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Mechanical Properties of Fluids

NCERT Class 11 · Physics Based on NCERT Class 11 Physics textbook · Free CBSE study kit

Chapter Notes

COMPREHENSIVE CHAPTER NOTES: MECHANICAL PROPERTIES OF FLUIDS

FLUIDS: DEFINITION AND CHARACTERISTICS

**Definition:** A fluid is any substance that can flow and has no definite shape of its own. Both liquids and gases are fluids.

**Key Distinguishing Properties of Fluids vs. Solids:**

  • **Solids:** Have definite shape and volume; resist shear stress significantly
  • **Liquids:** Have definite volume but no definite shape; nearly incompressible; occupy the shape of container
  • **Gases:** Have no definite shape or volume; fill entire container; highly compressible
  • **Shear Stress Resistance:** Fluids offer very little resistance to shear stress—approximately one millionth that of solids. Even minimal shear stress causes fluid flow, distinguishing fluids fundamentally from solids
  • **Common Properties:**

  • Compressibility varies: liquids have low compressibility (nearly incompressible), gases have high compressibility
  • Both liquids and gases can flow and transmit pressure
  • Force exerted by fluids at rest acts normal (perpendicular) to any surface in contact with them
  • ---

    PRESSURE IN FLUIDS

    Definition and Mathematical Expression

    **Pressure:** The normal force exerted per unit area of surface.

    **Mathematical Definition:**

  • Average pressure: **P_av = F/A** (equation 9.1)
  • Point pressure (limiting definition): **P = lim(ΔA→0) ΔF/ΔA** (equation 9.2)
  • Where:

  • F = magnitude of normal force (N)
  • A = area of surface (m²)
  • P = pressure (Pa)
  • **Important Note:** Pressure is a **scalar quantity**, not a vector. It represents the component of force normal to the surface, not the force vector itself.

    **Dimensions:** [ML⁻¹T⁻²]

    **SI Unit:** **Pascal (Pa) = N/m²**

    Named after French scientist Blaise Pascal (1623-1662) who pioneered fluid pressure studies.

    **Other Common Units:**

  • **1 atm (atmosphere) = 1.013 × 10⁵ Pa** (atmospheric pressure at sea level)
  • **1 bar = 10⁵ Pa**
  • **1 torr = 133 Pa** (used in medicine; 1 mm of Hg)
  • **760 mm Hg = 1 atm = 1.013 × 10⁵ Pa**
  • Density

    **Definition:** Mass per unit volume of a substance.

    **Formula:** **ρ = m/V** (equation 9.3)

    Where:

  • ρ = density (kg/m³)
  • m = mass (kg)
  • V = volume (m³)
  • **Dimensions:** [ML⁻³]

    **SI Unit:** kg/m³

    **Key Properties:**

  • Density is a positive scalar quantity
  • **Water density at 4°C (277 K) = 1.0 × 10³ kg/m³** (reference standard)
  • Liquids have nearly constant density across pressure ranges (nearly incompressible)
  • Gases show large density variations with pressure changes
  • **Relative Density:** Ratio of a substance's density to the density of water at 4°C; dimensionless.

  • Example: Relative density of aluminum = 2.7, so density of aluminum = 2.7 × 10³ kg/m³
  • Pressure Force Perpendicularity Principle

    When a fluid is at rest, the force it exerts on any surface is always perpendicular to that surface. **Why?** If a force component existed parallel to the surface, by Newton's third law, the fluid would push back with equal parallel force, causing the fluid to flow parallel to the surface. Since the fluid is at rest, this cannot occur; therefore, only perpendicular force exists.

    ---

    PASCAL'S LAW

    **Statement:** In a fluid at rest, pressure is the same at all points at the same height (same depth). Equivalently: any change in external pressure applied to a confined fluid is transmitted undiminished and equally in all directions.

    Proof (Using Prismatic Element)

    Consider a small right-angled prismatic element ABC-DEF inside a stationary fluid:

    **Given:** Pressures P_a, P_b, P_c act on faces with areas A_a, A_b, A_c respectively; forces F_a, F_b, F_c act normally on these faces.

