JEE Mains Physics · Mechanical Properties of Fluids

Continuity, Bernoulli's Equation and Efflux

In steady flow the volume passing each section per second is the same, so a liquid speeds up where a pipe narrows; Bernoulli's equation, P + ρgh + ½ρv² = constant, then says its pressure falls there, and it gives the speed of a jet leaving a tank as √(2gh).

All Mechanical Properties of Fluids notes JEE Mains Physics strategy

Why this matters

Twenty-three PYQs, ten of them numeric, and four from 2026. Twelve are pipes and tubes: continuity alone, the pressure difference across a narrowing, a venturi meter, a bent pipe or the correct form of Bernoulli's equation; five turn a pressure difference into a speed, for a valve gauge or an aircraft wing; six are liquid leaving a hole in a tank, for its speed, its range or the push it gives the tank.

Concept 1 of 3: Continuity and Bernoulli's equation in a pipe

A liquid cannot pile up inside a pipe, so the volume passing each section per second is the same: Av is constant. Where the pipe narrows, the liquid speeds up. It gains kinetic energy only if the liquid behind pushes harder than the liquid ahead, so the narrow part is at LOWER pressure. Bernoulli's equation is this energy balance written per unit volume.

Definition

Continuity: $A_1 v_1 = A_2 v_2 = Q$ , the volume flow rate. A hose feeding n holes of area a: $Av = n\,a\,v'$ . Radius halved: area ÷ 4, speed × 4.
A tank of area A emptied through a tap of area a at speed v: the level falls at $\dfrac{dh}{dt} = \dfrac{a v}{A}$ .
Bernoulli along a streamline: $P + \rho g h + \tfrac{1}{2}\rho v^{2} = \text{constant}$ . Each term is energy per unit volume, measured in pascals.
Horizontal pipe: $P_1 - P_2 = \tfrac{1}{2}\rho(v_2^{2} - v_1^{2})$ . If $A_2 = A_1/2$ , then $v_2 = 2v_1$ and $P_1 - P_2 = \tfrac{3}{2}\rho v_1^{2}$ .
Venturi meter: the difference h between the water columns gives $g h = \tfrac{1}{2}(v_2^{2} - v_1^{2})$ , with $v_2$ at the throat. Then $Q = A_1 v_1$ .
Pipe whose ends are at different heights: $P_1 - P_2 = \tfrac{1}{2}\rho(v_2^{2} - v_1^{2}) + \rho g(h_2 - h_1)$ . Read which end is higher from the figure.

Continuity and Bernoulli

A_1 v_1 = A_2 v_2, \qquad P + \rho g h + \tfrac{1}{2}\rho v^{2} = \text{constant}

Worked example

Water (

1000\ \text{kg/m}^{3}

) flows along a horizontal pipe. At X the area is

40\ \text{cm}^{2}

and the speed is 0.5 m/s; at Y the area is

10\ \text{cm}^{2}

. Find the speed at Y and

P_X - P_Y

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 8 Apr 2026 Shift 2 · Q8Moderate

Example 1 · Mechanical Properties of Fluids · Continuity, Bernoulli's Equation and Efflux

A liquid of density

600\text{ }kg/m^{3}

flowing steadily in a tube of varying cross-section. The cross-section at a point A is

1.0{\text{ }cm}^{2}

and that at B is

20{\text{ }mm}^{2}

. Both the points A and B are in same horizontal plane, the speed of the liquid at A is

10\text{ }cm/s

. The difference is pressures at A and B points is

\_\_\_\_

Pa.

Put both areas in the same unit

1 cm² = 100 mm². A ratio of 1 cm² to 20 mm² is 5, not 1/20. Convert before using continuity.

The narrow section is at the lower pressure

Faster flow means lower pressure. In a venturi meter with v₁ at the wide part, 2gh = v₂² − v₁². A statement written as v₁² − v₂² has the sign the wrong way round.

Square the speeds, then subtract

The pressure difference is ½ρ(v₂² − v₁²), not ½ρ(v₂ − v₁)². The second form is a common wrong option.

Drill 11 more on continuity and bernoulli's equation in a pipe

Concept 2 of 3: Speed from a pressure drop: gauges and wings

When liquid at rest starts to move, it pays for its speed with pressure: a fall of

\tfrac{1}{2}\rho v^{2}

in pressure buys speed v. A wing works the same way in air. The air over the top moves faster, so the pressure above is lower, and that difference times the wing area is the lift.

Definition

Valve closed, then opened, at the same height: $P_1 - P_2 = \tfrac{1}{2}\rho v^{2}$ . So $v = \sqrt{\dfrac{2(P_1 - P_2)}{\rho}}$ and $v \propto \sqrt{P_1 - P_2}$ .
Lift on a wing: $F = \tfrac{1}{2}\rho\left(v_{\text{top}}^{2} - v_{\text{bottom}}^{2}\right)A$ , with A the TOTAL area of the wings.
Level flight at constant speed: lift = weight, mg.
Two close speeds: $v_1^{2} - v_2^{2} = (v_1 + v_2)(v_1 - v_2)$ . This turns a pressure difference into a small speed difference.
Convert km/h to m/s by multiplying by 5/18.

