is halved and the pressure difference is doubled, what is the new flow rate compared to the original?” These require applying the Poiseuilles Law Formula in ratio form. Since \( Q \propto P r^4 \), the new flow rate is \( Q’ = Q \times (2) \times (1/2)^4 = Q \times 2 \times (1/16) = Q/8 \). Practise at least 10 such ratio problems before JEE Main.
Pattern 2 — Series and Parallel Pipe Networks (JEE Advanced)
JEE Advanced problems often involve networks of pipes with different radii and lengths. The approach mirrors Kirchhoff’s laws for circuits. For pipes in series: \( R_{total} = \sum R_i \). For pipes in parallel: \( 1/R_{total} = \sum (1/R_i) \). The Poiseuilles Law Formula provides the value of each \( R_i = 8\eta l_i / (\pi r_i^4) \). This analogy is explicitly mentioned in NCERT and is a high-yield topic for JEE Advanced.
Pattern 3 — Blood Flow Applications (NEET)
NEET questions on the human circulatory system often link directly to the Poiseuilles Law Formula. Questions ask about the effect of arterial narrowing (atherosclerosis) on blood flow, or compare blood flow in arteries of different radii. Since \( Q \propto r^4 \), even a small reduction in arterial radius causes a dramatic drop in blood flow. For example, reducing the radius by 20% (factor of 0.8) reduces flow to \( (0.8)^4 = 0.41 \) times the original — a 59% reduction. This is a powerful conceptual point for NEET Biology questions on cardiovascular physiology.
We recommend NEET students practise converting between medical units (mmHg, mL/min) and SI units (Pa, m³ s−1) when applying the Poiseuilles Law Formula to biological contexts.
FAQs on Poiseuilles Law Formula
For more related Physics formulas, explore our comprehensive guides on the Critical Velocity Formula, the Superposition Formula, and the Heat Load Formula. You can also browse the complete Physics Formulas hub for all Class 11 and Class 12 formula sheets. For the official NCERT syllabus reference, visit the NCERT official website.