PhysicsCBSE Class 11 · Chapter 9

Mechanical Properties of Fluids Class 11: Notes, Bernoulli and Surface Tension

Q: Which topics were removed from this chapter for 2026-27?

The Venturi meter, blood flow and heart attack, detergents and surface tension, and the appendix were removed. Bernoulli's theorem is kept with simple applications: Torricelli's law and dynamic lift.

Q: Is Reynolds number needed?

The CBSE content names streamline and turbulent flow and critical velocity, and the NCERT text explains that flow becomes turbulent beyond a critical speed. Learn this idea qualitatively; the current NCERT text does not develop a Reynolds-number formula.

Q: Why is the pressure inside a soap bubble higher than outside?

Surface tension tries to shrink the bubble. The extra pressure inside, 4S/r for a soap bubble with two surfaces, balances that pull.

Q: Why does a raindrop not keep accelerating?

Air drag grows with speed. When drag plus buoyancy equals the weight, the net force is zero and the drop falls at a steady terminal velocity.

Q: How is this chapter tested?

In the school annual exam through definitions and reasoning (Pascal's law, surface tension effects), derivations (Bernoulli, terminal velocity, excess pressure, capillary rise) and numericals. NEET often tests capillarity and terminal velocity.

Updated 14 October 2026 · Fact-checked

Mechanical Properties of Fluids is Chapter 9 of CBSE Class 11 Physics. It covers fluid pressure and Pascal's law with hydraulic machines, streamline and turbulent flow, Bernoulli's theorem with Torricelli's law and dynamic lift, viscosity, Stokes' law and terminal velocity, and surface tension with drops, bubbles and capillary rise.

What Mechanical Properties of Fluids is about

Liquids and gases are fluids: they flow and take the shape of their container, and they cannot resist a shearing force. This chapter studies what they do at rest and in motion.

At rest, a fluid presses equally in all directions, and the pressure grows with depth: $P = P_{a} + ρ g h$ . Pascal's law says pressure applied to an enclosed fluid is passed on undiminished to every part, which is how a hydraulic lift multiplies force and hydraulic brakes act equally on all wheels.

In motion, an ideal fluid in steady streamline flow obeys the equation of continuity ( $A v$ = constant) and Bernoulli's theorem ( $P + \frac{1}{2} ρ v^{2} + ρ g h$ = constant), which explain the speed of water from a hole in a tank (Torricelli's law) and the dynamic lift on aircraft wings and spinning balls. Real fluids have viscosity, internal friction between layers, which leads to Stokes' law and the terminal velocity of falling drops.

Finally, the surface of a liquid behaves like a stretched skin: surface tension. It explains why drops are round, why the pressure inside a drop or bubble is higher than outside, and why water rises in thin tubes (capillary rise). The Venturi meter, blood flow and detergents have been removed for 2026-27.

Weightage: This chapter is part of Unit VII, Properties of Bulk Matter. The syllabus groups Unit VII with Unit VIII, Thermodynamics and Unit IX, Behaviour of Perfect Gases and Kinetic Theory of Gases: together they carry 20 of the 70 theory marks in the school annual exam.Source: CBSE Physics (042) syllabus 2026-27, Class XI course structure, Unit VII Properties of Bulk Matter

Key concepts

Pressure and its variation with depth

NCERT section 9.2, 9.2.2 and 9.2.3

Pressure is the normal force per unit area: $P = \frac{F}{A}$ . It is a scalar; SI unit pascal (Pa = N m $^{- 2}$ ). A fluid at rest exerts pressure equally in all directions, always perpendicular to any surface.

Variation with depth: for a liquid of density ρ, the pressure at depth h below a surface where the pressure is $P_{a}$ is
$P = P_{a} + ρ g h$
It depends only on depth, not on the shape of the container (the hydrostatic paradox). $P - P_{a} = ρ g h$ is the gauge pressure.

Atmospheric pressure at sea level is about $1.013 \times 1 0^{5}$ Pa = 1 atm, and it holds up a column of mercury 76 cm high in a barometer ( $ρ g h$ with ρ = $13.6 \times 1 0^{3}$ kg m $^{- 3}$ ). This is why dams are made thicker at the bottom.

Worked example

Find the total pressure on a diver 25 m below the surface of a freshwater lake. ( $P_{a} = 1.013 \times 1 0^{5}$ Pa, ρ = 1000 kg m $^{- 3}$ , g = 9.8 m s $^{- 2}$ ) [1 mark]

Try it first, then show the solutionSolution

$P = 1.013 \times 1 0^{5} + 1000 \times 9.8 \times 25$
$P = 1.013 \times 1 0^{5} + 2.45 \times 1 0^{5} \approx 3.46 \times 1 0^{5} P a$ (about 3.4 atm)

Answer: About $3.5 \times 1 0^{5}$ Pa

Pascal's law and hydraulic machines

NCERT section 9.2.1 and 9.2.4

Pascal's law: the pressure in a fluid at rest is the same at all points at the same height, and any extra pressure you apply to a fluid shut inside a container reaches every part of it, and the container walls, without getting smaller.

Hydraulic lift: a small force $F_{1}$ on a small piston of area $A_{1}$ creates pressure $\frac{F _{1}}{A _{1}}$ , which acts on a large piston of area $A_{2}$ :
$F_{2} = F_{1} \frac{A _{2}}{A _{1}}$
The force is multiplied by $\frac{A _{2}}{A _{1}}$ , but the small piston must move farther, so the work done is the same (energy is not created).

