Physics · Mechanical Properties of Fluids

Bernoulli's Principle

Bernoulli's principle is the conservation of energy written for a flowing ideal fluid. NCERT §9.4 derives, from the work–energy theorem, that along a streamline the sum of pressure, kinetic energy per unit volume and potential energy per unit volume is constant: $P + \tfrac{1}{2}\rho v^2 + \rho g h = \text{constant}$. From that single line follow Torricelli's law of efflux, the Venturi-meter, and the dynamic lift on an aircraft wing or a spinning cricket ball. This deep-dive derives the equation, fixes its assumptions, and walks through every application NEET tests.

The Bernoulli equation

An incompressible, non-viscous fluid flows steadily through a pipe of varying cross-section that also varies in height. Its speed must change as the area changes, by the equation of continuity. A change in speed needs a net force, and that force comes from a difference in pressure between regions of the pipe. Bernoulli's equation is the general relation that ties these three quantities — pressure, speed and height — together for any two points on a streamline.

Bernoulli's principle (NCERT §9.4): As we move along a streamline, the sum of the pressure $P$, the kinetic energy per unit volume $\tfrac{1}{2}\rho v^2$, and the potential energy per unit volume $\rho g h$ remains a constant.

Written between two points 1 and 2 on the same streamline, and in its general form:

$$P_1 + \tfrac{1}{2}\rho v_1^2 + \rho g h_1 = P_2 + \tfrac{1}{2}\rho v_2^2 + \rho g h_2$$ $$P + \tfrac{1}{2}\rho v^2 + \rho g h = \text{constant}$$

Each term has the dimensions of pressure (energy per unit volume). $P$ is the pressure energy density, $\tfrac{1}{2}\rho v^2$ the kinetic energy density, and $\rho g h$ the gravitational potential energy density. The Swiss physicist Daniel Bernoulli established this relation in 1738.

Ideal fluid in a pipe of varying cross-section and height $h_1$ $h_2$ 1 $A_1,\,v_1,\,P_1$ 2 $A_2,\,v_2,\,P_2$ wide & low: slow, high pressure → narrow & high: fast, low pressure

Figure 1. Steady flow of an ideal fluid through a pipe of changing area and height. Where the bore narrows the fluid speeds up and its pressure drops; where it rises, potential energy increases. Bernoulli's equation keeps the three energy densities in balance.

Derivation from energy conservation

Take the fluid initially between sections B and D in Figure 1. In a small interval $\Delta t$ the fluid at B (speed $v_1$, area $A_1$, pressure $P_1$) moves a distance $v_1\Delta t$ to C; the fluid at D (speed $v_2$, area $A_2$, pressure $P_2$) moves $v_2\Delta t$ to E. By continuity the same volume $\Delta V$ crosses each section. The work done on the fluid at the left face is $W_1 = P_1 A_1 (v_1\Delta t) = P_1\Delta V$, and the work done by the fluid against the pressure at the right face is $W_2 = P_2 A_2 (v_2\Delta t) = P_2\Delta V$. The net work done on the fluid is therefore $(P_1 - P_2)\Delta V$.

This work changes the kinetic and potential energies of the displaced volume. With $\Delta m = \rho\,\Delta V$, the changes are $\Delta K = \tfrac{1}{2}\rho\,\Delta V\,(v_2^2 - v_1^2)$ and $\Delta U = \rho g\,\Delta V\,(h_2 - h_1)$. The work–energy theorem gives:

$$(P_1 - P_2)\,\Delta V = \tfrac{1}{2}\rho\,\Delta V\,(v_2^2 - v_1^2) + \rho g\,\Delta V\,(h_2 - h_1)$$

Dividing every term by $\Delta V$ and rearranging recovers the Bernoulli equation directly:

$$P_1 + \tfrac{1}{2}\rho v_1^2 + \rho g h_1 = P_2 + \tfrac{1}{2}\rho v_2^2 + \rho g h_2$$

For a fluid at rest, $v = 0$ everywhere and the equation collapses to $P_1 + \rho g h_1 = P_2 + \rho g h_2$, i.e. $P_1 - P_2 = \rho g (h_2 - h_1)$ — the hydrostatic pressure relation. Bernoulli's equation is thus the dynamic generalisation of the static pressure law you met under pressure in fluids.

Assumptions and limitations

Because the derivation used the work–energy theorem with no dissipative term, the equation holds only for an idealised fluid. NEET routinely tests whether you know the four conditions and where they break.

Assumption 1

Incompressible

Density $\rho$ is constant. The elastic energy stored in compression is not accounted for, so the relation fails for highly compressible gases at large pressure changes.

