Along a streamline of an ideal incompressible fluid: P + ½ ρ v² + ρ g h = constant. Higher speed → lower pressure. Basis of aerofoil lift, venturi meter, atomiser.
-- NCERT Class 11 Physics, Ch. 9, p. 188Bernoulli's Principle
8 MCQs2 revision cards9-step worked example
Source: NCERT Properties of Solids and LiquidsPYQ coverage: NEET 2023, 2026Official key: NTA-verifiedLast updated: 27 Sep 2026
Bernoulli's Principle, explained for NEET
The high-frequency trap with Bernoulli's equation is forgetting the height term. When a pipe changes both cross-section and elevation, aspirants who drop the ρgh term get a plausible-looking but wrong pressure or velocity — and lose marks to negative marking.
Bernoulli's principle states that for steady, incompressible, non-viscous flow along a streamline:
P + ½ρv² + ρgh = constant
This is energy conservation per unit volume. P is the pressure energy density, ½ρv² is the kinetic energy density, and ρgh is the gravitational potential energy density (NCERT Class 11 Physics Chapter 9, page 188).
Three conditions must hold for the equation to apply: (1) the flow is steady (no turbulence), (2) the fluid is incompressible (constant ρ), and (3) the fluid is non-viscous (no energy loss to friction). If any condition breaks — turbulent flow, compressible gas at high speed, or significant viscosity — Bernoulli's equation gives incorrect results.
How NEET uses this: Questions typically give two points along a pipe system (often a Venturi meter or a pipe with a constriction at a different height) and ask for the pressure difference or the velocity at one point. The continuity equation A₁v₁ = A₂v₂ is almost always needed alongside Bernoulli's equation to eliminate one unknown velocity.
Watch out: When both points are at the same height, the ρgh terms cancel and you get P₁ + ½ρv₁² = P₂ + ½ρv₂². Many aspirants memorise only this simplified form and then apply it to problems where heights differ — that is exactly the distractor NTA exploits.
Can you answer these Bernoulli's Principle MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
Bernoulli's theorem is a statement of conservation of which quantity for a fluid in streamline flow?
Show answer and why every option is right or wrong
Answer: A. Bernoulli's equation expresses conservation of energy per unit volume along a streamline — pressure energy + kinetic energy + potential energy = constant (NCERT Class 11 Physics Chapter 9, page 188).
Why B is wrong: B is wrong. Conservation of momentum for fluids relates to Newton's second law applied to fluid elements (Euler's equation), not Bernoulli's equation.
Why C is wrong: C is wrong. Conservation of mass in fluid flow is expressed by the continuity equation (A₁v₁ = A₂v₂), not by Bernoulli's equation.
Why D is wrong: D is wrong. Conservation of angular momentum applies to rotating systems and is not the basis of Bernoulli's theorem.
MCQ 2Easy RecallPractice
Which of the following is NOT a necessary condition for Bernoulli's equation to be valid?
Show answer and why every option is right or wrong
Answer: B. Bernoulli's equation requires steady, non-viscous, incompressible flow. Turbulent flow violates the steady-flow condition, so it is not a condition for validity — rather, its presence invalidates the equation (NCERT Class 11 Physics Chapter 9, page 188).
Why A is wrong: A is wrong. Incompressibility (constant ρ) is a required condition; without it, the derivation of Bernoulli's equation does not hold.
Why C is wrong: C is wrong. Non-viscous flow is required; viscosity causes energy dissipation that Bernoulli's equation does not account for.
Why D is wrong: D is wrong. Steady (streamline) flow is a required condition; the equation applies along a streamline under steady-state conditions.
MCQ 3Easy RecallPractice
In Bernoulli's equation P + ½ρv² + ρgh = constant, the term ½ρv² represents:
Show answer and why every option is right or wrong
Answer: C. ½ρv² has dimensions of energy per unit volume (J/m³ = Pa) and represents the kinetic energy density of the flowing fluid (NCERT Class 11 Physics Chapter 9, page 188).