Hydraulic brakes: pressing the pedal raises the pressure in the brake fluid, and this pressure reaches all four wheel cylinders equally, so all wheels are braked evenly.

Worked example

The pistons of a hydraulic press have radii 2 cm and 20 cm. What force on the small piston lifts a load of $1 0^{4}$ N on the large one? [1 mark]

Try it first, then show the solutionSolution

$F_{1} = F_{2} \frac{A _{1}}{A _{2}} = 1 0^{4} \times (\frac{2}{20})^{2}$
$F_{1} = 1 0^{4} \times \frac{1}{100} = 100 N$

Answer: 100 N

Streamline flow, turbulence and the equation of continuity

NCERT section 9.3

In steady (streamline) flow the velocity of the fluid at each point does not change with time. The path of a fluid particle is a streamline; streamlines never cross (otherwise a particle would have two velocities at one point). A bundle of streamlines forms a tube of flow.

Beyond a limiting speed, the critical speed, the flow becomes turbulent: irregular, with eddies, as in white water rapids. Low speeds, high viscosity and narrow pipes favour streamline flow.

Equation of continuity: for an incompressible fluid, the volume passing every cross-section per second is the same:
$A_{1} v_{1} = A_{2} v_{2} = constant$
The fluid speeds up where the pipe narrows. This is conservation of mass. The volume flow rate is $Q = A v$ (m $^{3}$ s $^{- 1}$ ).

Worked example

Water flows at 2 m s $^{- 1}$ in a pipe. The pipe narrows to half its radius. Find the speed in the narrow part. [1 mark]

Try it first, then show the solutionSolution

Area $\propto r^{2}$ , so the narrow part has one quarter of the area
$v_{2} = v_{1} \frac{A _{1}}{A _{2}} = 2 \times 4 = 8 m s^{- 1}$

Answer: 8 m s $^{- 1}$

Bernoulli's principle

NCERT section 9.4

For the steady flow of an incompressible, non-viscous fluid, along a streamline:
$P + \frac{1}{2} ρ v^{2} + ρ g h = constant$
The three terms are pressure energy, kinetic energy and potential energy per unit volume.

Derivation (work-energy): consider fluid between sections 1 and 2. Pressure forces do net work $(P_{1} - P_{2}) Δ V$ as volume ΔV moves through. This equals the gain in KE, $\frac{1}{2} ρ Δ V (v_{2}^{2} - v_{1}^{2})$ , plus the gain in PE, $ρ g Δ V (h_{2} - h_{1})$ . Rearranging gives $P_{1} + \frac{1}{2} ρ v_{1}^{2} + ρ g h_{1} = P_{2} + \frac{1}{2} ρ v_{2}^{2} + ρ g h_{2}$ .

Key consequence: in a horizontal pipe, where the speed is higher, the pressure is lower. For a fluid at rest, Bernoulli reduces to $P_{2} - P_{1} = ρ g (h_{1} - h_{2})$ , the hydrostatic result.

Worked example

Water flows through a horizontal pipe. Its speed is 2 m s $^{- 1}$ at one point and 6 m s $^{- 1}$ at another. Find the pressure difference between them. [2 marks]

Try it first, then show the solutionSolution

Same height: $P_{1} - P_{2} = \frac{1}{2} ρ (v_{2}^{2} - v_{1}^{2})$
$= \frac{1}{2} \times 1000 \times (36 - 4) = 16000 P a$ , higher where the water is slower

Answer: $1.6 \times 1 0^{4}$ Pa

Torricelli's law and dynamic lift

NCERT section 9.4.1 and 9.4.2

Speed of efflux (Torricelli's law): water flows out of a small hole at depth h below the free surface of a wide open tank. Both the surface and the jet are at atmospheric pressure, and the surface moves very slowly, so Bernoulli gives
$v = 2 g h$
the same speed as a body falling freely through h.

Dynamic lift is the upward force on a body moving through a fluid, caused by a pressure difference:

Aerofoil (aircraft wing): its shape makes air flow faster over the top than underneath. Faster flow means lower pressure on top, so there is a net upward force: lift = ΔP × wing area.
Spinning ball (Magnus effect): a spinning ball drags air round with it. On one side the air speed is higher and the pressure lower, so the ball curves (swing and spin in cricket and football).

Worked example

A small hole is made 5 m below the water surface in a large open tank. Find the speed of the water jet. (g = 9.8 m s $^{- 2}$ ) [1 mark]

Try it first, then show the solutionSolution

$v = 2 g h = 2 \times 9.8 \times 5$
$v \approx 9.9 m s^{- 1}$

Answer: About 9.9 m s $^{- 1}$

Viscosity

NCERT section 9.5

Viscosity is the internal friction of a fluid: adjacent layers moving at different speeds drag on each other. Honey is far more viscous than water.

For a fluid between a fixed plate and a moving plate, the force needed is proportional to the area and the velocity gradient:
$F = η A \frac{d v}{d x}$
where η is the coefficient of viscosity. SI unit: N s m $^{- 2}$ = Pa s (the poiseuille); the CGS unit is the poise (1 poise = 0.1 Pa s). Dimensions $[M L^{- 1} T^{- 1}]$ .

The viscosity of liquids falls as temperature rises (warm honey flows easily).
The viscosity of gases rises with temperature.
Viscosity resists relative motion of layers, unlike solid friction which acts between surfaces.

