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Mechanical Properties of Fluids

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Chapter Notes

Overview
This chapter explores the physical properties of fluids (liquids and gases), focusing on pressure, buoyancy, fluid dynamics, viscosity, and surface tension. It exists in the syllabus to bridge the gap between the mechanics of solids and the behavior of substances that can flow, which is essential for understanding natural phenomena and engineering applications. The one big idea a student must walk away with is that fluids at rest exert pressure uniformly in all directions (Pascal's law), and fluids in motion obey conservation of energy (Bernoulli's principle), with real-world implications from hydraulic lifts to airplane flight.
Key Concepts (18)
Fluid
A substance that can flow; liquids and gases are fluids because they offer very little resistance to shear stress.
e.g. Water flowing from a tap and air moving as wind are examples of fluids.
Pressure
The normal force acting per unit area. It is a scalar quantity defined as P = F/A, where F is the magnitude of the normal force on a surface of area A.
e.g. A sharp needle pierces the skin easily because the same force is applied over a very small area, producing high pressure.
Pascal's Law
The pressure in a fluid at rest is the same at all points if they are at the same height. Also, when external pressure is applied on any part of a fluid contained in a vessel, it is transmitted undiminished and equally in all directions.
e.g. In a hydraulic lift, a small force on a small piston creates a pressure that is transmitted undiminished to a larger piston, lifting a heavy car.
Gauge Pressure
The excess of pressure over atmospheric pressure, given by P_g = ρgh, where ρ is the density of the fluid, g is acceleration due to gravity, and h is the depth.
e.g. The pressure measured by a tyre gauge is the gauge pressure, i.e., the pressure inside the tyre above atmospheric pressure.
Atmospheric Pressure
The pressure exerted by the weight of a column of air of unit cross-sectional area extending from that point to the top of the atmosphere. At sea level, it is 1.013 × 10^5 Pa (1 atm).
e.g. A mercury barometer measures atmospheric pressure as the height of the mercury column, typically 76 cm at sea level.
Streamline Flow
A steady flow in which the velocity of each fluid particle at any given point is constant in time, and the paths of particles do not cross each other.
e.g. Water flowing smoothly from a tap at low speed is an example of streamline flow.
Equation of Continuity
For an incompressible fluid in steady flow, the product of the cross-sectional area and the fluid speed at any point along the pipe is constant: Av = constant.
e.g. When a river narrows, the water flows faster because the same volume of water must pass through a smaller area per unit time.
Bernoulli's Principle
For an incompressible, non-viscous fluid in steady flow, the sum of the pressure (P), kinetic energy per unit volume (½ρv²), and potential energy per unit volume (ρgh) remains constant along a streamline.
e.g. An airplane wing is shaped so that air flows faster over the top surface, creating lower pressure and providing lift.
Torricelli's Law
The speed of efflux of a fluid from an open tank is v = √(2gh), where h is the height of the fluid above the hole.
e.g. Water flowing out of a hole at the bottom of a bucket has a speed equal to that of a freely falling body from the same height.
Viscosity
The property of a fluid that resists relative motion between its layers, analogous to internal friction. It is defined as the ratio of shearing stress to the strain rate.
e.g. Honey is more viscous than water; it flows more slowly when poured.
Stokes' Law
The viscous drag force F on a small sphere of radius a moving with velocity v through a fluid of viscosity η is given by F = 6πηav.
e.g. A raindrop falling through air reaches a terminal velocity when the viscous drag (Stokes' law) plus buoyant force equals its weight.
Surface Tension
The property of a liquid surface that makes it behave like a stretched elastic membrane, arising from the extra energy of molecules at the surface compared to those in the bulk.
e.g. A needle can float on water if placed gently, because the water surface acts like a stretched film supporting it.
Surface Energy
The extra energy possessed by molecules at the surface of a liquid compared to those in the interior, due to unbalanced attractive forces.
e.g. To increase the surface area of a liquid, work must be done, which is stored as surface energy.