Why A is wrong: A is wrong. Pressure energy per unit volume is represented by the P term in Bernoulli's equation, not the ½ρv² term.
Why B is wrong: B is wrong. Potential energy per unit volume is represented by the ρgh term, not ½ρv².
Why D is wrong: D is wrong. Bernoulli's equation applies when no external work is added or removed from the fluid; ½ρv² is the kinetic energy density of the fluid itself.
MCQ 4Direct ApplicationPractice
Water flows through a horizontal pipe that narrows from cross-sectional area 4.0 × 10⁻² m² to 2.0 × 10⁻² m². If the speed of water in the wider section is 2.0 m/s, what is the speed in the narrower section?
Show answer and why every option is right or wrong
Answer: C. By the continuity equation A₁v₁ = A₂v₂: v₂ = (4.0 × 10⁻²)(2.0) / (2.0 × 10⁻²) = 4.0 m/s. Area halves, so speed doubles (NCERT Class 11 Physics Chapter 9, page 188).
Why A is wrong: A is wrong. This would be the result if you incorrectly computed v₂ = A₂v₁/A₁ (inverting the ratio), giving half the original speed instead of double.
Why B is wrong: B is wrong. Speed cannot remain the same when the cross-section changes — the continuity equation requires the product Av to stay constant.
Why D is wrong: D is wrong. This would result from squaring the area ratio (factor of 4) instead of using the correct linear ratio (factor of 2) from the continuity equation.
MCQ 5Direct ApplicationPractice
Water flows steadily through a horizontal pipe. At point 1, the pressure is 3.0 × 10⁵ Pa and the speed is 2.0 m/s. At point 2 (same height), the speed is 4.0 m/s. Taking ρ = 1.0 × 10³ kg/m³, the pressure at point 2 is:
Show answer and why every option is right or wrong
Answer: B. Same height, so ρgh cancels. P₂ = P₁ + ½ρ(v₁² − v₂²) = 3.0 × 10⁵ + ½(1.0 × 10³)(4.0 − 16.0) = 3.0 × 10⁵ − 6.0 × 10³ = 2.94 × 10⁵ Pa. Faster flow means lower pressure (NCERT Class 11 Physics Chapter 9, page 188).
Why A is wrong: A is wrong. This result (3.06 × 10⁵ Pa) comes from adding the kinetic energy difference instead of subtracting it — incorrectly assuming pressure increases where speed increases, which contradicts Bernoulli's principle.
Why C is wrong: C is wrong. This value (2.88 × 10⁵ Pa) drops the ½ in ½ρv²: ρ(v₂² − v₁²) = 1.2 × 10⁴ Pa, twice the true pressure drop.
Why D is wrong: D is wrong. This value (2.99 × 10⁵ Pa) comes from using the speeds without squaring them: ½ρ(v₂ − v₁) = ½(1.0 × 10³)(2.0) = 1.0 × 10³ Pa, far too small a drop.
MCQ 6Direct ApplicationPractice
A Venturi meter has a wide section of area 4.0 × 10⁻² m² and a constriction of area 1.0 × 10⁻² m², both at the same height. If the pressure difference between the two sections is 7.5 × 10³ Pa and ρ = 1.0 × 10³ kg/m³, the speed of flow in the wide section is:
Show answer and why every option is right or wrong
Answer: A. Continuity: v₂ = (A₁/A₂)v₁ = 4v₁. Bernoulli (same height): P₁ − P₂ = ½ρ(v₂² − v₁²) = ½ρ(16v₁² − v₁²) = ½ρ(15v₁²). So v₁² = 2(7.5 × 10³)/(1.0 × 10³ × 15) = 1.00 m²/s², giving v₁ = 1.0 m/s exactly (NCERT Class 11 Physics Chapter 9, page 188).
Why B is wrong: B is wrong. 0.71 m/s comes from dropping the ½ in Bernoulli's equation: ρ(15v₁²) = 7.5 × 10³ gives v₁² = 0.50, so v₁ ≈ 0.71 m/s.
