Mechanical Properties of Fluids ISC Class 11: Notes and Numericals

What Mechanical Properties of Fluids is about

Weightage: Mechanical Properties of Fluids belongs to unit 7 of the ISC Class XI Physics syllabus. CISCE weights units 7-9 together (Properties of Bulk Matter; Heat and Thermodynamics; Behaviour of Perfect Gases and Kinetic Theory of Gases) at 20 of the 70 marks in Paper I (Theory). Class XI is examined by the school, so your school sets the paper within this weightage.Source: CISCE ISC Physics syllabus, examination year 2028, Class XI unit weightage

Key concepts

Pressure in a fluid and the effect of gravity

Pressure is the normal force per unit area, $p=\frac{F}{A}$ (Pa; 1 atm = $1.013\times 10^5$ Pa). A fluid at rest exerts pressure equally in all directions at a point, and always perpendicular to any surface it touches. Considering a vertical column of fluid of height h: the pressure below must support the weight of the column above, so $p=p_0+\rho gh$, where $p_0$ is the pressure at the top (often atmospheric). The excess $p-p_0=\rho gh$ is the gauge pressure. Pressure depends only on depth, not on the shape of the container (hydrostatic paradox): connected vessels fill to the same level. This is why dam walls are thicker at the bottom.

Worked example

Find the absolute and gauge pressure on a diver 10 m below the surface of fresh water. ($p_0=1.013\times 10^5$ Pa, g = 9.8 m s⁻²)

Try it first, then show the solutionSolution

1. Gauge pressure = ρgh = 1000 × 9.8 × 10 = 9.8 × 10⁴ Pa
2. Absolute pressure = 1.013 × 10⁵ + 0.98 × 10⁵ = 1.99 × 10⁵ Pa

Answer: About 2 atm absolute; gauge pressure $9.8\times 10^4$ Pa

Pascal's law: hydraulic lift and brakes

Pascal's law: a change in pressure applied to an enclosed incompressible fluid is transmitted undiminished to every part of the fluid and to the walls of the container. Hydraulic lift (press): a small piston of area $a_1$ pushed with force $F_1$ creates pressure $\frac{F_1}{a_1}$, which acts on a large piston of area $a_2$, giving $F_2=F_1\frac{a_2}{a_1}$: a force multiplier. Energy is conserved: the small piston moves through a larger distance ($a_1{d}_1=a_2{d}_2$). Hydraulic brakes: pressing the brake pedal pushes a piston in the master cylinder; the pressure reaches all four wheel cylinders equally, whose wider pistons press the brake pads on the drums or discs with larger forces.

Worked example

In a hydraulic lift the pistons have areas 0.01 m² and 0.5 m². What force on the small piston lifts a 1200 kg car on the large one? (g = 9.8 m s⁻²)

Try it first, then show the solutionSolution

1. Weight of car = 1200 × 9.8 = 11 760 N
2. $F_1=F_2\frac{a_1}{a_2}=11760\times \frac{0.01}{0.5}=235N$

Answer: About 235 N, one-fiftieth of the car's weight

Buoyancy and Archimedes' principle

The pressure on the bottom of an immersed body is greater than on its top, so the fluid exerts a net upward force, the upthrust (buoyant force). Archimedes' principle: a body wholly or partly immersed in a fluid experiences an upthrust equal to the weight of the fluid it displaces: $U=V_{sub}\sigma g$, where σ is the fluid density. Apparent weight = true weight − upthrust. Flotation: a floating body displaces its own weight of fluid, so the fraction submerged is $\frac{{\rho }_{body}}{{\rho }_{fluid}}$. A body sinks if its average density exceeds that of the fluid. The upthrust acts at the centre of buoyancy (the centre of gravity of the displaced fluid).

Worked example

What fraction of an iceberg (density 917 kg m⁻³) lies below the surface of sea water (density 1025 kg m⁻³)?

