PhysicsISC Class 11 · Mechanical Properties of Fluids

Equation of Continuity and Bernoulli's Theorem with Applications

Updated 17 October 2026 · Fact-checked

For steady flow of an incompressible fluid, A₁v₁ = A₂v₂ (continuity). Bernoulli's theorem adds that p + ½ρv² + ρgh is constant along a streamline: where the speed is higher, the pressure is lower. Applications include the speed of efflux v = √(2gh), the Venturi tube and the lift on a wing.

In the narrow section the water moves faster (A₁v₁ = A₂v₂) and the pressure is lower, so the liquid column there stands lower.

Illustration drawn for QuestionBuddy.

Understand Continuity Equation and Bernoulli's Theorem

Streamline flow and continuity. In steady (streamline) flow each particle follows the same path as the one before it, and the velocity at a point does not change with time. For an incompressible fluid, the mass entering a pipe per second equals the mass leaving it: A₁v₁ = A₂v₂ = volume flow rate. A narrower section means faster flow.

Bernoulli's theorem. For a non-viscous, incompressible fluid in steady flow, p + ½ρv² + ρgh = constant along a streamline. The three terms are the pressure energy, kinetic energy and potential energy per unit volume. It follows from the work-energy theorem applied to a fluid element. Between two points: p₁ + ½ρv₁² + ρgh₁ = p₂ + ½ρv₂² + ρgh₂.

Speed of efflux (Torricelli). For a small hole at depth h below the open surface of a tank, both the surface and the jet are at atmospheric pressure, and the surface speed is negligible, so v = √(2gh). The jet leaves horizontally and lands at a horizontal distance v √(2H/g) when the hole is H above the ground.

Venturi tube. In a horizontal pipe with a constriction, A₁v₁ = A₂v₂ and p₁ − p₂ = ½ρ(v₂² − v₁²). The measured pressure difference gives the flow speed.

Dynamic lift. Air moving faster over the curved top of a wing creates lower pressure than the pressure below it, giving a net upward force. A spinning ball curves for a similar reason. Sprayers, carburettors and the bunsen burner also depend on low pressure where the speed is high.

Limits. The theorem ignores viscosity and assumes steady, incompressible flow along one streamline.

Continuity Equation and Bernoulli's Theorem: formulas and key facts

Continuity: $A_{1} v_{1} = A_{2} v_{2}$; Volume flow rate Q = Av.
Bernoulli: $p + ½ ρ v^{2} + ρ g h = co n s t an t$; Along a streamline in ideal flow.
Speed of efflux: $v = \sqrt (2 g h)$; h = depth of the hole below the free surface.
Venturi tube: $p_{1} - p_{2} = ½ ρ (v_{2}^{2} - v_{1}^{2})$; Horizontal pipe, so no height term.

Continuity Equation and Bernoulli's Theorem in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

A garden hose nozzle

Suppose water moves at 0.5 m s⁻¹ in a hose of internal diameter 2 cm, and the nozzle has a diameter 0.5 cm.

Find the exit speed, the flow rate, the height of a vertical jet (g = 10 m s⁻²) and the time to fill a 20 L bucket.

Try it first, then show the solutionSolution

Area ratio = (2/0.5)² = 16, so v₂ = 0.5 × 16 = 8 m s⁻¹.
Q = A₁v₁ = π(0.01)² × 0.5 = 1.57 × 10⁻⁴ m³ s⁻¹ = 0.157 L s⁻¹.
Jet height h = v²/2g = 64/20 = 3.2 m; time = 20/0.157 = 127 s.

Answer: 8 m s⁻¹; 0.157 L s⁻¹; 3.2 m; about 2.1 minutes.

Memory hook: Squeezing the nozzle raises the speed, not the flow rate.

Tricks to remember Continuity Equation and Bernoulli's Theorem

Short memory aids. Say each one aloud once, then use it on a question the same day.

Narrow means fast, fast means low pressure
Continuity gives speed, Bernoulli gives pressure.
Use it when: Qualitative questions on pipes and wings.
Torricelli is free fall
v = √(2gh) is the speed a body would reach falling through h.
Use it when: Remembering the speed of efflux.

Continuity Equation and Bernoulli's Theorem: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Equation of continuity · 1 mark · MCQBoard level

The equation of continuity expresses the conservation of:
(a) energy
(b) mass
(c) momentum
(d) angular momentum

Try it first, then show the solutionSolution

For incompressible flow the mass entering equals the mass leaving per second: ρA₁v₁ = ρA₂v₂.

