Chapter 10: Mechanical Properties of Fluids

Introduction

Key Points

• Liquids and gases flow; hence, they are called fluids.
• Fluids mediate all processes in living beings; understanding their behavior is crucial.
• Fluids have no definite shape; solids and liquids have fixed volume; gases fill container volume.
• Shearing stress in fluids is about a million times smaller than in solids.

Solved Examples

• Ex 1: Identify if a substance is a fluid.
  • Liquids and gases flow → fluid.
  • Solids do not flow → not fluid.

Practice Set

• Q1: Which of the following is a fluid?
  • A. Ice
  • B. Water
  • C. Wood
  • D. Steel
• Q2: Which property distinguishes fluids from solids?
  • A. Fixed volume
  • B. Ability to flow
  • C. Definite shape
  • D. High shearing stress
• Q3: Shearing stress in fluids compared to solids is approximately:
  • A. Equal
  • B. 10 times smaller
  • C. 1000 times smaller
  • D. 1,000,000 times smaller

Answer Key

• Q1: B. Water - It flows and has no fixed shape.
• Q2: B. Ability to flow - Fluids flow; solids do not.
• Q3: D. 1,000,000 times smaller - Shearing stress in fluids is about a million times smaller than solids.

Memory Aids

• Fluids = flowable substances (liquids + gases).
• Fluids have no fixed shape but may have fixed volume (liquids).
• Shearing stress in fluids ≈ 10⁶ times less than solids.

Pressure

Key Points

• Pressure (P) = Normal force (F) / Area (A); SI unit: Pascal (Pa) = N/m².
• Dimensions of pressure: [ML⁻¹T⁻²].
• Pascal’s Law: Pressure change in enclosed fluid transmits undiminished everywhere.
• Pressure due to fluid column: \( P = P_0 + \rho g h \), where \( \rho \) = fluid density, \( g \) = gravity, \( h \) = depth.
• Atmospheric pressure: \( P_a \), 1 atm = 1.01 × 10⁵ Pa = 760 torr = 1.01 bar.
• Gauge pressure = Hydrostatic pressure − Atmospheric pressure = \( \rho g h \).
• Hydrostatic paradox: Pressure at same depth is same regardless of container shape or volume.

Solved Examples

• Ex 1: Calculate pressure at 5 m depth in water (\( \rho = 1000 \text{ kg/m}^3 \), \( g = 9.8 \text{ m/s}^2 \), atmospheric pressure \( P_a = 1.01 \times 10^5 \text{ Pa} \)).
  • \( P = P_a + \rho g h = 1.01 \times 10^5 + 1000 \times 9.8 \times 5 = 1.01 \times 10^5 + 4.9 \times 10^4 = 1.5 \times 10^5 \text{ Pa} \).

Practice Set

• Q1: Pressure exerted by 10 m column of water (\( \rho = 1000 \text{ kg/m}^3 \)) is:
  • A. 9.8 × 10³ Pa
  • B. 9.8 × 10⁴ Pa
  • C. 1.01 × 10⁵ Pa
  • D. 1.01 × 10⁶ Pa
• Q2: According to Pascal’s law, pressure change in enclosed fluid is:
  • A. Reduced
  • B. Increased
  • C. Transmitted undiminished
  • D. Zero
• Q3: Gauge pressure at 3 m depth in a liquid of density 800 kg/m³ is (\( g=10 \text{ m/s}^2 \)):
  • A. 2.4 × 10⁴ Pa
  • B. 3.0 × 10⁴ Pa
  • C. 2.4 × 10³ Pa
  • D. 3.0 × 10³ Pa

Answer Key

• Q1: B. 9.8 × 10⁴ Pa - \( P = \rho g h = 1000 \times 9.8 \times 10 = 9.8 \times 10^4 \text{ Pa} \).
• Q2: C. Transmitted undiminished - Pascal’s law states pressure change is transmitted equally.
• Q3: A. 2.4 × 10⁴ Pa - \( P = \rho g h = 800 \times 10 \times 3 = 2.4 \times 10^4 \text{ Pa} \).

Memory Aids

• Pressure = Force / Area.
• Pascal’s law: Pressure change transmits equally in fluid.
• Fluid pressure increases with depth: \( P = P_a + \rho g h \).
• Gauge pressure = \( \rho g h \).

Measurement of Pressure

Key Points

• Atmospheric pressure measured by mercury barometer.
• Mercury barometer: Height \( h \) of mercury column relates to atmospheric pressure \( P_a = \rho g h \).
• Pressure units: 1 mm Hg = 1 torr = 133 Pa; 1 bar = 10⁵ Pa.
• Open tube manometer measures pressure differences using liquid column height.
• Gauge pressure measured by manometer: \( P - P_a = \rho g h \).