Dynamic Lift
The net upward force on a body moving through a fluid due to pressure differences caused by the shape or spin of the body.
e.g. A spinning cricket ball curves in the air due to the Magnus effect, a form of dynamic lift.
Magnus Effect
The phenomenon where a spinning ball moving through a fluid experiences a force perpendicular to its motion due to pressure differences caused by the spin.
e.g. In tennis, a topspin ball dips faster because the spin creates a pressure difference that pulls it downward.
Terminal Velocity
The constant velocity attained by a body falling through a viscous fluid when the net force (weight minus buoyant force minus viscous drag) becomes zero.
e.g. A skydiver reaches terminal velocity when air resistance balances their weight.
Hydrostatic Paradox
The phenomenon that the pressure at the bottom of a liquid depends only on the height of the liquid column and not on the shape or volume of the container.
e.g. Three vessels of different shapes but with the same base area and filled to the same height have the same pressure at the bottom, even though they hold different amounts of water.
Relative Density
The ratio of the density of a substance to the density of water at 4°C (1000 kg/m³). It is a dimensionless positive scalar.
e.g. The relative density of aluminium is 2.7, meaning its density is 2.7 times that of water.
Formulas & Equations (9)
P = F/A
Reads as: Pressure equals force divided by area
Variables: P = pressure (Pa or N/m²), F = normal force (N), A = area (m²)
Use when: Calculating pressure exerted by a force on a surface, or defining pressure.
⚠ Students use the total force vector instead of the normal component; pressure is scalar and uses only the normal force.
P = Pₐ + ρgh
Reads as: Pressure at depth equals atmospheric pressure plus rho g h
Variables: P = absolute pressure at depth (Pa), Pₐ = atmospheric pressure (Pa), ρ = density of fluid (kg/m³), g = acceleration due to gravity (m/s²), h = depth (m)
Use when: Finding pressure at a point below the surface of a liquid open to the atmosphere.
⚠ Forgetting to add atmospheric pressure; students often give only ρgh as the total pressure.
P₂ − P₁ = ρgh
Reads as: Pressure difference equals rho g h
Variables: P₂ − P₁ = pressure difference (Pa), ρ = density (kg/m³), g = gravity (m/s²), h = vertical height difference (m)
Use when: Calculating pressure difference between two points in a fluid at rest at different heights.
⚠ Using horizontal distance instead of vertical height difference.
A₁v₁ = A₂v₂
Reads as: A one v one equals A two v two
Variables: A = cross-sectional area (m²), v = fluid speed (m/s)
Use when: Applying the equation of continuity for incompressible fluids in steady flow.
⚠ Forgetting that this applies only to incompressible fluids; using it for gases without considering density changes.
P + ½ρv² + ρgh = constant
Reads as: P plus half rho v squared plus rho g h equals constant
Variables: P = pressure (Pa), ρ = density (kg/m³), v = fluid speed (m/s), g = gravity (m/s²), h = height (m)
Use when: Solving problems involving steady, non-viscous, incompressible flow along a streamline.
⚠ Applying Bernoulli's equation to viscous or turbulent flows where energy is lost due to friction.
v = √(2gh)
Reads as: v equals square root of 2 g h
Variables: v = speed of efflux (m/s), g = gravity (m/s²), h = height of fluid above the hole (m)
Use when: Finding the speed of fluid flowing out of a hole in an open tank (Torricelli's law).
⚠ Forgetting that this formula assumes the tank is open to the atmosphere and the hole is small compared to the tank area.
F = 6πηav
Reads as: F equals 6 pi eta a v
Variables: F = viscous drag force (N), η = coefficient of viscosity (Pa·s or N·s/m²), a = radius of sphere (m), v = velocity of sphere (m/s)
Use when: Calculating the viscous drag on a small sphere moving slowly through a fluid (Stokes' law).
⚠ Using diameter instead of radius; the formula uses radius a.
vₜ = [2a²(ρ − σ)g] / (9η)
Reads as: v sub t equals 2 a squared times (rho minus sigma) g divided by 9 eta
Variables: vₜ = terminal velocity (m/s), a = radius of sphere (m), ρ = density of sphere (kg/m³), σ = density of fluid (kg/m³), g = gravity (m/s²), η = viscosity (Pa·s)
Use when: Finding the terminal velocity of a sphere falling through a viscous fluid.
⚠ Forgetting to subtract the fluid density σ from the sphere density ρ; using ρ alone.
F₂ = (A₂/A₁) × F₁
Reads as: F two equals A two over A one times F one
Variables: F₂ = force on larger piston (N), A₂ = area of larger piston (m²), A₁ = area of smaller piston (m²), F₁ = force on smaller piston (N)