Worked example

A plate of area 0.1 m $^{2}$ moves at 0.1 m s $^{- 1}$ over a layer of liquid 0.5 mm thick resting on a fixed plate. If η = $1 0^{- 3}$ Pa s, find the force needed. [2 marks]

Try it first, then show the solutionSolution

Velocity gradient $= \frac{0.1}{0.5 \times 1 0 ^{- 3}} = 200 s^{- 1}$
$F = η A \frac{d v}{d x} = 1 0^{- 3} \times 0.1 \times 200 = 0.02 N$

Answer: 0.02 N

Stokes' law and terminal velocity

NCERT section 9.5.1

A small sphere of radius a moving slowly at speed v through a viscous fluid feels a drag force (Stokes' law):
$F = 6 π η a v$

A sphere falling through a fluid speeds up until the upward viscous drag plus buoyancy balances its weight. After that it falls at a constant terminal velocity:
$\frac{4}{3} π a^{3} ρ g = \frac{4}{3} π a^{3} σ g + 6 π η a v_{t} \Rightarrow v_{t} = \frac{2 a ^{2} ( ρ - σ ) g}{9 η}$
where ρ is the density of the sphere and σ that of the fluid.

$v_{t} \propto a^{2}$ : big raindrops fall faster than small ones; fine mist and dust stay suspended. If $σ > ρ$ (an air bubble in water), $v_{t}$ is negative: the bubble rises.

Worked example

Find the terminal velocity of a fine raindrop of radius 0.05 mm falling through air. (η of air = $1.8 \times 1 0^{- 5}$ Pa s, ρ of water = 1000 kg m $^{- 3}$ , density of air 1.2 kg m $^{- 3}$ , g = 9.8 m s $^{- 2}$ ) [2 marks]

Try it first, then show the solutionSolution

$v_{t} = \frac{2 a ^{2} ( ρ - σ ) g}{9 η} = \frac{2 \times ( 5 \times 1 0 ^{- 5} ) ^{2} \times 998.8 \times 9.8}{9 \times 1.8 \times 1 0 ^{- 5}}$
$v_{t} \approx 0.30 m s^{- 1}$

Answer: About 0.30 m s $^{- 1}$

Surface tension, surface energy and angle of contact

NCERT section 9.6, 9.6.1 to 9.6.3

Molecules inside a liquid are pulled equally in all directions, but molecules at the surface are pulled inwards. Bringing a molecule to the surface needs energy, so the surface has extra energy and tends to shrink: it behaves like a stretched membrane.

Surface tension S is the force per unit length acting along the surface, perpendicular to a line drawn in it: $S = \frac{F}{l}$ (N m $^{- 1}$ ). It equals the surface energy per unit area (J m $^{- 2}$ ): the work needed to increase the surface area by one unit.

A soap film has two surfaces, so pulling a wire of length l across it needs $F = 2 S l$ . Surface tension falls as temperature rises.

Angle of contact θ: the angle between the liquid surface and the solid surface, measured inside the liquid, where they meet. If θ < 90° the liquid wets the solid (water on clean glass, θ ≈ 0°); if θ > 90° it does not (mercury on glass, θ ≈ 140°). Liquid drops take a spherical shape because a sphere has the least surface for a given volume.

Worked example

How much work is needed to blow a soap bubble of radius 1 cm? (S = 0.03 N m $^{- 1}$ ) [2 marks]

Try it first, then show the solutionSolution

A bubble has two surfaces: total area $= 2 \times 4 π r^{2} = 8 π \times 1 0^{- 4} m^{2}$
$W = S \times Δ A = 0.03 \times 8 π \times 1 0^{- 4} \approx 7.5 \times 1 0^{- 5} J$

Answer: About $7.5 \times 1 0^{- 5}$ J

Excess pressure in drops and bubbles; capillary rise

NCERT section 9.6.4 and 9.6.5

Because of surface tension, the pressure on the concave side of a curved liquid surface is higher than on the convex side.

Drop of radius r (one surface): increasing the radius by Δr increases the surface energy by $8 π r Δ r S$ , which equals the work $(P_{i} - P_{o}) 4 π r^{2} Δ r$ , so
$P_{i} - P_{o} = \frac{2 S}{r}$

Air bubble inside a liquid (one surface): also $\frac{2 S}{r}$ .
Soap bubble in air (two surfaces): $\frac{4 S}{r}$ .

Smaller bubbles have a higher excess pressure.

Capillary rise: in a thin tube of radius a, a wetting liquid rises to a height
$h = \frac{2 S c o s θ}{ρ g a}$
Water (θ ≈ 0) rises; mercury (θ > 90°) is pushed down. Narrower tubes give greater rise. Examples: oil rising in a lamp wick, water rising in soil and in blotting paper.

Worked example

Find the height to which water rises in a glass capillary tube of radius 0.25 mm. (S = 0.073 N m $^{- 1}$ , θ = 0°, ρ = 1000 kg m $^{- 3}$ , g = 9.8 m s $^{- 2}$ ) [2 marks]

Try it first, then show the solutionSolution

$h = \frac{2 S c o s θ}{ρ g a} = \frac{2 \times 0.073 \times 1}{1000 \times 9.8 \times 2.5 \times 1 0 ^{- 4}}$
$h \approx 0.060 m = 6.0 c m$

Answer: About 6 cm

Physics in real life: Mechanical Properties of Fluids

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.