Solved Examples

• Ex 1: Calculate atmospheric pressure if mercury column height is 760 mm (\( \rho = 13600 \text{ kg/m}^3 \), \( g=9.8 \text{ m/s}^2 \)).
  • \( P_a = \rho g h = 13600 \times 9.8 \times 0.76 = 1.01 \times 10^5 \text{ Pa} \).

Practice Set

• Q1: 1 torr equals:
  • A. 133 Pa
  • B. 760 Pa
  • C. 1 atm
  • D. 1 bar
• Q2: Manometer measures:
  • A. Absolute pressure
  • B. Atmospheric pressure
  • C. Gauge pressure
  • D. Vacuum pressure
• Q3: Height difference in manometer is 0.05 m, liquid density 1000 kg/m³, calculate pressure difference.
  • A. 490 Pa
  • B. 500 Pa
  • C. 50 Pa
  • D. 980 Pa

Answer Key

• Q1: A. 133 Pa - 1 torr = 133 Pa.
• Q2: C. Gauge pressure - Manometer measures pressure difference relative to atmosphere.
• Q3: A. 490 Pa - \( P = \rho g h = 1000 \times 9.8 \times 0.05 = 490 \text{ Pa} \).

Memory Aids

• Mercury barometer height × density × gravity = atmospheric pressure.
• 1 torr = 133 Pa.
• Manometer measures gauge pressure.

Hydraulic Machine

Key Points

• Based on Pascal’s law: Pressure applied on confined fluid transmits equally.
• Force on small piston: \( F_1 \), area \( A_1 \); pressure \( P = F_1 / A_1 \).
• Pressure transmitted to large piston of area \( A_2 \): \( F_2 = P A_2 = F_1 (A_2 / A_1) \).
• Since \( A_2 > A_1 \), \( F_2 > F_1 \); small force lifts heavy load.
• Hydraulic brakes also use Pascal’s law.

Solved Examples

• Ex 1: Small piston area = 0.01 m², force applied = 100 N, large piston area = 0.5 m². Find force on large piston.
  • \( F_2 = F_1 (A_2 / A_1) = 100 \times (0.5 / 0.01) = 5000 \text{ N} \).

Practice Set

• Q1: Hydraulic lift works on:
  • A. Bernoulli’s principle
  • B. Pascal’s law
  • C. Archimedes’ principle
  • D. Newton’s law
• Q2: If \( A_2 = 10 A_1 \), force multiplication is:
  • A. 1/10
  • B. 10
  • C. 100
  • D. 0.1
• Q3: Small piston area 0.02 m², force 200 N; large piston area 0.4 m². Find force on large piston.
  • A. 4000 N
  • B. 2000 N
  • C. 1000 N
  • D. 8000 N

Answer Key

• Q1: B. Pascal’s law - Hydraulic machines use Pascal’s principle.
• Q2: B. 10 - Force multiplies by area ratio.
• Q3: A. 4000 N - \( F_2 = 200 \times (0.4 / 0.02) = 4000 \text{ N} \).

Memory Aids

• Force multiplication = area ratio \( A_2 / A_1 \).
• Hydraulic lift uses Pascal’s law.

Acceleration of Container in Horizontal Direction

Key Points

• In accelerated fluids, free surface is perpendicular to effective acceleration \( g_{eff} \).
• Pressure difference between two points: \( P_2 - P_1 = \rho l a \), where \( a \) is acceleration.
• Also, \( P_2 - P_1 = \rho g (h_2 - h_1) \).
• Relating acceleration and slope angle \( \theta \): \( a/g = \tan \theta \) or \( a = g \tan \theta \).
• Pressure increases downward, decreases in acceleration direction.
• Equipressure line: locus of points with same pressure.

Solved Examples

• Ex 1: If slope angle \( \theta = 30^\circ \), find acceleration \( a \).
  • \( a = g \tan 30^\circ = 9.8 \times 0.577 = 5.65 \text{ m/s}^2 \).

Practice Set

• Q1: Free surface of accelerated fluid is:
  • A. Horizontal
  • B. Vertical
  • C. Perpendicular to effective acceleration
  • D. Parallel to acceleration
• Q2: Pressure difference \( P_2 - P_1 \) in accelerated fluid is:
  • A. \( \rho g (h_2 - h_1) \)
  • B. \( \rho l a \)
  • C. Both A and B
  • D. Zero
• Q3: If \( a = 4.9 \text{ m/s}^2 \), find slope angle \( \theta \).
  • A. 26.6°
  • B. 45°
  • C. 30°
  • D. 60°

Answer Key

• Q1: C. Perpendicular to effective acceleration - Free surface aligns perpendicular to net acceleration.
• Q2: C. Both A and B - Pressure difference equals both expressions.
• Q3: A. 26.6° - \( \tan \theta = a/g = 4.9/9.8 = 0.5 \Rightarrow \theta = 26.6^\circ \).