Try it first, then show the solutionSolution

1. Floating: weight of ice = weight of water displaced: $V{\rho }_{ice}g=V_{sub}{\rho }_{sea}g$
2. $\frac{V_{sub}}{V}=\frac{917}{1025}=0.89$

Answer: About 89% is under water: only about a ninth shows

Streamline and turbulent flow; critical velocity and Reynolds number

In streamline (laminar) flow, the velocity at each point stays constant in time, and every particle passing a given point follows the same path, a streamline. Streamlines never cross (otherwise a particle would have two velocities at one point); where they crowd together the fluid moves faster. Above a certain speed, the critical velocity, flow becomes turbulent: irregular, with eddies, and energy is lost rapidly. The Reynolds number $R_e=\frac{\rho vD}{\eta }$ (ρ density, v speed, D pipe diameter, η viscosity) is dimensionless and indicates the type of flow: roughly below 1000 streamline, above 2000 turbulent, in between unsteady. Its significance: a high density and speed favour turbulence; a high viscosity favours streamline flow. So $v_c=\frac{R_e\eta }{\rho D}$.

Worked example

Water (η = 1.0 × 10⁻³ Pa s) flows at 0.05 m s⁻¹ through a pipe 2 cm in diameter. Find the Reynolds number and the speed at which the flow would become turbulent (take R_e = 2000).

Try it first, then show the solutionSolution

1. $R_e=\frac{1000\times 0.05\times 0.02}{1.0\times 10^{-3}}=1000$: streamline
2. $v_c=\frac{2000\times 10^{-3}}{1000\times 0.02}=0.1m{s}^{-1}$

Answer: 1000 (streamline); about 0.1 m s⁻¹

Equation of continuity

For the steady flow of an incompressible fluid through a pipe of varying cross-section, the mass entering per second must equal the mass leaving per second: $\rho a_1{v}_1=\rho a_2{v}_2$, so $a_1{v}_1=a_2{v}_2$ = constant, the volume flow rate Q (m³ s⁻¹). The speed is inversely proportional to the area: fluid speeds up where the pipe narrows. This is a statement of conservation of mass. It explains why a river flows faster through a narrow gorge and why covering part of a hose nozzle makes the water jet faster.

Worked example

Water flows at 1 m s⁻¹ in a hose of diameter 2 cm. Find its speed as it leaves a nozzle of diameter 0.5 cm, and the flow rate.

Try it first, then show the solutionSolution

1. $v_2=v_1\frac{a_1}{a_2}=1\times {\left(\frac{2}{0.5}\right)}^2=16m{s}^{-1}$
2. Q = π(0.01)² × 1 = 3.1 × 10⁻⁴ m³ s⁻¹

Answer: 16 m s⁻¹; about 0.31 litre per second

Bernoulli's theorem

Statement: for the steady, streamline flow of an ideal (incompressible, non-viscous) fluid, the sum of pressure energy, kinetic energy and potential energy per unit volume is constant along a streamline: $p+\frac{1}{2}\rho v^2+\rho gh$ = constant. Derivation: consider a tube of flow between sections 1 and 2. In time Δt, a volume ΔV enters at 1 and leaves at 2 (continuity). Work done by pressure = $(p_1-p_2)\Delta V$. This equals the gain in KE, $\frac{1}{2}\rho \Delta V(v_2^2-v_1^2)$, plus the gain in PE, $\rho \Delta Vg(h_2-h_1)$. Rearranging gives $p_1+\frac{1}{2}\rho v_1^2+\rho g{h}_1=p_2+\frac{1}{2}\rho v_2^2+\rho g{h}_2$. Assumptions: the fluid is incompressible and non-viscous, the flow is steady and streamline (irrotational), and no energy is lost to friction. In a horizontal pipe, where the speed is greater, the pressure is less. Torricelli's theorem (a special case): liquid flows out of a hole a depth h below the free surface at $v=\sqrt{2gh}$.

Worked example

Water flows through a horizontal pipe at 2 m s⁻¹ where the area is 4 cm² and the pressure is 2.0 × 10⁵ Pa. Find the speed and pressure where the area is 1 cm².