    **Equilibrium conditions (no acceleration):**

    Vertical force balance: **F_b sin θ = F_c**

    Horizontal force balance: **F_b cos θ = F_a**

    **Geometric relations:**

    **A_b sin θ = A_c** and **A_b cos θ = A_a**

    **Dividing force equations by area equations:**

    **F_a/A_a = F_b/A_b = F_c/A_c**

    Therefore: **P_a = P_b = P_c** (equation 9.4)

    **Conclusion:** Pressure is the same in all directions in a fluid at rest; it is not a vector but a scalar quantity.

    Application: Pressure at Same Horizontal Level

    For a horizontal fluid element in equilibrium, horizontal forces at both ends balance, so pressure is equal at all points in a horizontal plane within a stationary fluid.

    ---

    VARIATION OF PRESSURE WITH DEPTH

    Derivation of Pressure-Depth Relationship

    **Setup:** Consider a cylindrical column of fluid with uniform cross-sectional area A, with point 1 at height h above point 2.

    **Pressures:** P₁ (at top), P₂ (at bottom)

    **Force Balance (Vertical Equilibrium):**

    Upward force from below: **P₂A**

    Downward forces: **P₁A + mg** (where mg is weight of fluid column)

    Equilibrium condition: **P₂A = P₁A + mg**

    Rearranging: **(P₂ - P₁)A = mg**

    **Expressing mass in terms of density:**

    Mass of fluid: **m = ρV = ρhA**

    Substituting: **(P₂ - P₁)A = ρhAg**

    **Simplifying:** **P₂ - P₁ = ρgh** (equation 9.6)

    Absolute Pressure at Depth

    If point 1 is at the liquid surface open to atmosphere:

  • P₁ = P_a (atmospheric pressure)
  • P₂ = P (pressure at depth h)
  • **Formula for absolute pressure:** **P = P_a + ρgh** (equation 9.7)

    Where:

  • P_a = atmospheric pressure (1.013 × 10⁵ Pa)
  • ρ = density of fluid (kg/m³)
  • g = acceleration due to gravity (9.8 m/s²)
  • h = depth below surface (m)
  • Gauge Pressure

    **Definition:** Excess pressure above atmospheric pressure.

    **Formula:** **P_gauge = P - P_a = ρgh** (equation 9.8)

    **Important Observation:** Pressure depends only on depth h, not on the cross-sectional area or shape of the container (hydrostatic paradox).

    Hydrostatic Paradox

    Three vessels A, B, C of different shapes, connected at the bottom by a horizontal pipe, all maintain the same water level when filled to equal heights. Although vessels hold different amounts of water, the pressure at any depth is identical because pressure depends only on the vertical height of the fluid column, not on the total volume or container shape.

    ---

    ATMOSPHERIC PRESSURE AND MEASUREMENT

    Mercury Barometer

    **Device:** A long glass tube closed at one end, filled with mercury, inverted into a trough of mercury.

    **Principle:** The space above the mercury column contains only mercury vapor (negligible pressure ≈ 0 Pa).

    **Pressure Balance at Base of Column:**

  • Pressure at point A (top of column) = 0
  • Pressure at point B (bottom of column in tube) must equal pressure at point C (same horizontal level in trough, open to atmosphere)
  • At point C: P_C = P_a (atmospheric pressure)
  • **Formula:** **P_a = ρ_Hg × g × h** (equation 9.8)

    Where:

  • ρ_Hg = density of mercury ≈ 13,600 kg/m³
  • h = height of mercury column
  • **At sea level:** h ≈ 0.76 m = 76 cm = 760 mm

    Therefore: **1 atm = 760 mm Hg = 1.013 × 10⁵ Pa**

    Pressure Units and Conversions

  • **1 torr** = pressure equivalent of 1 mm Hg = 133 Pa (used in medicine/physiology)
  • **1 bar** = 10⁵ Pa (used in meteorology)
  • **1 millibar** = 100 Pa
  • Relationship: **1 atm ≈ 760 torr = 1.013 bar**
  • Open Tube Manometer

    **Function:** Measures pressure differences and gauge pressures.

    **Construction:** U-tube containing liquid (low-density for small differences, mercury for large differences); one end open to atmosphere, other connected to system under test.

    **Principle:** Pressure at same horizontal level on both sides of U-tube is equal. If pressure on closed end exceeds atmospheric, liquid rises on the open end.