Pressure at the base of a dam

Suppose a dam holds back a reservoir 50 m deep.

Find the gauge pressure on the dam wall at the bottom, and explain why the wall is made thicker there. (ρ = 1000 kg m $^{- 3}$ , g = 9.8 m s $^{- 2}$ )

Try it first, then show the solutionSolution

Gauge pressure $= ρ g h = 1000 \times 9.8 \times 50 = 4.9 \times 1 0^{5} P a$ (almost 5 atm)
Pressure grows linearly with depth, so the bottom of the wall must withstand the largest push.

Answer: $4.9 \times 1 0^{5}$ Pa; thicker at the base to withstand the higher pressure

Memory hook: Deeper water pushes harder: build thicker at the bottom.

Equation of continuity

Imagine watering plants with a garden hose and putting your thumb over the end so the opening becomes narrower.

Suppose the water leaves the open hose at 1.5 m s $^{- 1}$ , and your thumb reduces the opening to one fifth of its area. How fast does the water now come out?

Try it first, then show the solutionSolution

Same volume per second: $A_{1} v_{1} = A_{2} v_{2}$
$v_{2} = 1.5 \times 5 = 7.5 m s^{- 1}$

Answer: 7.5 m s $^{- 1}$ , so the jet reaches much farther

Memory hook: Narrow the gap, quicken the flow.

Hydraulic jack

Suppose a mechanic lifts a 1200 kg car with a hydraulic jack whose large piston has 20 times the radius of the small piston.

What force must be applied on the small piston? (g = 9.8 m s $^{- 2}$ )

Try it first, then show the solutionSolution

Area ratio $= 2 0^{2} = 400$
$F_{1} = \frac{1200 \times 9.8}{400} = 29.4 N$

Answer: About 29 N, roughly the weight of a 3 kg bag

Memory hook: Pascal multiplies force by the area ratio.

Topics in this chapter

Mechanical Properties of Fluids formula sheet

Pressure at depth: $P = P_{a} + ρ g h$
Hydraulic lift: $\frac{F _{1}}{A _{1}} = \frac{F _{2}}{A _{2}}$
Equation of continuity: $A_{1} v_{1} = A_{2} v_{2}, Q = A v$
Bernoulli's theorem: $P + \frac{1}{2} ρ v^{2} + ρ g h = constant$
Torricelli's law: $v = 2 g h$
Viscous force: $F = η A \frac{d v}{d x}$
Stokes' law and terminal velocity: $F = 6 π η a v, v_{t} = \frac{2 a ^{2} ( ρ - σ ) g}{9 η}$
Surface tension and surface energy: $S = \frac{F}{l} = \frac{W}{Δ A}$
Excess pressure: $drop or bubble in liquid : \frac{2 S}{r}; soap bubble : \frac{4 S}{r}$
Capillary rise: $h = \frac{2 S c o s θ}{ρ g a}$

How the school exam asks Mechanical Properties of Fluids

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Pressure and Pascal's law	Pressure at depth; gauge vs absolute; hydraulic lift forces; why dams are thick at the base; hydraulic brakes.	$P = P_{a} + ρ g h$ ; $F_{2} / F_{1} = A_{2} / A_{1} = (r_{2} / r_{1})^{2}$ .
Continuity and Bernoulli	Speed in a narrow section; pressure difference in a pipe; derivation of Bernoulli's theorem; lift on a wing.	Continuity first to get speeds, then Bernoulli for pressure; lift = ΔP × area.
Torricelli's law	Speed and rate of flow from a hole; derivation; where the jet lands.	$v = 2 g h$ with h measured from the free surface; $Q = A v$ .
Viscosity and Stokes' law	Viscous force on a plate; terminal velocity derivation and numericals; effect of temperature; why big drops fall faster.	Balance weight = buoyancy + drag at terminal velocity; $v_{t} \propto a^{2}$ .
Surface tension	Work to form drops or bubbles; energy released when drops merge; excess pressure derivation; capillary rise; angle of contact.	Count surfaces (two for a soap film); energy = S × change in area; use $2 S / r$ or $4 S / r$ .

Tricks to remember Mechanical Properties of Fluids

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

Fast flow, low pressure
Bernoulli in one line: where fluid speeds up (same height), its pressure drops.
Use it when: Wings, spinning balls, roofs blown off in storms.
Drop 2, soap bubble 4
One surface: 2S/r. Two surfaces (soap bubble in air): 4S/r.
Use it when: Excess pressure questions.
Terminal velocity grows as radius squared
$v_{t} \propto a^{2}$ : double the drop, four times the speed.
Use it when: Ratio MCQs on falling drops.
Thin tube, tall column
Capillary rise $h \propto \frac{1}{a}$ ; also $ha$ = constant for one liquid.
Use it when: Comparing tubes.
Hot liquids flow, hot gases resist
Viscosity of liquids falls with temperature; of gases rises.
Use it when: Assertion-reason items.

Mechanical Properties of Fluids: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Pressure in a soap bubble · 1 mark · MCQExam level

The excess pressure inside a soap bubble of radius r and surface tension S is:
(a) S/r
(b) 2S/r
(c) 4S/r
(d) 8S/r

Try it first, then show the solutionSolution

A soap bubble has two surfaces, each contributing 2S/r.

