Mechanical Properties of Fluids | ISC Class 11 Physics Notes

This note covers fluid pressure, buoyancy, hydraulic machines, continuity, Bernoulli’s theorem, fluid flow, viscosity, Poiseuille’s formula, terminal velocity, surface tension, drops, bubbles, angle of contact and capillary rise.

What are fluids, pressure and density?

Fluids are substances that can flow, including liquids and gases. They have no definite shape of their own. Liquids are largely incompressible, while gases show much greater changes in volume when pressure changes.

Shear stress means tangential force per unit area. Fluids offer very little resistance to shearing, so even a small tangential stress changes their shape. A liquid at rest cannot sustain a tangential force that would set it flowing.

Definition: Pressure is the normal force per unit area. A normal force acts perpendicular to the surface on which it acts.

Let P denote uniform pressure, F the magnitude of normal force and A the area receiving it. Then P = F/A. For non-uniform pressure, this ratio gives the average over the chosen area.

Pressure is a scalar, a quantity specified by magnitude without direction. The pressure force has a direction fixed by the surface orientation. Equal pressure on differently oriented surfaces does not mean that their force vectors point in the same direction.

How are these quantities measured?

The SI unit of pressure is the pascal, symbol Pa: 1 Pa = 1 N m⁻². SI means International System of Units; N denotes newton, m denotes metre, kg denotes kilogram and s denotes second.

Density, denoted by ρ, is mass per unit volume: ρ = m/V, where m is mass and V is volume. The SI unit of density is kg m⁻³. Here m in an equation denotes mass; m in a unit denotes metre.

Relative density is the ratio of a substance’s density to water’s density at 4 °C, where °C means degree Celsius. It has no unit because the density units cancel. Water at this temperature has density 1.0 × 10³ kg m⁻³.

Worked example 1. Two femurs, each of area 10 cm², support a body mass of 40 kg. Find their average pressure, taking gravitational acceleration g = 10 m s⁻². Here cm means centimetre.

Formula: F = mg; A = 2 × area of one femur; P = F/A.

Substitute: F = 40 × 10 = 400 N; A = 20 × 10⁻⁴ m².

Answer: P = 400/(20 × 10⁻⁴) = 200000 Pa.

How do gravity and buoyancy affect pressure in a liquid?

Hydrostatic pressure is pressure in a fluid at rest. Within a connected liquid of uniform density, points at the same horizontal level have equal pressure. A pressure difference along that level would produce an unbalanced force and initiate flow.

Derivation: Pressure due to a liquid column

Consider a vertical liquid cylinder of area A and height h. Let P₁ and P₂ be the pressures at its upper and lower faces. Its density is ρ and gravitational acceleration is g.

1. The downward pressure force is P₁A and the upward pressure force is P₂A.
2. The cylinder has volume Ah, mass ρAh and downward weight ρAhg.
3. Vertical equilibrium gives P₂A − P₁A = ρAhg. Dividing by A gives the pressure difference.

P₂ − P₁ = ρgh. If the upper face is the open liquid surface, its pressure is atmospheric pressure Pₐ, the pressure exerted by surrounding air. At depth h, P = Pₐ + ρgh, provided density remains constant.

Absolute pressure P includes atmospheric pressure. Gauge pressure is pressure relative to the atmosphere: Pɡ = P − Pₐ = ρgh, where Pɡ denotes gauge pressure. The container’s shape and base area do not appear in this expression.

A vertical cylinder is shown inside liquid. Its upper and lower faces are labelled 1 and 2, their separation is h, and the weight mg points downwards. Arrows show pressure forces on its surfaces.

See Fig. 9.3 in your NCERT textbook

What is Archimedes’ principle?

Buoyancy is the upward resultant force exerted by a fluid on an immersed body. The greater pressure on lower surfaces contributes to this upward force. Archimedes’ principle states that the buoyant force equals the weight of the displaced fluid.

If B is buoyant force and Vᵈ is displaced volume, B = ρgVᵈ. Apparent weight, the supporting force needed while immersed, is true weight minus buoyant force. For floating equilibrium without another vertical force, buoyancy balances weight.

Worked example 2. Find the absolute pressure on a swimmer 10 m below a lake surface. Use ρ = 1000 kg m⁻³, g = 10 m s⁻² and Pₐ = 1.01 × 10⁵ Pa.

Formula: Pɡ = ρgh; P = Pₐ + Pɡ.

Substitute: Pɡ = 1000 × 10 × 10 = 1.00 × 10⁵ Pa.