Answer: (b) mass

Example 2: Pipe of varying section · 3 marks · Short answerBoard level

Water flows at 1 m s⁻¹ through a pipe of radius 4 cm, which narrows to a radius of 2 cm. Find the speed in the narrow part and the volume flow rate.

Try it first, then show the solutionSolution

A₁/A₂ = (4/2)² = 4. (1 mark)
v₂ = 4 × 1 = 4 m s⁻¹. (1 mark)
Q = A₁v₁ = π(0.04)² × 1 = 5.03 × 10⁻³ m³ s⁻¹ ≈ 5.0 L s⁻¹. (1 mark)

Answer: v₂ = 4 m s⁻¹; Q = 5.0 L s⁻¹.

Example 3: Jet from a tank · 4 marks · Short answerHarder board level

Water stands in a large open tank. A small hole is 1.8 m below the surface and 3.2 m above the ground (g = 10 m s⁻²). Find the speed of the jet and the distance from the tank at which it lands.

Try it first, then show the solutionSolution

v = √(2gh) = √(2 × 10 × 1.8) = 6 m s⁻¹. (1 mark)
Fall time t = √(2H/g) = √(2 × 3.2/10) = 0.8 s. (1 mark)
Range = vt = 6 × 0.8. (1 mark)
= 4.8 m. (1 mark)

Answer: v = 6 m s⁻¹; range 4.8 m.

Example 4: Venturi tube · 4 marks · Short answerHarder board level

A horizontal Venturi pipe has wide and narrow sections of 40 cm² and 10 cm². The pressure difference between them is 12 kPa and the fluid is water (ρ = 1000 kg m⁻³). Find the speed in the wide part and the flow rate.

Try it first, then show the solutionSolution

v₂ = 4v₁ (area ratio 4). (1 mark)
p₁ − p₂ = ½ρ(v₂² − v₁²) = 500 × 15 v₁² = 7500 v₁². (1 mark)
12000 = 7500 v₁² → v₁² = 1.6, v₁ = 1.265 m s⁻¹. (1 mark)
Q = A₁v₁ = 40 × 10⁻⁴ × 1.265 = 5.06 × 10⁻³ m³ s⁻¹. (1 mark)

Answer: v₁ = 1.27 m s⁻¹; Q = 5.1 L s⁻¹.

Example 5: Pressure on an upper floor · 5 marks · Long answerHOTS / competency

In a building, water of density 1000 kg m⁻³ flows at 2 m s⁻¹ in a ground-floor pipe of diameter 4 cm at a gauge pressure of 3.0 × 10⁵ Pa. A pipe of diameter 2 cm takes it to a point 16 m higher (g = 10 m s⁻²). Find the speed and the gauge pressure at the upper point.

Try it first, then show the solutionSolution

v₂ = v₁ (d₁/d₂)² = 2 × 4 = 8 m s⁻¹. (1 mark)
p₂ = p₁ + ½ρ(v₁² − v₂²) − ρg(h₂ − h₁). (1 mark)
½ρ(v₁² − v₂²) = 500 × (4 − 64) = −30 000 Pa. (1 mark)
ρgh = 1000 × 10 × 16 = 160 000 Pa. (1 mark)
p₂ = 300 000 − 30 000 − 160 000 = 1.10 × 10⁵ Pa (gauge). (1 mark)

Answer: v₂ = 8 m s⁻¹; p₂ = 1.1 × 10⁵ Pa (gauge).

Where students lose marks in Continuity Equation and Bernoulli's Theorem

The slips examiners see most often, and what to write instead.

Mistake: Using the wrong area ratio
Using the ratio of diameters instead of the ratio of areas.
Do this: Speed ratio = inverse area ratio = (d₁/d₂)².
Usually costs: 1 to 2 marks.
Mistake: Dropping the height term in a vertical pipe
Writing p₁ + ½ρv₁² = p₂ + ½ρv₂² for pipes at different heights.
Do this: Include ρgh whenever the heights differ.
Usually costs: 1 to 2 marks.
Mistake: Mixing gauge and absolute pressure
Adding atmospheric pressure in some terms only.
Do this: Use either gauge or absolute pressure at both points.
Usually costs: 1 mark.

