Lesson 2: Mechanical Properties of Fluids

Link to earlier classes : The pressure, density and floating (buoyancy) ideas of classes 9-10 are extended here to flowing fluids.

In the figure : The point deep in the liquid with arrows in all directions; p grows as the depth h grows.

Rule 1: Definition of pressure — Force acting perpendicular to a surface ÷ its area : p = F ÷ A. Unit Pa = N/m²; dimensions [M L⁻¹ T⁻²]. Pressure is a scalar.

• Easy : F = 200 N, A = 0.05 m² : p = 200 ÷ 0.05 = 4000 Pa.
• Medium : A 60 kg person, each foot 200 cm² : on two feet p = 600 ÷ 0.04 = 15000 Pa; on one foot 30000 Pa.
• Exam (CET) : A 2 kg block 10 cm × 20 cm × 5 cm. Largest face down : p = 20 ÷ 0.02 = 1000 Pa; smallest face down : 4000 Pa. Ratio 1 : 4. (Options : 1) 1000 & 2000 2) 1000 & 4000 3) 500 & 2000 4) 2000 & 4000 → answer 2)

Rule 2: Pressure with depth — In a liquid at rest, at depth h : p = p₀ + ρ g h. Only ρ g h is the gauge pressure. Equal at all points on one level; independent of the vessel's shape.

• Easy : Water, h = 5 m : gauge pressure = 1000 × 10 × 5 = 50000 Pa.
• Medium : Absolute pressure at 20 m = 1.013 × 10⁵ + 2 × 10⁵ = 3.013 × 10⁵ Pa.
• Exam (CET) : At what depth in water is the absolute pressure double the atmospheric pressure? ρ g h = p₀ ⇒ h = 1.013 × 10⁵ ÷ 10⁴ = 10.13 m ≈ 10 m. (Options : 1) 5 m 2) 10.1 m 3) 20 m 4) 76 cm → answer 2)

Rule 3: Force on a wall — Pressure on a vertical wall grows with depth; average pressure = ½ ρ g H. So the force on a dam wall is F = ½ ρ g H² b (b = width of the wall).

• Easy : To get the average pressure on a wall take half of the bottom value (0 at the top, ρ g H at the bottom).
• Medium : H = 4 m, b = 10 m : F = ½ × 1000 × 10 × 16 × 10 = 800000 N.
• Exam (CET) : If the water depth is doubled, the force on the wall becomes 4 times (F ∝ H²). (Options : 1) 2× 2) 4× 3) 8× 4) same → answer 2)

Derivation : prove p = p₀ + ρ g h

• Step 1 : Imagine a vertical cylinder in the liquid : cross-section A, top on the surface, base at depth h.
• Step 2 : Air pressure p₀ gives a downward force p₀ A on the top; the liquid gives an upward force p A on the base.
• Step 3 : Weight of the liquid inside (downwards) = ρ A h g.
• Step 4 : Equilibrium : p A = p₀ A + ρ A h g ⇒ p = p₀ + ρ g h (A cancels, so the shape does not matter).

Spot the difference

In the figure : F₁ on the small piston A₁, F₂ on the large piston A₂; the pressure under both is equal.

Rule 1: Pascal's law — Pressure applied to a confined fluid is transmitted undiminished to every part of the fluid and to the walls of the container.

• Easy : Pressing the piston of a closed syringe raises the pressure equally everywhere.
• Medium : Raising the pressure by 10 N/cm² at the small piston raises it by the same amount below the large piston.
• Exam (CET) : Pascal's law treats the liquid as incompressible; in a gas it is less accurate because the gas compresses.

Rule 2: Hydraulic lift — The pressure under both pistons is equal : F₁ ÷ A₁ = F₂ ÷ A₂ ⇒ F₂ = F₁ (A₂ ÷ A₁). The force is multiplied but the work stays the same (energy conservation) : F₁ d₁ = F₂ d₂, so the large piston moves less.