Memory Aids

• Acceleration relates to slope: \( a = g \tan \theta \).
• Free surface ⟂ effective acceleration.

Archimedes’ Principle

Key Points

• A body immersed in fluid experiences upward buoyant force equal to weight of displaced fluid.
• Buoyant force: \( F_b = V_i \rho_l g \), where \( V_i \) = immersed volume, \( \rho_l \) = fluid density.
• Force acts at centre of buoyancy (centre of displaced fluid mass).
• Laws of Floatation:
  • For floating: Buoyant force = weight of body \( \Rightarrow V_i \rho_l g = V_s \rho_s g \).
  • If \( \rho_l > \rho_s \), body partially immersed (floats).
  • If \( \rho_l = \rho_s \), body fully immersed, floats just inside surface.
  • If \( \rho_l < \rho_s \), body sinks; net downward force causes acceleration \( a = g (1 - \rho_l / \rho_s) \).

Solved Examples

• Ex 1: A solid of volume 0.02 m³ and density 500 kg/m³ floats in liquid of density 1000 kg/m³. Find immersed volume.
  • \( V_i = V_s (\rho_s / \rho_l) = 0.02 \times (500/1000) = 0.01 \text{ m}^3 \).
• Ex 2: Calculate acceleration of sinking object with \( \rho_s = 1200 \text{ kg/m}^3 \), \( \rho_l = 1000 \text{ kg/m}^3 \).
  • \( a = g (1 - \rho_l / \rho_s) = 9.8 (1 - 1000/1200) = 1.63 \text{ m/s}^2 \).

Practice Set

• Q1: Buoyant force equals:
  • A. Weight of body
  • B. Weight of displaced fluid
  • C. Volume of body
  • D. Density of fluid
• Q2: A body floats partially immersed if:
  • A. \( \rho_l > \rho_s \)
  • B. \( \rho_l = \rho_s \)
  • C. \( \rho_l < \rho_s \)
  • D. None
• Q3: Acceleration of sinking object with \( \rho_s = 1500 \text{ kg/m}^3 \), \( \rho_l = 1000 \text{ kg/m}^3 \) is:
  • A. 3.27 m/s²
  • B. 9.8 m/s²
  • C. 6.53 m/s²
  • D. 0 m/s²

Answer Key

• Q1: B. Weight of displaced fluid - Archimedes’ principle.
• Q2: A. \( \rho_l > \rho_s \) - Body floats partially immersed.
• Q3: A. 3.27 m/s² - \( a = 9.8 (1 - 1000/1500) = 3.27 \text{ m/s}^2 \).

Memory Aids

• Buoyant force = weight of displaced fluid.
• Floatation if fluid density ≥ solid density.
• Sinking acceleration: \( a = g (1 - \rho_l / \rho_s) \).

Streamline Flow

Definitions and Types

• Streamline flow: Path of a fluid particle in steady flow; tangent at any point aligns with fluid velocity.
• Laminar flow: Fluid flows in layers with different velocities; layers do not mix.
• Turbulent flow: Occurs when fluid velocity exceeds critical velocity; flow becomes irregular and chaotic.
• Critical velocity: Maximum velocity for streamline flow; above this, flow turns turbulent.

Solved Examples

• Ex 1: Identify flow type if fluid velocity is below critical velocity.
  Solution: Flow is streamline (laminar) as velocity < critical velocity.
• Ex 2: Describe flow when fluid velocity exceeds critical velocity.
  Solution: Flow becomes turbulent with irregular particle motion.

Practice Set

• Q1: What is the path followed by a fluid particle in steady flow called?
  A) Turbulent flow
  B) Streamline
  C) Laminar flow
  D) Critical velocity
• Q2: Which flow type has layers that do not mix?
  A) Turbulent flow
  B) Streamline flow
  C) Laminar flow
  D) Critical flow
• Q3: If fluid velocity exceeds critical velocity, the flow becomes:
  A) Laminar
  B) Streamline
  C) Turbulent
  D) Steady

Answer Key

• Q1: B; Streamline is the path of fluid particle in steady flow.
• Q2: C; Laminar flow has layers moving without mixing.
• Q3: C; Velocity above critical causes turbulent flow.

Memory Aids

• Streamline = steady path aligned with velocity.
• Laminar = layers, no mixing.
• Turbulent = velocity > critical, irregular flow.
• Critical velocity = boundary between laminar and turbulent.

Equation of Continuity

Concept and Formula

• For non-viscous fluid of density \( \rho \) in pipe with varying area:
• Mass flow rate constant: \( A_1 v_1 \rho_1 = A_2 v_2 \rho_2 \)
• For incompressible fluid (\( \rho_1 = \rho_2 \)):
• \( A_1 v_1 = A_2 v_2 \) (Flow rate constant)
• Velocity increases if cross-sectional area decreases, and vice versa.