Exam tips

Write continuity first to find the speeds, then Bernoulli for pressure.
State the assumptions: steady, incompressible, non-viscous flow.
For tanks, mention that the top surface speed is neglected and both pressures are atmospheric.
Give the volume flow rate in m³ s⁻¹ or L s⁻¹.

Practice questions on Continuity Equation and Bernoulli's Theorem

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The lift on an aeroplane wing is mainly due to:
(a) higher pressure above the wing
(b) lower pressure above the wing than below
(c) the weight of the air
(d) the thrust of the engine
Hint
Air moves faster over the curved top surface.
Answer
(b) lower pressure above the wing than below
Q2 · 2 marks · Short answerBoard level
Water moves at 2 m s⁻¹ in a pipe of cross-section 5 cm². Find its speed where the section is 2 cm².
Hint
Use A₁v₁ = A₂v₂.
Answer
v₂ = 5 × 2/2 = 5 m s⁻¹.
Q3 · 3 marks · Short answerHarder board level
A horizontal pipe of diameter 10 cm carries water at 1.5 m s⁻¹. Find the flow rate in litres per second and the volume delivered in 10 minutes.
Hint
Q = Av.
Answer
Q = π(0.05)² × 1.5 = 1.178 × 10⁻² m³ s⁻¹ = 11.8 L s⁻¹. Volume = 11.8 × 600 = 7068 L ≈ 7.07 m³.
Q4 · 3 marks · Short answerHarder board level
A storm wind blows at 25 m s⁻¹ over a flat roof of area 80 m² while the air inside is still. Taking the air density as 1.2 kg m⁻³, find the pressure difference and the net upward force on the roof.
Hint
Δp = ½ρv² (same height).
Answer
Δp = ½ × 1.2 × 625 = 375 Pa. F = 375 × 80 = 30 000 N.
Q5 · 3 marks · Short answerHarder board level
A syringe has a piston of area 2 cm². A force of 4 N is applied and the fluid (density 1000 kg m⁻³) squirts out of a fine nozzle. Estimate the speed of the jet, ignoring the piston's own speed and gravity.
Hint
p = F/A, then ½ρv² = p.
Answer
p = 4/(2 × 10⁻⁴) = 2 × 10⁴ Pa. v = √(2p/ρ) = √40 = 6.3 m s⁻¹.
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a leak develops in a wide open-topped drum of water, 3.2 m below the water level (g = 10 m s⁻²).
(i) How fast does the water shoot out of the leak? (1 mark)
(ii) If the leak has an area of 1.0 cm², find the volume leaving per second. (1 mark)
(iii) The leak is at 12.8 m below the level instead. By what factor does the speed change? (1 mark)
(iv) Which hole gives the longer jet if the tank is on the ground and the holes are at heights 6 m and 2 m above the ground for the same water depth 8 m? (1 mark)
Hint
v = √(2gh) and the range depends on v and the fall time.
Answer
(i) √(2 × 10 × 3.2) = √64 = 8 m s⁻¹. (ii) 10⁻⁴ × 8 = 8 × 10⁻⁴ m³ s⁻¹. (iii) √(12.8/3.2) = 2 times. (iv) Hole at height 6 m: depth 2 m, v = 6.32 m s⁻¹, t = √(1.2) = 1.095 s, range 6.9 m. Hole at 2 m: depth 6 m, v = 10.95 m s⁻¹, t = 0.632 s, range 6.9 m. They give equal ranges (a hole at depth h and at height H − h reach the same distance).

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.State the equation of continuity.Show answer
A₁v₁ = A₂v₂ for incompressible steady flow.
2.Bernoulli's equation?Show answer
p + ½ρv² + ρgh = constant.
3.Speed of efflux?Show answer
√(2gh).
4.Where is the pressure low in a Venturi pipe?Show answer
In the narrow section, where the speed is high.

Continuity Equation and Bernoulli's Theorem: frequently asked questions

Does Bernoulli's theorem violate energy conservation?

No. It is the work-energy theorem for a fluid: the work done by pressure differences changes the kinetic and potential energy of the fluid.

Why does water speed up in a narrow pipe?

The same volume per second must pass each cross-section, so a smaller area needs a higher speed.

Does viscosity matter?

In real fluids viscosity causes energy loss, so the pressure falls along a pipe even if the section is constant; Bernoulli's theorem is for the ideal case.