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Poiseuille's Formula

NEET > Physics > Properties of Bulk Matter > Fluid Mechanics > Poiseuille's Formula

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Topic 13 of 14 • Chapter: Fluid Mechanics • Physics

Poiseuille's Formula – Complete Notes, Revision, Important Questions & Downloads

This topic starts with Poiseuille's Equation for laminar flow through a capillary tube, then extends that resistance idea into Series Combination of Capillary Tubes and Parallel Combination of Capillary Tubes. NEET usually tests the relation Q = (pi r^4 Delta P)/(8 eta l) through proportional reasoning: double the radius, compare two capillaries, or convert the flow law into the resistance form R = 8 eta l/(pi r^4). The point that decides most questions is the r^4 dependence, because a small change in radius overwhelms moderate changes in pressure, viscosity, or length. If a capillary radius becomes 2r while Delta P, eta, and l stay fixed, the volume flow rate becomes 16 times larger, and that same logic controls the series and parallel combination questions too.

⬇ Download Notes PDFView Important Questions →
r^4 DependenceCapillary ResistanceLaminar Flow
Expected QuestionsQ
1
Usually appears as one formula-driven MCQ on radius dependence, equivalent liquid resistance, or blood-flow analogy in a narrow vessel.
Time Required⏱
45 min
Roughly 15 minutes for the base law and assumptions, and 30 minutes for ratio questions, resistance combinations, and capillary-network numericals.
Difficulty⚡
Medium
The algebra is short, but students lose marks when they miss the fourth-power radius dependence or mix series and parallel resistance rules.
NRI USA Curriculum GapUS
Gap
Many U.S. courses mention laminar tube flow qualitatively, but NEET expects immediate use of the resistance form and direct comparison of capillary tubes connected in series or parallel.
10Subtopics
4Practice Questions
2Free Downloads
45 minPrep Time
⬇ Get Free Downloads

NEET Weightage - Poiseuille's Formula

Fluid Mechanics
NEET YearQuestions from this TopicBarMarks
NEET 20240
 
0 Q
0
NEET 20230
 
0 Q
0
NEET 20220
 
0 Q
0
NEET 20210
 
0 Q
0
NEET 20200
 
0 Q
0
NEET 20190
 
0 Q
0
Total (2019-2024)0 0
Poiseuille's law is the laminar-flow result for a long cylindrical tube, so the assumptions matter before the formula is used.
Flow rate is directly proportional to pressure difference and inversely proportional to viscosity and length.

Radius controls the result through the fourth power, so it is the fastest way to eliminate options in comparison questions.

Writing Q = Delta P / R with R = 8 eta l / (pi r^4) turns fluid-network questions into the same logic used for electrical resistance.
[AVG]
0.0
Avg Questions / Year
[MARKS]
0
Total Marks (6 yrs)
[PATTERN]
Direct
Pattern
[LEVEL]
Medium
Difficulty

Poiseuille Strategy for NEET

1

Check the flow conditions before touching the formula Use Poiseuille's law only when the question describes laminar flow through a long capillary-like tube. If turbulence or a non-capillary geometry is implied, the standard relation should not be used blindly.

2

Memorise which quantities are direct and inverse Flow rate rises with Delta P and r^4, but falls with eta and l. That map prevents sign mistakes when the question changes two parameters at once.

3

Treat radius change as the dominant effect The common trap is to compare a doubled radius with doubled length and call the net change twofold. Because radius enters as r^4, doubling r makes the flow sixteenfold before any other factor is applied.

4

Convert to resistance form when tubes are connected As soon as the question mentions two capillaries together, write R = 8 eta l / (pi r^4), then use R_eff = R1 + R2 for series or 1/R_eff = 1/R1 + 1/R2 for parallel.

5

Recognise the blood-vessel framing When the stem talks about a narrowed artery or cannula, look for the same r^4 law. A modest decrease in vessel radius can reduce flow sharply even if pressure difference changes only slightly.

Poiseuille's Formula Study Materials

PDF · Cheat Sheet · MCQ Set · PYQ
[NOTES]
Full Notes - Poiseuille's Formula
Topic notes that connect the capillary-flow equation with liquid resistance and then extend the same logic to series and parallel capillary combinations.
310 subtopicsResistance formCombination rules
Download Notes
[FORMULA]
Formula Sheet - Poiseuille's Formula
Quick sheet for Q = (pi r^4 Delta P)/(8 eta l), the liquid-resistance form, and the equivalent-resistance rules for capillary networks.
Q vs R formr^4 lawSeries and parallel
Download Formula Sheet
[MCQ]
MCQ Practice - Poiseuille's Formula
Applied MCQs focused on pressure change, capillary radius comparison, and effective resistance of connected tubes under the same laminar-flow assumption.
4 core MCQsCapillary networksFlow ratios
Download MCQ Set
[PYQ]
PYQ - Poiseuille's Formula
Revision download that groups the standard ratio tricks, radius-sensitivity checks, and capillary-combination patterns that resemble NEET objective questions.
Quick revisionr^4 trapsResistance analogy
Download PYQ Set

Poiseuille's Formula Subtopics

2-Column Table
Column AColumn B
Poiseuille's Equation↗
Series Combination of Capillary Tubes↗
Parallel Combination of Capillary Tubes↗
Increase in all directions↗
Never increases↗
Area of the bottom surface↗
Nature of the liquid↗
High density and high viscosity↗
Low density and low viscosity↗
Streamline flow↗

Rapid Revision - Poiseuille's Formula

Concept → Trap → Example

1) Poiseuille's Equation

Laminar capillary flow

Poiseuille's Equation gives the volume flow rate through a capillary as Q = (pi r^4 Delta P)/(8 eta l), so flow is directly proportional to pressure difference and fourth power of radius, and inversely proportional to viscosity and length.