• Easy : A₁ = 2 cm², A₂ = 200 cm², F₁ = 50 N : F₂ = 50 × 100 = 5000 N.
• Medium : To lift a 1200 kg car with A₂ ÷ A₁ = 50 : F₁ = 12000 ÷ 50 = 240 N.
• Exam (CET) : In the same machine, to raise the big piston by 1 cm the small piston must be pushed down 100 cm. (Options : 1) 1 cm 2) 10 cm 3) 100 cm 4) 50 cm → answer 3)

Rule 3: Measuring pressure — A barometer measures atmospheric pressure : p₀ = ρ g h, mercury column 76 cm. A manometer measures gauge pressure : p gauge = ρ g h, h = difference of the two levels.

• Easy : Mercury 76 cm : p₀ = 13600 × 10 × 0.76 ≈ 1.03 × 10⁵ Pa (with g = 9.8 it is 1.013 × 10⁵ Pa).
• Medium : Mercury levels in a manometer differ by 15 cm : gauge pressure = 13600 × 10 × 0.15 = 20400 Pa.
• Exam (CET) : With water instead of mercury the barometer height = 13.6 × 0.76 = 10.34 m ≈ 10.3 m. (Options : 1) 0.76 m 2) 1.03 m 3) 10.3 m 4) 103 m → answer 3)

Derivation : prove F₂ = F₁ (A₂ ÷ A₁)

• Step 1 : A force F₁ on the small piston creates a pressure p = F₁ ÷ A₁ in the liquid.
• Step 2 : By Pascal's law the same pressure acts under the large piston : p = F₂ ÷ A₂.
• Step 3 : Equate them : F₁ ÷ A₁ = F₂ ÷ A₂ ⇒ F₂ = F₁ (A₂ ÷ A₁). Work : F₁ d₁ = F₂ d₂ (incompressible, V = A₁d₁ = A₂d₂).

Spot the difference

In the figure : The liquid surface near a glass wall; θ is measured inside the liquid.

Rule 1: Surface tension — Force per unit length on a line in the free surface, tangential to the surface : S = F ÷ l (unit N/m; [M T⁻²]). Also S = W ÷ ΔA (J/m²), i.e. surface energy per unit area.

• Easy : A soap film frame of length l = 0.1 m, S = 0.03 N/m (two surfaces) : F = 2 S l = 0.006 N.
• Medium : Film area increased by 20 cm² (two surfaces) : W = S × 2ΔA = 0.03 × 2 × 20 × 10⁻⁴ = 1.2 × 10⁻⁴ J.
• Exam (CET) : S decreases when the temperature rises, and also when soap or detergent is added. (Options : 1) rises with T 2) falls with T 3) unchanged 4) doubles → answer 2)

Rule 2: Surface energy — Work done to increase the surface area is stored as surface energy : W = S ΔA. A liquid tries to have minimum area, so a free drop is spherical.

• Easy : S = 0.07 N/m, ΔA = 10⁻³ m² : W = 7 × 10⁻⁵ J.
• Medium : Drop of radius 2 mm : area 4πr² = 5.03 × 10⁻⁵ m²; surface energy = 0.07 × 5.03 × 10⁻⁵ = 3.52 × 10⁻⁶ J.
• Exam (CET) : For a soap bubble with two surfaces take total area 2 × 4πr². (Options : 1) 4πr² 2) 8πr² 3) 2πr² 4) πr² → answer 2)

Rule 3: Angle of contact — The angle θ measured inside the liquid between the liquid surface and the solid wall at their contact line. θ < 90° ⇒ the liquid wets the solid (water–glass); θ > 90° ⇒ it does not wet (mercury–glass).