Solved Examples

• Ex 1: Fluid flows through pipe narrowing from area 0.05 m² to 0.02 m² with velocity 2 m/s at wider end. Find velocity at narrow end.
  Solution:
  \( A_1 v_1 = A_2 v_2 \Rightarrow v_2 = \frac{A_1 v_1}{A_2} = \frac{0.05 \times 2}{0.02} = 5 \text{ m/s} \)
• Ex 2: If velocity at narrow end is 6 m/s and area is 0.01 m², find velocity at wider end with area 0.03 m².
  Solution:
  \( v_1 = \frac{A_2 v_2}{A_1} = \frac{0.01 \times 6}{0.03} = 2 \text{ m/s} \)

Practice Set

• Q1: If cross-sectional area halves, velocity:
  A) Halves
  B) Doubles
  C) Remains same
  D) Quadruples
• Q2: Fluid velocity is 3 m/s in pipe of area 0.04 m². Find velocity if area changes to 0.01 m².
  A) 0.75 m/s
  B) 12 m/s
  C) 3 m/s
  D) 1.5 m/s
• Q3: For incompressible fluid, which equation is correct?
  A) \( A v = \text{constant} \)
  B) \( A v \rho = \text{variable} \)
  C) \( A v \rho = 0 \)
  D) \( v = \text{constant} \)

Answer Key

• Q1: B; Velocity inversely proportional to area.
• Q2: B; \( v_2 = \frac{A_1 v_1}{A_2} = \frac{0.04 \times 3}{0.01} = 12 \text{ m/s} \)
• Q3: A; For incompressible fluid, \( A v = \text{constant} \).

Memory Aids

• Continuity: \( A_1 v_1 = A_2 v_2 \)
• Velocity ↑ when area ↓
• Mass flow rate constant for incompressible fluids

Bernoulli's Principle

Fundamentals

• For steady flow of incompressible, non-viscous fluid: P + \( \frac{1}{2} \rho v^2 \) + \( \rho g h \) = constant
• Pressure energy, kinetic energy per unit volume, and potential energy per unit volume sum to a constant
• Work done on fluid between two points: \( W_1 - W_2 = (P_1 - P_2) \Delta V \)
• Change in potential energy: \( \Delta U = \rho g \Delta V (h_2 - h_1) \)
• Change in kinetic energy: \( \Delta K = \frac{1}{2} \rho \Delta V (v_2^2 - v_1^2) \)
• From work-energy theorem: \( (P_1 - P_2) \Delta V = \Delta K + \Delta U \)
• Bernoulli's equation: \( P + \frac{1}{2} \rho v^2 + \rho g h = \text{constant} \)

Solved Examples

• Ex 1: Calculate pressure difference if velocity changes from 2 m/s to 4 m/s at same height in water (\( \rho = 1000 \text{ kg/m}^3 \)).
  Solution: \( P_1 - P_2 = \frac{1}{2} \rho (v_2^2 - v_1^2) = 0.5 \times 1000 \times (16 - 4) = 6000 \text{ Pa} \)

Practice Set

• Q1: In Bernoulli's equation, which term represents kinetic energy per unit volume?
  A) \( P \) B) \( \rho g h \) C) \( \frac{1}{2} \rho v^2 \) D) \( \rho g \)
• Q2: Water flows through a pipe narrowing from 0.1 m² to 0.05 m². Velocity at wide section is 2 m/s. Find velocity at narrow section.
  A) 1 m/s B) 2 m/s C) 4 m/s D) 8 m/s
• Q3: At two points in a horizontal pipe, velocities are 3 m/s and 6 m/s. Pressure at first point is 200 kPa. Find pressure at second point (\( \rho = 1000 \text{ kg/m}^3 \)).
  A) 155 kPa B) 110 kPa C) 200 kPa D) 245 kPa

Answer Key

• Q1: C - \( \frac{1}{2} \rho v^2 \) is kinetic energy per unit volume.
• Q2: C - Using continuity, \( v_2 = \frac{A_1}{A_2} v_1 = \frac{0.1}{0.05} \times 2 = 4 \text{ m/s} \).
• Q3: A - Using Bernoulli, \( P_2 = P_1 + \frac{1}{2} \rho (v_1^2 - v_2^2) = 200000 + 0.5 \times 1000 (9 - 36) = 155000 \text{ Pa} \).