Try it first, then show the solutionSolution

1. Continuity: v₂ = 2 × 4/1 = 8 m s⁻¹
2. Bernoulli: $p_2=p_1-\frac{1}{2}\rho (v_2^2-v_1^2)=2.0\times 10^5-500(64-4)$
3. p₂ = 2.0 × 10⁵ − 3.0 × 10⁴ = 1.7 × 10⁵ Pa

Answer: 8 m s⁻¹; $1.7\times 10^5$ Pa (faster flow, lower pressure)

Applications of Bernoulli's theorem

Venturimeter: a pipe with a narrow throat; the pressure difference between the wide part and the throat, read on a manometer, gives the flow speed: $v_1=\sqrt{\frac{2\Delta p}{\rho [(a_1/a_2{)}^2-1]}}$. Atomiser (sprayer): air blown fast across the top of a narrow tube lowers the pressure there; atmospheric pressure on the liquid pushes it up the tube, and the fast air breaks it into a spray (perfume sprays, paint guns, carburettors). Dynamic uplift on an aeroplane wing: the wing's shape (aerofoil) and angle make air flow faster over the top than below, so pressure above is less, giving an upward lift = (pressure difference) × (wing area). Magnus effect: a spinning ball drags air round with it; on one side this adds to the airflow and on the other it subtracts; the faster side has lower pressure, so the ball swerves (swing and spin in cricket and football). Also: roofs blown off in storms, and two boats moving side by side drawn together.

Worked example

Air (density 1.2 kg m⁻³) flows over the top of a wing of area 30 m² at 70 m s⁻¹ and under it at 63 m s⁻¹. Find the lift on the wing.

Try it first, then show the solutionSolution

1. $\Delta p=\frac{1}{2}\rho (v_1^2-v_2^2)=0.6(4900-3969)=559Pa$
2. Lift = Δp × A = 559 × 30 = 1.68 × 10⁴ N

Answer: About $1.7\times 10^4$ N (enough to hold up about 1.7 tonnes)

Viscosity and Poiseuille's formula

Viscosity is the property of a fluid by which it opposes relative motion between its layers (internal friction). When a liquid flows over a fixed surface, the layer in contact is at rest and the speed increases with distance, giving a velocity gradient $\frac{dv}{dx}$. Newton's law of viscosity: the viscous drag between layers is $F=\eta A\frac{dv}{dx}$, where η is the coefficient of viscosity: SI unit N s m⁻² (Pa s); CGS unit poise, 1 poise = 0.1 N s m⁻²; dimensions [M L⁻¹ T⁻¹]. Viscosity of liquids falls as temperature rises; that of gases rises. Poiseuille's formula for streamline flow through a narrow horizontal tube of radius r and length l under pressure difference p: $Q=\frac{\pi p{r}^4}{8\eta l}$. The r⁴ makes flow extremely sensitive to the radius: halving it cuts the flow to one-sixteenth.

Worked example

Water (η = 1.0 × 10⁻³ Pa s) flows through a tube 1 m long and 1 mm in radius under a pressure difference of 1000 Pa. Find the volume flowing per second.

Try it first, then show the solutionSolution

1. $Q=\frac{\pi p{r}^4}{8\eta l}=\frac{\pi \times 1000\times (10^{-3}{)}^4}{8\times 10^{-3}\times 1}$
2. Q = 3.9 × 10⁻⁷ m³ s⁻¹

Answer: About 0.39 cm³ per second

Stokes' law and terminal velocity

Stokes' law: a small sphere of radius r moving slowly at speed v through a viscous fluid experiences a drag $F=6\pi \eta rv$. A sphere of density ρ falling through a fluid of density σ is acted on by its weight $\frac{4}{3}\pi r^3\rho g$, the upthrust $\frac{4}{3}\pi r^3\sigma g$ and the drag. As it speeds up, the drag grows until the net force is zero; it then falls at a constant terminal velocity: $6\pi \eta r{v}_t=\frac{4}{3}\pi r^3(\rho -\sigma )g$, so $v_t=\frac{2r^2(\rho -\sigma )g}{9\eta }$. The v-t graph rises steeply at first, then curves and levels off at $v_t$. Since $v_t\propto r^2$, tiny fog droplets fall so slowly they seem to float, and larger raindrops fall faster. If σ > ρ (an air bubble in water), $v_t$ is negative: the bubble rises.

