NCERT Plus One (Class 11) Physics • Chapter 9

Mechanical Properties of Fluids

Q: State Pascal's law for transmission of fluid pressure. Explain how a hydraulic lift multiplies force without violating conservation of energy.

Pascal's Law: Pressure applied to an enclosed incompressible fluid is transmitted undiminished to every point of the fluid and walls of the vessel. In a hydraulic lift, F₂ = F₁ (A₂/A₁) multiplies force when A₂ > A₁ . However, fluid volume displaced is equal ( A₁ d₁ = A₂ d₂ ), so input work F₁ d₁ equals output work F₂ d₂ — energy is conserved.

Q: Three vessels of different shapes (wide, narrow, slanted) have the same base area and are filled with water to the exact same height h. Compare the pressure and force on their bases.

Both pressure AND force on the base are EQUAL in all three vessels! Pressure P = P_a + ρgh depends solely on vertical height h and density ρ , and since base area A is identical, force P × A is also identical.

Q: State Bernoulli's principle and write its mathematical expression. Mention two essential assumptions and two practical applications.

Statement: Along a streamline of steady, incompressible, non-viscous flow, the sum of pressure energy, kinetic energy per unit volume, and potential energy per unit volume remains constant: P + ½ρv² + ρgh = constant . Assumptions: Fluid is incompressible, non-viscous (zero friction), and flow is steady/irrotational. Applications: Venturi meter (flow speed measurement), dynamic aerofoil lift on aircraft wings, Magnus swing of spinning balls, and roof lift-off in storms.

Q: What is Torricelli's law of speed of efflux? At what depth should a hole be made in a tank of total height H for maximum horizontal range?

Speed of liquid flowing from a small orifice at depth h below the free surface is v = √(2gh) (equal to free-fall speed from height h ). Horizontal range R = 2√(h(H − h)) is maximum when h = H / 2 (halfway up the tank), giving R_max = H .

Q: State Stokes' law and derive the expression for terminal velocity v_t of a sphere of radius a and density ρ falling through a fluid of density σ and viscosity η.

Stokes' Law: Viscous drag on a small sphere moving at velocity v is F_v = 6 π η a v . At terminal velocity v_t , net downward force is zero: Weight = Buoyancy + Viscous Drag . (4/3)πa³ ρ g = (4/3)πa³ σ g + 6 π η a v_t 6 π η a v_t = (4/3)πa³ (ρ − σ) g ⇒ v_t = 2 a² (ρ − σ) g / (9 η) .

Q: Why is excess pressure inside a soap bubble twice that inside a liquid drop of the same radius R?

A liquid drop in air has only one liquid–air free surface, giving excess pressure ΔP_drop = 2S / R . A soap bubble has two free surfaces (inner wall and outer wall in contact with air), doubling the surface tension inward pull to ΔP_bubble = 4S / R .

Q: Derive the expression for capillary rise h of a liquid of surface tension S, density ρ, and contact angle θ inside a tube of radius r.

Concave meniscus has radius of curvature R = r / cos θ . Pressure just below the meniscus is lower than atmospheric pressure P₀ by ΔP = 2S / R = 2S cos θ / r . At equilibrium, liquid column of height h balances this deficit: h ρ g = 2S cos θ / r . Hence h = 2S cos θ / (r ρ g) .

Q: Calculate the terminal velocity of an air bubble of radius 0.5 mm rising in water (viscosity η = 1.0 × 10⁻³ Pa·s, ρ_water = 1000 kg/m³, g = 9.8 m/s², neglect air density).

Here a = 0.5 × 10⁻³ m , ρ_bubble ≈ 0 , σ_water = 1000 kg/m³ . Magnitude: |v_t| = 2 a² (σ − ρ) g / (9 η) = [2 × (0.5 × 10⁻³)² × 1000 × 9.8] / (9 × 10⁻³) = [2 × 2.5×10⁻⁷ × 9800] / (9 × 10⁻³) = 4.9×10⁻³ / (9×10⁻³) ≈ 0.54 m/s (directed upward!).

Pascal's law, hydraulic lift, hydrostatic pressure, barometer, continuity, Bernoulli's theorem, viscosity & surface tension.