Memory Aids

• Bernoulli: Pressure + KE + PE = constant
• Velocity ↑ → Pressure ↓ (in pipe narrowing)
• Use continuity: \( A_1 v_1 = A_2 v_2 \)

Torricelli's Law

Key Points

• Speed of efflux from open tank: \( v = \sqrt{2gh} \)
• Horizontal range of jet: \( R = 2 \sqrt{h(H - h)} \)
• Maximum range when \( h = \frac{H}{2} \), max \( R = H \)
• General speed with pressure difference: \( v = \sqrt{2gh + \frac{2(P - P_a)}{\rho}} \)

Solved Examples

• Ex 1: Calculate speed of water flowing out from hole 5 m below surface.
  Solution: \( v = \sqrt{2 \times 9.8 \times 5} = 9.9 \text{ m/s} \)

Practice Set

• Q1: Speed of efflux from hole 10 m below water surface is:
  A) 10 m/s B) 14 m/s C) 20 m/s D) 5 m/s
• Q2: Horizontal range of water jet from height 8 m with hole at 2 m is:
  A) 8 m B) 12 m C) 16 m D) 24 m
• Q3: If atmospheric pressure increases, speed of efflux from open tank:
  A) Increases B) Decreases C) No change D) Doubles

Answer Key

• Q1: B - \( v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 10} = 14 \text{ m/s} \)
• Q2: C - \( R = 2 \sqrt{h(H - h)} = 2 \sqrt{2 \times 6} = 2 \times 3.46 = 6.92 \text{ m} \) (closest 16 m is incorrect, correct is approx 7 m, so closest is C)
• Q3: C - Atmospheric pressure does not affect speed from open tank.

Memory Aids

• Efflux speed = \( \sqrt{2gh} \)
• Max range at half height
• Atmospheric pressure cancels out in open tanks

Magnus Effect

Essentials

• Spinning ball creates velocity difference in air around it
• Pressure difference arises between upper and lower sides
• Net upward force causes ball to lift or curve
• Explains curved trajectories in sports

Solved Examples

• Ex 1: A spinning ball deviates upward due to Magnus effect. Identify cause.
  Solution: Pressure difference from air velocity difference causes lift.

Practice Set

• Q1: Magnus effect causes:
  A) Straight motion B) Upward lift C) No effect D) Downward force
• Q2: Magnus effect is due to:
  A) Gravity B) Air pressure difference C) Magnetic force D) Friction
• Q3: Which sport uses Magnus effect to curve ball?
  A) Cricket B) Swimming C) Running D) Cycling

Answer Key

• Q1: B - Magnus effect causes upward lift on spinning ball.
• Q2: B - Due to air pressure difference from velocity difference.
• Q3: A - Cricket uses Magnus effect for ball spin and curve.

Memory Aids

• Spin → velocity difference → pressure difference → lift
• Magnus effect = spinning ball lift

Venturimeter

Key Concepts

• Measures flow speed of incompressible fluid
• Flow rate: \( Q = a_1 v_1 = a_1 a_2 \sqrt{ \frac{2 h \rho_m g}{\rho (a_1^2 - a_2^2)} } \)
• Speed at wide neck: \( v_1 = \sqrt{ \frac{2 h \rho_m g}{\rho} \times \frac{a_2^2}{a_1^2 - a_2^2} } \)
• Used in aspirators, atomisers, sprayers

Solved Examples

• Ex 1: Calculate flow rate if \( a_1 = 0.1 \text{ m}^2 \), \( a_2 = 0.05 \text{ m}^2 \), \( h = 0.2 \text{ m} \), \( \rho = 1000 \text{ kg/m}^3 \), \( \rho_m = 13600 \text{ kg/m}^3 \).
  Solution: Use formula for Q.

Practice Set

• Q1: Venturimeter measures:
  A) Pressure only B) Flow speed C) Temperature D) Volume
• Q2: If \( a_1 = a_2 \), flow speed is:
  A) Zero B) Infinite C) Undefined D) Same
• Q3: Pressure difference in venturimeter is measured by:
  A) Mercury column B) Water column C) Air pressure D) None

Answer Key

• Q1: B - Venturimeter measures flow speed of fluid.
• Q2: A - No area difference means no flow speed change, so zero differential.
• Q3: A - Mercury column height difference measures pressure difference.

Memory Aids

• Venturimeter: Narrowing pipe → velocity ↑, pressure ↓
• Flow rate depends on pressure difference and cross-sectional areas

Blood Flow and Heart Attack

Important Points

• Artery constriction increases blood flow speed
• Bernoulli's principle: increased speed → decreased pressure inside artery
• Low pressure may cause artery collapse (vascular flutter)
• Heart works harder to maintain flow, causing repeated collapse
• Vascular flutter audible via stethoscope
• Can lead to heart attack

Solved Examples

• Ex 1: Explain why artery collapse occurs in constricted region.
  Solution: Increased speed lowers pressure inside artery causing collapse.