Worked example

A steel ball of radius 1 mm (density 7800 kg m⁻³) falls through glycerine (density 1260 kg m⁻³, η = 1.5 Pa s). Find its terminal velocity. (g = 9.8 m s⁻²)

Try it first, then show the solutionSolution

1. $v_t=\frac{2\times (10^{-3}{)}^2\times (7800-1260)\times 9.8}{9\times 1.5}$
2. $v_t=\frac{0.128}{13.5}=9.5\times 10^{-3}m{s}^{-1}$

Answer: About 9.5 mm s⁻¹

Surface tension, surface energy and angle of contact

Molecular theory: a molecule inside a liquid is pulled equally in all directions by its neighbours (within its sphere of influence, of radius equal to the range of molecular forces). A molecule at the surface has a net inward pull, so the surface layer has extra energy and tends to contract like a stretched membrane. Surface tension S is the force per unit length acting perpendicular to an imaginary line drawn in the surface, tangential to it: $S=\frac{F}{l}$ (N m⁻¹). Surface energy is the work done to increase the surface area by unity at constant temperature: $W=S\Delta A$, so S also equals surface energy per unit area (J m⁻²). A soap film has two surfaces. S decreases as temperature rises and is changed by impurities (detergents lower it). The angle of contact θ is the angle between the liquid surface and the solid surface, measured inside the liquid, at the point of contact: acute for liquids that wet the solid (water on clean glass, θ ≈ 0°), obtuse for those that do not (mercury on glass, θ ≈ 140°). It depends on the balance of cohesive (liquid-liquid) and adhesive (liquid-solid) forces.

Worked example

Find the work done in blowing a soap bubble of radius 1 cm. (S = 0.03 N m⁻¹)

Try it first, then show the solutionSolution

1. A bubble has two surfaces: ΔA = 2 × 4πr² = 8π(0.01)² = 2.51 × 10⁻³ m²
2. W = SΔA = 0.03 × 2.51 × 10⁻³ = 7.5 × 10⁻⁵ J

Answer: $7.5\times 10^{-5}$ J

Excess pressure in drops and bubbles; capillary rise

Surface tension makes a curved surface press inward, so the pressure on the concave side is greater. Liquid drop (one surface): excess pressure $p=\frac{2S}{r}$ (balance the forces on one half: $p\pi r^2=2\pi rS$). Air bubble inside a liquid: also $\frac{2S}{r}$. Soap bubble in air (two surfaces): $p=\frac{4S}{r}$. Smaller bubbles have greater excess pressure. Capillary rise: in a narrow tube of radius r dipped in a wetting liquid, the curved meniscus has lower pressure under it, so liquid rises until the weight of the column balances the upward surface-tension force: $2\pi rS\cos \theta =\pi r^{2}h\rho g$, giving $h=\frac{2S\cos \theta }{r\rho g}$ (Jurin's law: h ∝ 1/r). For θ > 90° (mercury in glass), cos θ is negative and the liquid is depressed. Measuring h, r and ρ gives S (with θ ≈ 0 for water in clean glass). Examples: ink in a nib, oil in a wick, water in the soil and in plant xylem.

Worked example

How high does water rise in a clean glass capillary of radius 0.2 mm? (S = 0.073 N m⁻¹, θ = 0°, g = 9.8 m s⁻²) What is the excess pressure inside a water drop of radius 1 mm?

Try it first, then show the solutionSolution

1. $h=\frac{2\times 0.073\times 1}{2\times 10^{-4}\times 1000\times 9.8}=0.0745m$
2. Drop: $p=\frac{2S}{r}=\frac{2\times 0.073}{10^{-3}}=146Pa$

Answer: About 7.4 cm; 146 Pa

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.