Answer: (c) 4S/r

Example 2: Why drops are round · 1 mark · Assertion-reasonExam level

Assertion (A): A small free liquid drop takes a spherical shape.
Reason (R): Surface tension makes the surface area of a liquid as small as possible.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.

Try it first, then show the solutionSolution

For a given volume a sphere has the smallest surface area, so R explains A.

Answer: (a)

Example 3: Car on a hydraulic lift · 2 marks · Very short answerHarder exam level

A car of mass 1500 kg rests on a hydraulic lift whose large piston has area 0.5 m $^{2}$ . The small piston has area 25 cm $^{2}$ . What force on the small piston holds the car up? (g = 9.8 m s $^{- 2}$ ) [2 marks]

Try it first, then show the solutionSolution

$F_{1} = F_{2} \frac{A _{1}}{A _{2}} = 1500 \times 9.8 \times \frac{25 \times 1 0 ^{- 4}}{0.5}$
$F_{1} = 14700 \times 0.005 = 73.5 N$

Answer: 73.5 N

Example 4: Pressure drop in a narrowing pipe · 3 marks · Short answerHarder exam level

Water flows at 1 m s $^{- 1}$ through a horizontal pipe 4 cm in diameter, which narrows to 2 cm. Find the speed in the narrow part and the pressure difference between the two parts. [3 marks]

Try it first, then show the solutionSolution

Continuity: $v_{2} = v_{1} (\frac{d _{1}}{d _{2}})^{2} = 1 \times 4 = 4 m s^{- 1}$
Bernoulli (same height): $P_{1} - P_{2} = \frac{1}{2} ρ (v_{2}^{2} - v_{1}^{2})$
$= \frac{1}{2} \times 1000 \times (16 - 1) = 7500 P a$

Answer: 4 m s $^{- 1}$ ; 7500 Pa lower in the narrow part

Example 5: Ball falling through glycerine · 3 marks · Short answerHarder exam level

A steel ball of radius 1 mm falls through glycerine. Find its terminal velocity. (ρ of steel = 7800 kg m $^{- 3}$ , σ of glycerine = 1260 kg m $^{- 3}$ , η = 1.5 Pa s, g = 9.8 m s $^{- 2}$ ) [3 marks]

Try it first, then show the solutionSolution

At terminal velocity: weight = upthrust + Stokes' drag, giving $v_{t} = \frac{2 a ^{2} ( ρ - σ ) g}{9 η}$
$v_{t} = \frac{2 \times ( 1 0 ^{- 3} ) ^{2} \times 6540 \times 9.8}{9 \times 1.5}$
$v_{t} \approx 9.5 \times 1 0^{- 3} m s^{- 1}$ (about 1 cm per second)

Answer: About 9.5 mm s $^{- 1}$

Example 6: Capillary rise in full · 5 marks · Long answerHarder exam level

(a) Derive the expression for the height of capillary rise of a liquid in a tube of radius a. (b) Water rises 3 cm in a tube. How high will it rise in a tube of half the radius? (c) What happens if the tube is pushed down so that only 2 cm sticks out above the water: does water overflow? Explain. (d) Why does mercury show capillary depression? [5 marks]

Try it first, then show the solutionSolution

(a) The meniscus is a hemisphere of radius $a / cos θ$ ; the pressure just below it is lower than atmospheric by $\frac{2 S c o s θ}{a}$
This deficit is made up by the liquid column: $h ρ g = \frac{2 S c o s θ}{a} \Rightarrow h = \frac{2 S c o s θ}{ρ g a}$
(b) $h \propto \frac{1}{a}$ : half the radius gives 6 cm
(c) No overflow: the water reaches the top and the meniscus becomes flatter (larger radius of curvature), so the surface tension force supports only a 2 cm column
(d) For mercury and glass θ > 90°, so cos θ < 0 and h is negative: the level falls

Answer: $h = 2 S cos θ / ρ g a$ ; 6 cm; no overflow; θ > 90° for mercury

Example 7: Energy released when drops merge · 3 marks · Short answerHOTS / competency

Eight identical water drops, each of radius 1 mm, combine into one big drop. Find the radius of the big drop and the surface energy released. (S = 0.072 N m $^{- 1}$ ) [3 marks]

Try it first, then show the solutionSolution

Volume is conserved: $8 \times \frac{4}{3} π r^{3} = \frac{4}{3} π R^{3} \Rightarrow R = 2 r = 2$ mm
Area before $= 8 \times 4 π r^{2} = 32 π r^{2}$ ; after $= 4 π (2 r)^{2} = 16 π r^{2}$ ; decrease $= 16 π r^{2}$
Energy released $= S \times 16 π r^{2} = 0.072 \times 16 π \times 1 0^{- 6} \approx 3.6 \times 1 0^{- 6} J$ (it warms the drop slightly)

Answer: R = 2 mm; about $3.6 \times 1 0^{- 6}$ J

Example 8: Short capillary tube · 2 marks · Very short answerHOTS / competency

Water would rise about 3 cm in a glass tube of radius 0.5 mm. The tube sticks out only 2 cm above the water surface. Find the radius of curvature of the meniscus at the top. (S = 0.073 N m $^{- 1}$ , ρ = 1000 kg m $^{- 3}$ , g = 9.8 m s $^{- 2}$ ) [2 marks]

Try it first, then show the solutionSolution

The column is now 2 cm, and $h ρ g = \frac{2 S}{R}$ where R is the radius of curvature of the meniscus
$R = \frac{2 S}{ρ g h} = \frac{2 \times 0.073}{1000 \times 9.8 \times 0.02} \approx 7.4 \times 1 0^{- 4} m$ (larger than 0.5 mm: a flatter meniscus)

Answer: About 0.74 mm

Case-study practice (Section E style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 4-mark question split into 1 + 1 + 1 + 1 marks.