Assumption 2

Non-viscous

Zero viscosity. Real fluids lose kinetic energy to internal friction between layers, which becomes heat. So $P_2$ is in practice lower than Bernoulli predicts.

Assumption 3

Steady (streamline)

Velocity and pressure at each point are constant in time. The equation does not hold for turbulent or non-steady flow, where both fluctuate.

Assumption 4

Along one streamline

Points 1 and 2 must lie on the same streamline. The constant generally differs from one streamline to another.

In short, Bernoulli's equation is exact only for an ideal fluid and a good approximation for low-viscosity, incompressible, smoothly flowing real fluids. For a turbulent rapid in a river, or for blood squeezing through a constricted artery where viscosity dominates, it cannot be applied without correction.

The pressure–velocity inverse relation

Restrict the streamline to a horizontal pipe, so $h_1 = h_2$ and the $\rho g h$ term drops out. Bernoulli's equation reduces to its most-tested form:

$$P + \tfrac{1}{2}\rho v^2 = \text{constant}$$

The two terms trade against each other. Where the fluid moves faster, the kinetic-energy density $\tfrac{1}{2}\rho v^2$ is larger, so the pressure $P$ must be smaller. This is the single idea behind every Bernoulli application: high speed means low pressure, and low speed means high pressure. Combine it with continuity ($Av = \text{constant}$, so the fluid speeds up where the pipe narrows) and you can predict the pressure everywhere in a flow at a glance.

i
Prerequisite

Continuity sets the speed; Bernoulli sets the pressure. Revise the link first in streamline flow & the equation of continuity.

Torricelli's law of efflux

The word efflux means fluid outflow. Torricelli discovered that the speed at which liquid spurts from an open tank equals the speed of a freely falling body. Consider a tank of liquid of density $\rho$ with a small hole in its side at height $y_1$, the free surface being at height $y_2$, and air above the surface at pressure $P$. Let $A_1$ be the hole area and $A_2$ the tank's cross-section.

Torricelli efflux from a tank 2 (surface, $v_2\approx 0,\ P$) 1 (hole, $P_a$) $h = y_2-y_1$ $v_1=\sqrt{2gh}$

Figure 2. Efflux from a tank open to the atmosphere. With $A_2 \gg A_1$ the surface is essentially at rest, and Bernoulli between the surface and the hole gives $v_1 = \sqrt{2gh}$ — the free-fall speed.

By the equation of continuity $v_1 A_1 = v_2 A_2$. If the tank's area is much larger than the hole, $A_2 \gg A_1$, the surface descends so slowly that we take $v_2 \approx 0$. Applying Bernoulli between the surface (point 2) and the hole (point 1), where the pressure at the hole is atmospheric $P_a$:

$$P_a + \tfrac{1}{2}\rho v_1^2 + \rho g y_1 = P + \rho g y_2$$

Writing $h = y_2 - y_1$ and solving for the efflux speed:

$$v_1 = \sqrt{\,\dfrac{2(P - P_a)}{\rho} + 2gh\,}$$

Two limiting cases follow. When the tank is sealed and the air pressure is very large, $P \gg P_a$ and $2gh$ is negligible, so $v_1 \approx \sqrt{2(P-P_a)/\rho}$ — the efflux is set by the container pressure, as in rocket propulsion. When the tank is open to the atmosphere, $P = P_a$ and the pressure terms cancel, leaving Torricelli's law:

$$\boxed{\,v_1 = \sqrt{2gh}\,}$$

This is exactly the speed a body acquires falling freely through height $h$. Notice that for an open tank the density $\rho$ has cancelled out entirely — the efflux speed is independent of which liquid is in the tank.

The Venturi-meter

A Venturi-meter measures the rate of flow of a liquid through a pipe. It is a horizontal tube with a deliberate constriction; a manometer connects the wide section (area $A_1$, speed $v_1$, pressure $P_1$) to the narrow throat (area $A_2$, speed $v_2$, pressure $P_2$). Because the pipe is horizontal, the height term vanishes and only pressure and kinetic energy trade.

Venturi-meter with manometer $A_1,\,v_1,\,P_1$ $A_2,\,v_2,\,P_2$ $h$ narrow throat → faster → lower pressure

Figure 3. A Venturi-meter. The throat speeds the fluid up (continuity) and so lowers its pressure (Bernoulli). The manometer level difference $h$ measures $P_1 - P_2$, from which the flow speed and volume rate follow.