Topics in this chapter

• Continuity Equation and Bernoulli's Theorem
• Surface Tension and Capillary Rise

Mechanical Properties of Fluids formula sheet

Pressure at depth

$p=p_0+\rho gh$

Gauge pressure = ρgh.

Hydraulic lift

$\frac{F_1}{a_1}=\frac{F_2}{a_2}$

Pascal's law; $a_1{d}_1=a_2{d}_2$.

Upthrust

$U=V\sigma g$

Floating fraction = ${\rho }_{body}/{\rho }_{fluid}$.

Reynolds number

$R_e=\frac{\rho vD}{\eta }$

< 1000 streamline; > 2000 turbulent.

Equation of continuity

$a_1{v}_1=a_2{v}_2=Q$

Incompressible fluid.

Bernoulli's theorem

$p+\frac{1}{2}\rho v^2+\rho gh=\text{constant}$

Ideal fluid, streamline flow.

Torricelli's theorem

$v=\sqrt{2gh}$

h = depth of hole below the free surface.

Venturimeter

$v_1=\sqrt{\frac{2\Delta p}{\rho [(a_1/a_2{)}^2-1]}}$

Δp from the manometer.

Viscous force

$F=\eta A\frac{dv}{dx}$

1 poise = 0.1 N s m⁻².

Poiseuille's formula

$Q=\frac{\pi p{r}^4}{8\eta l}$

Q ∝ r⁴.

Stokes' law and terminal velocity

$F=6\pi \eta rv,v_t=\frac{2r^2(\rho -\sigma )g}{9\eta }$

$v_t\propto r^2$.

Surface tension and energy

$S=\frac{F}{l}=\frac{W}{\Delta A}$

Soap film: two surfaces.

Excess pressure

$\text{drop}\frac{2S}{r},\text{soap bubble}\frac{4S}{r}$

Air bubble in liquid: 2S/r.

Capillary rise

$h=\frac{2S\cos \theta }{r\rho g}$

Negative h (depression) for θ > 90°.

Mechanical Properties of Fluids: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the school exam asks Mechanical Properties of Fluids

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

Tricks to remember Mechanical Properties of Fluids

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

• Bernoulli's three Ps

  Pressure + ½ρv² + ρgh: 'Push, Pace, Peak' stay in balance; faster pace, less push.

  Use it when: Writing and applying Bernoulli's equation.

• Two for a drop, four for a bubble

  Drop 2S/r (one skin), soap bubble 4S/r (two skins).

  Use it when: Excess pressure questions.

• Poiseuille's fourth power

  Halve the radius, flow drops 16 times.

  Use it when: Comparing flow rates.

• Terminal velocity is r squared

  Double the radius, four times the speed.

  Use it when: Merging drops and fog questions.

• Wet means acute

  Wetting liquids: θ < 90°, rise; non-wetting (mercury): θ > 90°, fall.

  Use it when: Capillarity and angle of contact.

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.

Case-study practice (Section D style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 5-mark question split into 1 + 1 + 2 + 1 marks.

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.

Where students lose marks in Mechanical Properties of Fluids

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

• ✗Mistake: Using 2S/r for a soap bubble

  Students forget that a soap bubble has two surfaces.

  Wrong working

  Soap bubble of radius 2 cm: p = 2 × 0.03/0.02 = 3 Pa.

  Right working

  p = 4S/r = 6 Pa; work to blow it uses 2 × 4πr².

  ✓Do this: Count the surfaces before writing the formula.

  Usually costs: 1 mark.

• ✗Mistake: Applying Bernoulli without continuity

  In a narrowing pipe, students assume the speed is unchanged, or guess it, instead of using a₁v₁ = a₂v₂.

  Wrong working

  Pressure in the narrow part = pressure in the wide part, since no height changes.

  Right working

  First v₂ = v₁a₁/a₂, then p₂ = p₁ − ½ρ(v₂² − v₁²).

  ✓Do this: Always do continuity first, then Bernoulli.

  Usually costs: 2 marks.

• ✗Mistake: Using diameter for area ratios

  Area ∝ d², but students use the ratio of diameters.

