Hydrodynamic

Q: What is the equation of continuity?

It states that in steady flow of an incompressible fluid the volume flowing per second is the same at every cross-section, so A1 v1 equals A2 v2. It expresses conservation of mass: the fluid speeds up where a pipe narrows.

Q: What does Bernoulli's theorem state?

For steady, non-viscous, incompressible flow, the sum of pressure, rho g h and half rho v squared is constant along a streamline. It is conservation of energy per unit volume, so where the fluid moves faster its pressure is lower.

Q: What hydrodynamics questions come in JEE Main and JEE Advanced?

JEE Main asks continuity with pipe radii, Bernoulli between two points, Torricelli's speed and the venturimeter. JEE Advanced adds the range of jets and equal-range holes, time to empty a tank, thrust from a jet, pitot tubes and the height limit of a siphon.

Q: What is Torricelli's law of efflux?

The speed of liquid flowing out of a small hole in an open tank is root of 2 g h, where h is the depth of the hole below the free surface. It is the same speed a body gains by falling freely through height h.

Q: Where should a hole be made in a tank to get the maximum range?

At half the height of the liquid column above the ground. The range is 2 times root of h times H minus h, which is largest when the hole is at depth H over 2, and the maximum range then equals H.

Q: How does a venturimeter measure the speed of flow?

The pipe narrows to a throat where the fluid moves faster and its pressure drops. The pressure difference, read as a level difference h, combined with the ratio of areas gives the speed through continuity and Bernoulli's equation.

Q: How does an aeroplane wing produce lift?

The wing's shape and tilt make air flow faster over the top than underneath. By Bernoulli's principle the pressure above is lower, and the pressure difference multiplied by the wing area gives an upward dynamic lift.

Q: Which hydrodynamics topics are important for NEET?

NEET focuses on the equation of continuity, Bernoulli's principle and its applications such as dynamic lift and the blowing-off of roofs, Torricelli's speed of efflux and the venturimeter. Most questions are ratio or one-line numericals, for example speed in a narrowing pipe.

PhysicsFluid MechanicsFor JEE aspirants

Hydrodynamics is the study of fluids in motion. Two equations do almost all the work: the equation of continuity ( $A_{1} v_{1} = A_{2} v_{2}$ ), which conserves mass, and Bernoulli's equation ( $P + ρ g h + \frac{1}{2} ρ v^{2} = constant$ ), which conserves energy. Hydrodynamics questions in JEE Main, JEE Advanced and NEET apply them to Torricelli's efflux, venturimeters, pitot tubes, siphons, spinning balls and aeroplane lift, all with fully worked examples.

On this page1Types of flow2Continuity3Bernoulli's equation4Torricelli and range5Venturimeter6Pitot tube7Siphon8Lift and Magnus effect

Key Formulas - Quick Reference

★ Must learnVolume flow rate $Q = A v$ ; continuity $A_{1} v_{1} = A_{2} v_{2}$ (incompressible, steady)
Freely falling stream: $v = v_{0}^{2} + 2 g h$ , $a = \frac{A v _{0}}{v _{0}^{2} + 2 g h}$
★ Must learnBernoulli: $P + ρ g h + \frac{1}{2} ρ v^{2} = constant$ along a streamline
Heads: $\frac{P}{ρ g} + h + \frac{v ^{2}}{2 g} = constant$
★ Must learnTorricelli: $v = 2 g h$ ( $h$ = depth of hole below the free surface)
★ Must learnRange on the ground: $R = 2 h (H - h)$ ; $R_{m a x} = H$ at $h = \frac{H}{2}$
★ Must learnVenturimeter: $v_{1} = \frac{2 g h}{( A _{1} / A _{2} ) ^{2} - 1}$
Pitot tube: $\frac{1}{2} ρ_{g} v^{2} = (ρ - ρ_{g}) g h$
Siphon: $v = 2 g h_{1}$ ; works only if $h_{1} + h_{2} < \frac{P _{0}}{ρ g}$
Time to empty a tank through a small hole: $t = \frac{A}{a} \frac{2 H}{g}$

1. Types of Flow and the Ideal Fluid

When a fluid moves, the velocity of the fluid at each point can stay the same or change with time.