Answer: P = 201000 Pa, approximately twice atmospheric pressure.

How does Pascal’s law explain hydraulic machines?

Pascal’s law states that a pressure change applied to an enclosed fluid is transmitted undiminished throughout the fluid and to its container walls. In a fluid at rest, pressure at a point is also the same in every direction.

The transmitted quantity is pressure change. Different piston areas therefore experience different force changes. Existing differences in pressure caused by height are not removed when an additional pressure is applied.

How does a hydraulic lift multiply force?

Let A₁ and A₂ denote small and large piston areas, and F₁ and F₂ their corresponding forces. For an ideal hydraulic arrangement at the same level, F₁/A₁ = F₂/A₂, hence F₂ = F₁A₂/A₁.

The mechanical advantage, output force divided by input force, is A₂/A₁. If the liquid is incompressible, equal volumes are displaced. With L₁ and L₂ denoting piston displacements, A₁L₁ = A₂L₂. A larger output force accompanies a smaller output displacement.

Connected liquid-filled cylinders have a small piston labelled A₁ and a larger piston labelled A₂ carrying a car. F₁ points downwards at the small piston, while F₂ points upwards beneath the large piston.

See Fig. 9.6b in your NCERT textbook

In hydraulic brakes, the pedal moves a master piston. Pressure passes through brake oil to the wheel cylinders. The larger piston areas produce large forces that press the brake shoes against the brake lining, opposing wheel motion.

Worked example 3. Water-filled syringes have piston diameters 1.0 cm and 3.0 cm. A 10 N input force pushes the small piston through 6.0 cm. Find the large piston’s force and displacement, treating water as incompressible.

Formula: A₂/A₁ = (D₂/D₁)²; F₂ = F₁A₂/A₁; L₂ = L₁A₁/A₂. D₁ and D₂ denote the piston diameters.

Substitute: A₂/A₁ = (3.0/1.0)² = 9; F₂ = 10 × 9; L₂ = 6.0/9.

Answer: F₂ = 90 N and L₂ ≈ 0.67 cm.

What are streamlines and the equation of continuity?

Steady flow means that fluid velocity at a fixed point remains constant with time. It does not require equal velocities at different points. A particle can accelerate as it moves through a steady flow with changing cross-section.

A streamline is a curve whose tangent gives the fluid’s velocity direction at that point. In steady flow, it is also the path followed by a particle. Two streamlines cannot intersect because the velocity at an intersection would have two directions.

A tube of flow is a region bounded by streamlines. Fluid travels along its boundaries rather than crossing them. Conservation of mass connects what enters and leaves a section of such a tube.

How is continuity obtained?

1. Choose two cross-sections with areas A₁ and A₂ and fluid speeds v₁ and v₂.
2. In a short time Δt, meaning a small time interval, the fluid travels distances v₁Δt and v₂Δt.
3. At constant density ρ, the masses passing are ρA₁v₁Δt and ρA₂v₂Δt.
4. For steady flow with no leakage or accumulation, equate these masses and cancel the common factors.

A₁v₁ = A₂v₂. The product Av is the volume flow rate Q, the volume crossing a section per unit time: Q = Av. The SI unit of volume flow rate is m³ s⁻¹.

A narrower nozzle at a hose outlet increases water speed for a given volume flow rate. The same volume must pass through a smaller area in the same time. This follows from mass conservation, before any pressure calculation is made.

When streamlines represent equal portions of the flow, closer spacing indicates greater speed. For a compressible fluid, density can change, so the more general steady-flow relation is ρ₁A₁v₁ = ρ₂A₂v₂, where ρ₁ and ρ₂ are the densities at the two sections.

What does Bernoulli’s theorem state and how is it derived?

Bernoulli’s theorem expresses mechanical energy conservation in fluid flow. Use an incompressible, non-viscous fluid in steady flow along a streamline. Incompressible means constant density; non-viscous means no internal friction. The ideal-liquid model also assumes irrotational flow, with no local spinning of fluid elements.

Let P be pressure, ρ density, v speed and y height above a chosen horizontal reference. Then P + ½ρv² + ρgy = constant along the streamline. The three terms represent pressure, kinetic energy per unit volume and gravitational potential energy per unit volume.

Derivation: Bernoulli’s equation

Choose two sections labelled 1 and 2. Let ΔV be the small volume passing each section in the same time; its mass is ρΔV. Subscripts identify each section’s pressure, speed and height.