• Easy : Water–glass : θ ≈ 0° (acute), concave meniscus.
• Medium : Mercury–glass : θ ≈ 135°, convex meniscus.
• Exam (CET) : Adhesion larger ⇒ small θ; cohesion larger ⇒ large θ. (Options : 1) mercury–glass : adhesion > cohesion 2) water–glass : adhesion > cohesion 3) both equal 4) no relation → answer 2)

Derivation : show that S = F ÷ l and S = W ÷ ΔA are the same

• Step 1 : A soap film on a U-shaped wire with a sliding wire of length l; the film has two surfaces ⇒ pulling force F = 2 S l.
• Step 2 : Pull the wire a distance x : work W = F x = 2 S l x.
• Step 3 : Increase in area (two surfaces) ΔA = 2 l x.
• Step 4 : So W ÷ ΔA = 2 S l x ÷ 2 l x = S. The two definitions are identical.

Spot the difference

In the figure : Water rises to a height h in a narrow tube; the meniscus is concave.

Rule 1: Excess pressure — The pressure inside a curved surface exceeds the outside pressure. Drop : ΔP = 2S ÷ r; soap bubble (two surfaces) : ΔP = 4S ÷ r; air bubble inside a liquid : ΔP = 2S ÷ r.

• Easy : Water drop r = 1 mm, S = 0.07 N/m : ΔP = 2 × 0.07 ÷ 0.001 = 140 Pa.
• Medium : Soap bubble r = 1 cm, S = 0.03 N/m : ΔP = 4 × 0.03 ÷ 0.01 = 12 Pa.
• Exam (CET) : Two soap bubbles of radii 1 : 2 have excess pressures in the ratio 2 : 1 (ΔP ∝ 1/r). The smaller bubble has the higher pressure. (Options : 1) 1 : 2 2) 2 : 1 3) 1 : 4 4) 4 : 1 → answer 2)

Rule 2: Capillary action — In a narrow tube a liquid rises or falls : h = 2 S cos θ ÷ (r ρ g). θ < 90° ⇒ rise; θ > 90° ⇒ depression. h ∝ 1/r : the narrower the tube the larger the effect.

• Easy : Water, r = 0.5 mm, S = 0.07, θ = 0° : h = 2 × 0.07 ÷ (0.0005 × 1000 × 10) = 2.8 cm.
• Medium : Mercury, r = 1 mm, S = 0.465, θ = 135°, ρ = 13600 : h = -4.8 mm (negative = down).
• Exam (CET) : Halving the radius doubles h. (Options : 1) half 2) double 3) four times 4) same → answer 2)

Rule 3: Splitting and merging drops — When a big drop splits into n small drops the area increases, so energy is spent : W = S · 4πR² (n^{1/3} − 1). Merging releases energy.

• Easy : Volume is conserved : n r³ = R³ ⇒ r = R ÷ n^{1/3}.
• Medium : A drop of R = 2 mm splits into 8 drops (r = 1 mm) : W = 0.07 × 4π(0.002)² × (2 − 1) = 3.52 × 10⁻⁶ J.
• Exam (CET) : For n = 1000, n^{1/3} = 10 ⇒ area becomes 10 times. (Options : 1) 10× 2) 100× 3) 1000× 4) same → answer 1)

Derivation : capillary rise formula h = 2S cos θ ÷ (r ρ g)

• Step 1 : Along the circumference (2πr) the surface-tension force S pulls at angle θ; its vertical component is S cos θ.
• Step 2 : Total upward force = (2πr) S cos θ.
• Step 3 : Weight of the raised liquid = π r² h ρ g.
• Step 4 : Balance : π r² h ρ g = 2π r S cos θ ⇒ h = 2 S cos θ ÷ (r ρ g).

Spot the difference

In the figure : Where the pipe narrows the streamlines crowd together and the speed v₂ is larger.

Rule 1: Streamline and turbulent flow — In streamline flow every particle follows a fixed path and paths never cross. Beyond a critical speed the flow becomes turbulent — eddies form.