Type of flow	What it means	Example
Steady (streamline) flow	Velocity at any given point does not change with time; it may differ from point to point	Slow flow of water in a smooth pipe
Laminar flow	Fluid moves in parallel layers that slide over each other without mixing	Honey poured slowly
Turbulent flow	Velocity at a point changes irregularly; eddies and mixing appear above a critical speed	Water from a fully open tap, rapids

A streamline is the path followed by a fluid particle in steady flow; the tangent at any point gives the velocity there. Two streamlines never cross (otherwise one point would have two velocities). A bundle of streamlines forms a tube of flow, and fluid never crosses its wall. Where streamlines crowd together the fluid moves faster.

Figure 1: (a) In steady flow the velocity at every point is tangent to the streamline, streamlines never cross, and they crowd together where the fluid moves faster (arrows at P, Q, R drawn with

v \propto \frac{1}{r ^{2}}

). (b) Above a critical speed the motion becomes irregular with eddies: turbulent flow.

Streamline (laminar) flow

Speed below the critical speed. Each particle follows the path of the one before it; velocity at a point is fixed; layers do not mix.

Turbulent flow

Speed above the critical speed. Velocity at a point changes irregularly; eddies form and the fluid mixes, wasting energy as heat.

Ideal fluid (the model behind continuity and Bernoulli):

Incompressible: density stays the same everywhere.
Non-viscous: no internal friction between layers.
Irrotational: a small body released in the flow does not spin about its centre of mass.
Steady: velocity at a point does not change with time (it may vary from point to point).

Real fluids are viscous and become turbulent at high speeds; viscosity, Stokes' law and the Reynolds number are covered in Properties of Fluids.

2. Equation of Continuity

Consider steady flow through a tube whose cross-section changes from $A_{1}$ to $A_{2}$ , with speeds $v_{1}$ and $v_{2}$ . No fluid is created or lost between the two sections, so in steady flow the mass entering per second equals the mass leaving per second.

In time $d t$ the fluid at section 1 moves $v_{1} d t$ , so volume $A_{1} v_{1} d t$ and mass $ρ_{1} A_{1} v_{1} d t$ enter.
Similarly mass $ρ_{2} A_{2} v_{2} d t$ leaves at section 2.
Steady flow: $ρ_{1} A_{1} v_{1} = ρ_{2} A_{2} v_{2}$ . For an incompressible fluid $ρ_{1} = ρ_{2}$ :
$A_{1} v_{1} = A_{2} v_{2} = Q = constant$

$Q = A v$ is the volume flow rate (unit $m^{3} s^{- 1}$ ). The equation of continuity is conservation of mass: a fluid speeds up where the pipe narrows and slows down where it widens, $v \propto \frac{1}{A}$ . For a circular pipe, $v \propto \frac{1}{r ^{2}}$ .

Figure 2: Continuity. Equal volumes cross every section per second, so

A_{1} v_{1} = A_{2} v_{2}

. Here the radius halves roughly (

r_{1} / r_{2} = 1.92

), so

v_{2} = 3.7 v_{1}

, drawn to scale.

2.1 Freely falling liquid

Water leaving a tap speeds up as it falls, so by continuity its stream gets thinner. If it leaves a mouth of area $A$ with speed $v_{0}$ , then at depth $h$ below the mouth

v^{2} = v_{0}^{2} + 2 g h and v_{0} A = a v_{0}^{2} + 2 g h \Rightarrow a = \frac{v _{0} A}{v _{0}^{2} + 2 g h}

Figure 3: Falling stream.

v = v_{0}^{2} + 2 g h

and

a = \frac{A v _{0}}{v _{0}^{2} + 2 g h}

. Shape computed for

v_{0} = 1 m s^{- 1}

: at

h = 0.3 m

v = 2.62 m s^{- 1}

and

a = 0.38 A

Key idea

Continuity is conservation of mass:

A_{1} v_{1} = A_{2} v_{2}

. Narrow section, fast flow; for a round pipe

v \propto \frac{1}{r ^{2}}

3. Bernoulli's Equation

Bernoulli's equation is the work-energy theorem applied to an ideal fluid in steady flow. Take the fluid between section S (area $A_{1}$ , pressure $P_{1}$ , speed $v_{1}$ , height $h_{1}$ ) and section T (area $A_{2}$ , $P_{2}$ , $v_{2}$ , $h_{2}$ ).