Rooftop water tank

Suppose a large open water tank on a roof has water 1.8 m deep. A tap of cross-section 2 cm $^{2}$ near the bottom is opened. Take g = 10 m s $^{- 2}$ and ρ = 1000 kg m $^{- 3}$ .

(i) · 1 mark
What is the gauge pressure at the tap when it is closed?
Answer
Answer: $1.8 \times 1 0^{4}$ Pa
$ρ g h = 1000 \times 10 \times 1.8$
(ii) · 1 mark
With what speed does water come out of the open tap?
Answer
Answer: 6 m s $^{- 1}$
$2 \times 10 \times 1.8 = 36$
(iii) · 1 mark
How much water flows out per second?
Answer
Answer: $1.2 \times 1 0^{- 3}$ m $^{3}$ (1.2 L)
$2 \times 1 0^{- 4} \times 6$
(iv) · 1 mark
How does the speed change as the tank empties?
Answer
Answer: It decreases
$v = 2 g h$ falls as h falls.

Raindrops and mist

Suppose two water droplets fall through still air: a drizzle drop of radius 0.1 mm and a fog droplet of radius 0.01 mm. Take η of air = $1.8 \times 1 0^{- 5}$ Pa s, and ignore the density of air compared with water (1000 kg m $^{- 3}$ ), g = 9.8 m s $^{- 2}$ .

(i) · 1 mark
Which law gives the drag on each droplet?
Answer
Answer: Stokes' law, $F = 6 π η a v$
Small spheres moving slowly through a viscous fluid.
(ii) · 1 mark
Find the terminal velocity of the drizzle drop.
Answer
Answer: About 1.2 m s $^{- 1}$
$\frac{2 \times 1 0 ^{- 8} \times 1000 \times 9.8}{9 \times 1.8 \times 1 0 ^{- 5}}$
(iii) · 1 mark
Find the terminal velocity of the fog droplet.
Answer
Answer: About 1.2 cm s $^{- 1}$
$v_{t} \propto a^{2}$ : one hundredth of 1.2 m s $^{- 1}$
(iv) · 1 mark
Why does fog seem to hang in the air?
Answer
Answer: Its droplets fall extremely slowly
At about 1 cm per second, slight air currents keep them up.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Derive Torricelli's law for the speed of efflux of a liquid from a small hole in a tank. [2 marks] (2 marks)

Step	Marks
Apply Bernoulli between the free surface (pressure $P_{a}$ , speed ≈ 0, height h above the hole) and the hole (pressure $P_{a}$ , speed v)	1
$P_{a} + ρ g h = P_{a} + \frac{1}{2} ρ v^{2} \Rightarrow v = 2 g h$	1

Derive the excess pressure inside a liquid drop of radius r in terms of its surface tension S. [3 marks] (3 marks)

Step	Marks
If the radius increases by Δr, the surface area increases by $8 π r Δ r$ , so the surface energy rises by $8 π r Δ r S$	1
Work done by the excess pressure: $(P_{i} - P_{o}) \times 4 π r^{2} Δ r$	1
Equating: $P_{i} - P_{o} = \frac{2 S}{r}$	1

(a) State and derive Bernoulli's theorem. (b) Water stands 2 m deep in a large open tank. A hole of area 1 cm $^{2}$ is made at the bottom. Find the speed of efflux and the volume of water leaving per second. (g = 9.8 m s $^{- 2}$ ) [5 marks] (5 marks)

Step	Marks
(a) Statement: for steady, incompressible, non-viscous flow, $P + \frac{1}{2} ρ v^{2} + ρ g h$ is constant along a streamline	1
Work done by pressure forces as volume ΔV passes: $(P_{1} - P_{2}) Δ V$	1
Equals ΔKE + ΔPE: $\frac{1}{2} ρ Δ V (v_{2}^{2} - v_{1}^{2}) + ρ g Δ V (h_{2} - h_{1})$ ; rearranged to the constant form	1
(b) $v = 2 g h = 2 \times 9.8 \times 2 \approx 6.26 m s^{- 1}$	1
$Q = A v = 1 0^{- 4} \times 6.26 \approx 6.3 \times 1 0^{- 4} m^{3} s^{- 1}$ (0.63 L per second)	1

Where students lose marks in Mechanical Properties of Fluids

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

Mistake: Using one surface for a soap bubble
Forgetting the film has an inside and an outside surface.
Wrong working
Excess pressure in a soap bubble = 2S/r
Right working
4S/r
Do this: Count surfaces first: a soap film always has two.
Usually costs: 1-2 marks.
Mistake: Measuring h in Torricelli's law from the bottom of the tank
Misreading the diagram.
Do this: h is the depth of the hole below the free surface of the liquid.
Usually costs: 1 mark.
Mistake: Forgetting buoyancy in the terminal velocity
Using ρ instead of (ρ − σ).
Do this: Net downward force is weight minus upthrust: $\frac{4}{3} π a^{3} (ρ - σ) g$ .
Usually costs: 1 mark.
Mistake: Using radii instead of areas in the hydraulic lift
Rushing.
Do this: Force ratio = area ratio = square of the radius ratio.
Usually costs: 1 mark.
Mistake: Writing about the Venturi meter or blood flow
Old notes.
Do this: These were removed for 2026-27; the syllabus keeps only Torricelli's law and dynamic lift as applications of Bernoulli's theorem.
Usually costs: Time.