• Easy : Honey pouring slowly in a smooth stream = streamline.
• Medium : Foaming water under a waterfall = turbulent.
• Exam (CET) : Turbulent flow wastes more energy; the streamline idea applies only to streamline flow. (Options : 1) both 2) only streamline 3) only turbulent 4) neither → answer 2)

Rule 2: Equation of continuity — For an incompressible fluid the volume flowing per second is constant : A₁ v₁ = A₂ v₂ (= Q, volume flow rate, m³/s). Where the cross-section is smaller the speed is larger.

• Easy : Pipe diameter 4 cm → 2 cm, v₁ = 1.5 m/s : v₂ = 1.5 × (4 ÷ 2)² = 6 m/s.
• Medium : Pipe diameter 4 cm, v = 1.5 m/s : Q = π(0.02)² × 1.5 = 1.88 × 10⁻³ m³/s ≈ 1.88 litre/s.
• Exam (CET) : Halving the diameter makes the speed 4 times (v ∝ 1/d²). (Options : 1) 2× 2) 4× 3) 8× 4) half → answer 2)

Rule 3: Reynolds number — Rₑ = ρ v d ÷ η (a pure number, no unit). Below about 1000 ⇒ streamline; above about 2000 ⇒ turbulent; in between unstable (check the limits against your textbook).

• Easy : Water, d = 2 cm, v = 0.05 m/s, η = 10⁻³ Pa·s : Rₑ = 1000 × 0.05 × 0.02 ÷ 10⁻³ = 1000.
• Medium : Doubling the speed doubles Rₑ; so the critical speed v_c = Rₑ η ÷ (ρ d).
• Exam (CET) : Large η (honey, oil) ⇒ small Rₑ ⇒ flow stays streamline. (Options : 1) honey turbulent easily 2) honey streamline easily 3) no difference 4) only gases → answer 2)

Derivation : continuity equation A₁v₁ = A₂v₂

• Step 1 : In a pipe the speed is v₁ at cross-section A₁; volume entering in Δt = A₁ v₁ Δt.
• Step 2 : Volume leaving through A₂ in Δt = A₂ v₂ Δt.
• Step 3 : The fluid is incompressible (ρ constant) and nothing accumulates ⇒ mass in = mass out.
• Step 4 : Hence A₁ v₁ = A₂ v₂.

Spot the difference

In the figure : On the ball W acts down, B and the drag act up; at terminal speed W = B + F.

Rule 1: Viscosity — The property that opposes relative motion between layers of a fluid. Viscous force F = η A (dv ÷ dx). η = coefficient of viscosity; unit Pa·s = N s/m²; [M L⁻¹ T⁻¹]. 1 poise = 0.1 Pa·s.

• Easy : 2 poise = 0.2 Pa·s.
• Medium : Plate A = 0.1 m², gap d = 2 mm, speed 0.5 m/s, η = 0.1 Pa·s : F = 0.1 × 0.1 × 0.5 ÷ 0.002 = 2.5 N.
• Exam (CET) : On heating, η of a liquid decreases while η of a gas increases. (Options : 1) both rise 2) liquid falls, gas rises 3) liquid rises, gas falls 4) both fall → answer 2)

Rule 2: Stokes' law — The viscous drag on a sphere of radius r moving at speed v through a fluid : F = 6π η r v.

• Easy : η = 0.5 Pa·s, r = 2 mm, v = 0.1 m/s : F = 6π × 0.5 × 0.002 × 0.1 = 1.88 × 10⁻³ N.
• Medium : Doubling the speed doubles F (F ∝ v).
• Exam (CET) : Doubling both radius and speed ⇒ F becomes 4 times. (Options : 1) 2× 2) 3× 3) 4× 4) 8× → answer 3)

Rule 3: Terminal velocity — When weight = buoyancy + drag the acceleration is zero and the speed is constant : v_t = 2 r² (ρ − σ) g ÷ (9 η) (ρ of the sphere, σ of the liquid).