  Wrong working

  Diameter halves, so speed doubles.

  Right working

  Area falls to a quarter, so speed becomes four times.

  ✓Do this: Square the diameter ratio.

  Usually costs: 1 mark.

• ✗Mistake: Forgetting upthrust in terminal velocity

  Students balance weight against drag only.

  Wrong working

  $v_t=\frac{2r^2\rho g}{9\eta }$ for a steel ball in glycerine.

  Right working

  $v_t=\frac{2r^2(\rho -\sigma )g}{9\eta }$: subtract the fluid density.

  ✓Do this: List all three forces: weight, upthrust and drag.

  Usually costs: 1 mark.

• ✗Mistake: Explaining dynamic lift or the Magnus effect without saying where the flow is faster

  Students write 'by Bernoulli's principle the ball moves' without the pressure argument.

  Wrong working

  The spinning ball curves because of Bernoulli's theorem.

  Right working

  On the side where the spin adds to the airflow, air moves faster and pressure is lower; the higher pressure on the other side pushes the ball towards the low-pressure side.

  ✓Do this: Name the faster side, the lower-pressure side and the direction of the net force.

  Usually costs: 1-2 marks.

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.

1. Q1 · 1 mark · MCQExam level

   The SI unit of the coefficient of viscosity is:
   (a) N m⁻²
   (b) N s m⁻²
   (c) N m s⁻¹
   (d) poise

   Hint

   F = ηA dv/dx.

   Answer

   (b) N s m⁻² (Pa s); poise is the CGS unit

2. Q2 · 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

   Hint

   Two surfaces.

   Answer

   (c)

3. Q3 · 1 mark · Very short answerExam level

   State Pascal's law.

   Hint

   Enclosed fluid.

   Answer

   A change of pressure applied to an enclosed fluid is transmitted undiminished to every part of the fluid and the walls of the container.

4. Q4 · 1 mark · Very short answerExam level

   Why does mercury not wet glass?

   Hint

   Cohesion versus adhesion.

   Answer

   Cohesive forces between mercury atoms are stronger than adhesive forces between mercury and glass, so the angle of contact is obtuse.

5. Q5 · 2 marks · Short answerExam level

   State Archimedes' principle. A 0.5 kg body of volume 2 × 10⁻⁴ m³ is fully immersed in water. Find its apparent weight. (g = 9.8 m s⁻²)

   Hint

   Upthrust = Vρg.

   Answer

   A body immersed in a fluid experiences an upthrust equal to the weight of fluid displaced. Upthrust = 2 × 10⁻⁴ × 1000 × 9.8 = 1.96 N; apparent weight = 4.9 − 1.96 = 2.94 N.

6. Q6 · 2 marks · Short answerHarder exam level

   Water rises 3 cm in a capillary of radius 0.5 mm. How high will it rise in a capillary of radius 1 mm?

   Hint

   h ∝ 1/r.

   Answer

   1.5 cm.

7. Q7 · 2 marks · Short answerHarder exam level

   Blood (density 1060 kg m⁻³, η = 3.5 × 10⁻³ Pa s) flows at 0.3 m s⁻¹ through an artery of diameter 2.5 cm. Find the Reynolds number and comment on the flow.

   Hint

   R_e = ρvD/η.

   Answer

   R_e = 1060 × 0.3 × 0.025/3.5 × 10⁻³ = 2270: near the limit, so the flow may become turbulent.

8. Q8 · 2 marks · Short answerHarder exam level

   A plate of area 0.05 m² is moved at 0.1 m s⁻¹ over a fixed plate, separated by an oil film 1 mm thick (η = 0.8 Pa s). Find the force needed.

   Hint

   Velocity gradient = 0.1/0.001.

   Answer

   F = 0.8 × 0.05 × 100 = 4 N.

9. Q9 · 2 marks · Short answerHarder exam level

   By what factor does the rate of flow through a narrow tube change if its radius is halved and the pressure difference is doubled?

   Hint

   Q ∝ pr⁴.