📄 Official NCERT Textbook PDF

Topic 9.2.1: Pascal's Law & Hydraulic Lift / Brakes

Visit a busy service station in Thrissur and watch a mechanic press one small lever to hoist an entire 1500 kg SUV smoothly into the air! How can a finger-sized push lift an elephant-sized car? Not by creating free energy — by trapping incompressible oil inside connected pistons under Pascal's Law!

PRESSURE (P)P = F_normalA | SCALAR! | SI unit: Pa = N/m²

PASCAL'S LAWExtra pressure applied to an enclosed fluid transmits undiminished to every point & wall!

HYDRAULIC LIFTF₁A₁ = F₂A₂ ⇒ F₂ = F₁ (A₂A₁) = F₁ (r₂r₁)²

🧠 Memory Trick: A hydraulic lift is a FORCE multiplier, NOT a WORK multiplier — F₂ grows by A₂/A₁, while stroke shrinks by A₁/A₂ so F₁d₁ = F₂d₂!

💡 Simple Kid-Check: Pressure is like crowding inside a sealed water bottle: push on the tiny cap with 10 Pa, and EVERY square centimetre of the bottle walls suddenly feels an extra 10 Pa push! Put a wide 100 cm² lid on the other side, and those 100 little pushes team up to lift 100× more weight!

1
Pressure is a Scalar: Even though force is a vector, fluid pressure at a point pushes perpendicular to ANY surface placed there with the same magnitude — so P has no fixed direction.
2
Same-Height Rule & Transmission: In a fluid at rest, pressure is equal at all points at the same horizontal level, and any added pressure ΔP spreads equally throughout the closed vessel.
3
Hydraulic Brakes & Lifts: Pressing the brake pedal pushes master-cylinder fluid (small A₁) to wide wheel cylinders (large A₂), gripping all four tyres equally at once!

P = F₁A₁ = F₂A₂ | Volume pushed: A₁ d₁ = A₂ d₂ ⇒ Work: F₁ d₁ = F₂ d₂

🧒 Super-Easy Example: A hydraulic lift has pistons of radii 2 cm and 20 cm. To lift a 1500 kg car (F₂ ≈ 15,000 N):
A₂A₁ = (202)² = 10² = 100.
Required push F₁ = 15,000 / 100 = 150 N (just 15 kg weight)! To lift the car 10 cm, push the small piston 100 × 10 cm = 10 m!

Watch two classic 1-mark traps! First, Pressure is a SCALAR, never a vector (many students pick vector just because Force is in F/A!). Second, a hydraulic lift does NOT violate energy conservation — it multiplies force, while the small piston must travel proportionally farther so Work_in = Work_out!

Topic 9.2.2 – 9.2.3: Pressure with Depth, Manometer & Torricelli Barometer

Dive down to the bottom of a deep Kerala temple pond and your eardrums feel the squeeze — whether the pond is 10 metres wide or 10 kilometres wide! Long ago, Evangelista Torricelli flipped a 1-metre mercury tube into a dish to measure how heavy Earth's entire ocean of air really is!

DEPTH FORMULAP = P_a + ρ g h | Gauge Pressure: P_g = P − P_a = ρ g h

1 ATMOSPHERE1 atm = 1.013 × 10⁵ Pa = 76 cm Hg ≈ 10.3 m water | 1 bar = 10⁵ Pa

HYDROSTATIC PARADOXBottom pressure depends ONLY on vertical depth h & density ρ — NOT container shape!

🧠 Memory Trick: Water column for 1 atm = 10.3 m (3-storey building!), Mercury (13.6× denser!) = just 0.76 m (76 cm) — that's why barometers use Mercury!

💡 Simple Kid-Check: Picture a vertical water column of depth h and area A pressing down from above: its weight is mg = (ρ A h) g. Divide by A, add air pressing on the top surface (P_a), and boom — total pressure at depth h is P = P_a + ρ g h!