Use when: Calculating the output force in a hydraulic machine based on Pascal's law.
⚠ Confusing area ratio; students sometimes invert the ratio.
How It Works
Proof of Pascal's Law using a prismatic fluid element
  1. Consider a small right-angled prismatic element ABC-DEF inside a fluid at rest.
  2. The fluid exerts normal forces Fₐ, F_b, F_c on the three rectangular faces.
  3. By equilibrium: F_b sinθ = F_c and F_b cosθ = Fₐ.
  4. By geometry: A_b sinθ = A_c and A_b cosθ = Aₐ.
  5. Dividing force by area gives Pₐ = P_b = P_c, proving pressure is same in all directions.
💡 This proof assumes the element is small enough that gravity can be ignored; the forces are purely due to fluid pressure.
Variation of Pressure with Depth
  1. Consider a cylindrical fluid element of area A and height h at rest.
  2. Forces: P₁A downward at top, P₂A upward at bottom, weight mg downward.
  3. Equilibrium: (P₂ − P₁)A = mg.
  4. Mass m = ρV = ρhA, so (P₂ − P₁)A = ρhAg.
  5. Cancel A: P₂ − P₁ = ρgh.
💡 The pressure difference depends only on the vertical height h, not on the shape or cross-sectional area of the container.
Working of a Hydraulic Lift
  1. A small force F₁ is applied on a piston of small area A₁.
  2. This creates a pressure P = F₁/A₁ in the fluid.
  3. According to Pascal's law, this pressure is transmitted undiminished throughout the fluid.
  4. The same pressure acts on a larger piston of area A₂, producing an upward force F₂ = P × A₂ = (A₂/A₁) × F₁.
  5. The mechanical advantage is A₂/A₁, allowing a small force to lift a heavy load.
💡 The work done (force × distance) is the same on both pistons; the smaller piston moves a larger distance than the larger piston.
Derivation of Bernoulli's Equation
  1. Consider an incompressible, non-viscous fluid in steady flow through a pipe of varying cross-section and height.
  2. In time Δt, fluid at region 1 moves distance v₁Δt, and at region 2 moves v₂Δt.
  3. Work done on fluid at left end: W₁ = P₁A₁(v₁Δt) = P₁ΔV.
  4. Work done by fluid at right end: W₂ = P₂A₂(v₂Δt) = P₂ΔV; net work on fluid = (P₁ − P₂)ΔV.
  5. Change in kinetic energy: ΔK = ½ρΔV(v₂² − v₁²).
  6. Change in potential energy: ΔU = ρgΔV(h₂ − h₁).
  7. Apply work-energy theorem: (P₁ − P₂)ΔV = ½ρΔV(v₂² − v₁²) + ρgΔV(h₂ − h₁).
  8. Divide by ΔV and rearrange: P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂.
💡 Bernoulli's equation assumes no energy loss due to viscosity (friction) and applies only to steady, incompressible flow along a streamline.
Determination of Terminal Velocity using Stokes' Law
  1. A sphere of radius a and density ρ falls through a fluid of density σ and viscosity η.
  2. Forces acting: weight (4/3πa³ρg) downward, buoyant force (4/3πa³σg) upward, viscous drag (6πηav) upward.
  3. Initially, net force downward causes acceleration; as speed increases, drag increases.
  4. When net force becomes zero, the sphere moves with constant terminal velocity vₜ.
  5. Set weight = buoyant force + drag: (4/3)πa³ρg = (4/3)πa³σg + 6πηavₜ.
  6. Solve for vₜ: vₜ = [2a²(ρ − σ)g] / (9η).
💡 Terminal velocity is proportional to the square of the radius; larger spheres fall faster in the same fluid.
Magnus Effect (Dynamic Lift on a Spinning Ball)
  1. A ball moving through air spins, dragging a layer of air with it due to viscosity.
  2. Relative to the ball, air moves faster on the side where spin direction and airflow direction are the same.
  3. On the opposite side, the spin opposes the airflow, reducing relative speed.
  4. According to Bernoulli's principle, higher speed means lower pressure.
  5. The pressure difference across the ball creates a net force perpendicular to the motion (dynamic lift).
💡 The Magnus effect explains curveballs in cricket, tennis, and baseball; the direction of spin determines the direction of the curve.
Exam Traps — Don't Make These Mistakes
⚠ Pressure is a vector
✗ Pressure is a vector quantity because force is a vector.
→ Pressure is defined as the normal force per unit area, but it is a scalar. The force has a direction, but pressure itself has no direction; it acts equally in all directions in a fluid at rest.
✓ Pressure is a scalar quantity. It is the magnitude of the normal force per unit area, and it acts equally in all directions in a fluid at rest.
⚠ Using horizontal distance in pressure-depth formula
✗ P₂ − P₁ = ρg × (horizontal distance between points).
→ The pressure difference in a fluid at rest depends only on the vertical height difference, not the horizontal separation. Horizontal distance does not affect pressure if the points are at the same depth.