In time $d t$ a slab of volume $d V = A_{1} d x_{1}$ enters at S. By continuity, the same volume $d V = A_{2} d x_{2}$ leaves at T; its mass is $d m = ρ d V$ .
Work done by the pressure behind: $W_{1} = P_{1} A_{1} d x_{1} = P_{1} d V$ . Work done against the pressure ahead: $W_{2} = - P_{2} d V$ .
Net effect: the slab $d m$ has moved from S to T, gaining kinetic energy $\frac{1}{2} d m (v_{2}^{2} - v_{1}^{2})$ and potential energy $d m g (h_{2} - h_{1})$ .
Work-energy theorem:
$(P_{1} - P_{2}) d V = \frac{1}{2} ρ d V (v_{2}^{2} - v_{1}^{2}) + ρ g d V (h_{2} - h_{1})$
Rearranging:
$P_{1} + ρ g h_{1} + \frac{1}{2} ρ v_{1}^{2} = P_{2} + ρ g h_{2} + \frac{1}{2} ρ v_{2}^{2}$

Figure 4: In time

d t

equal volumes

d V = A_{1} d x_{1} = A_{2} d x_{2}

enter at S and leave at T. Net work by pressure

(P_{1} - P_{2}) d V

equals the gain in kinetic and potential energy.

Bernoulli's equation: along a streamline in steady flow of an ideal fluid,

P + ρ g h + \frac{1}{2} ρ v^{2} = constant

Each term is an energy per unit volume: pressure energy, potential energy and kinetic energy.

Dividing by $ρ g$ gives the same law in terms of heads (each has the unit of length):

\frac{P}{ρ g} + h + \frac{v ^{2}}{2 g} = constant

Head	Term	Meaning
Pressure head	$\frac{P}{ρ g}$	Height of liquid column the pressure could support
Gravitational (elevation) head	$h$	Height above the reference level
Velocity head	$\frac{v ^{2}}{2 g}$	Height the fluid could rise to by using up its speed

Special cases. Fluid at rest ( $v = 0$ ): $P + ρ g h = constant$ , which is hydrostatics, $P = P_{0} + ρ g h$ . Horizontal flow ( $h_{1} = h_{2}$ ): $P + \frac{1}{2} ρ v^{2} = constant$ , so where speed is high, pressure is low.

Exam Trick

Continuity first, Bernoulli second. Almost every flow problem is: use $A_{1} v_{1} = A_{2} v_{2}$ to get the unknown speed, then Bernoulli between the same two points for the pressure. Pick points where you know the most: free surfaces ( $P = P_{0}$ , $v \approx 0$ for a wide tank) and open jets ( $P = P_{0}$ ).

Limitations: Bernoulli's equation ignores viscous energy loss, so real pressure drops are larger; it holds along one streamline in steady flow; and it does not apply to turbulent flow or to compressible gases at high speed.

Key idea

Bernoulli is conservation of energy per unit volume along a streamline:

P + ρ g h + \frac{1}{2} ρ v^{2}

stays constant, so at one height faster fluid has lower pressure.

Quick Recall: tap to check

A pipe's radius halves. What happens to the speed?

It becomes

4

times:

v \propto \frac{1}{r ^{2}}

Name the three heads in Bernoulli's equation.

Pressure head

\frac{P}{ρ g}

, elevation head

h

, velocity head

\frac{v ^{2}}{2 g}

Can Bernoulli be used between points on different streamlines?

Only if the flow is irrotational; in general it holds along one streamline in steady, non-viscous flow.

4. Applications of Bernoulli's Principle

4.1 Torricelli's law of efflux

A wide tank is open to air and has a small hole at depth $h$ below the free surface. Apply Bernoulli between the surface A and the hole B. Both are at pressure $P_{0}$ . By continuity $A_{tank} v_{A} = a v_{B}$ , and since the tank is much wider than the hole, $v_{A} ≪ v_{B}$ , so $v_{A} \approx 0$ :

P_{0} + ρ g h + 0 = P_{0} + 0 + \frac{1}{2} ρ v^{2} \Rightarrow v = 2 g h

The liquid leaves with the speed of a body falling freely through height $h$ .

Range on the ground. Let the liquid stand at height $H$ above the ground, with the hole at depth $h$ , i.e. at height $H - h$ . The jet leaves horizontally with $v = 2 g h$ :

Time to fall $H - h$ : $t = \frac{2 ( H - h )}{g}$ .
Range: $R = v t = 2 g h \frac{2 ( H - h )}{g} = 2 h (H - h)$ .
Swapping $h$ and $H - h$ leaves $R$ unchanged: holes at depths $h$ and $H - h$ throw water equally far.
For the maximum, $\frac{d}{d h} [4 (H h - h^{2})] = 4 (H - 2 h) = 0$ , so $h = \frac{H}{2}$ and $R_{m a x} = 2 \frac{H}{2} \cdot \frac{H}{2} = H$ .