Why C is wrong: C is wrong. 2.2 m/s comes from taking the speed ratio as the square root of the area ratio, v₂ = 2v₁: then ½ρ(3v₁²) = 7.5 × 10³ gives v₁ = √5 ≈ 2.2 m/s. Continuity makes speed inversely proportional to area, so v₂ = 4v₁.
Why D is wrong: D is wrong. Getting 4.0 m/s conflates the speed at the constriction (v₂ = 4v₁) with the speed at the wide section. The question asks for the wide-section speed, which is the smaller value.
MCQ 7Concept TrapPractice
An ideal fluid flows through a pipe that rises vertically by height h while maintaining the same cross-sectional area throughout. Compared to the bottom, the pressure at the top is:
Show answer and why every option is right or wrong
Answer: D. Same area means same speed (continuity). Bernoulli's equation gives P₁ + ρgh₁ = P₂ + ρgh₂. Since h₂ > h₁, P₂ = P₁ − ρg(h₂ − h₁) = P₁ − ρgh. Pressure drops by ρgh, identical to static fluid pressure variation (NCERT Class 11 Physics Chapter 9, page 188).
Why A is wrong: A is wrong. Pressure increases with depth, not with height. At a higher elevation, the fluid has more potential energy and correspondingly less pressure energy.
Why B is wrong: B is wrong. The ½ρv² term does not change here because the speed is the same at both points (same cross-section). The pressure difference comes entirely from the ρgh term.
Why C is wrong: C is wrong. Equal area ensures equal speed, but it does not ensure equal pressure. The ρgh term in Bernoulli's equation still applies when height changes — pressure decreases with increasing height even at constant speed.
MCQ 8CalculationPractice
Water (ρ = 1.0 × 10³ kg/m³) flows through a pipe that narrows from area 8.0 × 10⁻² m² to 4.0 × 10⁻² m² while rising 5.0 m. The speed in the wider lower section is 2.0 m/s and the pressure there is 2.0 × 10⁵ Pa. Taking g = 10 m/s², the pressure in the narrower upper section is:
Show answer and why every option is right or wrong
Answer: A. Step 1 — Continuity: v₂ = (8.0 × 10⁻²/4.0 × 10⁻²)(2.0) = 4.0 m/s. Step 2 — Bernoulli with height change: P₂ = P₁ + ½ρ(v₁² − v₂²) − ρgΔh = 2.0 × 10⁵ + ½(10³)(4 − 16) − (10³)(10)(5.0) = 2.0 × 10⁵ − 6.0 × 10³ − 5.0 × 10⁴ = 2.0 × 10⁵ − 5.6 × 10⁴ = 1.44 × 10⁵ Pa (NCERT Class 11 Physics Chapter 9, page 188).
Why B is wrong: B is wrong. This value (1.50 × 10⁵ Pa) ignores the speed change, as if the speed stayed 2.0 m/s, and subtracts only ρgΔh = 5.0 × 10⁴ Pa. Continuity doubles the speed in the narrower section, which costs a further 6.0 × 10³ Pa.
Why C is wrong: C is wrong. This value (2.53 × 10⁵ Pa) results from adding both the kinetic and height terms instead of subtracting them — incorrectly assuming pressure increases at the higher, narrower section. Both increased speed and increased height reduce the pressure.
Why D is wrong: D is wrong. This value (0.94 × 10⁵ Pa) results from doubling the ρgΔh contribution (perhaps using Δh = 10 m instead of 5.0 m or applying an incorrect factor), leading to an overcorrected pressure drop.
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Bernoulli's Principle: quick recall before you leave
How do you solve a Bernoulli's Principle question? A worked example
- 1
Given
A pipe carries water (ρ = 1.00 × 10³ kg/m³) from ground level to a point 3.0 m higher. At the ground-level section, the pipe area is 6.0 × 10⁻² m², the speed is 3.0 m/s, and the gauge pressure is 2.50 × 10⁵ Pa. At the upper section the pipe area is 3.0 × 10⁻² m². Take g = 10 m/s² (exact, problem-defined).