Continuity gives $A_1 v_1 = A_2 v_2$, so $v_2 > v_1$ at the throat. Bernoulli for the horizontal tube gives:

$$P_1 - P_2 = \tfrac{1}{2}\rho\,(v_2^2 - v_1^2)$$

Eliminating $v_2$ using continuity and reading $P_1 - P_2 = \rho_m g h$ from the manometer (level difference $h$) lets us solve for the inlet speed:

$$v_1 = \sqrt{\dfrac{2(P_1 - P_2)}{\rho\left[(A_1/A_2)^2 - 1\right]}}$$

Since every term except $h$ is fixed for a given meter, $v_1 \propto \sqrt{h}$, and the volume flow rate is $Q = A_1 v_1 = K'\sqrt{h}$. The pressure is least where the bore is narrowest — Venturi's principle — which is precisely the pressure–velocity inverse relation in instrument form. NEET 2023 asked the bare fact that the Venturi-meter works on Bernoulli's principle.

Dynamic lift — aerofoil and Magnus effect

Dynamic lift is the force on a body — an aircraft wing, a hydrofoil, a spinning ball — arising from its motion through a fluid. It is the most spectacular consequence of the pressure–velocity relation, and NCERT §9.4.2 treats three cases.

Case (i)

Ball without spin

Streamlines around a non-spinning ball are symmetric top and bottom. The air speed at corresponding points is equal, so there is zero pressure difference and no upward or downward force.

Case (ii)

Spinning ball — Magnus effect

A spinning ball drags air with it. Air speed becomes larger on one side and smaller on the other; streamlines crowd on the fast side and rarefy on the slow side. The pressure difference produces a net sideways force that deflects the ball — the Magnus effect seen in cricket and tennis swing.

Case (iii)

Aerofoil — wing lift

A wing is shaped and angled so streamlines crowd more above than below. Flow speed on top is higher, so pressure on top is lower. The excess pressure underneath gives an upward lift that balances the aircraft's weight.

Streamlines past an aerofoil producing lift fast flow → low pressure slow flow → high pressure Lift

Figure 4. Air flowing past an aerofoil. Streamlines crowd above the wing, so the flow there is faster and the pressure lower; the higher pressure below produces a net upward dynamic lift.

NCERT Example 9.7

A fully loaded Boeing aircraft has a mass of $3.3\times10^5$ kg and total wing area $500~\text{m}^2$. It is in level flight at $960~\text{km/h}$. Estimate (a) the pressure difference between the lower and upper wing surfaces, and (b) the fractional increase in air speed over the upper surface. Take $\rho_{\text{air}} = 1.2~\text{kg m}^{-3}$.

(a) The weight is balanced by the lift force $\Delta P \times A$: $\;\Delta P = \dfrac{(3.3\times10^5)(9.8)}{500} = 6.5\times10^3~\text{N m}^{-2}$.

(b) Ignoring the height difference, $\Delta P = \tfrac{1}{2}\rho(v_2^2 - v_1^2) = \tfrac{1}{2}\rho(v_2 - v_1)(v_2 + v_1)$. With average speed $v_{av} = (v_1+v_2)/2 = 960~\text{km/h} = 267~\text{m s}^{-1}$, the fractional increase $\dfrac{v_2 - v_1}{v_{av}} = \dfrac{\Delta P}{\rho\,v_{av}^2} \approx 0.08$. The air over the wing need be only about 8% faster than the air beneath.

Atomiser, sprayer and blood flow

The same low-pressure-where-fast logic runs through a family of everyday devices and one important physiological caution.

Atomiser

Scent / paint sprayer

Squeezing the bulb blows air fast through a narrow orifice, dropping the pressure there. Atmospheric pressure pushes liquid up the tube into the low-pressure stream, which shatters it into a fine spray.

Sprayer

Spray gun / Bunsen burner

A piston (or gas jet) forces air through a small hole at high speed and low pressure, sucking insecticide — or, in a Bunsen burner, air through a side hole to mix with the gas before combustion.

Blood flow

Arterial constriction

Where an artery narrows (a plaque), blood speeds up and the lateral pressure falls. The vessel can be squeezed shut and reopen repeatedly, fluttering. Note blood is viscous, so Bernoulli applies only qualitatively here.

Quick recap

Bernoulli's principle in one breath

  • $P + \tfrac{1}{2}\rho v^2 + \rho g h = \text{constant}$ along a streamline — energy conservation per unit volume.
  • Assumptions: incompressible, non-viscous, steady flow, single streamline. Fails for viscous, turbulent or compressible flow.
  • Horizontal pipe: $P + \tfrac{1}{2}\rho v^2 = \text{constant}$ → high speed means low pressure.
  • Torricelli's efflux (open tank): $v = \sqrt{2gh}$, equal to free-fall speed and independent of density.
  • Venturi-meter measures flow rate from the throat pressure drop; $Q \propto \sqrt{h}$.
  • Dynamic lift: faster, lower-pressure flow above an aerofoil (or one side of a spinning ball) gives lift / Magnus deflection.