    **Measurement:** **P_gauge = ρ_liquid × g × h** where h is height difference of liquid columns.

    ---

    HYDRAULIC MACHINES (APPLICATIONS OF PASCAL'S LAW)

    Principle

    Any external pressure applied to a confined fluid is transmitted undiminished throughout the fluid in all directions. This principle enables **force multiplication** through **area difference**.

    Hydraulic Lift

    **Components:** Two pistons of different cross-sections (A₁ and A₂) separated by a fluid-filled space.

    **Operation:**

  • Small force F₁ applied to piston of area A₁
  • Pressure generated: **P = F₁/A₁**
  • This pressure acts on larger piston of area A₂
  • Upward force on large piston: **F₂ = P × A₂ = F₁ × (A₂/A₁)**
  • **Mechanical Advantage:** **MA = A₂/A₁**

    The output force is amplified by the ratio of piston areas. By controlling F₁, the platform can be raised or lowered.

    **Real-world example:** Car lifts in service stations, wheelchair lifts, hydraulic jacks.

    Hydraulic Brakes in Automobiles

    **Operation:**

  • Driver applies small force to brake pedal
  • Master piston in master cylinder pressurizes brake fluid
  • Pressure transmits equally to all four wheel cylinders (critical advantage)
  • Larger pistons in wheel cylinders experience amplified force
  • Force expands brake shoes against brake lining, creating braking action
  • **Advantages:**

  • Small pedal force produces large braking force
  • Pressure distributed equally to all wheels ensures uniform braking
  • Fail-safe: if one circuit fails, others still function
  • ---

    STREAMLINE FLOW

    Definition and Characteristics

    **Streamline Flow (Laminar Flow):** Flow in which at any given point in space, the velocity of each fluid particle passing through remains constant in time.

    **Key characteristics:**

  • **Steady state:** Velocity field at each point doesn't change with time
  • **Path lines and streamlines coincide:** Particles follow definite paths
  • **No turbulence:** Fluid layers slide smoothly past each other without mixing
  • **Orderly motion:** Particles at the same point always have the same velocity
  • **Occurs when:** Flow velocity is low; observed in viscous fluids; flow through narrow channels.

    Equation of Continuity

    **Statement:** For incompressible fluid in streamline flow, the mass flow rate is constant along a streamline.

    **Derivation:** Consider fluid flowing through a tube of varying cross-section.

    Consider two points with cross-sectional areas A₁ and A₂, velocities v₁ and v₂.

    **Volume flowing past point 1 in time t:** V₁ = A₁v₁t

    **Volume flowing past point 2 in time t:** V₂ = A₂v₂t

    For incompressible fluid (density ρ constant): **V₁ = V₂**

    Therefore: **A₁v₁t = A₂v₂t**

    **Equation of Continuity:** **A₁v₁ = A₂v₂ = (constant)** (equation 9.9)

    Or in terms of mass flow rate: **ρA₁v₁ = ρA₂v₂**

    **Physical meaning:** Where the tube narrows (smaller A), fluid velocity increases (larger v) to maintain constant flow rate. Where the tube widens, velocity decreases.

    **Real-world observation:** Water from a tap flows fastest where the stream is thinnest; this is why a narrower section forms as water falls.

    ---

    BERNOULLI'S PRINCIPLE

    Statement and Principle

    **Bernoulli's Principle:** For streamline flow of an incompressible, non-viscous fluid under gravity, the sum of pressure energy, kinetic energy, and gravitational potential energy per unit volume remains constant along a streamline.

    Mathematical Form

    **Bernoulli's Equation:**

    **P + ½ρv² + ρgh = constant** (equation 9.10)

    Or between any two points on a streamline:

    **P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂**

    Where:

  • **P** = pressure (Pa)
  • **ρ** = density (kg/m³)
  • **v** = velocity of fluid (m/s)
  • **g** = acceleration due to gravity (9.8 m/s²)
  • **h** = height above reference level (m)
  • **Energy Interpretation (per unit volume):**

  • **P** = pressure energy per unit volume (N/m²)
  • **½ρv²** = kinetic energy per unit volume (N/m²)
  • **ρgh** = gravitational potential energy per unit volume (N/m²)
  • Derivation Using Work-Energy Theorem

    **Setup:** Consider fluid element of mass dm = ρA₁dx₁ moving through tube section from point 1 (area A₁, velocity v₁, pressure P₁, height h₁) to point 2 (area A₂, velocity v₂, pressure P₂, height h₂).