NCERT exercise guide: Mechanical Properties of Fluids

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Exercises 9.1-9.4 (reasoning)

What it tests: Everyday effects of pressure, surface tension, angle of contact and viscosity.

How to approach it: Name the principle (P = F/A, ρgh, surface tension minimising area, Bernoulli) and apply it in two sentences.

Same kind as Exercise 9.1

Explain why (a) a sharp knife cuts better than a blunt one, (b) small drops of mercury are nearly spherical.

Show the solutionSolution

(a) The same force on a smaller area gives a much larger pressure ( $P = F / A$ ).
(b) Surface tension tries to make the surface area as small as possible, and for a given volume a sphere has the least area; for small drops gravity is too weak to flatten them.

Answer: (a) larger pressure (b) surface tension minimises area

Exercises 9.5-9.16 (pressure and flow)

What it tests: High heels, wine barometers, off-shore structures, hydraulic lifts, U-tubes, Bernoulli questions, glycerine flow, wind-tunnel lift and spray pumps.

How to approach it: Pressure balance at equal heights in U-tubes; $F_{2} / F_{1} = A_{2} / A_{1}$ ; continuity then Bernoulli; lift = ΔP × area.

Same kind as Exercise 9.14

Air flows over the top of an aircraft wing at 80 m s $^{- 1}$ and under it at 70 m s $^{- 1}$ . The wing area is 30 m $^{2}$ and the air density 1.2 kg m $^{- 3}$ . Find the lift and the mass it can support. (g = 9.8 m s $^{- 2}$ )

Show the solutionSolution

$Δ P = \frac{1}{2} \times 1.2 \times (8 0^{2} - 7 0^{2}) = 0.6 \times 1500 = 900 P a$
Lift $= 900 \times 30 = 2.7 \times 1 0^{4} N$
Mass supported $= \frac{2.7 \times 1 0 ^{4}}{9.8} \approx 2.76 \times 1 0^{3} k g$

Answer: $2.7 \times 1 0^{4}$ N; about 2760 kg

Exercises 9.17-9.20 (surface tension)

What it tests: Soap films on wires, liquid films supporting weights, and excess pressure inside mercury drops and soap bubbles.

How to approach it: Film force = 2Sl (two surfaces); drop 2S/r; soap bubble 4S/r; add the hydrostatic pressure if the bubble is under a liquid.

Same kind as Exercise 9.19

Find the excess pressure inside a mercury drop of radius 2.00 mm. (S of mercury = 0.465 N m $^{- 1}$ )

Show the solutionSolution

$P_{i} - P_{o} = \frac{2 S}{r} = \frac{2 \times 0.465}{2.00 \times 1 0 ^{- 3}}$
$= 465 P a$

Answer: 465 Pa

Practice questions on Mechanical Properties of Fluids

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 · MCQExam level
The SI unit of the coefficient of viscosity is:
(a) N m $^{- 1}$
(b) Pa s
(c) N s
(d) J m $^{- 2}$
Hint
F = ηA dv/dx.
Answer
(b) Pa s
Q2 · 1 mark · MCQExam level
Bernoulli's theorem is a consequence of the conservation of:
(a) mass
(b) momentum
(c) energy
(d) charge
Hint
Derived from the work-energy theorem.
Answer
(c) energy
Q3 · 1 mark · Assertion-reasonExam level
Assertion (A): Hydraulic brakes apply equal braking force on all the wheels.
Reason (R): By Pascal's law, the pressure applied to the brake fluid is transmitted undiminished throughout it.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
Same pressure, same piston sizes.
Answer
(a)
Q4 · 2 marks · Very short answerExam level
Why does the viscosity of a liquid decrease when it is heated, while that of a gas increases?
Hint
Cohesion vs collisions.
Answer
In a liquid, viscosity comes mainly from cohesive forces between molecules, which weaken as they move faster on heating. In a gas, it comes from molecules carrying momentum between layers, which increases as they move faster.
Q5 · 2 marks · Very short answerHarder exam level
Find the total pressure on a diver 30 m below the surface of the sea. (ρ of sea water = 1030 kg m $^{- 3}$ , $P_{a} = 1.013 \times 1 0^{5}$ Pa, g = 9.8 m s $^{- 2}$ )
Hint
$P_{a} + ρ g h$ .
Answer
$P = 1.013 \times 1 0^{5} + 1030 \times 9.8 \times 30 \approx 4.04 \times 1 0^{5}$ Pa, about 4 atm.
Q6 · 2 marks · Very short answerHarder exam level
Water flows at 3 L s $^{- 1}$ through a pipe of radius 2 cm. Find its speed. What is the speed where the radius is 1 cm?
Hint
v = Q/A.
Answer
$v = \frac{3 \times 1 0 ^{- 3}}{π \times ( 0.02 ) ^{2}} \approx 2.39$ m s $^{- 1}$ ; with half the radius, four times: about 9.55 m s $^{- 1}$ .
Q7 · 1 mark · Very short answerHarder exam level
Two capillary tubes of radii r and 2r are dipped in water. Find the ratio of the heights to which water rises in them.
Hint
$h \propto 1/ a$ .
Answer
2 : 1 (the narrower tube has twice the rise).
Q8 · 2 marks · Very short answerHarder exam level
A metal plate of area 0.05 m $^{2}$ rests on a layer of oil 1 mm thick. A force of 0.002 N moves it at a steady 0.2 m s $^{- 1}$ . Find the coefficient of viscosity of the oil.
Hint
η = F d/(A v).
Answer
$η = \frac{0.002 \times 1 0 ^{- 3}}{0.05 \times 0.2} = 2 \times 1 0^{- 4}$ Pa s.
Q9 · 3 marks · Short answerHarder exam level
Derive the expression for the terminal velocity of a sphere falling through a viscous fluid, and explain why an air bubble rises in water.
Hint
Weight = upthrust + drag.
Answer
At terminal velocity, $\frac{4}{3} π a^{3} ρ g = \frac{4}{3} π a^{3} σ g + 6 π η a v_{t}$ , so $v_{t} = \frac{2 a ^{2} ( ρ - σ ) g}{9 η}$ . For an air bubble ρ < σ, so $v_{t}$ is negative: it moves upwards at a steady speed.
Q10 · 2 marks · Very short answerHOTS / competency
Two soap bubbles of radii 2 cm and 4 cm are joined by a narrow tube. In which direction does air flow? Explain using excess pressure.
Hint
4S/r is larger for small r.
Answer
Excess pressure 4S/r is twice as large in the 2 cm bubble, so air flows from the smaller bubble into the larger one: the small bubble shrinks and the large one grows.
Q11 · 3 marks · Short answerHOTS / competency
During a storm, wind blows at 30 m s $^{- 1}$ over a flat roof of area 100 m $^{2}$ . Estimate the upward force on the roof. (density of air = 1.2 kg m $^{- 3}$ ; still air inside)
Hint
Bernoulli between inside and outside.
Answer
$Δ P = \frac{1}{2} \times 1.2 \times 3 0^{2} = 540$ Pa (lower pressure outside, above the roof). Upward force $= 540 \times 100 = 5.4 \times 1 0^{4}$ N, enough to lift a poorly fixed roof.