Figure 5: Efflux speed

v = 2 g h

(

h

= depth of the hole). Holes at depths

h

and

H - h

give the same range

R = 2 h (H - h)

; the hole at

H /2

gives

R_{m a x} = H

(dashed). Jets computed for

h = H /4

and

3 H /4

Figure 6:

R = 2 h (H - h)

(exact curve: half of an ellipse). Symmetric about

h = H /2

, where

R_{m a x} = H

Exam Trick

Efflux speed = free-fall speed from the surface. The jet leaves with the speed a stone would gain falling from the free surface to the hole, whatever the liquid. Holes at depths $h$ and $H - h$ throw equally far; the middle hole throws farthest, $R_{m a x} = H$ .

JEE Advanced

Time to empty a tank. With liquid height $y$ above a hole of area $a$ in a tank of area $A$ : $- A \frac{d y}{d t} = a 2 g y$ . Integrating from $H_{1}$ to $H_{2}$ :

t = \frac{A}{a} \frac{2}{g} (H_{1} - H_{2}), t_{empty} = \frac{A}{a} \frac{2 H}{g}

The level falls fast at first and slowly later. Pressurised tank: if the gas above the liquid is at $P$ , then $v = 2 (\frac{P - P _{0}}{ρ} + g h)$ . Thrust on the tank: the jet carries momentum $ρ a v$ per second times $v$ , so the tank feels a backward force $F = ρ a v^{2} = 2 ρ a g h$ .

Figure 7: Tank emptying:

h

falls linearly with time, so

h = H (1 - \frac{t}{T})^{2}

with

T = \frac{A}{a} \frac{2 H}{g}

. The level reaches

\frac{H}{4}

\frac{T}{2}

: the last quarter of the height takes half the time. For

A = 0.5 m^{2}

a = 1 cm^{2}

H = 2 m

T = 3194 s

4.2 Venturimeter

A venturimeter measures the flow speed in a pipe. The pipe narrows from area $A_{1}$ to a throat of area $A_{2}$ , both at the same height. Bernoulli with $h_{1} = h_{2}$ and continuity $v_{2} = \frac{A _{1} v _{1}}{A _{2}}$ give

P_{1} - P_{2} = \frac{1}{2} ρ (v_{2}^{2} - v_{1}^{2}) = \frac{1}{2} ρ v_{1}^{2} [(\frac{A _{1}}{A _{2}})^{2} - 1]

Since $A_{1} > A_{2}$ , $v_{2} > v_{1}$ and $P_{2} < P_{1}$ . The fluid is pushed forward (accelerated) as it enters the throat and slowed as it leaves. If vertical tubes at the two sections show a level difference $h$ , then $P_{1} - P_{2} = ρ g h$ and

v_{1} = \frac{2 g h}{( A _{1} / A _{2} ) ^{2} - 1}

Figure 8: Venturimeter. Faster flow in the throat means lower pressure, so the throat column stands

h

lower.

v_{1} = \frac{2 g h}{( A _{1} / A _{2} ) ^{2} - 1}

4.3 Pitot tube

A pitot tube measures the speed of a flowing fluid (and the airspeed of aircraft). It is a U-tube holding a liquid of density $ρ$ that does not mix with the flowing fluid. Opening A faces the flow: the fluid there is brought to rest, so its pressure rises. Opening B lies parallel to the flow and feels only the static pressure. Bernoulli between a point just upstream (speed $v$ ) and A (speed $0$ ):

P_{A} = P_{B} + \frac{1}{2} ρ_{g} v^{2}

The liquid levels differ by $h$ , so $P_{A} - P_{B} = (ρ - ρ_{g}) g h$ , which is $\approx ρ g h$ for a gas. Hence $v = \frac{2 ( ρ - ρ _{g} ) g h}{ρ _{g}}$ . On an aeroplane the pitot tube gives its speed relative to the air.

Figure 9: Pitot tube. Gas stops at opening A, so

P_{A} - P_{B} = \frac{1}{2} ρ_{g} v^{2} = (ρ - ρ_{g}) g h \approx ρ g h

Venturimeter

Measures the flow speed in a pipe. Compares the pressure at a wide section and a narrow throat: $P_{1} - P_{2} = \frac{1}{2} ρ v_{1}^{2} [(\frac{A _{1}}{A _{2}})^{2} - 1]$ .

Pitot tube

Measures the speed of a free stream (air speed of a plane). Compares stagnation and static pressure: $P_{A} - P_{B} = \frac{1}{2} ρ_{g} v^{2}$ .