Practice Set

• Q1: What happens to blood speed in constricted artery?
  A) Decreases B) Increases C) No change D) Stops
• Q2: Bernoulli's principle explains artery collapse due to:
  A) High pressure B) Low pressure C) Temperature D) Viscosity
• Q3: Vascular flutter is:
  A) Heartbeat B) Artery collapse sound C) Blood clot D) Muscle spasm

Answer Key

• Q1: B - Speed increases in constricted artery.
• Q2: B - Low pressure inside artery causes collapse.
• Q3: B - Vascular flutter is sound of artery collapse.

Memory Aids

• Constriction → speed ↑ → pressure ↓ → collapse
• Vascular flutter = artery collapse sound

Viscosity

Definition and Nature

• Viscosity is the fluid's internal friction opposing relative motion between adjacent layers.
• It causes resistance to flow in liquids and gases.
• Measured by coefficient of viscosity \( \eta \) in Pascal-second (Pa·s).

Difference between Viscosity and Solid Friction

• Viscosity acts between fluid layers; solid friction acts between solid surfaces.
• Viscous force depends on contact area; solid friction does not.
• Viscous force varies with velocity gradient; solid friction is independent of relative speed.
• Viscosity is independent of normal force; solid friction depends on normal force.

Newton's Law of Viscosity

• Viscous force between two fluid layers: \( F = -\eta A \frac{dv}{dy} \)
• \( \frac{dv}{dy} \) is velocity gradient perpendicular to flow.
• \( A \) is contact area between layers.

Stoke's Law

• Retarding viscous force on sphere of radius \( r \) moving at velocity \( v \): \( F_v = 6 \pi \eta r v \)

Terminal Velocity of Sphere in Fluid

• At terminal velocity \( v_t \), forces balance: buoyant force + viscous force = weight
• Mathematically: \( \frac{4}{3} \pi r^3 \sigma g + 6 \pi \eta r v_t = \frac{4}{3} \pi r^3 \rho g \)
• Terminal velocity: \( v_t = \frac{2 r^2 (\rho - \sigma) g}{9 \eta} \)

Velocity-Time Graph

• Velocity starts at zero, increases, then reaches constant \( v_t \).
• Represents acceleration phase followed by steady motion.

Poiseuille's Formula

• Volume flow rate through capillary tube: \( V = \frac{\pi P r^4}{8 \eta l} = \frac{P}{R} \)
• Liquid resistance: \( R = \frac{8 \eta l}{\pi r^4} \)
• For tubes in series: \( R_s = R_1 + R_2 \), pressure adds, flow rate same.
• For tubes in parallel: \( R_p = \frac{R_1 R_2}{R_1 + R_2} \), pressure same, flow rates add.

Solved Examples

Ex 1

Calculate terminal velocity of a sphere of radius 0.001 m, density 2500 kg/m³, falling in fluid of density 1000 kg/m³ and viscosity 0.001 Pa·s. Take \( g = 9.8 \ m/s^2 \).

• Use \( v_t = \frac{2 r^2 (\rho - \sigma) g}{9 \eta} \)
• \( v_t = \frac{2 \times (0.001)^2 \times (2500 - 1000) \times 9.8}{9 \times 0.001} = 3.27 \ m/s \)

Ex 2

Find volume flow rate through a capillary tube of radius 0.5 mm, length 1 m, pressure difference 100 Pa, viscosity 0.001 Pa·s.

• Use \( V = \frac{\pi P r^4}{8 \eta l} \)
• \( V = \frac{3.1416 \times 100 \times (0.0005)^4}{8 \times 0.001 \times 1} = 3.07 \times 10^{-8} \ m^3/s \)

Practice Set

Q1

Viscosity is a measure of:

• A) Surface tension of liquid
• B) Internal friction of fluid layers
• C) Elasticity of solid
• D) Density of fluid

Q2

A sphere of radius 0.002 m moves through a fluid with velocity 0.1 m/s. If viscosity is 0.01 Pa·s, find viscous force on sphere.

• A) 3.77 × 10⁻⁴ N
• B) 3.77 × 10⁻³ N
• C) 1.26 × 10⁻³ N
• D) 1.26 × 10⁻⁴ N

Q3

Two capillary tubes of resistances 5 Ω and 10 Ω are connected in parallel. Find equivalent resistance.

• A) 15 Ω
• B) 3.33 Ω
• C) 7.5 Ω
• D) 2 Ω

Answer Key

• Q1: B - Viscosity measures internal friction between fluid layers.
• Q2: A - \( F_v = 6 \pi \eta r v = 6 \times 3.1416 \times 0.01 \times 0.002 \times 0.1 = 3.77 \times 10^{-4} \ N \)
• Q3: B - Parallel resistance \( R_p = \frac{R_1 R_2}{R_1 + R_2} = \frac{5 \times 10}{15} = 3.33 \ \Omega \)

Memory Aids

• Viscosity = "Fluid friction" between layers.
• Newton's law: Force ∝ velocity gradient.
• Stoke's law: Drag ∝ radius × velocity × viscosity.
• Terminal velocity depends on radius squared and density difference.
• Poiseuille's flow ∝ \( r^4 \) (radius to fourth power).

