Mechanical Properties of Fluids | AAI ATC Physics

💧

Section 9.2

Pressure in Fluids

Pressure is defined as the normal force per unit area exerted by a fluid. When an object is submerged in a fluid at rest, the fluid exerts a force perpendicular to every surface of the object.

P = F / A

Pav = Average Pressure | F = Normal Force (N) | A = Area (m²) | Unit: Pascal (Pa) = N m⁻²

Pressure is a scalar quantity. The SI unit is Pascal (Pa), named after Blaise Pascal (1623–1662). Other units:

1 atm = 1.013 × 10⁵ Pa (pressure at sea level)
1 bar = 10⁵ Pa
1 torr = 133 Pa = 1 mm of Hg

🖼️ Insert Diagram: Fluid pressure on submerged object (force perpendicular to surfaces) Recommended: Fig 9.1(a) from NCERT — Normal forces on submerged object

Density (ρ) is another key quantity:

ρ = m / V

ρ = Density (kg m⁻³) | m = Mass (kg) | V = Volume (m³)

Fluid	Density ρ (kg m⁻³)
Water (4°C)	1.00 × 10³
Sea Water	1.03 × 10³
Mercury	13.6 × 10³
Air (STP)	1.29
Whole Blood	1.06 × 10³

🎯 Practice MCQs — Pressure

AAI ATC Style

1 A force of 500 N acts on a piston of area 0.025 m². What is the pressure exerted?

A 1.25 × 10³ Pa

B 500 Pa

C 2 × 10⁴ Pa

D 1.25 × 10⁴ Pa

✅ P = F/A = 500 / 0.025 = 20,000 Pa = 2 × 10⁴ Pa. Pressure = Normal Force ÷ Area. This is a direct application of the basic pressure formula.

2 The density of mercury is 13.6 × 10³ kg/m³. What is the pressure at the bottom of a mercury column of height 0.76 m? (g = 9.8 m/s²)

A 9.8 × 10³ Pa

B 1.01 × 10⁵ Pa

C 13.6 × 10³ Pa

D 7.6 × 10⁴ Pa

✅ P = ρgh = 13600 × 9.8 × 0.76 ≈ 1.01 × 10⁵ Pa = 1 atm. This is the standard atmosphere — exactly how Torricelli's barometer is calibrated.

3 Two femur bones, each with cross-sectional area 10 cm², support a body mass of 40 kg. The average pressure sustained by the femurs is: (g = 10 m/s²)

A 4 × 10⁴ Pa

B 1 × 10⁵ Pa

C 2 × 10⁵ Pa

D 4 × 10⁵ Pa

✅ Total area = 2 × 10 cm² = 20 × 10⁻⁴ m². F = 40 × 10 = 400 N. P = 400 / 20×10⁻⁴ = 2 × 10⁵ Pa. Note: both femurs together support the load, so total area = 2 × 10 cm².

⚖️

Section 9.2.1 & 9.2.2

Pascal's Law & Pressure with Depth

Pascal's Law: Pressure applied to an enclosed fluid is transmitted undiminished and equally in all directions to every point of the fluid and the walls of the container.

P = Pₐ + ρgh

P = Absolute Pressure | Pₐ = Atmospheric Pressure | ρ = Density | g = 9.8 m/s² | h = Depth

The pressure difference depends on vertical distance h, fluid density ρ, and g. It does NOT depend on the shape or cross-section of the container — this is the hydrostatic paradox.

Gauge Pressure = P − Pₐ = ρgh (excess over atmospheric pressure).

🖼️ Insert Diagram: Hydrostatic Paradox — Three vessels A, B, C of different shapes with equal liquid levels Recommended: Fig 9.4 from NCERT

      🔑 AAI ATC Exam Tip: Atmospheric pressure at sea level = 1.013 × 10⁵ Pa ≈ 1 atm ≈ 76 cm of Hg. The mercury barometer (Torricelli) uses this principle. A 10 mm drop in Hg level = sign of approaching storm.
    