4.4 Siphon

A siphon drains liquid over the rim of a tank to a lower level through a pipe that first rises and then falls. Take the pipe uniform (area $A$ ), the free surface P at height $0$ , the top of the pipe Q at height $h_{2}$ above it, and the outlet R at depth $h_{1}$ below it. The tank is wide, so $v_{P} \approx 0$ .

Bernoulli from P to R (both at $P_{0}$ ): $P_{0} + 0 + 0 = P_{0} - ρ g h_{1} + \frac{1}{2} ρ v^{2} \Rightarrow v = 2 g h_{1}$ . The outlet must be below the free surface ( $h_{1} > 0$ ).
The pipe is uniform, so by continuity the speed is the same at Q and R.
Bernoulli from Q to R: $P_{Q} + ρ g h_{2} = P_{0} - ρ g h_{1}$ , so $P_{Q} = P_{0} - ρ g (h_{1} + h_{2})$ .
Pressure cannot be negative (the liquid column breaks), so $P_{Q} > 0$ :
$h_{1} + h_{2} < \frac{P _{0}}{ρ g} (\approx 10.3 m for water)$

Figure 10: Siphon. Outflow speed

v = 2 g h_{1}

needs

h_{1} > 0

; the top works only if

P_{Q} = P_{0} - ρ g (h_{1} + h_{2}) > 0

4.5 Dynamic lift, Magnus effect and sprayers

Aeroplane wing. A wing (aerofoil) is shaped and tilted so that air flows faster over its upper surface than under it. By Bernoulli the pressure on top is lower, and the pressure difference times the wing area is an upward force called dynamic lift. When lift exceeds the plane's weight, the plane rises.

Magnus effect. A ball moving to the right with speed $V$ sees air moving left past it. If the ball also spins (back-spin), the air above it moves at $V + r ω$ and below at $V - r ω$ relative to the ball. Lower pressure above gives an upward force: the ball stays up longer and curves less sharply. Top-spin gives a downward force and a dipping ball; spin about a vertical axis gives sideways swing (in-swing or out-swing) in cricket, tennis and football.

Figure 11: Bernoulli in action. Wing (NACA 2412 profile, computed): faster air on top gives lift

= Δ P \times

wing area. Spinning ball: speeds

V + r ω

above and

V - r ω

below give an upward Magnus force.

Atomiser, sprayer, carburettor. A piston drives fast air across the open top of a thin tube dipped in the liquid. The fast air lowers the pressure above the tube, the liquid (scent, paint, insecticide, petrol) is pushed up, and the air stream breaks it into fine droplets.

Key idea

Every device here is Bernoulli plus continuity: fast flow at a throat, over a wing or above a spinning ball means low pressure there.

Quick Recall: tap to check

Water in a wide open tank stands

5 m

above a small hole. Efflux speed?

2 \times 9.8 \times 5 \approx 9.9 m s^{- 1}

A tank takes

T

to empty. When is the level at

\frac{H}{4}

\frac{T}{2}

, because

h = H (1 - \frac{t}{T})^{2}

Which way does a ball with back-spin deflect?

Upward (it floats), because the air above it moves faster relative to the ball.

Figure 12: Most flow problems are continuity first, then Bernoulli between the same two points. Torricelli and the horizontal-pipe form

P_{1} - P_{2} = \frac{1}{2} ρ (v_{2}^{2} - v_{1}^{2})

are just short cuts of Bernoulli.

Figure 13: Revision map of hydrodynamics: continuity conserves mass, Bernoulli conserves energy, and every device on this page is one of the two.

5. Solved Examples

Solved Example 1

Liquid flows steadily through a horizontal pipe from a wide section A (pressure

P_{1}

, speed

v_{1}

) to a narrow section B (pressure

P_{2}

, speed

v_{2}

). Compare

P_{1}

and

P_{2}

Solution:

Continuity: $A_{1} v_{1} = A_{2} v_{2}$ with $A_{1} > A_{2}$ , so $v_{1} < v_{2}$ .

Bernoulli (same height): $P_{1} + \frac{1}{2} ρ v_{1}^{2} = P_{2} + \frac{1}{2} ρ v_{2}^{2}$ . The smaller speed goes with the larger pressure.

Answer: $P_{1} > P_{2}$ .