1. Pressure does work P₁ΔV at entry. The fluid does work P₂ΔV at exit. Net pressure work is (P₁ − P₂)ΔV.
2. The change in kinetic energy is ½ρΔV(v₂² − v₁²).
3. The change in gravitational potential energy is ρgΔV(y₂ − y₁).
4. Equate net pressure work to the sum of these energy changes, then divide by ΔV and rearrange.

P₁ + ½ρv₁² + ρgy₁ = P₂ + ½ρv₂² + ρgy₂. This form shows how pressure work can change both speed and elevation.

The pipe rises and widens from its left section to its right section. Labels show pressure forces, cross-sectional areas, heights and the distances travelled during the same time interval.

See Fig. 9.9 in your NCERT textbook

What are the limitations?

Real fluids have viscosity, so some mechanical energy becomes heat. The simple equation is therefore an approximation for suitable flows. It does not directly describe turbulent flow, in which pressure and velocity fluctuate with time.

Note: Greater speed implies lower pressure when comparing points at the same height under Bernoulli’s assumptions. Where heights differ, the gravitational term must also be included.

How is Bernoulli’s theorem applied?

How do a Venturimeter and an atomiser work?

A Venturimeter measures flow using a pipe with a narrow throat. For steady incompressible flow, continuity gives higher speed at the throat. In a horizontal instrument, Bernoulli’s equation then gives lower throat pressure.

If the inlet and throat areas are A₁ and A₂, respectively, and ΔP = P₁ − P₂ is their pressure difference, Q = A₂√[2ΔP/{ρ(1 − (A₂/A₁)²)}]. Q is volume flow rate. This ideal expression follows by combining continuity with Bernoulli’s equation.

An atomiser uses a rapid air stream across the mouth of a tube dipping into liquid. The reduced pressure near the mouth allows atmospheric pressure on the reservoir liquid to drive it upwards. The air stream disperses the emerging liquid into droplets.

What is dynamic lift?

Dynamic lift is a force on a body arising from motion through a fluid. In the idealised aerofoil description, air moves faster above the wing than below it. An aerofoil is a shaped section designed to produce lift.

Neglecting the small height difference, lower pressure above and higher pressure below give an upward resultant. If vᵤ and vₗ are upper and lower air speeds, Pₗ − Pᵤ = ½ρ(vᵤ² − vₗ²), where Pₗ and Pᵤ are the corresponding pressures.

The Magnus effect is dynamic lift associated with a spinning body moving through a fluid. A spinning ball drags surrounding air, producing unequal speeds and pressures on opposite sides. Its departure from a parabolic path can be partly explained using Bernoulli’s principle.

What is Torricelli’s law?

Efflux means fluid outflow. For an open tank with a small hole a depth h below the free surface, v = √(2gh). The surface and outlet are both at atmospheric pressure, and the tank is broad enough to neglect surface speed.

The result follows by converting the decrease in gravitational potential energy per unit volume into kinetic energy per unit volume. Retain the small-hole and ideal-flow assumptions; a pressurised tank requires an additional pressure term.

What is viscosity and how is it measured?

Viscosity is internal friction associated with relative motion between fluid layers. A faster layer tends to pull a slower neighbouring layer forwards, while the slower layer retards the faster one. Viscous drag opposes relative motion.

Laminar flow is orderly flow in layers. Between a stationary plate and a parallel moving plate, the fluid next to each plate takes that plate’s velocity. For the simple steady arrangement, speed increases uniformly across the gap.

What is Newton’s formula for viscosity?

The velocity gradient dv/dx is the change in flow velocity v per unit distance x measured perpendicular to the layers. Newton’s viscosity relation, using force magnitude, is F = ηA |dv/dx|, where η is the coefficient of viscosity and A is layer area.

Thus η is shear stress divided by velocity gradient. The SI unit of coefficient of viscosity is Pa s, equivalent to N s m⁻². Its dimensions are [ML⁻¹T⁻¹], where M, L and T represent mass, length and time dimensions.

The centimetre-gram-second unit is the poise, equal to one dyne second per square centimetre; a dyne is that system’s force unit. 1 poise = 0.1 N s m⁻². Do not confuse viscosity with density: they describe different properties.

One sketch shows a fixed lower plate and a right-moving upper plate with velocity arrows between them. The pipe sketch shows longer arrows near the axis and shorter arrows near the walls.