📌 Key Points — Pascal's Law

Pressure in a fluid at rest is same at all points at the same horizontal level.
Gauge pressure = P − Pa = ρgh. This is what tyre gauges and BP gauges measure.
1 torr = 133 Pa (used in medicine/physiology); 1 bar = 10⁵ Pa (used in meteorology).
Mercury barometer: Pa = ρgh → h ≈ 76 cm at sea level.
Open-tube manometer measures gauge pressure proportional to height h.

🎯 Practice MCQs — Pascal's Law & Pressure with Depth

AAI ATC Style

1 A swimmer is 10 m below the surface of a lake (ρ_water = 1000 kg/m³, g = 10 m/s²). The absolute pressure at that depth is:

A 1.01 × 10⁵ Pa

B 2.01 × 10⁵ Pa

C 1.00 × 10⁵ Pa

D 3.01 × 10⁵ Pa

✅ P = Pa + ρgh = 1.01×10⁵ + 1000×10×10 = 1.01×10⁵ + 10⁵ = 2.01×10⁵ Pa ≈ 2 atm. This is a 100% increase in pressure from sea level!

2 Pascal's law states that an externally applied pressure to an enclosed fluid is transmitted:

A Only to the nearest wall of the container

B Upward only due to buoyancy

C Undiminished and equally in all directions

D Only downward due to gravity

✅ Pascal's Law — pressure is transmitted undiminished in all directions. This is the operating principle behind hydraulic lifts, hydraulic brakes, and all hydraulic machines.

3 At a depth of 1000 m in an ocean (ρ_sea = 1.03 × 10³ kg/m³, g = 10 m/s²), the gauge pressure is approximately:

A 1 atm

B 50 atm

C 103 atm

D 104 atm

✅ Pg = ρgh = 1.03×10³ × 10 × 1000 = 103 × 10⁵ Pa ≈ 103 atm. Submarines must withstand such enormous pressures. This is why submarine hulls use thick high-tensile steel.

🔧

Section 9.2.4

Hydraulic Machines

Based on Pascal's law, hydraulic machines use fluid to multiply force. In a hydraulic lift, a small force on a small piston creates a large force on a larger piston.

F₂ = F₁ × (A₂ / A₁)

Mechanical Advantage = A₂/A₁ | F₁ = Input force | F₂ = Output force | A₁, A₂ = Piston areas

Hydraulic Brakes also work on this principle — a small force on the brake pedal generates large braking force equally on all four wheels.

🖼️ Insert Diagram: Hydraulic Lift Principle — Small piston A₁ & Large piston A₂ Recommended: Fig 9.6(b) from NCERT — Schematic of hydraulic lift

🎯 Practice MCQs — Hydraulic Machines

AAI ATC Style

1 In a hydraulic lift, pistons have areas A₁ = 5 cm² and A₂ = 50 cm². A force of 20 N is applied on the smaller piston. The force on the larger piston is:

A 20 N

B 100 N

C 200 N

D 500 N

✅ F₂ = F₁ × (A₂/A₁) = 20 × (50/5) = 20 × 10 = 200 N. Mechanical advantage = 10. The hydraulic lift multiplies force by the ratio of areas.

2 Two syringes connected by a tube have piston diameters 1 cm and 3 cm. A 10 N force on the smaller piston produces what force on the larger piston?

A 10 N

B 30 N

C 90 N

D 270 N

✅ F₂ = F₁ × (r₂/r₁)² = 10 × (3/1)² = 10 × 9 = 90 N. Since area ∝ r², the force ratio = (r₂/r₁)² = 9. Answer = 90 N.

3 In a car lift, a compressed air force F₁ acts on a small piston of radius 5 cm. The pressure is transmitted to a piston of radius 15 cm lifting a car of mass 1350 kg (g = 9.8 m/s²). The value of F₁ is approximately:

A 1470 N ≈ 1.5 × 10³ N

B 13230 N

C 4900 N

D 735 N

✅ F₁ = F₂ × (A₁/A₂) = 1350×9.8 × (5/15)² = 13230 × (1/9) ≈ 1470 N. The small piston area is 1/9th of large piston area, so input force is 1/9th of the weight to be lifted.

🌊

Section 9.3

Streamline Flow & Equation of Continuity

A flow is steady (streamline) when every fluid particle at a given point has the same velocity at all times. A streamline is a curve whose tangent at any point is in the direction of the fluid velocity.