Solved Example 2

A cylindrical tank of radius

1 m

rests on a platform

5 m

high. It is filled with water to a height of

5 m

. A plug of area

1 0^{- 4} m^{2}

is removed from an orifice in the side at the bottom. Find (a) the initial speed of efflux, (b) the initial speed with which the water strikes the ground, (c) the time to empty the tank to half its original volume, and (d) whether this time depends on the height of the stand. (

g = 10 m s^{- 2}

)

Solution:

(a) $v_{H} = 2 g h = 2 (10) (5) = 10 m s^{- 1}$ (horizontal).

(b) Falling $5 m$ from rest vertically: $v_{V} = 2 (10) (5) = 10 m s^{- 1}$ . Speed at the ground $= v_{H}^{2} + v_{V}^{2} = 102 = 14.1 m s^{- 1}$ .

t = \frac{A}{a} \frac{2}{g} (H - H^{'}) = \frac{π ( 1 ) ^{2}}{1 0 ^{- 4}} 0.2 (5 - 2.5)

$t \approx 9.2 \times 1 0^{3} s \approx 2.6 h$ .

(d) No: the expression for $t$ does not contain the height of the stand.

Answer: (a) $10 m s^{- 1}$ ; (b) $14.1 m s^{- 1}$ ; (c) $9.2 \times 1 0^{3} s$ ; (d) No.

Solved Example 3

Water flows at

2 m s^{- 1}

through a horizontal pipe of radius

2 cm

that narrows to radius

1 cm

. Find the speed in the narrow part and the pressure difference between the two parts.

Solution:

Continuity: $v_{2} = v_{1} (\frac{r _{1}}{r _{2}})^{2} = 2 \times 4 = 8 m s^{- 1}$ .

Bernoulli: $P_{1} - P_{2} = \frac{1}{2} ρ (v_{2}^{2} - v_{1}^{2}) = \frac{1}{2} (1000) (64 - 4)$ .

Answer: $8 m s^{- 1}$ ; $3.0 \times 1 0^{4} Pa$ (the wide part is at higher pressure).

Solved Example 4

Water leaves a tap of cross-section

1 cm^{2}

1 m s^{- 1}

. Find the cross-section of the stream

15 cm

below the tap. (

g = 10 m s^{- 2}

)

Solution:

$v = v_{0}^{2} + 2 g h = 1 + 2 (10) (0.15) = 4 = 2 m s^{- 1}$ .

Continuity: $a = \frac{A v _{0}}{v} = \frac{( 1 ) ( 1 )}{2}$ .

Answer: $a = 0.5 cm^{2}$ .

Solved Example 5

In a venturimeter the area of the main pipe is twice that of the throat. The vertical tubes show a level difference of

5 cm

of the flowing water. Find the speed of water in the main pipe. (

g = 9.8 m s^{- 2}

)

Solution:

$\frac{A _{1}}{A _{2}} = 2$ , so $v_{1} = \frac{2 g h}{( A _{1} / A _{2} ) ^{2} - 1} = \frac{2 ( 9.8 ) ( 0.05 )}{4 - 1} = 0.327$ .

Answer: $v_{1} \approx 0.57 m s^{- 1}$ (and $v_{2} = 1.14 m s^{- 1}$ in the throat).

Solved Example 6

A pitot tube in an air stream shows a water level difference of

1.0 cm

. Density of air

1.2 kg m^{- 3}

, water

1000 kg m^{- 3}

g = 9.8 m s^{- 2}

. Find the air speed.

Solution:

$\frac{1}{2} ρ_{air} v^{2} = (ρ_{w} - ρ_{air}) g h \approx ρ_{w} g h$ .

$v = \frac{2 ρ _{w} g h}{ρ _{air}} = \frac{2 ( 1000 ) ( 9.8 ) ( 0.01 )}{1.2} = 163$ .

Answer: $v \approx 12.8 m s^{- 1}$ .

Solved Example 7

Water stands

2 m

deep in a tank resting on the ground. Where should a small hole be made in its side to throw the water farthest on the ground, and how far does it go?
(A)

0.5 m

below the surface,

1.7 m

(B)

1.0 m

below the surface,

2.0 m

(C)

1.5 m

below the surface,

1.7 m

(D) at the bottom,

2.0 m

Solution:

Answer: (B). $R = 2 h (H - h)$ is largest at $h = \frac{H}{2} = 1 m$ , giving $R_{m a x} = H = 2 m$ . Options (A) and (C) give equal but smaller ranges ( $2 0.75 = 1.73 m$ ); a hole at the bottom gives $R = 0$ .