• Easy : Steel ball r = 1 mm (ρ = 7800) in glycerine (σ = 1260, η = 0.8) : v_t = 1.82 cm/s.
• Medium : Raindrop r = 0.1 mm in air (η = 1.8 × 10⁻⁵) : v_t ≈ 1.23 m/s.
• Exam (CET) : Doubling the radius ⇒ v_t becomes 4 times (v_t ∝ r²). (Options : 1) 2× 2) 4× 3) 8× 4) half → answer 2)

Derivation : the terminal velocity formula

• Step 1 : Weight of the sphere W = (4/3)π r³ ρ g (down).
• Step 2 : Buoyancy B = (4/3)π r³ σ g (up); drag F = 6π η r v (up).
• Step 3 : At terminal speed W = B + F ⇒ (4/3)π r³ (ρ − σ) g = 6π η r v_t.
• Step 4 : Solving : v_t = 2 r² (ρ − σ) g ÷ (9 η).

Spot the difference

In the figure : The speed v₂ is higher in the narrow part; the lower liquid level shows the lower pressure p₂.

Rule 1: Bernoulli's equation — Along a streamline of an ideal fluid in steady flow : p + ½ ρ v² + ρ g h = constant (energy conservation). Three terms : pressure, kinetic and potential energy per unit volume.

• Easy : In a horizontal pipe (h constant) : where speed rises, pressure falls.
• Medium : Horizontal pipe, A₁ = 20 cm², A₂ = 10 cm², v₁ = 2 m/s ⇒ v₂ = 4 m/s : p₁ − p₂ = ½ × 1000 × (16 − 4) = 6000 Pa.
• Exam (CET) : Bernoulli holds for non-viscous, incompressible, steady flow; viscosity wastes energy. (Options : 1) all flows 2) only ideal fluids 3) only gases 4) only turbulent → answer 2)

Rule 2: Torricelli's law — Speed of efflux from a hole at depth h below the surface : v = √(2 g h) (same as free fall).

• Easy : h = 1.8 m : v = √(2 × 10 × 1.8) = 6 m/s.
• Medium : Tank 5 m high; hole 1.25 m below the surface : v = 5 m/s; fall time √(2 × 3.75 ÷ 10) = 0.866 s; range = 4.33 m.
• Exam (CET) : Range x = 2√[h (H − h)]; maximum at h = H ÷ 2 (x = H). (Options : 1) h = H 2) h = H ÷ 2 3) h = H ÷ 4 4) h = 0 → answer 2)

Rule 3: Applications of Bernoulli — Aeroplane wing : higher speed above ⇒ lower pressure ⇒ lift. Atomizer : fast air jet lowers the pressure so liquid rises. Venturimeter : pressure drop in the narrow part gives the flow rate. Swinging ball (Magnus effect).

• Easy : Air over the wing 80 m/s, under it 70 m/s (ρ = 1.2) : Δp = ½ × 1.2 × (6400 − 4900) = 900 Pa.
• Medium : Wing area 20 m² : lift = 900 × 20 = 18000 N.
• Exam (CET) : In a strong wind a roof lifts off because pressure above is lower than inside. (Options : 1) higher pressure above 2) lower pressure above 3) equal 4) density rises → answer 2)

Derivation : Bernoulli's equation (from the work–energy theorem)

• Step 1 : Take a small mass of fluid between section 1 (A₁, p₁, v₁, h₁) and section 2 (A₂, p₂, v₂, h₂). In time Δt the same volume ΔV passes both.
• Step 2 : Net work done by pressure = (p₁ − p₂) ΔV.
• Step 3 : Gravity does work − ρ ΔV g (h₂ − h₁); change in kinetic energy = ½ ρ ΔV (v₂² − v₁²).
• Step 4 : Work–energy theorem : (p₁ − p₂) ΔV − ρ ΔV g (h₂ − h₁) = ½ ρ ΔV (v₂² − v₁²).
• Step 5 : Divide by ΔV and rearrange : p₁ + ½ρv₁² + ρgh₁ = p₂ + ½ρv₂² + ρgh₂.

Spot the difference