See Fig. 9.12 in your NCERT textbook

Generally, thin liquids such as water are less viscous than thick liquids such as glycerine. Viscosity decreases with increasing temperature for liquids and increases for gases. A more mobile liquid flows more readily under comparable conditions.

Worked example 4. A block of area 0.10 m² moves at constant speed 0.085 m s⁻¹ on a liquid film 0.30 mm thick. A hanging mass 0.010 kg pulls it through a massless, frictionless pulley. Find η using g = 9.8 m s⁻²; mm means millimetre.

Formula: F = mg; η = Fd/(Av), where d is film thickness and v is block speed.

Substitute: F = 0.098 N; d = 3.0 × 10⁻⁴ m; η = (0.098 × 3.0 × 10⁻⁴)/(0.10 × 0.085).

Answer: η = 0.00346 Pa s.

How do Reynolds number and Poiseuille’s formula describe pipe flow?

Turbulent flow contains irregular fluctuations and eddies, meaning swirling regions of fluid. A gently opened tap can produce smooth flow; increasing the outflow speed can destroy that smoothness. Rapid streams encountering rocks provide another example of disturbed flow.

Critical velocity is the limiting speed beyond which orderly flow loses stability under the given conditions. The transition depends on fluid properties and the flow arrangement, so speed alone does not provide a universal criterion.

What is the significance of Reynolds number?

Reynolds number, denoted by Re, compares inertial and viscous effects. Inertia is resistance to a change in motion. For pipe flow, Re = ρvD/η, where ρ is fluid density, v mean flow speed, D internal pipe diameter and η viscosity. It is dimensionless.

Small Reynolds number indicates stronger viscous control and favours laminar flow. Large Reynolds number indicates stronger inertial effects and a greater tendency towards turbulence. The expression also shows why a change in pipe size can affect the flow regime.

What does Poiseuille’s formula predict?

For steady laminar flow of an incompressible viscous liquid through a long, horizontal, uniform circular tube, Q = πΔPr⁴/(8ηl). Here Q is volume flow rate, ΔP the pressure difference between the ends, r internal radius, l tube length and π the circle constant.

The formula assumes that liquid touching the tube wall is stationary. At constant pressure difference, viscosity and length, Q is proportional to r⁴. Narrowing a tube therefore strongly reduces flow. At constant radius, increasing pressure difference increases flow proportionally.

Unlike the simple non-viscous Bernoulli model, this formula includes the pressure drop required to overcome viscous resistance. In a horizontal uniform tube, the mean speed can remain constant while pressure decreases along its length.

Worked example 5. Glycerine flows through a horizontal tube of length 1.5 m and radius 1.0 cm. Its mass flow rate is 4.0 × 10⁻³ kg s⁻¹, density 1.3 × 10³ kg m⁻³ and viscosity 0.83 Pa s. Assuming laminar flow, find ΔP.

Formula: Q = ṁ/ρ; ΔP = 8ηlQ/(πr⁴). The symbol ṁ denotes mass passing per second.

Substitute: Q = (4.0 × 10⁻³)/(1.3 × 10³) = 3.08 × 10⁻⁶ m³ s⁻¹; r = 0.010 m.

Answer: ΔP ≈ 980 Pa.

How do Stokes’ law and terminal velocity describe moving bodies?

Stokes’ law gives the viscous drag on a sphere moving slowly through a fluid: Fᵥ = 6πηrv. Here Fᵥ is drag magnitude, η fluid viscosity, r sphere radius and v speed relative to the fluid. Drag acts opposite to the relative motion.

Use the expression for slow, low-Reynolds-number motion where the sphere’s flow is laminar and boundary effects are negligible. It should not be applied automatically to every body moving through air or liquid.

Derivation: Terminal velocity of a falling sphere

Let ρₛ denote sphere density and ρ𝒻 fluid density, with ρₛ greater than ρ𝒻. Terminal velocity vₜ is the constant velocity reached when the resultant force becomes zero.

1. The sphere’s downward weight is (4πr³/3)ρₛg.
2. Its upward buoyant force is (4πr³/3)ρ𝒻g.
3. At terminal speed, upward drag 6πηrvₜ plus buoyancy equals weight.
4. Rearrange 6πηrvₜ = (4πr³/3)(ρₛ − ρ𝒻)g to isolate vₜ.

vₜ = 2r²(ρₛ − ρ𝒻)g/(9η). For fixed densities and gravitational acceleration, terminal speed increases with the square of radius and decreases with viscosity.