Solved Example 8

A closed tank has water

2 m

deep and air above it at a gauge pressure of

3 \times 1 0^{5} Pa

. Find the speed of efflux from a small hole at the bottom. (

g = 9.8 m s^{- 2}

)

Solution:

Bernoulli from the surface to the hole: $P + ρ g h = P_{0} + \frac{1}{2} ρ v^{2}$ , so

v = 2 (\frac{P - P _{0}}{ρ} + g h) = 2 (300 + 19.6)

Answer: $v \approx 25.3 m s^{- 1}$ (compared with only $6.3 m s^{- 1}$ for an open tank).

Solved Example 9

Water flows out of a hole of area

1 cm^{2}

at a depth of

1.25 m

in a tank on a frictionless cart. Find the thrust on the tank. (

g = 10 m s^{- 2}

)

Solution:

$v = 2 g h = 25 = 5 m s^{- 1}$ . Momentum carried away per second $= (ρ a v) v$ :

$F = ρ a v^{2} = (1000) (1 0^{- 4}) (25)$ .

Answer: $F = 2.5 N$ , acting on the tank opposite to the jet.

Solved Example 10

A water siphon has its outlet

0.5 m

below the water surface in the tank. Find the speed of outflow and the greatest height of the siphon's top above the water surface at which it can still work. (

P_{0} = 1.013 \times 1 0^{5} Pa

g = 9.8 m s^{- 2}

)

Solution:

$v = 2 g h_{1} = 2 (9.8) (0.5) = 3.13 m s^{- 1}$ .

Condition: $h_{1} + h_{2} < \frac{P _{0}}{ρ g} = \frac{1.013 \times 1 0 ^{5}}{9800} = 10.34 m$ , so $h_{2} < 10.34 - 0.5$ .

Answer: $3.13 m s^{- 1}$ ; $h_{2} < 9.8 m$ (in practice less, because water boils at low pressure).

Solved Example 11

Air flows over the top of an aeroplane wing at

70 m s^{- 1}

and under it at

60 m s^{- 1}

. The wing area is

25 m^{2}

and the density of air is

1.3 kg m^{- 3}

. Find the lift.

Solution:

$Δ P = \frac{1}{2} ρ (v_{top}^{2} - v_{bottom}^{2}) = \frac{1}{2} (1.3) (4900 - 3600) = 845 Pa$ .

Lift $= Δ P \times A = 845 \times 25$ .

Answer: lift $\approx 2.1 \times 1 0^{4} N$ , enough to support about $2150 kg$ .

Solved Example 12

An open tank of cross-section

0.5 m^{2}

holds water

2 m

deep. How long does it take to empty through a hole of area

1 cm^{2}

in the bottom? (

g = 9.8 m s^{- 2}

)

Solution:

$t = \frac{A}{a} \frac{2 H}{g} = \frac{0.5}{1 0 ^{- 4}} \frac{4}{9.8} = 5000 \times 0.639$ .

Answer: $t \approx 3.2 \times 1 0^{3} s \approx 53 min$ .

Solved Example 13

Water flows through a horizontal pipe whose radius at section A is

3

times its radius at section B. The ratio of the speeds

v_{B} : v_{A}

is
(A)

3

(B)

\frac{1}{3}

(C)

9

(D)

\frac{1}{9}

Solution:

Answer: (C). Continuity: $π r_{A}^{2} v_{A} = π r_{B}^{2} v_{B}$ , so $\frac{v _{B}}{v _{A}} = (\frac{r _{A}}{r _{B}})^{2} = 9$ . The speed depends on the area, i.e. on $r^{2}$ , not on $r$ .

Solved Example 14

A tank draining through a small hole in its bottom takes

10 min

for the water level to fall from

H

\frac{H}{4}

. How long does it take from

\frac{H}{4}

to empty?
(A)

2.5 min

(B)

5 min

(C)

10 min

(D)

20 min

Solution:

Answer: (C). $t \propto H_{1} - H_{2}$ . From $H$ to $\frac{H}{4}$ : $H - \frac{H}{2} = \frac{H}{2}$ . From $\frac{H}{4}$ to $0$ : also $\frac{H}{2}$ . The two stages take equal times, $10 min$ each.

Solved Example 15

Air (density

1.2 kg m^{- 3}

) blows horizontally over a flat roof of area

40 m^{2}

30 m s^{- 1}

, while the air inside is still. Find the net upward force on the roof.

Solution:

Bernoulli at one height: pressure outside is lower by $Δ P = \frac{1}{2} ρ v^{2} = \frac{1}{2} (1.2) (30)^{2} = 540 Pa$ .

Force $= Δ P \times A = 540 \times 40 = 21600 N$ .