Lift on a wing

F = \tfrac{1}{2}\rho\left(v_{\text{top}}^{2} - v_{\text{bottom}}^{2}\right)A

Worked example

Air (

1.2\ \text{kg/m}^{3}

) flows over a glider's wings at 144 km/h above and 108 km/h below. The total wing area is

25\ \text{m}^{2}

. Find the lift and the largest mass the glider can have in level flight. (

g = 10\ \text{m/s}^{2}

)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 1 February 2024 · Q28Moderate

Example 2 · Mechanical Properties of Fluids · Continuity, Bernoulli's Equation and Efflux

A plane is in level flight at constant speed and each of its two wings has an area of

40{\text{ }m}^{2}

. If the speed of the air is

180\text{ }km/h

over the lower wing surface and

252\text{ }km/h

over the upper wing surface, the mass of the plane is ______ kg. (Take air density to be

1\text{ }kg{\text{ }m}^{- 3}

and

g = 10{\text{ }ms}^{- 2}

)

Two wings: add both areas

'Each of its two wings has an area A' means a total of 2A. Using one wing's area halves the lift and the mass.

Convert km/h before squaring

Square the speeds in m/s. Converting after squaring needs (5/18)², and mixing the two orders is a common slip.

Drill 4 more on speed from a pressure drop: gauges and wings

Concept 3 of 3: Liquid leaving a hole: Torricelli's law

Liquid leaving a small hole at depth h below the free surface comes out as fast as a body that has fallen freely through h:

v = \sqrt{2gh}

. Anything that adds pressure on the surface, such as a load or a piston, adds to the push. Once out, the jet is a projectile launched horizontally.

Definition

$v = \sqrt{2gh}$ , with h measured DOWN from the free surface to the hole. The tank is wide, so the speed of the top surface is neglected.
A load or piston of weight W on the surface of area A: $\tfrac{1}{2}\rho v^{2} = \rho g h + \dfrac{W}{A}$ . The atmosphere acts on both the surface and the jet, so it cancels.
Range on the floor from a hole at height y above the base, depth h below the surface: $x = v\sqrt{2y/g} = 2\sqrt{h y}$ .
For water of total height H, the range is greatest for a hole at the middle, $h = H/2$ , and then $x = H$ .
If the tank stands on a block, the jet falls through the block's height as well.
The jet pushes the tank back with force $\rho a v^{2} = 2\rho a g h$ . To hold a massless tank by friction, $\mu \ge \dfrac{2a}{A}$ .

Torricelli's law and the range

v = \sqrt{2gh}, \qquad x = 2\sqrt{h\,y}

Worked example

A wide tank holds water 1.25 m deep. A small hole is made 0.45 m above the base. Find the speed of the jet and how far from the tank it lands. Where should the hole be for the longest range? (

g = 10\ \text{m/s}^{2}

)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 3 Apr 2025 · Q2Moderate

Example 3 · Mechanical Properties of Fluids · Continuity, Bernoulli's Equation and Efflux

Consider a completely full cylindrical water tank of height 1.6 m and cross-sectional area

0.5{\text{ }m}^{2}

. It has a small hole in its side at a height 90 cm from the bottom. Assume, the cross-sectional area of the hole to be negligibly small as compared to that of the water tank. If a load 50 kg is applied at the top surface of the water in the tank then the velocity of the water coming out at the instant when the hole is opened is :

\left( g= 10\text{ }m/s^{2} \right)

Depth is measured from the free surface

h in √(2gh) is the depth of the hole below the water surface: the water's height minus the hole's height. Using the hole's height above the base gives the wrong speed.

A load adds W/A, the atmosphere adds nothing

Air presses on the surface and on the jet alike, so P₀ cancels. Only the extra pressure of the load, its weight over the tank's area, joins ρgh.

Drill 5 more on liquid leaving a hole: torricelli's law

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Continuity and Bernoulli's equation in a pipe
Continuity and Bernoulli
$A_1 v_1 = A_2 v_2, \qquad P + \rho g h + \tfrac{1}{2}\rho v^{2} = \text{constant}$
Speed from a pressure drop: gauges and wings
Lift on a wing
$F = \tfrac{1}{2}\rho\left(v_{\text{top}}^{2} - v_{\text{bottom}}^{2}\right)A$
Liquid leaving a hole: Torricelli's law
Torricelli's law and the range
$v = \sqrt{2gh}, \qquad x = 2\sqrt{h\,y}$

Watch out for (7)

Put both areas in the same unit
→ Continuity and Bernoulli's equation in a pipe
The narrow section is at the lower pressure
→ Continuity and Bernoulli's equation in a pipe
Square the speeds, then subtract
→ Continuity and Bernoulli's equation in a pipe
Two wings: add both areas
→ Speed from a pressure drop: gauges and wings
Convert km/h before squaring
→ Speed from a pressure drop: gauges and wings
Depth is measured from the free surface
→ Liquid leaving a hole: Torricelli's law
A load adds W/A, the atmosphere adds nothing
→ Liquid leaving a hole: Torricelli's law

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