Initially, drag is small and the sphere accelerates. Increasing speed increases drag, reducing the acceleration. At terminal speed, forces balance; they have not disappeared. The sphere continues moving with zero acceleration.

Put time t on the horizontal axis and downward speed v on the vertical axis. For release from rest, draw a rising curve with decreasing slope approaching a horizontal level labelled vₜ.

How do rising spheres and parachutes compare?

A rigid hollow sphere whose average density is below the liquid density rises. Average density here means total sphere mass divided by its external volume. At terminal rise, upward buoyancy balances downward weight and drag; the speed is 2r²(ρ𝒻 − ρₛ)g/(9η).

A parachute increases air resistance and lowers the eventual descent speed. Its terminal motion also involves force balance, but its shape and flow conditions do not justify using the sphere’s Stokes formula.

Worked example 6. A copper sphere of radius 2.0 mm falls through oil at terminal speed 6.5 cm s⁻¹ at 20 °C. Copper and oil densities are 8.9 × 10³ and 1.5 × 10³ kg m⁻³. Find η using g = 9.8 m s⁻².

Formula: η = 2r²(ρₛ − ρ𝒻)g/(9vₜ).

Substitute: r = 2.0 × 10⁻³ m; vₜ = 6.5 × 10⁻² m s⁻¹; ρₛ − ρ𝒻 = 7.4 × 10³ kg m⁻³.

Answer: η ≈ 9.9 × 10⁻¹ kg m⁻¹ s⁻¹, equivalent to 0.99 Pa s.

Why do liquid surfaces possess energy and tension?

Surface energy is the excess energy associated with molecules at an interface compared with molecules inside the liquid. An interface is the boundary separating two substances. Molecules in the interior have neighbouring liquid molecules around them; surface molecules have a different surrounding arrangement.

Creating more liquid surface requires energy. The liquid therefore tends towards the least surface area permitted by external conditions. Surface energy depends on both materials at the interface, rather than on the liquid alone.

How is surface tension defined?

Surface tension S is force per unit length acting in the plane of an interface, perpendicular to a line drawn on that interface. It is also the surface energy per unit area. The SI unit of surface tension is N m⁻¹.

The SI unit of surface energy is joule, symbol J. Energy per unit area has unit J m⁻², equivalent to N m⁻¹. If ΔE is the energy required to create extra surface area ΔA, then S = ΔE/ΔA at constant temperature.

How much work stretches a soap film?

Consider a movable bar of length l supporting a soap film. The film has two liquid-air surfaces, so the total pull is F = 2Sl. If the bar moves a distance d, the total added surface area is 2ld.

The work W supplied is W = Fd = 2Sld = SΔA. Count both surfaces when finding ΔA. A single liquid surface and a thin film do not require the same area factor.

The two sketches show a film between parallel guides ending at a movable bar of length l. Opposed arrows represent forces on the bar; the second sketch marks its additional displacement d.

See Fig. 9.15 in your NCERT textbook

Surface tension usually falls with temperature. Surface tension concerns the interface, while viscosity concerns resistance associated with relative fluid motion.

Worked example 7. A soap film on a U-shaped wire supports a total weight of 1.5 × 10⁻² N on a slider of length 30 cm. Find its surface tension.

Formula: S = F/(2l).

Substitute: S = (1.5 × 10⁻²)/(2 × 0.30).

Answer: S = 2.5 × 10⁻² N m⁻¹. Both film surfaces pull on the slider.

How do angle of contact and curvature affect drops and bubbles?

The angle of contact, θ, is measured inside the liquid between the tangent to its surface at contact and the solid surface. It depends on the pair of materials meeting at the boundary.

Water on clean glass has an acute contact angle and tends to wet it. Mercury on glass has an obtuse contact angle and does not wet it. Acute means less than a right angle; obtuse means greater than a right angle but less than a straight angle.

A meniscus is the curved liquid surface near a wall. Water in a clean glass capillary forms a concave meniscus, curving upwards at its edges. Mercury forms a convex meniscus, raised in the middle. Wetting agents make the contact angle smaller.

Why is pressure higher inside a spherical drop?

Free drops and bubbles are spherical if effects of gravity can be neglected. A sphere has the least surface area for a given volume, so this shape reduces surface energy when competing effects are negligible.

Let r be a drop’s radius and Δr a very small increase in radius. Its area increases approximately by 8πrΔr and its volume by 4πr²Δr. Pressure work supplies the extra surface energy.

