PhysicsCBSE Class 11 · Mechanical Properties of Fluids

Equation of Continuity and Bernoulli's Principle Class 11

Updated 17 October 2026 · Fact-checked

For steady flow of an incompressible fluid, A₁v₁ = A₂v₂ (continuity). Bernoulli's principle states P + ½ρv² + ρgh is constant along a streamline, an energy conservation law per unit volume. It gives Torricelli's speed of efflux v = √(2gh), the pressure drop where a fluid speeds up, and the dynamic lift on a wing.

Where the pipe narrows, the fluid speeds up (A₁v₁ = A₂v₂) and the pressure falls, so the tube there shows a lower liquid level.

Illustration drawn for QuestionBuddy.

Understand Equation of Continuity and Bernoulli's Principle

Pressure at depth. In a fluid at rest, $P = P_{a} + ρ g h$ ; the extra part $ρ g h$ is the gauge pressure. Pressure is transmitted equally (Pascal's law).

Equation of continuity. For the steady, streamline flow of an incompressible fluid, the volume rate of flow is the same at every cross-section: $A_{1} v_{1} = A_{2} v_{2}$ . Where the pipe narrows, the fluid speeds up.

Bernoulli's principle. For a non-viscous, incompressible fluid in steady flow, along a streamline
$ $P + \frac{1}{2} ρ v^{2} + ρ g h = constant .$ $
It is the work-energy theorem for the fluid: pressure energy, kinetic energy and potential energy per unit volume. For a horizontal flow, $P_{1} - P_{2} = \frac{1}{2} ρ (v_{2}^{2} - v_{1}^{2})$ : where speed is greater, pressure is lower.

Applications.

*Speed of efflux (Torricelli):* from a hole at depth $h$ below the open surface of a wide tank, $v = 2 g h$ (the speed of free fall through $h$ ). If the tank has gauge pressure $Δ P$ above the liquid, $v = 2 (Δ P / ρ + g h)$ .
*Range of the jet:* a hole at depth $h$ and height $H$ above the ground gives a horizontal range $x = v 2 H / g = 2 h H$ . Holes at depths $h$ and $H$ (swapped) give equal ranges; the range is greatest when the hole is halfway down.
*Venturimeter:* in a pipe with throat, $P_{1} - P_{2} = \frac{1}{2} ρ (v_{2}^{2} - v_{1}^{2})$ with $A_{1} v_{1} = A_{2} v_{2}$ gives the flow speed from a pressure difference.
*Dynamic lift:* an aerofoil makes air move faster above it than below; the pressure difference times the wing area is the lift. Spinning balls curve for a similar reason.
Two ships or buses passing close are drawn together because the fluid between them moves faster, so has lower pressure.

Limits. Bernoulli's principle neglects viscosity and turbulence; it applies only along a streamline in steady flow.

Equation of Continuity and Bernoulli's Principle: formulas and key facts

Pressure at depth: $P = P_{a} + ρ g h$; Static fluid.
Continuity: $A_{1} v_{1} = A_{2} v_{2}$; Incompressible, steady flow.
Bernoulli: $P + \frac{1}{2} ρ v^{2} + ρ g h = constant$; Along a streamline.
Efflux speed: $v = 2 g h$; Torricelli, open tank.
Range of a jet: $x = 2 h H$; Depth h, height H above ground.
Pressure drop (horizontal): $P_{1} - P_{2} = \frac{1}{2} ρ (v_{2}^{2} - v_{1}^{2})$; Venturi and lift.

How to solve Equation of Continuity and Bernoulli's Principle questions

Use continuity to find the new speed, then Bernoulli to find the pressure change between two points on the same streamline.

1Draw the streamline and mark the two points of interest.
2Use A₁v₁ = A₂v₂ to find the unknown speed.
3Write Bernoulli's equation between the points; cancel the ρgh terms if the flow is horizontal.
4For a tank hole at atmospheric pressure, take v at the top as zero and P equal at both ends.
5Substitute SI units (area in m², pressure in Pa).
6For jets, find the speed first, then the time of fall, then the range.

Equation of Continuity and Bernoulli's Principle 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 a hose of cross-section 4.0 cm² carries water at 1.5 m/s and ends in a nozzle of cross-section 1.0 cm².

Find the speed of the water leaving the nozzle.

Try it first, then show the solutionSolution

Q = A₁v₁ = 4.0 × 10⁻⁴ × 1.5 = 6.0 × 10⁻⁴ m³/s.
v₂ = Q/A₂ = 6.0 × 10⁻⁴/1.0 × 10⁻⁴ = 6.0 m/s.

Answer: 6.0 m/s.