1
Hydrostatic Paradox: Pour water to the exact same height h inside a wide funnel, a zig-zag pipe, and a narrow straw. Bottom pressure P_a + ρgh is identical in all three because tilted walls push sideways/upward to carry extra fluid weight!
2
Torricelli's Mercury Barometer: Air pressure outside pushes mercury up an evacuated glass tube to height h = 76 cm: P_a = ρ_Hg g h = 13600 × 9.8 × 0.76 ≈ 1.013 × 10⁵ Pa. (Also 1 torr = 1 mm Hg ≈ 133 Pa.)
3
Open-Tube Manometer: Measures gauge pressure P − P_a = ρ g h from the U-tube liquid level difference h (used in blood pressure cuffs and gas cylinders!).

Absolute: P = P_a + ρ g h | Gauge: P_g = ρ g h | Buoyancy (Archimedes): F_B = V_sub ρ_fluid g

🧒 Super-Easy Example: At 20 m depth in a lake (ρ = 1000 kg/m³, g = 10 m/s², P_a = 10⁵ Pa):
• Gauge pressure = 1000 × 10 × 20 = 2 × 10⁵ Pa (2 atm).
• Total (Absolute) pressure = 10⁵ + 2 × 10⁵ = 3 × 10⁵ Pa (3 atm)!

Watch three board-exam traps: (1) Blood pressure & tyre gauges report Gauge pressure (P − P_a), NOT absolute pressure! (2) Even if you tilt a mercury barometer tube slightly or use a wider tube, the vertical height h = 76 cm stays unchanged (slant length grows, vertical height does not)! (3) Dam walls are made thick at the bottom because pressure ρgh increases linearly with depth!

Topic 9.3: Streamline Flow & Equation of Continuity

Watering your garden banana plants with a rubber hose, you squeeze half of the pipe opening with your thumb — instantly, the gentle trickle blasts out into a fast jet reaching twice as far! Why? Because every litre entering the hose per second must exit just as fast through the narrow gate!

STREAMLINE FLOWEvery fluid particle follows the exact path of the one ahead; two streamlines NEVER cross!

CONTINUITY EQUATIONA₁ v₁ = A₂ v₂ = constant (Volume flow rate Q = A v)

CONSERVATION LAWEquation of Continuity is simply Conservation of Mass for incompressible flow!

🧠 Memory Trick: Narrower pipe = Faster flow (v ∝ 1/A ∝ 1/d²)! Halving pipe diameter makes water sprint 4× FASTER!

💡 Simple Kid-Check: Picture a crowd walking through a wide school corridor (A₁) entering a narrow classroom doorway (A₂). If nobody piles up inside (incompressible fluid!), everyone MUST walk much faster (v₂) through the narrow door so the same number of kids pass every second!

1
Steady vs Turbulent Flow: At low speeds fluid layers glide smoothly along fixed curves called streamlines (tangent = velocity at that point). Above a critical speed, eddies swirl chaotically (turbulent flow, like fast rapids or smoke curling up!).
2
Mass Conservation Derivation: In time Δt, mass entering area A₁ is ρ A₁ v₁ Δt, and mass exiting area A₂ is ρ A₂ v₂ Δt. For an incompressible liquid (ρ constant), A₁ v₁ = A₂ v₂!
3
Streamline Spacing Rule: Closely packed streamlines mean high speed; wide-spaced streamlines mean low speed!

A₁ v₁ = A₂ v₂ ⇒ Circular pipe: π r₁² v₁ = π r₂² v₂ ⇒ v₂ = v₁ (d₁d₂)²

🧒 Super-Easy Example: Water flows at 1.5 m/s through a 2 cm diameter pipe that narrows to a 1 cm nozzle.
Since diameter halved (d₁/d₂ = 2), speed becomes v₂ = 1.5 × 2² = 1.5 × 4 = 6.0 m/s!

Top 1-mark exam question: Why can two streamlines NEVER intersect? Because if two streamlines crossed at a point, a fluid particle reaching that intersection would have two different velocity directions at the exact same instant — impossible in steady flow! Also, remember Continuity (Av = const) expresses Conservation of Mass (for incompressible fluids).

Topic 9.4: Bernoulli's Principle, Venturi Meter, Efflux & Aerofoil Lift

During a heavy monsoon squall, why do tin or coconut-leaf roofs sometimes lift straight **upward** off sheds instead of blowing sideways? And how does a heavy airplane wing lift hundreds of passengers off Cochin airport? Fast-moving air overhead drops the pressure, letting higher pressure below push **UP**!