✓ P₂ − P₁ = ρgh, where h is the vertical height difference between the two points.
⚠ Applying Bernoulli's equation to viscous or turbulent flow
✗ Using P + ½ρv² + ρgh = constant for any fluid flow problem.
→ Bernoulli's equation assumes the fluid is non-viscous (no internal friction) and the flow is steady. In real viscous fluids, energy is lost due to friction, so the sum is not constant.
✓ Bernoulli's equation applies only to ideal fluids: incompressible, non-viscous, and in steady streamline flow. For real fluids, energy losses due to viscosity must be accounted for.
⚠ Confusing gauge pressure with absolute pressure
✗ The pressure at depth h is ρgh.
→ ρgh gives the gauge pressure (excess over atmospheric pressure). The absolute pressure at depth h is Pₐ + ρgh, where Pₐ is atmospheric pressure.
✓ The absolute pressure at depth h below the surface of a liquid open to the atmosphere is P = Pₐ + ρgh. The gauge pressure is P − Pₐ = ρgh.
⚠ Using diameter instead of radius in Stokes' law
✗ F = 6πηdv, where d is the diameter of the sphere.
→ Stokes' law uses the radius a of the sphere, not the diameter. The formula is F = 6πηav.
✓ The viscous drag force on a sphere is F = 6πηav, where a is the radius of the sphere.
⚠ Forgetting that Torricelli's law assumes a large tank
✗ v = √(2gh) for any hole in any container.
→ Torricelli's law assumes the cross-sectional area of the tank is much larger than the area of the hole, so the velocity of the fluid surface is approximately zero. If the tank is small, the surface velocity cannot be ignored.
✓ Torricelli's law v = √(2gh) applies when the tank's cross-sectional area is much larger than the hole's area, so the fluid surface is nearly stationary.
Compare & Contrast
Definition and reference point
Absolute Pressure
Gauge Pressure
Absolute pressure is the total pressure measured relative to a perfect vacuum, while gauge pressure is the excess pressure over atmospheric pressure (P_g = P_abs − P_atm).
Nature of flow and velocity
Streamline Flow
Turbulent Flow
In streamline flow, fluid particles move along smooth, non-crossing paths with constant velocity at a given point; in turbulent flow, the velocity fluctuates randomly and eddies form, causing energy loss.
Effect of temperature
Viscosity of Liquids
Viscosity of Gases
The viscosity of liquids decreases with an increase in temperature, whereas the viscosity of gases increases with an increase in temperature.
Exam Focus
Marks pattern: Typically 2-3 MCQs (1 mark each), 1-2 short answer questions (2-3 marks), 1 long answer/numerical (5 marks). Total weightage: 7-10 marks.
Most asked: Pascal's law and hydraulic machines (numericals) · Bernoulli's principle and its applications (lift, Torricelli's law) · Equation of continuity · Variation of pressure with depth · Stokes' law and terminal velocity · Surface tension and surface energy
Don't skip: Numerical on hydraulic lift using Pascal's law; Numerical on Bernoulli's equation (e.g., speed of efflux, pressure difference on airplane wing); Derivation of terminal velocity using Stokes' law; Proof of Pascal's law or derivation of Bernoulli's equation
One-Line Revision (read aloud before the exam)
  1. Fluids are substances that can flow; they offer very little resistance to shear stress.
  2. Pressure is a scalar quantity defined as normal force per unit area (P = F/A).
  3. Pascal's law: Pressure in a fluid at rest is transmitted undiminished in all directions.
  4. Pressure at depth h in a liquid: P = Pₐ + ρgh.
  5. Gauge pressure is the excess pressure over atmospheric pressure: P_g = ρgh.
  6. The equation of continuity for incompressible fluids: A₁v₁ = A₂v₂.
  7. Bernoulli's equation: P + ½ρv² + ρgh = constant along a streamline.
  8. Torricelli's law: Speed of efflux from an open tank is v = √(2gh).
  9. Viscosity is the internal friction in fluids; it opposes relative motion between layers.
  10. Stokes' law: Viscous drag on a sphere is F = 6πηav.
  11. Terminal velocity of a sphere in a fluid: vₜ = [2a²(ρ − σ)g] / (9η).
  12. Surface tension arises because molecules at the surface have extra energy compared to those in the bulk.
  13. Dynamic lift on an airplane wing is due to higher air speed (lower pressure) above the wing than below.
  14. The Magnus effect explains the curved path of a spinning ball in sports.
  15. Hydraulic machines use Pascal's law to multiply force: F₂ = (A₂/A₁) × F₁.
  16. Atmospheric pressure at sea level is 1.013 × 10⁵ Pa or 76 cm of Hg.
  17. Relative density is the ratio of a substance's density to that of water at 4°C.