Memory hook: A narrower exit gives a faster jet.

A leaking water tank

Suppose a rooftop tank has a small hole 1.25 m below the water surface.

Find the speed of the escaping water (g = 10 m/s²).

Try it first, then show the solutionSolution

v = √(2gh) = √(2 × 10 × 1.25) = √25 = 5 m/s.

Answer: 5 m/s.

Memory hook: Same speed as a body falling through h.

Tricks to remember Equation of Continuity and Bernoulli's Principle

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

Narrow means fast
A₁v₁ = A₂v₂; the narrower section has the larger speed.
Use it when: Continuity questions.
Fast means low pressure
Bernoulli: increase in v means decrease in P.
Use it when: Qualitative questions.
2 root hH
Range = 2√(hH); swap h and H for equal range.
Use it when: Jet problems.
Free-fall speed
v = √(2gh) is the speed of free fall through h.
Use it when: Torricelli.

Equation of Continuity and Bernoulli's Principle: solved examples

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

Example 1: Continuity in a pipe · 2 marks · Very short answerBoard level

Water flows at 0.5 m/s through a pipe of cross-section 8 cm². Find the volume rate of flow and the speed in a narrower part of the pipe of cross-section 2 cm².

Try it first, then show the solutionSolution

Q = Av = 8 × 10⁻⁴ × 0.5 = 4 × 10⁻⁴ m³/s (0.4 L/s). (1)
v₂ = Q/A₂ = 4 × 10⁻⁴/2 × 10⁻⁴ = 2 m/s. (1)

Answer: 0.4 L/s; 2 m/s.

Example 2: Pressure drop in a horizontal pipe · 3 marks · Short answerBoard level

Water (1000 kg/m³) flows through a horizontal pipe. The speed rises from 1 m/s to 4 m/s as the pipe narrows, and the pressure at the wide part is 1.5 × 10⁵ Pa. Find the pressure at the narrow part.

Try it first, then show the solutionSolution

Horizontal, so P₁ + ½ρv₁² = P₂ + ½ρv₂². (1)
P₁ − P₂ = ½ × 1000 × (16 − 1) = 7500 Pa. (1)
P₂ = 1.5 × 10⁵ − 7.5 × 10³ = 1.425 × 10⁵ Pa. (1)

Answer: 1.425 × 10⁵ Pa.

Example 3: Venturimeter · 3 marks · Short answerHarder board level

A venturimeter has a wide section of area 20 cm² and a throat of area 5 cm². The pressure difference between them is 1.5 × 10⁴ Pa. Find the flow speed in the wide section and the volume rate of flow of water.

Try it first, then show the solutionSolution

v₂ = 4v₁ (area ratio 4). ΔP = ½ρ(16v₁² − v₁²) = 7500 v₁². (1)
v₁² = 1.5 × 10⁴/7500 = 2, so v₁ = 1.41 m/s. (1)
Q = A₁v₁ = 20 × 10⁻⁴ × 1.41 = 2.8 × 10⁻³ m³/s. (1)

Answer: 1.41 m/s; 2.8 × 10⁻³ m³/s.

Example 4: Range of a jet from a tank · 4 marks · Long answerHarder board level

A tall tank stands on level ground and water stands 2.6 m above the ground. A small hole A is made 0.8 m below the surface. (a) Find the speed of the water leaving A and its range on the ground. (b) Show that a hole B at 1.8 m below the surface gives the same range. (c) At what depth is the range greatest and what is it? (g = 10 m/s²)

Try it first, then show the solutionSolution

(a) v = √(2 × 10 × 0.8) = 4 m/s; height of A above ground = 1.8 m, so t = √(2 × 1.8/10) = 0.6 s and range = 4 × 0.6 = 2.4 m. (1)
(b) For B: h = 1.8 m, H = 0.8 m, range = 2√(hH) = 2√(1.8 × 0.8) = 2.4 m, the same. (1)
(c) Range is greatest when h = H = 1.3 m (halfway down). (1)
R_max = 2√(1.3 × 1.3) = 2.6 m. (1)

Answer: (a) 4 m/s, 2.4 m; (b) shown; (c) 1.3 m below the surface, 2.6 m.

Example 5: Lift on a wing · 4 marks · Long answerHOTS / competency

Air flows over a wing so that its speed is 70 m/s over the upper surface and 60 m/s under the lower surface (density of air 1.2 kg/m³). The wing area is 20 m². Find the pressure difference, the lift force and the mass the lift can support (g = 10 m/s²). Explain why the pressure is lower on top.