Answer: about $2.2 \times 1 0^{4} N$ upward, which is why storms lift roofs off.

Practice Questions

A garden hose of internal radius $1 cm$ carries water at $3 m s^{- 1}$ . The nozzle has radius $0.5 cm$ . Find the speed at the nozzle.Answer: $12 m s^{- 1}$
Find the speed of efflux from a hole $20 cm$ below the water surface of a wide open tank. ( $g = 10 m s^{- 2}$ )Answer: $2 m s^{- 1}$
Holes at depths $h$ and $H - h$ in a tank give equal ranges. Show this, and find the range for $H = 1.6 m$ , $h = 0.4 m$ .Answer: $R = 2 0.4 \times 1.2 \approx 1.39 m$
Water flows through a horizontal pipe; at a point where the speed is $1 m s^{- 1}$ the pressure is $2 \times 1 0^{5} Pa$ . What is the pressure where the speed is $3 m s^{- 1}$ ?Answer: $1.96 \times 1 0^{5} Pa$
Why does a strong wind blow off a tin roof instead of pushing it down?Answer: Fast air above lowers the pressure on top; the still air inside pushes up (Bernoulli)
How long does it take to drain a tank from $4 m$ to $1 m$ depth if it takes $t_{0}$ to drain from $4 m$ to empty?Answer: $\frac{t _{0}}{2}$ (since $t \propto H_{1} - H_{2}$ )

Common Mistakes to Avoid

Watch out

Using $h$ as the height of the hole above the ground in $v = 2 g h$ . In Torricelli's law $h$ is the depth below the free surface.
Assuming high speed means high pressure. In horizontal flow the fast region has lower pressure.
Forgetting that $v \propto \frac{1}{r ^{2}}$ (not $\frac{1}{r}$ ) when a circular pipe narrows.
Taking the surface speed of a wide tank as the efflux speed: in a wide tank $v_{surface} \approx 0$ .
Writing $P_{0}$ at one end and gauge pressure at the other in the same Bernoulli equation. Use absolute pressures throughout or gauge throughout.
Applying Bernoulli between points on different streamlines in rotating or viscous flow, or in turbulent flow, where it does not hold.
Using $ρ g h$ of the flowing gas instead of the manometer liquid in pitot-tube and venturimeter questions.

Frequently Asked Questions

What is the equation of continuity?

It states that in steady flow of an incompressible fluid the volume flowing per second is the same at every cross-section, so A1 v1 equals A2 v2. It expresses conservation of mass: the fluid speeds up where a pipe narrows.

What does Bernoulli's theorem state?

For steady, non-viscous, incompressible flow, the sum of pressure, rho g h and half rho v squared is constant along a streamline. It is conservation of energy per unit volume, so where the fluid moves faster its pressure is lower.

What hydrodynamics questions come in JEE Main and JEE Advanced?

JEE Main asks continuity with pipe radii, Bernoulli between two points, Torricelli's speed and the venturimeter. JEE Advanced adds the range of jets and equal-range holes, time to empty a tank, thrust from a jet, pitot tubes and the height limit of a siphon.

What is Torricelli's law of efflux?

The speed of liquid flowing out of a small hole in an open tank is root of 2 g h, where h is the depth of the hole below the free surface. It is the same speed a body gains by falling freely through height h.

Where should a hole be made in a tank to get the maximum range?

At half the height of the liquid column above the ground. The range is 2 times root of h times H minus h, which is largest when the hole is at depth H over 2, and the maximum range then equals H.

How does a venturimeter measure the speed of flow?

The pipe narrows to a throat where the fluid moves faster and its pressure drops. The pressure difference, read as a level difference h, combined with the ratio of areas gives the speed through continuity and Bernoulli's equation.

How does an aeroplane wing produce lift?

The wing's shape and tilt make air flow faster over the top than underneath. By Bernoulli's principle the pressure above is lower, and the pressure difference multiplied by the wing area gives an upward dynamic lift.

Which hydrodynamics topics are important for NEET?

NEET focuses on the equation of continuity, Bernoulli's principle and its applications such as dynamic lift and the blowing-off of roofs, Torricelli's speed of efflux and the venturimeter. Most questions are ratio or one-line numericals, for example speed in a narrowing pipe.

Previous year questions on Hydrodynamic

23 questions from past papers, each with a step-by-step solution.

Show all 23 questions

Ready to master Fluid Mechanics?

Take a full mock test, practice concept-by-concept, and get an AI-powered rank prediction — all on Fundamenthol.

Practice questions →Take a mock test