No two streamlines can cross. When flow speed exceeds the critical speed, flow becomes turbulent.

Av = constant | A₁v₁ = A₂v₂

Equation of Continuity | A = Cross-section area | v = Velocity | (Conservation of Mass for incompressible fluids)

At narrower portions (smaller A), velocity is greater. At wider portions, velocity is smaller. This is why water speeds up through a narrow nozzle.

🖼️ Insert Diagram: Streamlines in a pipe — velocity increases at constriction Recommended: Fig 9.7(b) from NCERT — Region of streamline flow with three cross-sections

🎯 Practice MCQs — Streamline Flow

AAI ATC Style

1 Water flows through a pipe of cross-section 8 cm² at 2 m/s. It enters a narrow section of cross-section 2 cm². The velocity in the narrow section is:

A 2 m/s

B 4 m/s

C 8 m/s

D 16 m/s

✅ A₁v₁ = A₂v₂ → 8×2 = 2×v₂ → v₂ = 16/2 = 8 m/s. By the equation of continuity, flow rate (Av) is constant. Smaller area = higher velocity.

2 The spray pump tube has cross-section 8 cm² with 40 holes each of diameter 1 mm. If liquid flows inside at 1.5 m/min, the ejection speed through holes is:

A 0.25 m/s

B 0.5 m/s

C 6.37 m/s

D 1.5 m/s

✅ A_tube = 8 cm² = 8×10⁻⁴ m². Each hole area = π(0.5×10⁻³)² ≈ 7.85×10⁻⁷ m². Total hole area = 40 × 7.85×10⁻⁷ ≈ 3.14×10⁻⁵ m². v_tube = 1.5/60 = 0.025 m/s. v₂ = A₁v₁/A₂ ≈ 6.37 m/s.

3 Two streamlines in steady flow of a fluid:

A Can cross at high flow speeds

B Always run parallel to each other

C Never cross each other

D Cross only in turbulent flow

✅ Streamlines never cross. If they did, a fluid particle at the crossing point would have two different velocities simultaneously, which is impossible in steady flow.

✈️

Section 9.4

Bernoulli's Principle

Bernoulli's Principle is an expression of energy conservation for fluid flow. Developed by Swiss physicist Daniel Bernoulli (1738).

P + ½ρv² + ρgh = constant

P = Pressure | ρ = Fluid density | v = Velocity | h = Height | (For non-viscous, incompressible, steady flow)

As flow speed increases, pressure decreases — and vice versa. This principle explains aircraft lift, the Magnus effect (spinning ball), Torricelli's law, and many other phenomena.

      ✈️ Aircraft Lift (Aerofoil): Air flowing over the curved upper surface of a wing travels faster than air below → lower pressure above → net upward lift force. For a Boeing 747: ΔP ≈ 6.5 × 10³ Pa with air speed ≈ 960 km/h supports 3.3 × 10⁵ kg.
    

📌 Applications of Bernoulli's Principle

Torricelli's Law: Speed of efflux from an open tank: v = √(2gh)
Magnus Effect: Spinning ball deviates from parabolic path (cricket/tennis swing)
Aerofoil / Aircraft Lift: Higher speed over wing = lower pressure = upward lift
Venturimeter: Flow speed measurement using pressure difference at constriction
Limitations: Not valid for viscous fluids, turbulent flow, or compressible fluids

🖼️ Insert Diagram: Aerofoil with streamlines — crowded above wing, sparse below; lift force shown Recommended: Fig 9.11(c) from NCERT — Air flowing past aerofoil

🎯 Practice MCQs — Bernoulli's Principle

AAI ATC Style

1 In a test on an aeroplane model, flow speeds on the upper and lower wing surfaces are 70 m/s and 63 m/s respectively. Wing area = 2.5 m², air density = 1.3 kg/m³. The lift force on the wing is:

A 1500 N

B 1.505 × 10³ N

C 3500 N

D 546 N

✅ ΔP = ½ρ(v₂²−v₁²) = ½×1.3×(70²−63²) = ½×1.3×(4900−3969) = ½×1.3×931 = 605.15 Pa. Lift = ΔP × A = 605.15 × 2.5 ≈ 1513 N ≈ 1.5 × 10³ N.