If ΔP = Pᵢ − Pₒ is inside pressure minus outside pressure, then ΔP(4πr²Δr) = S(8πrΔr). Hence ΔP = 2S/r. The pressure difference is called excess pressure.

For a thin soap bubble, the two surfaces have approximately the same radius. Their combined surface energy gives ΔP = 4S/r. At constant surface tension, a smaller spherical drop or bubble has greater excess pressure of its corresponding type.

Three circular cross-sections show a liquid drop, a gas cavity in liquid and a thin liquid bubble. The labels identify inside pressure Pᵢ, outside pressure Pₒ and radius r.

See Fig. 9.18 in your NCERT textbook

How does capillary rise measure surface tension?

Capillary rise is the rise of a liquid above the surrounding level inside a narrow tube. For a wetting liquid in a clean glass tube, the concave surface produces a pressure difference that supports a raised liquid column.

How is the capillary formula obtained?

Let a be the internal radius of a uniform circular capillary, h the rise, ρ liquid density, S surface tension, θ contact angle and g gravitational acceleration. The meniscus radius of curvature R satisfies R = a/cos θ, where cos θ is the cosine of θ.

The curved surface gives pressure difference 2S/R = 2S cos θ/a. The raised liquid column requires hydrostatic pressure difference ρgh. Equating these gives h = 2S cos θ/(ρga).

For the same liquid, temperature and contact angle, a narrower tube gives a greater rise. For a non-wetting liquid with an obtuse contact angle, cos θ is negative. The formula then gives a depression below the external liquid level.

A narrow vertical tube is immersed in water, with water raised inside it by height h. The enlarged sketch shows the curved meniscus, tube radius a, radius of curvature r and contact angle θ.

See Fig. 9.19 in your NCERT textbook

How is surface tension determined experimentally?

1. Use a clean capillary of uniform bore, meaning uniform internal diameter, held vertically in the liquid.
2. Allow the liquid to reach equilibrium and measure the vertical rise h above the outside liquid level.
3. Measure the internal radius a and use the liquid density ρ and contact angle θ at the experimental temperature.
4. Calculate S = ρgah/(2 cos θ). For clean water and glass, the approximation θ ≈ 0 makes cos θ ≈ 1.

This elementary calculation neglects the small correction for liquid in the curved meniscus. Keep the tube radius distinct from the meniscus radius of curvature. The angle is measured inside the liquid, which determines the sign of the result.

Worked example 8. Calculate water’s rise in a capillary of radius 0.05 cm. Use S = 0.073 N m⁻¹, ρ = 10³ kg m⁻³, g = 9.8 m s⁻² and θ = 0.

Formula: h = 2S/(ρga).

Substitute: a = 5 × 10⁻⁴ m; h = (2 × 0.073)/(10³ × 9.8 × 5 × 10⁻⁴).

Answer: h = 2.98 × 10⁻² m = 2.98 cm.

Glossary

• Fluid — A substance that can flow and takes the shape of its container.
• Pressure — The normal force acting per unit area on a surface.
• Density — The mass contained per unit volume of a substance.
• Gauge pressure — The difference between a system’s absolute pressure and the surrounding atmospheric pressure.
• Buoyancy — The upward resultant force a fluid exerts on a partly or wholly immersed body.
• Streamline — A curve whose tangent gives the direction of fluid velocity at each point.
• Volume flow rate — The volume of fluid passing through a chosen cross-section per unit time.
• Viscosity — Internal friction associated with relative motion between neighbouring layers of a fluid.
• Velocity gradient — The change in flow velocity per unit perpendicular distance between fluid layers.
• Reynolds number — A dimensionless quantity comparing inertial and viscous effects in fluid flow.
• Terminal velocity — The constant velocity reached when the resultant force on a moving body becomes zero.
• Surface tension — Force per unit length acting tangentially to an interface and perpendicular to a chosen line.
• Angle of contact — The angle inside a liquid between the solid surface and the liquid-surface tangent at contact.
• Excess pressure — The pressure on the inside minus the pressure outside a drop or bubble.
• Capillary rise — The elevation of a liquid inside a narrow tube above the surrounding liquid level.

