What this chapter is about
This chapter extends your study of matter by examining how fluids — liquids and gases — behave under various conditions. While solids have definite shape and resist deformation, fluids flow and take the shape of their container. You will learn how pressure acts within fluids, why objects float or sink, and how fluids move through pipes and around obstacles.
The chapter builds on your understanding of force and energy from earlier work. You will meet fundamental principles like Pascal's law (used in hydraulic machines), Archimedes' principle (explaining buoyancy), and Bernoulli's principle (explaining how aeroplanes generate lift). These ideas connect directly to everyday observations: why a steel ship floats, how blood flows through arteries, and why a fast-moving train creates a dangerous pull on people standing nearby.
After studying this chapter, you should be able to calculate pressure at various depths in a fluid, predict whether an object will float, apply the equation of continuity to flowing liquids, and use Bernoulli's equation to analyse fluid flow. You will also understand viscosity and surface tension — properties that affect everything from oil flow in machines to water drops on a lotus leaf.
Key ideas
- Pressure in a fluid acts equally in all directions at a point and increases with depth. At depth h below the surface, pressure P = P₀ + ρgh, where P₀ is atmospheric pressure, ρ is fluid density, and g is acceleration due to gravity.
- Pascal's law states that pressure applied to an enclosed fluid is transmitted undiminished to every part of the fluid and to the walls of the container. This principle enables hydraulic lifts and brakes.
- Archimedes' principle states that when a body is wholly or partially immersed in a fluid, it experiences an upward buoyant force equal to the weight of the fluid displaced. An object floats when the buoyant force equals its weight.
- Equation of continuity for incompressible fluid flow: A₁v₁ = A₂v₂, where A is cross-sectional area and v is flow speed. This means fluid speeds up when passing through a narrower section.
- Bernoulli's principle states that for steady, non-viscous, incompressible flow, the sum P + ½ρv² + ρgh remains constant along a streamline. Where speed increases, pressure decreases.
- Viscosity is the internal friction in a fluid that opposes relative motion between its layers. The coefficient of viscosity η has SI unit Pa·s (pascal-second).
- Surface tension is the property of a liquid surface that makes it behave like a stretched elastic membrane. It arises from unbalanced molecular forces at the surface and has SI unit N/m.
- Capillary rise occurs when adhesive forces between liquid and tube walls exceed cohesive forces within the liquid, causing the liquid to rise in a narrow tube. The height of rise h = 2S cos θ / (ρgr), where S is surface tension, θ is contact angle, and r is tube radius.
Formulas and facts to remember
Pressure at depth h: P = P₀ + ρgh (P₀ is surface pressure, ρ is fluid density)
Pascal's law application: F₁/A₁ = F₂/A₂ (force multiplication in hydraulic systems)
Buoyant force: F_b = ρ_fluid × V_displaced × g (Archimedes' principle)
Equation of continuity: A₁v₁ = A₂v₂ (for incompressible fluids)
Bernoulli's equation: P + ½ρv² + ρgh = constant (along a streamline)
Stokes' law for viscous drag: F = 6πηrv (on a sphere of radius r moving at speed v through fluid of viscosity η)
Terminal velocity of a sphere: v_t = 2r²(ρ_sphere − ρ_fluid)g / 9η
Capillary rise: h = 2S cos θ / (ρgr) (S is surface tension, θ is contact angle, r is tube radius)
Worked examples
Example 1: Pressure at depth in a lake
Problem: A diver swims 15 m below the surface of a freshwater lake. Calculate the total pressure on the diver. Take atmospheric pressure P₀ = 1.01 × 10⁵ Pa, density of water ρ = 1000 kg/m³, and g = 10 m/s².
Solution: Pressure due to water column = ρgh = 1000 × 10 × 15 = 1.5 × 10⁵ Pa
Total pressure = P₀ + ρgh = 1.01 × 10⁵ + 1.5 × 10⁵ = 2.51 × 10⁵ Pa
The diver experiences about 2.5 times atmospheric pressure at this depth.
Example 2: Hydraulic lift
Problem: In a hydraulic car lift, the small piston has area 5 cm² and the large piston has area 200 cm². What force must be applied on the small piston to lift a car weighing 8000 N?
Solution: By Pascal's law, pressure is transmitted equally: F₁/A₁ = F₂/A₂
Here F₂ = 8000 N (weight to be lifted), A₁ = 5 cm², A₂ = 200 cm²
F₁ = F₂ × (A₁/A₂) = 8000 × (5/200) = 8000 × 0.025 = 200 N
A force of only 200 N on the small piston lifts 8000 N — a mechanical advantage of 40.
Example 3: Speed of water through a pipe
Problem: Water flows through a horizontal pipe that narrows from diameter 4 cm to diameter 2 cm. If the water speed in the wider section is 1 m/s, find the speed in the narrower section and the pressure difference between the two sections. Take ρ = 1000 kg/m³.
Solution: Using equation of continuity: A₁v₁ = A₂v₂
A₁ = π(0.02)² = 4π × 10⁻⁴ m², A₂ = π(0.01)² = π × 10⁻⁴ m²
v₂ = v₁ × (A₁/A₂) = 1 × (4π × 10⁻⁴)/(π × 10⁻⁴) = 4 m/s
For pressure difference, use Bernoulli's equation (horizontal pipe, so height terms cancel): P₁ + ½ρv₁² = P₂ + ½ρv₂²
P₁ − P₂ = ½ρ(v₂² − v₁²) = ½ × 1000 × (16 − 1) = 500 × 15 = 7500 Pa
The water speeds up to 4 m/s in the narrow section, and pressure drops by 7500 Pa there.
Common mistakes
- Forgetting to add atmospheric pressure when calculating total pressure at depth → The formula P = P₀ + ρgh gives absolute pressure; gauge pressure is just ρgh.
- Thinking buoyant force depends on the object's density → Buoyant force depends only on the volume of fluid displaced and the fluid's density, not the object's material.
- Applying Bernoulli's equation to viscous or turbulent flow → Bernoulli's principle applies only to ideal, steady, non-viscous, incompressible flow along a streamline.
- Confusing speed increase with pressure increase in flowing fluids → In horizontal flow, where speed increases, pressure decreases (not increases).
- Using diameter instead of radius in capillary rise formula → The formula h = 2S cos θ / (ρgr) requires radius r, so divide the given diameter by 2.
Quick revision
- Pressure in a fluid at depth h: P = P₀ + ρgh; acts equally in all directions.
- Buoyant force equals weight of displaced fluid; object floats when this equals object's weight.
- Continuity: fluid speeds up in narrow sections (A₁v₁ = A₂v₂).
- Bernoulli: fast-moving fluid has lower pressure than slow-moving fluid at same height.
- Viscosity is internal friction in fluids; terminal velocity is reached when drag equals net weight.
- Surface tension causes liquids to minimise surface area; capillary rise occurs in narrow tubes.