NEET PYQ Snapshot — Bernoulli's Principle

Bernoulli and efflux are heavily tested. Each of these reduces to the same constant-energy line.

NEET 2023

The Venturi-meter works on:

  1. The principle of perpendicular axes
  2. Huygen's principle
  3. Bernoulli's principle
  4. The principle of parallel axes
Answer: (3) Bernoulli's principle

Direct recall. The Venturi-meter uses a constriction: continuity speeds the fluid up in the throat, Bernoulli's principle drops the pressure there, and the manometer reads the difference. A pure Bernoulli device.

Concept · Efflux

A large open tank is filled with water to a depth $H$. A small hole is made in the side wall at a depth $h$ below the free surface. The speed of the water emerging from the hole is:

  1. $\sqrt{2gH}$
  2. $\sqrt{2gh}$
  3. $\sqrt{2g(H-h)}$
  4. $\sqrt{gh}$
Answer: (2) √(2gh)

Torricelli's law. For an open tank the efflux speed depends only on the depth of the hole below the free surface, $v=\sqrt{2gh}$, identical to a free fall through $h$. The total depth $H$ and the water's density are irrelevant.

Concept · Pressure–velocity

Water flows steadily through a horizontal pipe of non-uniform cross-section. At a point where the cross-sectional area is halved, the speed doubles. Compared with the wider section, the pressure at the narrow section is:

  1. Higher
  2. Lower
  3. Unchanged
  4. Zero
Answer: (2) Lower

Continuity + Bernoulli. Halving the area doubles the speed ($Av=$ const). On a horizontal pipe $P+\tfrac{1}{2}\rho v^2=$ const, so the rise in $\tfrac{1}{2}\rho v^2$ forces $P$ down. The narrow, fast section is always the low-pressure section.

FAQs — Bernoulli's Principle

Short answers to the Bernoulli and efflux questions NEET aspirants get wrong most often.

What are the assumptions behind Bernoulli's equation?
Bernoulli's equation applies to an ideal fluid: incompressible, non-viscous (zero viscosity), in steady (streamline) flow, and the relation holds only between two points lying on the same streamline. It is energy conservation per unit volume, so it assumes no energy is lost to internal friction. Real fluids have viscosity, so the result is only approximate; it fails for turbulent or non-steady flow because pressure and velocity then fluctuate in time.
Why does pressure fall where the fluid speed is high?
Along a horizontal streamline ρgh is constant, so P + ½ρv² = constant. If the speed v rises (as it must where a pipe narrows, by the equation of continuity), the kinetic-energy term ½ρv² rises and the pressure P must fall to keep the sum constant. High speed therefore means low pressure. This inverse relation drives the Venturi-meter, atomiser, aerofoil lift and the Magnus effect.
Does the speed of efflux depend on the density of the liquid?
No. For a tank open to the atmosphere, Torricelli's law gives v = √(2gh), where h is the depth of the hole below the free surface. Density cancels out of the Bernoulli equation when the tank is open, so a hole at depth h squirts water and mercury at the same speed. Density only re-enters when the space above the liquid is sealed and pressurised to a pressure P different from atmospheric.
What is the difference between Torricelli's law and a freely falling body?
They give the same speed. Torricelli's law states that the efflux speed v = √(2gh) is identical to the speed a body would acquire falling freely through the same height h. The energy interpretation is the same: pressure energy at depth h converts entirely to kinetic energy at the hole, just as gravitational potential energy mgh converts to ½mv² in free fall.
On what principle does the Venturi-meter work?
The Venturi-meter works on Bernoulli's principle. It is a tube with a constriction; by the equation of continuity the fluid speeds up in the narrow throat, and by Bernoulli's principle the pressure there drops. A manometer reads the pressure difference between the wide and narrow sections, and from that difference the flow speed and volume rate are computed. NEET has asked this exact fact (2023).
How does Bernoulli's principle explain the lift on an aircraft wing?
An aerofoil is shaped and angled so that streamlines crowd together above the wing more than below it. Crowded streamlines mean higher air speed on top, and by Bernoulli's principle higher speed means lower pressure. The pressure on the lower surface therefore exceeds that on top, giving a net upward force — the dynamic lift that balances the aircraft's weight. The same speed-pressure logic explains the Magnus deflection of a spinning ball.