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 Pascal's law.Show answer
Pressure applied to an enclosed fluid is transmitted undiminished to every part of the fluid and the container walls.
2.Write the equation of continuity.Show answer
$A_{1} v_{1} = A_{2} v_{2}$ for an incompressible fluid in steady flow.
3.State Bernoulli's theorem.Show answer
Along a streamline, $P + \frac{1}{2} ρ v^{2} + ρ g h$ is constant for steady, non-viscous, incompressible flow.
4.Write Torricelli's law.Show answer
Speed of efflux $v = 2 g h$ .
5.Write Stokes' law.Show answer
$F = 6 π η a v$ .
6.What is the SI unit of viscosity?Show answer
Pa s (N s m $^{- 2}$ ), also called the poiseuille.
7.What is the excess pressure inside a soap bubble?Show answer
$4 S / r$ .
8.Why does mercury fall in a glass capillary?Show answer
Its angle of contact with glass is more than 90°, so cos θ is negative.

Best order to study it

Learn pressure, its variation with depth, atmospheric and gauge pressure.
Learn Pascal's law and hydraulic machines.
Learn streamline and turbulent flow and the equation of continuity.
Derive Bernoulli's theorem; apply it to Torricelli's law and dynamic lift.
Learn viscosity, Stokes' law and terminal velocity.
Learn surface tension, surface energy and angle of contact.
Derive the excess pressure in drops and bubbles; finish with capillary rise.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you find the pressure at a depth in a liquid?
Can you explain the hydrostatic paradox?
Can you solve a hydraulic lift problem?
Can you use continuity to find speeds in a pipe?
Can you derive Bernoulli's theorem?
Can you derive Torricelli's law?
Can you explain lift on an aircraft wing?
Can you derive the terminal velocity of a sphere?
Can you derive the excess pressure inside a drop?
Can you derive and use the capillary-rise formula?

Mechanical Properties of Fluids: frequently asked questions

Which topics were removed from this chapter for 2026-27?

The Venturi meter, blood flow and heart attack, detergents and surface tension, and the appendix were removed. Bernoulli's theorem is kept with simple applications: Torricelli's law and dynamic lift.

Is Reynolds number needed?

The CBSE content names streamline and turbulent flow and critical velocity, and the NCERT text explains that flow becomes turbulent beyond a critical speed. Learn this idea qualitatively; the current NCERT text does not develop a Reynolds-number formula.

Why is the pressure inside a soap bubble higher than outside?

Surface tension tries to shrink the bubble. The extra pressure inside, 4S/r for a soap bubble with two surfaces, balances that pull.

Why does a raindrop not keep accelerating?

Air drag grows with speed. When drag plus buoyancy equals the weight, the net force is zero and the drop falls at a steady terminal velocity.

How is this chapter tested?

In the school annual exam through definitions and reasoning (Pascal's law, surface tension effects), derivations (Bernoulli, terminal velocity, excess pressure, capillary rise) and numericals. NEET often tests capillarity and terminal velocity.

Based on: CBSE Physics (042) syllabus 2026-27 (Classes XI-XII); NCERT Physics Part II, Textbook for Class XI, Chapter 9 (Reprint 2026-27)