- 2
Required
Find the gauge pressure at the upper section.
- 3
Concept
Bernoulli's equation along a streamline connects pressure, speed, and height. The continuity equation connects speeds at different cross-sections.
- 4
Formula
Continuity: A₁v₁ = A₂v₂
Bernoulli: P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂ - 5
Substitution
From continuity: v₂ = (6.0 × 10⁻² / 3.0 × 10⁻²) × 3.0 = 6.0 m/s
Set h₁ = 0, h₂ = 3.0 m.
P₂ = P₁ + ½ρ(v₁² − v₂²) − ρgΔh
P₂ = 2.50 × 10⁵ + ½(1.00 × 10³)(9.0 − 36.0) − (1.00 × 10³)(10)(3.0) - 6
Calculation
½(1.00 × 10³)(−27.0) = −1.35 × 10⁴ Pa
(1.00 × 10³)(10)(3.0) = 3.00 × 10⁴ Pa
P₂ = 2.50 × 10⁵ − 1.35 × 10⁴ − 3.00 × 10⁴
P₂ = 2.50 × 10⁵ − 4.35 × 10⁴ = 2.065 × 10⁵ Pa
Note: g = 10 m/s² is stated as exact (problem-defined constant) and the integer 2 in the area ratio is a counting ratio. Neither limits significant figures. The given data (three significant figures) governs the final answer. - 7
Final answer
P₂ = 2.07 × 10⁵ Pa (to three significant figures).
The pressure drops because both the increased speed and the increased height drain pressure energy from the fluid. - 8
Common trap
Forgetting the ρgΔh term. If you drop it, you get P₂ = 2.50 × 10⁵ − 1.35 × 10⁴ = 2.365 × 10⁵ Pa — a plausible-looking but wrong answer. This is the distractor NTA builds into the options (per pattern description: "drops ρgh term").
- 9
Similar NEET-style question
Oil (ρ = 8.0 × 10² kg/m³) flows through a horizontal pipe that widens from area 2.0 × 10⁻² m² to 8.0 × 10⁻² m². If the speed in the narrow section is 8.0 m/s and the pressure there is 1.0 × 10⁵ Pa, find the pressure in the wide section. (Answer: Apply continuity to get v₂ = 2.0 m/s, then Bernoulli at constant height gives P₂ = 1.0 × 10⁵ + ½(800)(64 − 4) = 1.0 × 10⁵ + 2.4 × 10⁴ = 1.24 × 10⁵ Pa.)
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What to remember before solving Bernoulli's Principle questions
Which Bernoulli's Principle formulas do you need for NEET?
1 formula — click to collapse
Bernoulli's equation
Conservation of energy along a streamline of incompressible non-viscous flow.
| Symbol | Quantity | SI Unit |
|---|---|---|
| P | pressure | Pa |
| rho | density | kg/m^3 |
| v | speed | m/s |
| g | gravity | m/s^2 |
| h | height | m |
Valid when
- Steady, non-viscous, incompressible flow
- Along a single streamline
- No work added/removed
More in Properties of Bulk Matter: 5 exam traps and mistakes · 11 formulas · 4 question patterns from its other lessons.
Bernoulli's Principle questions from past NEET papers
2 questions from NEET 2023, 2026. Answers verified against NTA official keys. — click to collapse
NEET 2026
1100
2200
3300
4400
NTA Answer: Option 4(final)
NEET 2023
1Bernoulli’s principle
2The principle of parallel axes
3The principle of perpendicular axes
4Huygen’s principle
NTA Answer: Option 1(final)
All 17 past-paper questions from Properties of Bulk Matter →
How does NEET ask about Bernoulli's Principle?
1 recurring pattern from past papers — click to collapse
Apply Bernoulli's equation to flow through pipes of varying cross-section / heights / Venturi-like geometries.
InterpretationMedium
Common distractors
forgets height term
Drops rho*g*h term
equates pressures incorrectly
Picks wrong reference points
Sources
NCERT refs: Class 11 Physics Chapter 9, p.188
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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