2 According to Torricelli's law, the speed of efflux of a liquid from a hole at depth h below the open surface is:

A v = gh

B v = √(gh)

C v = √(2gh)

D v = 2gh

✅ v = √(2gh) — This is Torricelli's Law. The efflux speed equals the speed of a freely falling body from height h. Derived from Bernoulli's equation with the tank open to atmosphere.

3 A fully loaded Boeing aircraft (mass 3.3 × 10⁵ kg, wing area 500 m²) flies at constant altitude. The pressure difference between lower and upper wing surfaces is: (g = 9.8 m/s²)

A 1.3 × 10³ Pa

B 3.3 × 10³ Pa

C 6.47 × 10³ Pa

D 9.8 × 10³ Pa

✅ ΔP × A = mg → ΔP = mg/A = (3.3×10⁵ × 9.8)/500 = 3.234×10⁶/500 = 6468 Pa ≈ 6.47 × 10³ Pa. At level flight, lift = weight. ΔP × wing area = aircraft weight.

🍯

Section 9.5

Viscosity & Stokes' Law

Viscosity is the property of fluids that offers resistance to flow — it is the internal friction between fluid layers. The coefficient of viscosity η (eta) is defined as:

η = (F/A) / (v/l) = Fl / vA

η = Coefficient of viscosity (Pa·s or Poiseuille) | F/A = Shear stress | v/l = Strain rate

Stokes' Law: The viscous drag on a sphere of radius a moving at velocity v through fluid of viscosity η:

F = 6πηav

Terminal velocity: vₜ = 2a²(ρ−σ)g / 9η

📌 Key Facts about Viscosity

Liquids: viscosity DECREASES with temperature (molecules become more mobile).
Gases: viscosity INCREASES with temperature (more random atomic collisions).
Blood is more viscous than water; relative viscosity of blood is constant 0°C–37°C.
Terminal velocity ∝ a² (square of radius) and ∝ 1/η (inversely to viscosity).
SI unit of viscosity: Pa·s (poiseuille, Pl). Dimensions: [ML⁻¹T⁻¹].

🖼️ Insert Diagram: Velocity profile for viscous flow in a pipe — max at centre, zero at walls Recommended: Fig 9.12(b) from NCERT — Velocity distribution in pipe

🎯 Practice MCQs — Viscosity & Stokes' Law

AAI ATC Style

1 A metal block of area 0.10 m² on a liquid film of thickness 0.30 mm moves at 0.085 m/s under a force equal to a 0.010 kg mass (g = 9.8 m/s²). The coefficient of viscosity of the liquid is:

A 1.5 × 10⁻³ Pa·s

C 3.46 × 10⁻³ Pa·s

B 9.8 × 10⁻³ Pa·s

D 6.5 × 10⁻³ Pa·s

✅ F = mg = 0.01×9.8 = 0.098 N. Shear stress = F/A = 0.098/0.10 = 0.98 N/m². Strain rate = v/l = 0.085/(0.30×10⁻³) = 283.3 s⁻¹. η = stress/strain rate = 0.98/283.3 ≈ 3.46×10⁻³ Pa·s.

2 The terminal velocity of a copper ball (radius 2 mm) falling in oil at 20°C is 6.5 cm/s. Density of oil = 1.5×10³ kg/m³, density of copper = 8.9×10³ kg/m³. The viscosity of oil is approximately:

A 0.1 Pa·s

B 0.99 Pa·s

C 9.9 Pa·s

D 1.99 Pa·s

✅ η = 2a²(ρ−σ)g / 9vₜ = 2×(2×10⁻³)²×(8900−1500)×9.8 / (9×0.065) = 2×4×10⁻⁶×7400×9.8/0.585 ≈ 0.99 Pa·s.

3 How does the viscosity of a liquid change with increasing temperature?

A Increases

B Decreases

C Remains constant

D First increases then decreases

✅ Viscosity of liquids DECREASES with temperature. Higher temperature → more mobile molecules → less internal friction. (Opposite for gases — gas viscosity increases with temperature.)