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.

Assertion-Reason Questions for Mechanical Properties of Fluids

Q1. Assertion (A):

Pressure is a scalar quantity even though force is a vector.

Reason (R):

Pressure is defined as the normal component of force per unit area, and the direction-dependent aspect of force is removed in the averaging process over area.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both the assertion and reason are true. Pressure, being force per unit area with no directional component, is a scalar. The reason correctly explains why despite force being a vector, pressure becomes a scalar—the normal component per unit area removes directional dependence.

Q2. Assertion (A):

Liquids are less compressible than gases because liquids have much stronger intermolecular forces and molecules are closer together.

Reason (R):

Greater intermolecular distances in liquids allow more room for compression compared to gases.

Answer: C — A is true but R is false — The assertion is true—liquids are indeed less compressible than gases due to strong intermolecular cohesion and close molecular packing. However, the reason is false and contradictory: it states liquids have greater intermolecular distances, which is opposite to reality. Gases have larger molecular separations, making them compressible.

Q3. Assertion (A):

In a fluid at rest, the pressure at a point is the same in all directions.

Reason (R):

This is because the fluid element at that point is in static equilibrium and the net force on it must be zero.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both assertion and reason are correct. Isotropic pressure (same in all directions) at a point in a static fluid is a consequence of equilibrium. If pressure varied directionally, there would be a net force causing fluid motion, contradicting the 'at rest' condition.

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Case-Based Questions (CBQ) for Mechanical Properties of Fluids

Case Study:

A deep-sea diving operation is being conducted off the coast of Goa. Divers descend to a depth of 200 m where the pressure is approximately 2.0 × 10⁷ Pa. The atmospheric pressure at sea level is 1.0 × 10⁵ Pa. The density of seawater is approximately 1025 kg/m³. The diving suit must be pressurised to balance the external pressure. Engineers are designing a spherical observation window for a submarine with diameter 0.5 m.

Q1. What is the gauge pressure (pressure above atmospheric) at 200 m depth?

  • A. 2.0 × 10⁷ Pa
  • B. 1.9 × 10⁷ Pa ✓
  • C. 1.0 × 10⁵ Pa
  • D. 1.99 × 10⁷ Pa

Answer: B

Q2. Using P = P_atm + ρgh, what depth corresponds to a total pressure of 3.0 × 10⁶ Pa?

  • A. Approximately 290 m
  • B. Approximately 30 m ✓
  • C. Approximately 290 cm
  • D. Approximately 3000 m

Answer: B

Q3. The observation window experiences force F = P × A. If the total pressure at 200 m is 2.0 × 10⁷ Pa and the window's cross-sectional area is π(0.25)² m², what is the approximate force on the window?

  • A. 3.9 × 10⁶ N ✓
  • B. 1.96 × 10⁶ N
  • C. 2.0 × 10⁷ N
  • D. 5.0 × 10⁶ N

Answer: A

Q4. Why is the diving suit pressurised to match the external water pressure?

  • A. To prevent the diver's body from being crushed by the net pressure difference ✓
  • B. To reduce the buoyancy force
  • C. To increase the diver's weight
  • D. To make the suit more rigid

Answer: A

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Class 11 Physics Formula Sheet

Every formula from all 14 chapters on one page, checked against the NCERT textbook and the CBSE 2026-27 syllabus. Free PDF download.

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