BERNOULLI'S EQUATIONP + 12 ρ v² + ρ g h = constant (Energy per unit volume!)

HIGH SPEED = LOW PAt level height (h₁ = h₂): where speed v is HIGH, pressure P is LOW!

TORRICELLI EFFLUXTank hole depth h below surface: speed v = √(2gh)! Max range at h = H/2!

🧠 Memory Trick: Bernoulli is just Conservation of Energy in disguise: Pressure Energy (P) + Kinetic (½ρv²) + Potential (ρgh) = Constant!

💡 Simple Kid-Check: Blow air fast BETWEEN two hanging paper strips or balloons — instead of flying apart, they slam TOGETHER! Fast air between them creates a low-pressure zone, so calm outer air shoves them inward!

1
Venturi Meter: Narrows a pipe from area A₁ to throat A₂. By continuity v₂ > v₁, so Bernoulli makes throat pressure drop (P₁ − P₂ = ½ ρ (v₂² − v₁²)), measured cleanly by a U-tube manometer to calculate pipe speed!
2
Torricelli's Law (Speed of Efflux): Water shooting from a small tank hole at depth h below the free surface has speed v = √(2gh) — identical to a stone dropped freely through height h!
3
Dynamic Lift (Aerofoil & Magnus Effect): Curved airplane wings speed up air on top (low P) relative to bottom (high P) → net upward lift! A spinning cricket/football drags air faster on one side → Magnus swing!

P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂ | Efflux: v = √(2gh), Range: R = 2√(h(H−h))

🧒 Super-Easy Example: A horizontal water pipe narrows so water speeds up from 2 m/s to 6 m/s (ρ = 1000 kg/m³). Find the pressure drop:
P₁ − P₂ = ½ ρ (v₂² − v₁²) = ½ × 1000 × (36 − 4) = 500 × 32 = 16,000 Pa!

Watch three golden exam traps! (1) Bernoulli holds strictly for steady, incompressible, non-viscous (zero friction), irrotational flow — in real viscous pipes, pressure still drops along a flat uniform pipe! (2) In Torricelli's efflux v = √(2gh), h is depth below the open top surface, NOT height from the floor! (3) Why is blood flow through narrowed arteries dangerous (heart attack risk)? High blood speed inside constricted plaque lowers internal pressure, letting external pressure collapse the artery temporarily!

Topic 9.5: Viscosity, Stokes' Law & Terminal Velocity

Tilt a spoonful of water and a spoonful of thick Kerala wild honey: water splashes down instantly while honey oozes layer after sliding layer! Now think about monsoon raindrops falling from 2 kilometres up — without air viscosity slowing them via terminal velocity, raindrops would smash down like bullets!

VISCOSITY (η)Internal friction between fluid layers: η = F / Av / l | SI: Pa·s (1 poise = 0.1 Pa·s)

STOKES' DRAGF_v = 6 π η a v (Viscous drag on sphere of radius a at speed v!)

TERMINAL VELOCITYv_t = 2 a² (ρ − σ) g9 η (ρ = sphere, σ = fluid; v_t ∝ a²!)

🧠 Memory Trick: Heat makes Liquids RUNNIER (η_liquid ↓) and Gases STICKIER (η_gas ↑)! Warm coconut oil pours fast; hot air resists more!

💡 Simple Kid-Check: Picture fluid layers as playing cards sliding on each other: bottom layer sticks to the floor (v = 0), top layer moves fastest, and honey molecules tug hard between layers! When a steel ball drops into oil, viscous drag grows with speed until Upward (Drag + Buoyancy) = Downward Weight — then acceleration stops and it glides at steady terminal velocity v_t!

1
Coefficient of Viscosity (η): Ratio of shearing stress (F/A) to velocity gradient (dv/dx). SI unit: poiseuille (Pl) = N·s/m² = Pa·s; CGS unit: poise (1 Pl = 10 poise).
2
Stokes' Law: A small smooth sphere of radius a moving at velocity v experiences viscous drag F_v = 6 π η a v opposite to motion.
3
Terminal Velocity Derivation: At equilibrium:
6 π η a v_t + 43πa³ σ g = 43πa³ ρ g ⇒ v_t = 2 a² (ρ − σ) g9 η!