🫧

Section 9.6

Surface Tension

Surface tension (S) is the force per unit length (or surface energy per unit area) acting in the plane of a liquid surface. It arises because surface molecules have fewer neighbours than interior molecules → extra potential energy.

S = F / 2l | S = W / 2ΔA

S = Surface tension (N/m) | F = Force | l = Length | Unit: N m⁻¹

Key consequences: Spherical drops & bubbles (minimum surface area for given volume); capillary rise/fall; pressure difference across curved surfaces.

Excess pressure inside drops and bubbles:

Surface	Excess Pressure Formula	Reason
Liquid Drop	ΔP = 2S/r	1 liquid-air interface
Air Bubble in Liquid	ΔP = 2S/r	1 liquid-air interface
Soap Bubble	ΔP = 4S/r	2 liquid-air interfaces

Capillary Rise: h = 2S cosθ / (ρga) — the rise height is inversely proportional to capillary radius a.

🖼️ Insert Diagram: Angle of contact — water on glass (acute) vs mercury on glass (obtuse) Recommended: Fig 9.17 from NCERT — Interfacial tensions and angle of contact

📌 Key Points — Surface Tension

Surface tension of liquids generally decreases with temperature.
Water on glass: acute contact angle (water wets glass). Mercury on glass: obtuse angle (does not wet).
Soaps/detergents reduce angle of contact → improve wetting and penetration.
Water-proofing agents create large contact angle → water forms droplets (lotus effect).
Capillary rise: h ∝ 1/a (smaller tube = greater rise).

🎯 Practice MCQs — Surface Tension

AAI ATC Style

1 The excess pressure inside a soap bubble of radius 5 mm is: (Surface tension of soap solution = 2.50 × 10⁻² N/m)

A 5 Pa

B 10 Pa

C 20 Pa

D 40 Pa

✅ Soap bubble has 2 surfaces: ΔP = 4S/r = 4 × 0.025 / 0.005 = 0.1/0.005 = 20 Pa. Remember: bubble (4S/r) vs drop (2S/r). The factor of 4 comes from two interfaces.

2 A soap film is stretched on a U-shaped wire of slider length 30 cm and supports a weight of 1.5 × 10⁻² N. The surface tension of the film is:

A 0.01 N/m

B 0.015 N/m

C 0.025 N/m

D 0.05 N/m

✅ S = F/(2l) = 0.015/(2×0.30) = 0.015/0.60 = 0.025 N/m. The factor 2l accounts for the two sides of the soap film. S = 2.5 × 10⁻² N/m.

3 The pressure inside a mercury drop of radius 3 mm is: (S_mercury = 4.65 × 10⁻¹ N/m, Pₐ = 1.01 × 10⁵ Pa)

A 1.01 × 10⁵ Pa

B 1.013 × 10⁵ Pa

C 1.013 × 10⁵ Pa + 310 Pa = 1.041 × 10⁵ Pa

D 2 × 10⁵ Pa

✅ Excess pressure = 2S/r = 2×0.465/0.003 = 310 Pa. Total pressure = Pa + 310 = 1.01×10⁵ + 310 ≈ 1.013×10⁵ Pa. For a drop (not bubble), only 1 surface → 2S/r formula.

Mechanical Propertiesof Fluids

Pressure in Fluids

🎯 Practice MCQs — Pressure

Pascal's Law & Pressure with Depth

📌 Key Points — Pascal's Law

🎯 Practice MCQs — Pascal's Law & Pressure with Depth

Hydraulic Machines

🎯 Practice MCQs — Hydraulic Machines

Streamline Flow & Equation of Continuity

🎯 Practice MCQs — Streamline Flow

Bernoulli's Principle

📌 Applications of Bernoulli's Principle

🎯 Practice MCQs — Bernoulli's Principle

Viscosity & Stokes' Law

📌 Key Facts about Viscosity

🎯 Practice MCQs — Viscosity & Stokes' Law

Surface Tension

📌 Key Points — Surface Tension

🎯 Practice MCQs — Surface Tension

🚀 Ace Your AAI ATC Exam!

Mechanical Properties
of Fluids