   Answer

   Q ∝ pr⁴: factor = 2 × (1/2)⁴ = 1/8.

10. Q10 · 3 marks · Short answerHarder exam level

    A water tank stands with its free surface 4.9 m above a small hole in its side, and the hole is 1.25 m above the ground. Find the speed of efflux and how far from the tank the jet hits the ground. (g = 9.8 m s⁻²)

    Hint

    Torricelli, then projectile.

    Answer

    v = √(2 × 9.8 × 4.9) = 9.8 m s⁻¹; t = √(2 × 1.25/9.8) = 0.505 s; distance = 4.95 m.

11. Q11 · 3 marks · Short answerHarder exam level

    Compare the excess pressure inside an air bubble of radius 2 mm just below a water surface (S = 0.073 N m⁻¹) with that inside a soap bubble of the same radius (S = 0.03 N m⁻¹).

    Hint

    One surface versus two.

    Answer

    Air bubble in water: 2S/r = 2 × 0.073/0.002 = 73 Pa. Soap bubble: 4S/r = 4 × 0.03/0.002 = 60 Pa.

12. Q12 · 5 marks · Long answerHarder exam level

    (a) State the equation of continuity and explain why a river flows faster through a narrow gorge. [2]
    (b) Explain, using Bernoulli's theorem, how an atomiser works. [2]
    (c) Why are storm winds able to lift roofs off houses? [1]

    Hint

    Speed up, pressure down.

    Answer

    (a) a₁v₁ = a₂v₂ for an incompressible fluid; in the gorge the cross-section is smaller, so the same volume per second must flow faster. (b) Air blown fast across the top of a narrow tube lowers the pressure there; atmospheric pressure on the liquid surface pushes liquid up the tube, and the fast air breaks it into fine droplets. (c) Fast wind over the roof lowers the pressure above it; the still air inside is at higher pressure, so a net upward force lifts the roof.

13. Q13 · 3 marks · Short answerHOTS / competency

    During a cyclone, wind blows at 30 m s⁻¹ across a flat roof of area 120 m², while the air inside the house is still. Find the net upward force on the roof, and the least mass the roof must have (if it is not fastened down) to stay in place. (air density 1.2 kg m⁻³, g = 9.8 m s⁻²)

    Hint

    Inside, v ≈ 0; outside, v = 30 m s⁻¹.

    Answer

    Δp = ½ρv² = ½ × 1.2 × 30² = 540 Pa; upward force = 540 × 120 = 6.48 × 10⁴ N; the roof stays only if mg ≥ 6.48 × 10⁴ N, i.e. m ≥ 6610 kg (about 6.6 tonnes), which is why roofs are bolted down.

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. 1.Pressure at depth h in a liquid?Show answer

   p = p₀ + ρgh.

2. 2.Equation of continuity?Show answer

   a₁v₁ = a₂v₂.

3. 3.Bernoulli's equation?Show answer

   p + ½ρv² + ρgh = constant along a streamline.

4. 4.1 poise in SI units?Show answer

   0.1 N s m⁻².

5. 5.Stokes' law?Show answer

   F = 6πηrv.

6. 6.Terminal velocity formula?Show answer

   v_t = 2r²(ρ − σ)g/9η.

7. 7.Capillary rise formula?Show answer

   h = 2S cos θ/(rρg).

8. 8.What does the Reynolds number indicate?Show answer

   Whether flow is streamline (low) or turbulent (high).

Quick revision checklist

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

• ☐Can you find pressure at a depth and explain the hydraulic lift?
• ☐Can you use Archimedes' principle for floating and immersed bodies?
• ☐Can you use continuity and Bernoulli in a narrowing pipe?
• ☐Can you derive Bernoulli's theorem and list its assumptions?
• ☐Can you explain the venturimeter, atomiser, lift and Magnus effect?
• ☐Can you use Poiseuille's formula and Stokes' law?
• ☐Can you derive the terminal velocity and draw the v-t graph?
• ☐Can you find excess pressure and capillary rise?

Mechanical Properties of Fluids: frequently asked questions