F = 6 π η a v | v_t = 2 a² (ρ − σ) g9 η | [η] = [M L⁻¹ T⁻¹]

🧒 Super-Easy Example: Two spherical raindrops have radii in ratio 1 : 2 falling through air. Find the ratio of their terminal velocities:
Since v_t ∝ a², doubling radius quadruples terminal speed: v₁ : v₂ = 1² : 2² = 1 : 4! (That is why tiny mist droplets practically float!)

Watch two huge board-exam traps! First, temperature affects liquid vs gas viscosity oppositely: heating lowers η in liquids (weakening cohesion) but INCREASES η in gases (faster random molecular collisions across layers)! Second, if fluid density σ > sphere density ρ (like an air bubble inside water!), v_t flips negative — the bubble shoots UPWARD!

Topic 9.6: Surface Tension, Excess Pressure in Drops & Capillary Rise

Why do morning dewdrops on a colocasia leaf curl into shiny little spheres, why can pond-skater insects walk effortlessly on water, and how does lamp oil climb all the way up a cotton wick without any motor? Welcome to the stretched invisible skin of liquids: Surface Tension!

SURFACE TENSION (S)S = Fl = Surface EnergyΔA | SI: N/m or J/m²

EXCESS PRESSURE (ΔP)Liquid drop / air bubble in water: 2SR | Soap bubble (2 surfaces!): 4SR

CAPILLARY RISE (h)h = 2S cos θr ρ g (θ < 90° rises; Mercury θ > 90° depresses!)

🧠 Memory Trick: Drop = 1 skin = 2S/R; Soap bubble = 2 skins (inside + outside) = 4S/R! Smaller bubble has BIGGER pressure (1/R)!

💡 Simple Kid-Check: Molecules inside water get hugged equally from all sides, whereas surface molecules miss top neighbours — so they cling extra tight inward, shrinking the surface area like a stretched rubber balloon (making spheres, the smallest area for a given volume!).

1
Surface Energy Work: Increasing liquid surface area by ΔA costs work W = S ΔA (for a soap film with two faces, use W = 2 S ΔA!). Adding soap lowers water's surface tension S so clothes clean easily!
2
Angle of Contact (θ): Angle inside the liquid between tangent and solid wall. Water–clean glass has acute θ (concave meniscus, wets glass, cos θ > 0). Mercury–glass has obtuse θ ≈ 140° (convex meniscus, doesn't wet, cos θ < 0!).
3
Capillary Rise (Jurin's Law): Concave meniscus has lower pressure just below it by 2S cos θ / r, pulling liquid up height h = 2S cos θ / (r ρ g) (notice h ∝ 1/r: thinner straw = higher climb!).

Drop: ΔP = 2SR | Soap Bubble: ΔP = 4SR | Capillary: h = 2S cos θr ρ g

🧒 NCERT-Style Example: Water (S = 0.073 N/m, θ ≈ 0°, ρ = 1000 kg/m³) enters a capillary tube of radius r = 0.5 mm = 5 × 10⁻⁴ m (g = 9.8 m/s²).
Rise h = 2 × 0.073 × 1(5 × 10⁻⁴) × 1000 × 9.8 = 0.1464.9 ≈ 0.0298 m ≈ 3.0 cm!

Two legendary board-exam traps! (1) Connect a small soap bubble to a large soap bubble via a straw — which way does air flow? Since ΔP = 4S/R ∝ 1/R, the SMALLER bubble has HIGHER internal pressure, so air rushes from SMALL to LARGE (small shrinks, large swells)! (2) A soap film/bubble has TWO free surfaces, so always use 4S/R for excess pressure and 2ΔA for surface energy!

Solved Previous Year Questions (Kerala DHSE & NCERT)

Q1 (Kerala HSE March • 2 Marks): State Pascal's law for transmission of fluid pressure. Explain how a hydraulic lift multiplies force without violating conservation of energy.