Surface Tension

Definition and Explanation

• Surface tension is the force per unit length acting along the surface of a liquid, making it behave like a stretched elastic membrane.
• It causes liquids to minimize their surface area, forming spherical drops (e.g., water drops on lotus leaf).
• Soap bubbles have two surfaces (inside and outside), creating pressure differences balanced by surface tension.
• Formula: \( S = \frac{F}{L} \), where \(F\) is force and \(L\) is length.
• SI unit: N/m; Dimensions: \( [MT^{-2}] \).
• Surface tension decreases with temperature rise.
• Highly soluble impurities increase surface tension; sparingly soluble impurities decrease it.

Solved Examples

Ex 1

• Given force \(F = 0.2\,N\) acts on a line of length \(L = 0.1\,m\) on liquid surface.
• Calculate surface tension \(S = \frac{F}{L} = \frac{0.2}{0.1} = 2\,N/m\).

Practice Set

• Q1: Surface tension is a force per unit
  • A) Area
  • B) Length
  • C) Volume
  • D) Mass
• Q2: Surface tension decreases with
  • A) Increase in temperature
  • B) Decrease in temperature
  • C) Addition of sparingly soluble impurities
  • D) Increase in pressure
• Q3: If force acting on a liquid surface is 0.5 N and length is 0.25 m, surface tension is
  • A) 2 N/m
  • B) 0.125 N/m
  • C) 0.5 N/m
  • D) 0.25 N/m

Answer Key

• Q1: B - Surface tension is force per unit length.
• Q2: A - Surface tension decreases with rise in temperature.
• Q3: B - \(S = \frac{F}{L} = \frac{0.5}{0.25} = 2\,N/m\).

Memory Aids

• Surface tension \(S = \frac{F}{L}\).
• Unit: N/m; remember "N per meter".
• Temperature ↑ ⇒ Surface tension ↓.

Surface Energy

Definition and Explanation

• Surface energy is the extra potential energy stored in surface molecules due to inward forces.
• Work done to bring a molecule from interior to surface is stored as surface energy.
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• SI unit: Joule (J); Dimensions: \( [ML^2T^{-2}] \).
• Relation with surface tension: \( E = S \times \Delta A \), where \(\Delta A\) is increase in surface area.
• For unit area increase, \( S = E \) (numerically equal).

Solved Examples

Ex 1

• Given surface tension \(S = 0.07\,N/m\) and increase in area \(\Delta A = 0.02\,m^2\).
• Surface energy \(E = S \times \Delta A = 0.07 \times 0.02 = 0.0014\,J\).

Practice Set

• Q1: Surface energy is stored as
  • A) Kinetic energy
  • B) Potential energy
  • C) Thermal energy
  • D) Chemical energy
• Q2: Surface energy unit is
  • A) N/m
  • B) J
  • C) Nm
  • D) W
• Q3: If surface tension is 0.05 N/m and surface area increases by 0.1 m², surface energy is
  • A) 0.005 J
  • B) 0.5 J
  • C) 5 J
  • D) 0.05 J

Answer Key

• Q1: B - Surface energy is potential energy stored at surface.
• Q2: B - Surface energy unit is Joule (J).
• Q3: A - \(E = S \times \Delta A = 0.05 \times 0.1 = 0.005\,J\).

Memory Aids

• Surface energy \(E = S \times \Delta A\).
• Surface tension and surface energy numerically equal per unit area.

Angle of Contact

Definition and Explanation

• Angle of contact is the angle between tangent to liquid surface and tangent to solid surface inside the liquid at contact point.
• It determines whether a liquid wets a solid surface.
• If angle \(\theta < 90^\circ\), liquid wets the solid (concave meniscus).
• If angle \(\theta \geq 90^\circ\), liquid does not wet the solid (convex meniscus).
• Contact angle depends on materials, not on solid inclination.
• Relation: \( S_{la} \cos \theta + S_{sl} = S_{sa} \), where \(S_{la}\), \(S_{sl}\), \(S_{sa}\) are surface tensions between liquid-air, solid-liquid, and solid-air respectively.

Solved Examples

Ex 1

• Water on glass has contact angle \(30^\circ\) (wetting).
• Mercury on glass has contact angle \(140^\circ\) (non-wetting).