    **Work-Energy Analysis:**

    Net work done by pressure forces: **W_pressure = P₁A₁dx₁ - P₂A₂dx₂**

    Using continuity: **A₁v₁ = A₂v₂**, so **A₁dx₁ = A₂dx₂ = dV** (volume element)

    **W_pressure = (P₁ - P₂)dV**

    Work against gravity: **W_gravity = ρg(h₂ - h₁)dV**

    **Net work = Change in kinetic energy:**

    **(P₁ - P₂)dV - ρg(h₂ - h₁)dV = ½ρ(v₂² - v₁²)dV**

    Dividing by dV:

    **P₁ - P₂ = ½ρ(v₂² - v₁²) + ρg(h₂ - h₁)**

    Rearranging:

    **P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂**

    This proves Bernoulli's equation.

    Special Cases of Bernoulli's Equation

    **Case 1: Horizontal Flow (h₁ = h₂)**

    **P₁ + ½ρv₁² = P₂ + ½ρv₂²**

    As velocity increases, pressure decreases and vice versa. This explains why water from a narrower section has lower pressure.

    **Case 2: Stationary Fluid (v₁ = v₂ = 0)**

    **P₁ + ρgh₁ = P₂ + ρgh₂**

    This reduces to the hydrostatic pressure formula: **P₂ - P₁ = ρg(h₂ - h₁)**

    **Case 3: At Same Height with Same Pressure (P₁ = P₂, h₁ = h₂)**

    **v₁ = v₂** (velocity remains constant)

    Applications of Bernoulli's Principle

    **Venturimeter:** Measures fluid flow rate using pressure difference in narrowed section.

    **Carburettor in vehicles:** Low pressure in narrow section draws fuel into air stream.

    **Aerofoil/Airplane wing:** Lower pressure above wing (faster flow) creates lift force.

    **Atomizer/Perfume spray:** Fast-moving air creates low pressure, drawing liquid up tube.

    **Spinning ball in cricket/tennis:** Spin creates pressure difference, curving the trajectory (Magnus effect).

    ---

    VISCOSITY

    Definition and Concept

    **Viscosity:** The property of a fluid that opposes relative motion between its layers; a measure of fluid's internal resistance to flow.

    **Physical Cause:** Molecular friction between fluid layers moving at different velocities.

    **Analogy:** Like friction between solid surfaces, viscosity acts between fluid layers.

    Coefficient of Viscosity

    **Definition:** Coefficient of viscosity (η) quantifies the viscous force between layers.

    **Newton's Law of Viscous Flow:**

    **F = -η A (dv/dy)** (equation 9.11)

    Where:

  • **F** = viscous force (N)
  • **η** = coefficient of viscosity (Pa·s or N·s/m²)
  • **A** = area of contact between layers (m²)
  • **dv/dy** = velocity gradient perpendicular to flow (s⁻¹)
  • Negative sign indicates force opposes relative motion
  • **Dimensions of η:** [ML⁻¹T⁻¹]

    **SI Unit:** **Pascal-second (Pa·s)** = **N·s/m²** = **kg/(m·s)**

    **Other common units:**

  • **Poise (P) = 0.1 Pa·s** (CGS unit; used historically)
  • **Centipoise (cP) = 0.01 P = 0.001 Pa·s**
  • Viscosity Values and Temperature Dependence

    **At 20°C (293 K):**

  • Water: η ≈ 1.0 × 10⁻³ Pa·s = 1 centipoise
  • Glycerin: η ≈ 1.5 Pa·s (highly viscous)
  • Air: η ≈ 1.8 × 10⁻⁵ Pa·s (very low viscosity)
  • **Temperature effects:**

  • **Liquids:** Viscosity DECREASES with increasing temperature (molecules move faster, overcome intermolecular attractions)
  • **Gases:** Viscosity INCREASES with increasing temperature (molecular collisions increase)
  • Stokes' Law

    **Statement:** The viscous drag force on a small sphere moving through a viscous fluid is directly proportional to the sphere's radius, the fluid's viscosity, and the sphere's velocity.