Pascal's Law: Pressure applied to an enclosed incompressible fluid is transmitted undiminished to every point of the fluid and walls of the vessel.
In a hydraulic lift, F₂ = F₁ (A₂/A₁) multiplies force when A₂ > A₁. However, fluid volume displaced is equal (A₁ d₁ = A₂ d₂), so input work F₁ d₁ equals output work F₂ d₂ — energy is conserved.

Q2 (Kerala HSE Model • 1 Marks): Three vessels of different shapes (wide, narrow, slanted) have the same base area and are filled with water to the exact same height h. Compare the pressure and force on their bases.

Both pressure AND force on the base are EQUAL in all three vessels! Pressure P = P_a + ρgh depends solely on vertical height h and density ρ, and since base area A is identical, force P × A is also identical.

Q3 (Kerala HSE March • 3 Marks): State Bernoulli's principle and write its mathematical expression. Mention two essential assumptions and two practical applications.

Statement: Along a streamline of steady, incompressible, non-viscous flow, the sum of pressure energy, kinetic energy per unit volume, and potential energy per unit volume remains constant:
P + ½ρv² + ρgh = constant.
Assumptions: Fluid is incompressible, non-viscous (zero friction), and flow is steady/irrotational.
Applications: Venturi meter (flow speed measurement), dynamic aerofoil lift on aircraft wings, Magnus swing of spinning balls, and roof lift-off in storms.

Q4 (Kerala HSE Say • 2 Marks): What is Torricelli's law of speed of efflux? At what depth should a hole be made in a tank of total height H for maximum horizontal range?

Speed of liquid flowing from a small orifice at depth h below the free surface is v = √(2gh) (equal to free-fall speed from height h).
Horizontal range R = 2√(h(H − h)) is maximum when h = H / 2 (halfway up the tank), giving R_max = H.

Q5 (Kerala HSE March • 3 Marks): State Stokes' law and derive the expression for terminal velocity v_t of a sphere of radius a and density ρ falling through a fluid of density σ and viscosity η.

Stokes' Law: Viscous drag on a small sphere moving at velocity v is F_v = 6 π η a v.
At terminal velocity v_t, net downward force is zero: Weight = Buoyancy + Viscous Drag.
(4/3)πa³ ρ g = (4/3)πa³ σ g + 6 π η a v_t
6 π η a v_t = (4/3)πa³ (ρ − σ) g ⇒ v_t = 2 a² (ρ − σ) g / (9 η).

Q6 (Kerala HSE March • 2 Marks): Why is excess pressure inside a soap bubble twice that inside a liquid drop of the same radius R?

A liquid drop in air has only one liquid–air free surface, giving excess pressure ΔP_drop = 2S / R.
A soap bubble has two free surfaces (inner wall and outer wall in contact with air), doubling the surface tension inward pull to ΔP_bubble = 4S / R.

Q7 (Kerala HSE Say • 3 Marks): Derive the expression for capillary rise h of a liquid of surface tension S, density ρ, and contact angle θ inside a tube of radius r.

Concave meniscus has radius of curvature R = r / cos θ.
Pressure just below the meniscus is lower than atmospheric pressure P₀ by ΔP = 2S / R = 2S cos θ / r.
At equilibrium, liquid column of height h balances this deficit: h ρ g = 2S cos θ / r.
Hence h = 2S cos θ / (r ρ g).

Q8 (NCERT / DHSE Model • 2 Marks): Calculate the terminal velocity of an air bubble of radius 0.5 mm rising in water (viscosity η = 1.0 × 10⁻³ Pa·s, ρ_water = 1000 kg/m³, g = 9.8 m/s², neglect air density).

Here a = 0.5 × 10⁻³ m, ρ_bubble ≈ 0, σ_water = 1000 kg/m³.
Magnitude: |v_t| = 2 a² (σ − ρ) g / (9 η) = [2 × (0.5 × 10⁻³)² × 1000 × 9.8] / (9 × 10⁻³)
= [2 × 2.5×10⁻⁷ × 9800] / (9 × 10⁻³) = 4.9×10⁻³ / (9×10⁻³) ≈ 0.54 m/s (directed upward!).