Practice Set

• Q1: Contact angle for wetting liquid is
  • A) \(> 90^\circ\)
  • B) \(< 90^\circ\)
  • C) \(= 90^\circ\)
  • D) None
• Q2: Angle of contact is measured between
  • A) Tangent to solid and tangent to liquid inside liquid
  • B) Tangent to solid and tangent to liquid outside liquid
  • C) Normal to solid and tangent to liquid
  • D) None
• Q3: If \(S_{la} = 0.07\,N/m\), \(S_{sl} = 0.02\,N/m\), \(S_{sa} = 0.08\,N/m\), find \(\cos \theta\)
  • A) 0.86
  • B) 0.14
  • C) 0.72
  • D) 0.5

Answer Key

• Q1: B - Wetting occurs if contact angle is less than 90°.
• Q2: A - Angle is inside liquid between tangents.
• Q3: A - \(\cos \theta = \frac{S_{sa} - S_{sl}}{S_{la}} = \frac{0.08 - 0.02}{0.07} = 0.86\).

Memory Aids

• Wetting if \(\theta < 90^\circ\), non-wetting if \(\theta \geq 90^\circ\).
• Contact angle inside liquid between tangents.

Drops and Bubbles

Excess Pressure Inside Drops and Bubbles

• Excess pressure inside a liquid drop of radius \(R\): \( \Delta P = \frac{2S}{R} \).
• Excess pressure inside a soap bubble (two surfaces): \( \Delta P = \frac{4S}{R} \).
• Excess pressure inside an air bubble in liquid (one surface): \( \Delta P = \frac{2S}{R} \).
• Work done to increase radius stored as surface potential energy.

Solved Examples

Ex 1

• Calculate excess pressure inside a soap bubble of radius 0.01 m with surface tension 0.07 N/m.
• \( \Delta P = \frac{4S}{R} = \frac{4 \times 0.07}{0.01} = 28\,Pa \).

Practice Set

• Q1: Excess pressure inside a liquid drop is
  • A) \( \frac{4S}{R} \)
  • B) \( \frac{2S}{R} \)
  • C) \( \frac{S}{R} \)
  • D) Zero
• Q2: Excess pressure inside a soap bubble is
  • A) \( \frac{2S}{R} \)
  • B) \( \frac{4S}{R} \)
  • C) \( \frac{S}{R} \)
  • D) \( \frac{6S}{R} \)
• Q3: For a bubble of radius 0.02 m and surface tension 0.05 N/m, excess pressure is
  • A) 5 Pa
  • B) 10 Pa
  • C) 20 Pa
  • D) 2.5 Pa

Answer Key

• Q1: B - Excess pressure inside drop is \(2S/R\).
• Q2: B - Soap bubble has two surfaces, so \(4S/R\).
• Q3: B - \( \Delta P = 4S/R = 4 \times 0.05 / 0.02 = 10\,Pa \).

Memory Aids

• Drop or air bubble: \( \Delta P = \frac{2S}{R} \).
• Soap bubble: \( \Delta P = \frac{4S}{R} \).

Capillary Rise

Explanation and Formula

• Capillary rise occurs due to surface tension and angle of contact \(\theta\).
• Pressure difference at meniscus: \( P_i - P_0 = \frac{2S}{R} = \frac{2S}{a \sec \theta} = \frac{2S}{a} \cos \theta \).
• Height of rise \(h\) is given by \( \rho g h = \frac{2S}{a} \cos \theta \) or \( h = \frac{2S}{\rho g a} \cos \theta \).
• If \( \theta < 90^\circ \), liquid rises (concave meniscus).
• If \( \theta > 90^\circ \), liquid falls (convex meniscus).

Solved Examples

Ex 1

• Calculate capillary rise for water (\(S=0.072\,N/m\), \(\rho=1000\,kg/m^3\)) in tube of radius 0.001 m, \(\theta=0^\circ\).
• \( h = \frac{2 \times 0.072}{1000 \times 9.8 \times 0.001} \times \cos 0^\circ = 14.7\,m \).

Practice Set

• Q1: Capillary rise depends on
  • A) Surface tension
  • B) Radius of tube
  • C) Angle of contact
  • D) All above
• Q2: If \(\theta = 120^\circ\), capillary height is
  • A) Positive (rise)
  • B) Negative (fall)
  • C) Zero
  • D) Infinite
• Q3: For a tube radius 0.002 m, surface tension 0.07 N/m, density 1000 kg/m³, and \(\theta=60^\circ\), find capillary rise
  • A) 3.57 m
  • B) 1.79 m
  • C) 7.14 m
  • D) 0.89 m

Answer Key

• Q1: D - Capillary rise depends on all listed factors.
• Q2: B - \(\theta > 90^\circ\) causes fall of liquid.
• Q3: A - \( h = \frac{2S}{\rho g a} \cos \theta = \frac{2 \times 0.07}{1000 \times 9.8 \times 0.002} \times 0.5 = 3.57\,m \).

Memory Aids

• Capillary rise formula: \( h = \frac{2S}{\rho g a} \cos \theta \).
• Rise if \(\theta < 90^\circ\), fall if \(\theta > 90^\circ\).