Common errors and misconceptions

• Misconception: Pressure is a vector because force is a vector. Correct: Pressure is scalar; the surface orientation determines the pressure force’s direction.
• Misconception: Pascal’s law makes pressure identical at every depth. Correct: Applied pressure changes are transmitted undiminished; hydrostatic differences remain.
• Misconception: Steady flow requires the same speed everywhere. Correct: Velocity stays constant with time at each fixed point but can differ between points.
• Misconception: Bernoulli’s equation applies unchanged to every flow. Correct: The simple form requires steady, incompressible, non-viscous flow along a streamline.
• Misconception: Terminal velocity means that no forces act. Correct: Weight, buoyancy and drag balance, giving zero resultant force and zero acceleration.
• Misconception: Every bubble has excess pressure 4S/r. Correct: An air bubble in liquid has one interface and excess pressure 2S/r; a thin soap bubble has two.
• Misconception: Surface tension always decreases with temperature. Correct: Surface tension usually falls with temperature; retain that qualification.
• Misconception: Capillary rise depends on tube diameter in the radius formula. Correct: Use internal radius a in h = 2S cos θ/(ρga).

Exam-style questions with model answers

Q1. Define pressure and explain why it is a scalar quantity. [2 marks]

1. Pressure is the magnitude of normal force per unit area: P = F/A, where F is normal force and A is area.
2. Pressure has no assigned direction. The direction of its associated force depends on the orientation of the surface.

Q2. A swimmer is 10 m below an open lake surface. Take water density as 1000 kg m⁻³, gravitational acceleration as 10 m s⁻² and atmospheric pressure as 1.01 × 10⁵ Pa. Calculate gauge and absolute pressure, distinguishing them. [3 marks]

1. Gauge pressure is the pressure above atmospheric pressure. With ρ denoting water density, g gravitational acceleration and h depth, it is Pɡ = ρgh.
2. Substitution gives Pɡ = 1000 × 10 × 10 = 1.00 × 10⁵ Pa for the pressure contributed by the liquid column.
3. Absolute pressure includes atmospheric pressure Pₐ: P = Pₐ + Pɡ = 1.01 × 10⁵ + 1.00 × 10⁵ = 2.01 × 10⁵ Pa.

Q3. Connected water-filled syringes have piston diameters 1.0 cm and 3.0 cm. A force of 10 N moves the smaller piston inwards by 6.0 cm. Assuming an ideal incompressible system at the same level, calculate output force and displacement and name the principle used. [4 marks]

1. Pascal’s law transmits the applied pressure change undiminished through the enclosed water. Let subscripts 1 and 2 identify the smaller and larger pistons.
2. Piston area is proportional to diameter squared. Therefore the area ratio is A₂/A₁ = (3.0/1.0)² = 9.
3. Equal applied pressure gives F₂ = F₁A₂/A₁ = 10 × 9 = 90 N, where F₁ and F₂ denote input and output forces.
4. Incompressibility gives A₁L₁ = A₂L₂ for piston displacements L₁ and L₂. Thus L₂ = 6.0/9 ≈ 0.67 cm outwards.

Q4. Derive Bernoulli’s equation for steady incompressible non-viscous flow along a streamline, stating the ideal-liquid assumptions and defining the quantities used. [5 marks]

1. Assume constant density ρ, steady flow, no viscous energy loss and an irrotational ideal liquid. Compare sections 1 and 2 with pressures P₁ and P₂, speeds v₁ and v₂, and heights y₁ and y₂.
2. During the same short interval, continuity makes the passing volume ΔV equal at both sections. The corresponding fluid mass is ρΔV.
3. Net pressure work on that volume is (P₁ − P₂)ΔV: entry pressure supplies work, while work is done against the exit pressure.
4. The kinetic energy increase is ½ρΔV(v₂² − v₁²), and the gravitational potential energy increase is ρgΔV(y₂ − y₁), where g is gravitational acceleration.
5. Equate pressure work to the total energy increase, divide by ΔV and rearrange: P₁ + ½ρv₁² + ρgy₁ = P₂ + ½ρv₂² + ρgy₂.

Q5. Glycerine flows steadily and laminarly through a horizontal tube of length 1.5 m and radius 1.0 cm. Its mass flow rate is 4.0 × 10⁻³ kg s⁻¹, density 1.3 × 10³ kg m⁻³ and viscosity 0.83 Pa s. Use Poiseuille’s formula to find the pressure difference. [3 marks]

1. Convert mass flow rate ṁ to volume flow rate Q using density ρ: Q = ṁ/ρ = (4.0 × 10⁻³)/(1.3 × 10³) = 3.08 × 10⁻⁶ m³ s⁻¹.
2. Poiseuille’s formula rearranges to ΔP = 8ηlQ/(πr⁴), where η is viscosity, l tube length and r tube radius. Here r = 0.010 m.
3. Substituting gives ΔP = [8 × 0.83 × 1.5 × 3.08 × 10⁻⁶]/[π × (0.010)⁴] ≈ 9.8 × 10² Pa.