    **Formula:**

    **F_drag = 6πηrv** (equation 9.12)

    Where:

  • **F_drag** = viscous drag force (N)
  • **η** = coefficient of viscosity (Pa·s)
  • **r** = radius of sphere (m)
  • **v** = velocity of sphere relative to fluid (m/s)
  • **6π ≈ 18.85** (constant factor)
  • **Applicability:** Valid for laminar flow around sphere at low velocities (low Reynolds number); breakdown occurs at high speeds where turbulence develops.

    Terminal Velocity

    **Definition:** Maximum constant velocity attained by an object falling through a viscous fluid when drag force equals gravitational force.

    **Derivation:**

    For a sphere of radius r, density ρ_s, falling through fluid of density ρ_f and viscosity η:

    **Forces acting:**

  • Gravitational force (downward): **F_g = mg = ⁴⁄₃πr³ρ_s g**
  • Buoyant force (upward): **F_b = ⁴⁄₃πr³ρ_f g**
  • Viscous drag (upward): **F_drag = 6πηrv_t**
  • **At terminal velocity (acceleration = 0):** **F_g = F_b + F_drag**

    **⁴⁄₃πr³(ρ_s - ρ_f)g = 6πηrv_t**

    **Terminal Velocity Formula:**

    **v_t = (2/9) × (r²/η) × (ρ_s - ρ_f)g** (equation 9.13)

    Or simplified: **v_t = 2r²(ρ_s - ρ_f)g / 9η**

    **Key observations:**

  • Terminal velocity ∝ r² (larger spheres fall faster)
  • Terminal velocity ∝ 1/η (lower viscosity → higher terminal velocity)
  • Terminal velocity ∝ (ρ_s - ρ_f) (greater density difference → higher terminal velocity)
  • **Real-world applications:**

  • Raindrops reach terminal velocity of ~9 m/s (why heavy rain doesn't hurt more)
  • Sedimentation of particles in centrifuges
  • Blood cell motion through blood vessels
  • Oil drop experiment (Millikan) to measure electron charge
  • Energy Dissipation Due to Viscosity

    In viscous flow, mechanical energy is continuously converted to heat due to internal friction. This is why:

  • Moving fluid gradually slows down without external forces
  • Highly viscous fluids generate heat during rapid flow
  • Viscous damping removes mechanical energy from oscillating systems
  • ---

    SURFACE TENSION

    Definition and Cause

    **Surface Tension:** The property of a liquid surface due to which it behaves like a stretched elastic membrane. It results from unbalanced intermolecular forces at the liquid-air interface.

    **Molecular Explanation:**

  • In the bulk of liquid, each molecule is surrounded by others in all directions, experiencing balanced intermolecular attractive forces
  • At the surface, molecules have neighbors only on one side (below), resulting in a net inward pull
  • This creates a "skin-like" layer with excess energy, causing the surface to contract minimizing surface area
  • Surface molecules require energy to move to the surface (surface energy)
  • Coefficient of Surface Tension

    **Definition:** Surface tension (T or σ) is the force per unit length acting perpendicular to an imaginary line on the liquid surface, trying to minimize surface area.

    **Formula:**

    **T = F/l** (equation 9.14)

    Where:

  • **T** = surface tension (N/m)
  • **F** = force acting on the surface (N)
  • **l** = length of the line over which force acts (m)
  • **Alternative definition:** Surface tension = surface energy per unit area = **T = E/A** (J/m² = N/m)

    **Dimensions:** [MT⁻²]

    **SI Unit:** **N/m** (Newton per meter) or **J/m²** (Joule per square meter) or **kg/s²**

    Surface Tension Values (at 20°C, in contact with air)

  • **Water:** T ≈ 0.073 N/m
  • **Mercury:** T ≈ 0.486 N/m (highest among common liquids)
  • **Glycerin:** T ≈ 0.063 N/m
  • **Soap solution:** T ≈ 0.025 N/m (reduced due to soap)
  • **Alcohol:** T ≈ 0.022 N/m
  • Temperature Dependence of Surface Tension

    **Effect:** Surface tension **decreases with increasing temperature**

    **Reason:** Higher temperature increases molecular kinetic energy, reducing intermolecular attractive forces and the inward pull at the surface.