Try it first, then show the solutionSolution

ΔP = ½ρ(v_top² − v_bottom²) = ½ × 1.2 × (4900 − 3600) = 780 Pa. (1)
Lift = ΔP × A = 780 × 20 = 15600 N. (1)
Mass = 15600/10 = 1560 kg. (1)
Faster flow over the upper surface means lower pressure there (Bernoulli), so the net force on the wing acts upward. (1)

Answer: 780 Pa; 15600 N; 1560 kg.

Where students lose marks in Equation of Continuity and Bernoulli's Principle

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

Mistake: Using the diameter ratio as the area ratio.
Forgetting to square.
Do this: A ∝ d², so halving d divides A by 4.
Usually costs: The answer mark.
Mistake: Dropping ρgh terms for a vertical pipe.
Treating all flow as horizontal.
Do this: Keep ρgh when the heights differ.
Usually costs: Method mark.
Mistake: Using Torricelli's v = √(2gh) with h measured from the bottom.
Wrong reference point.
Do this: h is the depth below the free surface.
Usually costs: The answer mark.
Mistake: Forgetting to convert cm² to m².
Unit slip.
Do this: 1 cm² = 10⁻⁴ m².
Usually costs: The answer mark.

Exam tips

State continuity and Bernoulli with their conditions: steady, incompressible, non-viscous, along a streamline.
Draw the streamline and name the two points.
Give the sentence "greater speed, lower pressure" for qualitative items.
For jets, split the answer into speed, time of fall and range.
Quote densities and g in the first line.

Practice questions on Equation of Continuity and Bernoulli's Principle

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 · MCQHarder board level
A horizontal pipe narrows to half its diameter. The kinetic energy per unit volume of the water in the narrow part is greater by a factor of
(a) 2 (b) 4 (c) 8 (d) 16
Hint
Area falls to 1/4, so speed rises to 4 times.
Answer
(d) 16.
Q2 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Two ships sailing close and parallel to each other tend to be drawn together.
Reason (R): The water between them moves faster, so its pressure is lower than on the outer sides.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Bernoulli.
Answer
(a).
Q3 · 2 marks · Very short answerBoard level
A tap fills a 10 litre bucket in 40 s. Find the flow rate and the speed of water in a pipe of cross-section 1 cm² supplying the tap.
Hint
Q = V/t.
Answer
Q = 10/40 = 0.25 L/s = 2.5 × 10⁻⁴ m³/s; v = Q/A = 2.5 × 10⁻⁴/10⁻⁴ = 2.5 m/s.
Q4 · 3 marks · Short answerHarder board level
Water stands 1.8 m deep in a closed tank whose air above the water is at an extra gauge pressure of 2.0 × 10⁵ Pa. Find the speed of water leaving a small hole at the bottom (ρ = 1000 kg/m³, g = 10 m/s²).
Hint
v = √(2(ΔP/ρ + gh)).
Answer
v = √(2(200 + 18)) = √436 ≈ 20.9 m/s. (Without the extra pressure it would be 6 m/s.)
Q5 · 4 marks · Long answerHOTS / competency
A wide open tank of cross-section 1.0 m² has a small hole of area 1.0 cm² at its base. The water is initially 1.8 m deep (g = 10 m/s²). (a) Find the initial speed and rate of outflow. (b) Show that the time to drain completely is t = (A/a)√(2H/g) and find it.
Hint
dh/dt = −(a/A)√(2gh).
Answer
(a) v = √(2 × 10 × 1.8) = 6 m/s; Q = 6 × 10⁻⁴ m³/s. (b) A dh/dt = −a√(2gh); integrating, √H − √h = (a/A)√(g/2) t, so t = (A/a)√(2H/g) = 10⁴ × √0.36 = 6000 s ≈ 100 min.

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.Equation of continuity?Show answer
A₁v₁ = A₂v₂.
2.Bernoulli's equation?Show answer
P + ½ρv² + ρgh = constant.
3.Speed of efflux?Show answer
√(2gh).
4.Range of a jet?Show answer
2√(hH).
5.Cause of dynamic lift?Show answer
Pressure difference from different air speeds.
6.Gauge pressure at depth h?Show answer
ρgh.

Equation of Continuity and Bernoulli's Principle: frequently asked questions

Why does water speed up in a narrow pipe?

The same volume must pass each second, so a smaller area needs a greater speed.

Is Bernoulli's principle a conservation law?

Yes, it expresses conservation of energy for the flowing fluid.

Why does a roof lift off in a storm?

Fast wind over the roof lowers the pressure outside, so the higher pressure inside pushes the roof up.

Does Torricelli's result depend on the liquid?

No; the density cancels, so the speed depends only on the depth.

When does Bernoulli's principle fail?

For viscous, turbulent or compressible flow, or between points not on the same streamline.