  • Use it for laminar flow through a long cylindrical tube of uniform radius.
  • The same relation can be written as Q = Delta P / R, where R = 8 eta l / (pi r^4).
  • The trap is to compare radius changes linearly when the law actually depends on r^4.
Example (NEET-style)If one capillary has radius 1 mm and another has radius 2 mm with the same Delta P, eta, and l, the second capillary carries 2^4 = 16 times the flow rate of the first.

2) Series Combination of Capillary Tubes

Same flow, added resistance

Series Combination of Capillary Tubes keeps the same flow rate through each tube, so the effective liquid resistance is R_eff = R1 + R2 and the total pressure drop is shared across the two capillaries.

  • The same volume per second must pass through both capillaries in a series path.
  • Equivalent resistance increases, so for the same applied Delta P the total flow decreases compared with a single tube.
  • The trap is to use the parallel rule in a series question just because two tubes are mentioned together.
Example (NEET-style)If two capillaries have liquid resistances 3 x 10^8 and 5 x 10^8 SI units in series, the effective resistance is 8 x 10^8 and the flow becomes Delta P/(8 x 10^8).

3) Parallel Combination of Capillary Tubes

Same pressure, split flow

Parallel Combination of Capillary Tubes keeps the same pressure difference across each branch, while the total flow divides among them, so 1/R_eff = 1/R1 + 1/R2.

  • Each branch experiences the same Delta P between the common inlet and outlet points.
  • Equivalent resistance decreases, which makes the total flow larger than the flow through any single branch alone.
  • The trap is to assume the flow is the same in each branch even when the branch resistances are different.
Example (NEET-style)If two identical capillaries of resistance R are connected in parallel, the effective resistance becomes R/2, so the same pressure difference drives twice the flow obtained through one capillary alone.

US Curriculum Gaps - Poiseuille's Formula

What U.S. Students Usually Miss

AP Physics 2 often treats tube-flow resistance qualitatively rather than as a fast ratio tool

NEET expects immediate use of Q proportional to r^4 and R proportional to 1/r^4 in one-step elimination. Students trained only on qualitative pressure-flow language often miss how violently the flow changes when radius changes even slightly.

  • Memorise that doubling radius multiplies flow by 16 under fixed Delta P, eta, and l.
  • Translate artery narrowing or cannula size into the same capillary-flow ratio without re-deriving it.

Honors Physics rarely packages series and parallel capillary tubes as a circuit-style resistance problem

This topic does not stop at the single-capillary law. NEET regularly expects the resistance analogy itself: same flow in series, same pressure difference in parallel, and the correct effective-resistance formula chosen without hesitation.

  • Write the liquid resistance first when two or more capillaries are connected.
  • Choose R_eff = R1 + R2 for series and 1/R_eff = 1/R1 + 1/R2 for parallel before substituting numbers.

Concept IQ Check - Poiseuille's Formula

4 NEET-style MCQs with Answers
1A capillary tube carries a liquid under laminar flow. If its radius is doubled while pressure difference, length, and viscosity stay unchanged, the flow rate becomesPoiseuille's Equation
2 times
4 times
8 times
16 times
Poiseuille's law gives Q proportional to r^4. So if radius becomes 2r, the new flow is 2^4 = 16 times the original flow. The smaller options come from forgetting that radius enters to the fourth power, not linearly or quadratically.
2Two capillary tubes of liquid resistances 3R and 5R are connected in series. Their effective resistance isSeries Combination of Capillary Tubes
8R
15R/8
2R
R/8
In series, the same flow passes through each capillary and the pressure drops add, so the effective liquid resistance is the direct sum: R_eff = 3R + 5R = 8R. The reciprocal formula belongs to parallel combination, not series combination.
3Two identical capillaries are connected in parallel between the same pressure points. Compared with one capillary alone, the total flow isParallel Combination of Capillary Tubes
halved
unchanged
doubled
quadrupled
Two identical resistances R in parallel give R_eff = R/2. Since total flow is Q = Delta P / R_eff, the same pressure difference produces twice the original single-capillary flow. The pressure difference stays the same across both branches, while the total current of liquid splits between them.
4A liquid flows through a capillary under fixed pressure difference. Which change cuts the flow to half?Poiseuille's Equation
Doubling the radius
Halving the length
Doubling the viscosity
Doubling the pressure difference
Poiseuille's law gives Q proportional to Delta P and inversely proportional to eta and l. Doubling viscosity halves the flow. Doubling radius would increase flow sixteenfold, halving length would double it, and doubling pressure difference would also double it.