    **Critical point:** At critical temperature, surface tension becomes zero (liquid-gas distinction disappears).

    Surface Tension Phenomena and Applications

    **Capillarity:**

    **Definition:** The ability of a liquid to flow in narrow tubes (capillaries) against or with gravity.

    **Capillary rise (or depression) formula:**

    **h = (2T cos θ) / (ρgr)** (equation 9.15)

    Where:

  • **h** = height of capillary rise (m)
  • **T** = surface tension (N/m)
  • **θ** = contact angle between liquid and capillary wall
  • **ρ** = density of liquid (kg/m³)
  • **g** = acceleration due to gravity (9.8 m/s²)
  • **r** = radius of capillary tube (m)
  • **Interpretation:**

  • If θ < 90° (cos θ > 0): liquid wets surface, **capillary rise occurs** (liquid rises in tube)
  • Example: water in glass capillary (θ ≈ 0°)
  • If θ > 90° (cos θ < 0): liquid doesn't wet surface, **capillary depression occurs** (liquid level drops in tube)
  • Example: mercury in glass capillary (θ ≈ 140°)
  • **Key observations:**

  • Capillary rise ∝ T and ∝ 1/r (narrower tubes show greater rise)
  • Capillary rise ∝ 1/ρ (lighter liquids rise higher)
  • **Real-world applications:**

  • Water rises in soil and plant roots through capillary action (essential for plant nutrition)
  • Capillary action in lamp wicks draws oil upward
  • Blood flow through capillaries in human circulatory system
  • Ink absorption in paper and textiles
  • **Droplet and Bubble Formation:**

    **Droplets:** Spherical shape minimizes surface area for given volume, reducing surface energy (why water beads on hydrophobic surfaces).

    **Bubbles:** Similar principle; pressure inside bubble exceeds external pressure due to surface tension.

    **Excess pressure inside bubble:**

    **ΔP = 4T/r** (double-surfaced bubble like soap bubble)

    **ΔP = 2T/r** (single-surfaced droplet)

    ---

    KEY FORMULAS SUMMARY

    | Concept | Formula | Variables |

    |---------|---------|-----------|

    | Pressure | P = F/A | F (force), A (area) |

    | Density | ρ = m/V | m (mass), V (volume) |

    | Pressure-depth | P = P_a + ρgh | P_a (atmospheric), h (depth) |

    | Gauge pressure | P_g = ρgh | ρ, g, h |

    | Pascal's law | P₁ = P₂ = ... | Equal throughout fluid |

    | Continuity equation | A₁v₁ = A₂v₂ | A (area), v (velocity) |

    | Bernoulli's equation | P + ½ρv² + ρgh = const | P, ρ, v, h |

    | Viscous force | F = ηA(dv/dy) | η (viscosity), A, dv/dy |

    | Stokes' law | F_drag = 6πηrv | η, r (radius), v |

    | Terminal velocity | v_t = 2r²(ρ_s - ρ_f)g/9η | r, ρ, g, η |

    | Surface tension | T = F/l | F (force), l (length) |

    | Capillary rise | h = 2T cos θ/(ρgr) | T, θ, ρ, g, r |

    ---

    IMPORTANT EXAM-FOCUSED POINTS

    1. **Pressure is scalar**, not vector; always acts perpendicular to surfaces

    2. **Pascal's law** enables hydraulic machines with mechanical advantage = A₂/A₁

    3. **Equation of continuity** applies only to incompressible fluids; use A₁v₁ = A₂v₂

    4. **Bernoulli's equation** assumes frictionless (non-viscous) flow; always includes all three terms or show which are zero

    5. **Viscosity** causes energy dissipation; different temperature dependence for liquids vs. gases

    6. **Surface tension** acts to minimize surface area; causes capillary rise/depression depending on contact angle

    7. **Terminal velocity** is independent of time; reached when net force = zero

    8. **Manometer reading** gives gauge pressure directly; absolute pressure requires adding P_a

    9. **Hydrostatic paradox:** Pressure at depth depends only on height, not container shape or total volume

    10. Numerical problems require careful unit conversions (cm→m, g→kg, atm→Pa)

    MCQs — 10 Questions with Answers

    Q1. The SI unit of pressure is pascal (Pa). Which of the following is the correct relationship?