Q6. Derive the terminal speed of a sphere of radius r and density ρₛ falling through a fluid of density ρ𝒻 and viscosity η, where ρₛ > ρ𝒻. Take gravitational acceleration as g and assume Stokes’ law applies. Explain the terminal condition. [5 marks]

1. The sphere has volume 4πr³/3, so its downward gravitational force is (4πr³/3)ρₛg. The symbol π denotes the circle constant.
2. Archimedes’ principle gives upward buoyancy equal to displaced fluid weight, namely (4πr³/3)ρ𝒻g. The fluid therefore reduces the downward resultant before drag is included.
3. At speed v relative to the liquid, Stokes’ law gives upward drag 6πηrv. As the falling sphere accelerates, increasing speed increases this opposing force.
4. At terminal speed vₜ, acceleration becomes zero. Weight equals buoyancy plus drag, giving 6πηrvₜ = (4πr³/3)(ρₛ − ρ𝒻)g.
5. Rearranging gives vₜ = 2r²(ρₛ − ρ𝒻)g/(9η). Motion continues at this speed because the forces balance, rather than because the forces cease to act.

Q7. Compare the excess pressure in a spherical liquid drop and a thin soap bubble, each of radius r and surface tension S. Explain the difference. [2 marks]

1. A liquid drop has one interface, so its inside pressure exceeds outside pressure by 2S/r.
2. A thin soap bubble has two interfaces of approximately equal radius, doubling the excess pressure to 4S/r.

Q8. Water rises in a uniform capillary of internal radius 0.05 cm. Given surface tension 0.073 N m⁻¹, density 10³ kg m⁻³, gravitational acceleration 9.8 m s⁻² and contact angle zero, calculate the rise using the elementary capillary formula. [3 marks]

1. Use h = 2S cos θ/(ρga), where h is rise, S surface tension, θ contact angle, ρ density, g gravitational acceleration and a internal radius.
2. The given zero contact angle makes cos θ = 1. Convert the radius to metres: a = 0.05 cm = 5 × 10⁻⁴ m.
3. Substitution gives h = (2 × 0.073)/(10³ × 9.8 × 5 × 10⁻⁴) = 2.98 × 10⁻² m, or 2.98 cm above the outside water level.

Key takeaways

• Fluid pressure acts normally to a surface, but pressure itself is scalar and is measured in pascals.
• Hydrostatic pressure increases with depth; buoyant force equals the weight of fluid displaced by an immersed body.
• Pascal’s law transmits pressure changes through enclosed fluid, allowing hydraulic systems to produce larger forces on larger piston areas.
• Continuity conserves mass, while Bernoulli’s equation relates pressure, speed and height under its ideal-flow assumptions.
• Viscosity opposes relative fluid motion; Poiseuille’s formula shows the strong dependence of laminar tube flow on radius.
• Terminal velocity occurs when weight, buoyancy and drag balance, leaving zero acceleration while motion continues.
• Surface tension is force per unit length and energy per unit area; count interfaces carefully for films and bubbles.
• Capillary rise depends on surface tension, contact angle, density and internal tube radius, with non-wetting liquids showing depression.

Test yourself

Why can two differently shaped connected vessels have equal pressure at the same depth?

For a resting liquid of uniform density, pressure depends on depth and surface pressure, not container shape.

What does steady flow mean at a fixed point?

The velocity of fluid passing that point remains constant with time, although velocity may differ elsewhere.

Why does water speed up in a narrower nozzle?

For steady incompressible flow, the same volume must pass through a smaller area in the same time.

Why is Bernoulli’s simple equation inadequate when viscous losses matter?

Some mechanical energy becomes heat, so pressure, kinetic energy and gravitational energy terms alone do not account for the loss.

What does Reynolds number compare?

It compares inertial and viscous effects, helping indicate whether a flow tends to remain laminar or become turbulent.

How can a hollow sphere rise at constant speed?

Its upward buoyancy can balance downward weight and drag, producing zero resultant force at terminal rise.

Why must both surfaces of a soap film be counted?

Stretching the film increases two liquid-air interfacial areas, and both surfaces contribute energy and force.

Why is mercury depressed in a glass capillary?

Its obtuse contact angle makes the cosine negative, giving a convex meniscus and capillary depression.