Practice Questions - Poiseuille's Formula

Click "Reveal Answer" after attempting
1Water flows through a capillary of radius r with flow rate Q. A second capillary has the same pressure difference, viscosity, and length, but radius 1.5r. What is the new flow rate?
1.5Q
2.25Q
5.06Q
6.75Q
👁 Reveal Answer
Correct option: 3. Since Q is proportional to r^4, the new flow is (1.5)^4 Q = 5.0625Q, which is about 5.06Q. This is the standard NEET comparison trap: even a modest radius increase produces a much larger effect than students expect.
2Two capillaries of the same radius and viscosity have lengths 10 cm and 15 cm. They are connected in series. If their individual liquid resistances are proportional only to length, what is the ratio R_eff/R1, where R1 is the resistance of the 10 cm tube?
1.5
2.0
2.5
3.0
👁 Reveal Answer
Correct option: 3. With equal radius and viscosity, resistance is proportional to length. So R2/R1 = 15/10 = 1.5, hence R_eff = R1 + R2 = 2.5R1. The same flow must pass through both tubes because they are in series.
3Three identical capillaries, each of resistance R, are connected in parallel. Under the same pressure difference Delta P, what is the total flow compared with the flow through one capillary alone?
one-third
same
three times
nine times
👁 Reveal Answer
Correct option: 3. For three identical parallel capillaries, 1/R_eff = 1/R + 1/R + 1/R = 3/R, so R_eff = R/3. Therefore total flow is Delta P/(R/3) = 3 Delta P/R, which is three times the flow through one capillary.
4A narrowed blood vessel has the same pressure difference and length as before, but its effective radius becomes 80% of the original value. Assuming laminar flow and unchanged viscosity, what fraction of the original flow remains?
0.80
0.64
0.41
0.25
👁 Reveal Answer
Correct option: 3. Flow is proportional to r^4, so the new fraction is (0.8)^4 = 0.4096, about 0.41. This is why Poiseuille's law is so useful in blood-flow analogies: a small reduction in vessel radius can reduce flow much more than intuition suggests.

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Poiseuille's Formula FAQs

Notes · Downloads · Revision · Important Questions
What does Poiseuille's formula state?
It states that the volume flow rate through a capillary tube under laminar flow is directly proportional to the pressure difference and the fourth power of the radius, and inversely proportional to viscosity and length. In compact form, Q = (pi r^4 Delta P)/(8 eta l).
Why is the radius term so important in this topic?
Because radius enters as r^4, not as r or r^2. That makes radius the most sensitive factor in most comparison questions. Doubling radius increases flow sixteen times, while halving radius reduces flow to one-sixteenth under the same pressure difference, length, and viscosity.
What is liquid resistance in Poiseuille's formula?
Liquid resistance is the factor R = 8 eta l / (pi r^4) that opposes volume flow. Writing the law as Q = Delta P / R makes capillary-flow problems look like resistance problems in circuits, which is especially useful for series and parallel combinations.
How do I know whether two capillary tubes are in series or in parallel?
If the same stream of liquid must pass through one tube and then the next, they are in series, so the same flow passes through each and resistances add. If the liquid can split into branches and then recombine, they are in parallel, so each branch has the same pressure difference and the reciprocal resistance rule applies.
Why is the same flow rate maintained in series combination?
Because the liquid has only one path. Whatever volume per second enters the first capillary must emerge from it and then pass through the second capillary. So the pressure drops may differ, but the flow rate through each element in the series path remains the same.
Why is the same pressure difference maintained in parallel combination?
Because each branch begins and ends at the same two junctions. Those junctions fix the same inlet and outlet pressures for every branch, so each branch experiences the same Delta P even though the branch flow rates can be different.
Can Poiseuille's law be used for any fluid flow question in a pipe?
No. It is the laminar-flow result for a long cylindrical tube under the usual ideal assumptions for this school-level treatment. If the question implies turbulent flow or a strongly non-capillary geometry, the standard Poiseuille relation should not be applied without checking the assumptions.
How is Poiseuille's formula relevant to blood flow?
It gives the first-order idea that blood flow in a narrow vessel is extremely sensitive to vessel radius. Even when a question is framed biologically, the NEET-style inference is often the same: a slight narrowing can produce a very large reduction in flow because of the fourth-power radius dependence.
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Poiseuille's Equation

Series Combination of Capillary Tubes

Parallel Combination of Capillary Tubes

Increase in all directions

Never increases

Area of the bottom surface

Nature of the liquid

High density and high viscosity

Low density and low viscosity

Streamline flow

Subtopics

Poiseuille's Equation

Series Combination of Capillary Tubes

Parallel Combination of Capillary Tubes

Increase in all directions

Never increases

Area of the bottom surface

Nature of the liquid

High density and high viscosity

Low density and low viscosity

Streamline flow

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Poiseuille's Formula > Streamline flow > Streamline flow
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Poiseuille's Equation

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