    • A. 1 Pa = 1 N/m² ✓
    • B. 1 Pa = 1 dyne/cm²
    • C. 1 Pa = 10 N/cm²
    • D. 1 Pa = 1 kg/m²

    Answer: A — By definition, pressure = force/area, so 1 Pa = 1 N/m² is the correct SI unit relationship.

    Q2. The two femur bones of a 40 kg person have a combined cross-sectional area of 20 cm². What is the average pressure sustained by the femurs? (Take g = 10 m/s²)

    • A. 2.0 × 10⁴ Pa
    • B. 2.0 × 10⁵ Pa ✓
    • C. 5.0 × 10⁴ Pa
    • D. 2.0 × 10⁶ Pa

    Answer: B — F = 40 × 10 = 400 N; A = 20 cm² = 20 × 10⁻⁴ m²; P = F/A = 400 / (20 × 10⁻⁴) = 2.0 × 10⁵ Pa.

    Q3. The relative density of aluminium is 2.7. What is the density of aluminium in kg/m³?

    • A. 2.7 kg/m³
    • B. 2.7 × 10² kg/m³
    • C. 2.7 × 10³ kg/m³ ✓
    • D. 2.7 × 10⁶ kg/m³

    Answer: C — Relative density = ρ_substance / ρ_water; 2.7 = ρ_Al / (1.0 × 10³), so ρ_Al = 2.7 × 10³ kg/m³.

    Q4. A force of 200 N is applied to a liquid using two different pistons: Piston A has area 10 cm² and Piston B has area 100 cm². Which statement is correct?

    • A. Pressure from Piston A is 10 times greater than Piston B ✓
    • B. Pressure from Piston B is 10 times greater than Piston A
    • C. Both pistons exert equal pressure
    • D. Pressure depends on the type of liquid, not the piston area

    Answer: A — Pressure P = F/A; P_A = 200/(10 × 10⁻⁴) = 2 × 10⁵ Pa and P_B = 200/(100 × 10⁻⁴) = 2 × 10⁴ Pa, so P_A = 10 × P_B.

    Q5. Which of the following correctly explains why a needle pierces skin while a blunt object does not, even with the same applied force?

    • A. The needle has greater force
    • B. The needle concentrates the same force over a much smaller area, creating higher pressure that exceeds the skin's strength ✓
    • C. The skin has different material properties for sharp and blunt objects
    • D. The blunt object distributes the force, but the needle breaks the skin's bonds chemically

    Answer: B — Pressure = F/A; a smaller area (needle point) means higher pressure, which exceeds the skin's tensile strength and causes piercing.

    Q6. For a fluid element in the form of a right-angled prism at rest, the forces on three faces have areas A_a, A_b, A_c and pressures P_a, P_b, P_c respectively. Which relationship is correct according to Pascal's Law?

    • A. P_a > P_b > P_c (increases with depth)
    • B. P_a = P_b = P_c (equal at same height) ✓
    • C. P_a + P_b = P_c
    • D. P_a · A_a = P_b · A_b = P_c · A_c always

    Answer: B — At the same height in a resting fluid, Pascal's Law states pressure is equal in all directions; P_a = P_b = P_c.

    Q7. Why must the force exerted by a resting fluid on an object's surface be perpendicular (normal) to that surface?

    • A. Because fluids are less dense than solids
    • B. Because a parallel component would cause the fluid to flow parallel to the surface, contradicting the rest condition ✓
    • C. Because pressure is defined only perpendicular to surfaces
    • D. Because the object's surface is always perpendicular to gravity

    Answer: B — If a parallel force component existed, the fluid would flow by Newton's third law, violating the equilibrium condition; only normal forces can exist in a resting fluid.

    Q8. A liquid in a horizontal container is at rest. Which of the following statements is NOT correct?

    • A. Pressure is the same at all points in a horizontal plane at the same depth
    • B. The pressure at the bottom is greater than at the surface due to the weight of the liquid above
    • C. Pressure is a vector quantity with a direction perpendicular to any surface ✓
    • D. In the absence of flow, horizontal forces on a fluid element must be balanced

    Answer: C — Pressure is a scalar quantity, not a vector; it has no direction assigned to it, though the force it exerts is perpendicular to surfaces.

    Q9. Two containers with different fluids at 4°C are connected to a U-tube manometer. Container A has water (ρ = 1.0 × 10³ kg/m³) and Container B has an unknown fluid. If the height difference in the manometer is the same for equal pressures, what can be inferred about the unknown fluid's density?

    • A. The unknown fluid must have the same density as water ✓
    • B. The unknown fluid must be more dense than water
    • C. The unknown fluid must be less dense than water
    • D. Density cannot be determined from height difference alone

    Answer: A — In a manometer at equal pressure, ρ_A × h_A = ρ_B × h_B; if heights are equal, then ρ_A = ρ_B, so the unknown fluid has the same density as water.

    Q10. A circus performer survives having an elephant step on a wooden plank placed across his chest (rather than the elephant stepping directly on him). This is because the plank:

    • A. reduces the elephant's weight through leverage
    • B. distributes the elephant's weight over a larger area of the chest, reducing pressure and preventing rib fracture ✓
    • C. reflects the elephant's force backward
    • D. increases the elephant's weight capacity by material strength

    Answer: B — The plank increases the contact area (A), and since P = F/A, a larger area reduces pressure on the chest, preventing injury.

    Flashcards

    What is pressure in fluids, and what are its dimensions?

    Pressure P = F/A (normal force per unit area), with dimensions [ML⁻¹T⁻²] and SI unit pascal (Pa) or N/m².

    Why does a needle pierce skin but a spoon doesn't, even with same force?

    The needle has a much smaller contact area, so the pressure (force/area) is higher, exceeding the material's strength.

    State Pascal's Law for fluids at rest.

    The pressure in a fluid at rest is the same at all points at the same height, and is transmitted equally in all directions.

    What is the difference between a solid and a fluid in terms of shear stress?

    Fluids offer negligible resistance to shear stress (about million times less than solids) and change shape with very small stress.

    Define density and relative density with their units.

    Density ρ = m/V (kg/m³); relative density is the ratio of substance density to water density at 4°C and is dimensionless.

    Why must the force exerted by a resting fluid on an object be perpendicular to the surface?

    If there were a parallel component, the fluid would flow parallel to the surface by Newton's third law, contradicting the rest condition.

    Convert 1 atm to pascal and state the relationship.

    1 atm = 1.013 × 10⁵ Pa; this is the atmospheric pressure at sea level.

    What does it mean that liquids are 'largely incompressible' unlike gases?

    Liquids have much lower compressibility; their density and volume remain nearly constant even with large changes in external pressure.

    A prismatic fluid element has three faces: why are pressures Pa, Pb, Pc equal?

    By force balance (Fb sin θ = Fc and Fb cos θ = Fa) and geometry, ratios of forces to areas are equal, so Pa = Pb = Pc.

    What is the practical implication of pressure being a scalar, not a vector?

    Pressure has no direction; the force is always perpendicular to any surface regardless of surface orientation, simplifying fluid statics calculations.

    Important Board Questions

    Define pressure and state its SI unit. If a force of 500 N acts on an area of 25 cm², calculate the pressure exerted. [2 marks]

    Pressure is normal force per unit area (P = F/A); convert 25 cm² to m² (25 × 10⁻⁴ m²) and compute P in pascals.

    Explain Pascal's Law using the equilibrium of a prismatic fluid element at rest. Show using force balance and geometry that pressures on all faces are equal. [5 marks]

    Draw a right-angled prism with forces F_a, F_b, F_c on faces with areas A_a, A_b, A_c; apply equilibrium conditions (Fb sin θ = Fc, Fb cos θ = F_a) and relate areas by geometry (Ab sin θ = Ac, Ab cos θ = A_a) to prove Pa = Pb = Pc.

    A hydraulic lift uses Pascal's Law. Small piston A (area 5 cm²) applies force to raise large piston B (area 500 cm²). If a person of mass 60 kg stands on piston A, what mass can piston B lift? (Take g = 10 m/s²). Explain why this simple machine is so effective. [6 marks]

    Use Pascal's Law (Pa = Pb); calculate F_A = 60 × 10 = 600 N; then P = 600 / (5 × 10⁻⁴) = 1.2 × 10⁷ Pa; find F_B = P × (500 × 10⁻⁴); solve for mass at B. Explain that equal pressure on different areas produces greatly different forces, enabling mechanical advantage.

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