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Physics — Second Secondary

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Physics — Second Secondary 2026/2027

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Text lesson

Chapter 5: Properties of Hydrostatics

Physics — Second Secondary

Chapter 5: Properties of Hydrostatics

Overview

Overview

(Unit Three→Properties of Matter — Second Secondary, Term 1)

🌳 Concept tree

Chapter 5→Properties of Hydrostatics

├── What you will learn

├── Overview

├── Lesson 1→Pressure

│ ├── 1.1 Definition

│ ├── 1.2 Enrichment: heel against elephant

│ └── 1.3 Worked examples

├── Lesson 2→Pressure at a Point Inside a Liquid

│ ├── 2.1 Observations

│ ├── 2.2 Deriving P = Pa + ρgh

│ ├── 2.3 Consequences

│ ├── 2.4 Worked examples

  1. │ └── 2.5 Balance of two immiscible liquids in a U-tube
  2. Atmospheric Pressure and the Barometer

│ ├── 3.1 Atmospheric pressure

│ ├── 3.2 The mercury barometer (Torricelli)

│ ├── 3.3 Units of pressure

  • │ ├── 3.4 Factors that do not affect the barometer height

│ ├── 3.5 Worked examples

│ └── 3.6 Enrichment: the ear at heights

├── Lesson 4→The Manometer and Applications of Pressure

│ ├── 4.1 The manometer

│ ├── 4.2 Worked examples

│ ├── 4.3 Applications of pressure

│ └── 4.4 More worked examples

├── Lesson 5→Pascal's Principle and the Hydraulic Press

│ ├── 5.1 Pascal's principle

│ ├── 5.2 The hydraulic press

│ ├── 5.3 Applications

│ └── 5.4 Worked examples

  • ├── Short answers to typical discussion questions

└── Summary

Chapter 5: Properties of Hydrostatics · 1 / 15
What you will learn

What you will learn

  • Pressure at a point and its unit; why a pointed heel can press more than an elephant’s foot.
  • Pressure at a point inside a liquid P = Pa + ρgh, with its consequences (connecting vessels, dams, divers).
  • Balance of immiscible liquids in a U-tube; finding the relative density of oil.
  • Atmospheric pressure, the mercury barometer (Torricelli), and units of pressure.
  • The manometer.
  • Applications of pressure (blood pressure, tyre pressure).
  • Pascal’s principle and the hydraulic press, mechanical advantage, hydraulic brakes and lifts.
Chapter 5: Properties of Hydrostatics · 2 / 15
Overview

Overview

  • After studying fluids in motion (hydrodynamics), we study fluids at rest (hydrostatics): how a liquid exerts pressure at every point inside it, how we measure the pressure of the atmosphere and of gases, and how pressure applied to a liquid is transmitted.
Chapter 5: Properties of Hydrostatics · 3 / 15
Lesson 1: Pressure

Lesson 1: Pressure

1.1 Definition

  • Pressure at a point is the average force acting perpendicularly on a unit area surrounding that point.
  • If a force F acts perpendicularly on a surface of area A, the pressure is
  • P = F / A
  • Unit newton per square metre, N/m2, called the pascal (Pa): 1 Pa = 1 N/m2.
  • Dimensional formula ML−1T−2.
  • Pressure is a scalar quantity (the force it produces on a surface is always perpendicular to that surface).
  • Pressure increases when the force increases or when the area decreases.
  • Which presses harder on the ground, an elephant’s foot or a woman’s pointed heel?
  • Because the area of a pointed heel is tiny, the pressure under it can be much greater than under an elephant’s broad foot, even though the elephant’s weight is far larger.

1.3 Worked examples

Comparison aspectMain ideaKeywords
Example 1.
  • A 60 kg person stands on both feet
  • each of contact area 200 cm2. (g = 9.8 m/s2) Weight = 60 × 9.8 = 588 N
  • total area = 400 cm2 = 0.04 m2. P = 588/0.04 = 14 700 N/m2. Standing on one foot doubles the pressure to 29 400 N/m2
personstandsbothfeet
Example 2.A heel of area 1 cm2 carries 300 N: P = 300/10−4 = 3 × 106 N/m2. A foot of area 0.04 m2 carrying 5000 N gives only 1.25 × 105 N/m2heelareacarriesfoot
Example 3 (a block).→A solid block 4 cm × 10 cm × 25 cm has density 2500 kg/m3 (take g = 10 m/s2). Its volume is 1000 cm3 = 10−3 m3; mass = 2.5 kg; weight = 25 N.
Comparison aspectMain ideaKeywords
Largest pressure (smallest face 4 × 10 cm2 = 4 × 10−3 m2)P = 25 / 4 × 10−3 = 6250 N/m26250
•Smallest pressure (largest face 10 × 25 = 250 cm2 = 0.025 m2): P = 25/0.025 = 1000 N/m2Smallestpressurelargestface
Why do we use a sharp knife, and wide tracks on a tractor?Sharp edge → small area → large pressure (cuts easily). Wide tracks → large area → small pressure (does not sink in soft ground)edgesmallarealarge
Chapter 5: Properties of Hydrostatics · 4 / 15
2.1 Observations

2.1 Observations

  • Push a cork under water and release it it rises back to the surface.
  • So the water exerts an upward force on the submerged cork; this force comes from the difference in pressure acting on the bottom and the top of the cork.
  • At any point inside a liquid the pressure acts in all directions.
  • The force produced by liquid pressure on a surface is always perpendicular to that surface.
  • The pressure increases with depth.
Chapter 5: Properties of Hydrostatics · 5 / 15
2.2 Deriving P = Pa + ρgh

2.2 Deriving P = Pa + ρgh

  • Imagine a horizontal plate of area A at depth h below the free surface of a liquid of density ρ.
  • The plate carries a vertical column of liquid above it, of height h and cross-section A.
  1. Volume of the column = A h
  2. Mass of the column = ρ A h
  3. Weight of the column = Fg = ρ A h g
  1. Because the liquid is at rest and incompressible
  2. the force on the plate due to the liquid equals the weight of this column.

So the pressure due to the liquid alone is

P = F/A = ρ A h g / A → P = ρ g h (N/m2)

  • If the liquid surface is open to the atmosphere (atmospheric pressure Pa), this pressure is added, so the total (absolute) pressure at depth h is

P = Pa + ρ g h

  • The quantity ρgh is the gauge pressure — the pressure due to the liquid only.
  • The pressure at a point depends on:

the depth h (greater depth → greater pressure),

  • the density of the liquid ρ (greater density → greater pressure at the same depth),
  • g and the pressure at the surface Pa.
    It does not depend on the shape or width of the container nor on the area of its base, nor on the amount of liquid.
Chapter 5: Properties of Hydrostatics · 6 / 15
2.3 Consequences

2.3 Consequences

Comparison aspectMain ideaKeywords
•All points at the same horizontal level inside the same liquid at rest have the same pressurepointssamehorizontallevel
Connecting vessels
  • a liquid poured into vessels connected at the bottom rises to the same level in all of them
  • whatever their shapes (the pressure at their common base level must be the same). So the sea level is the same for all connected seas
liquidpouredconnectedbottom
Damsare built thicker at the base to withstand the greater pressure at greater depthsbuiltthickerbasewithstand
Divers
  • wear protective suits. Near the surface
  • a diver blows air into the sinuses to equalise the pressure. At great depths (hundreds of metres) a pressurised diving suit and helmet protect him from the huge external pressure
wearprotectivesuitsNear
Chapter 5: Properties of Hydrostatics · 7 / 15
2.4 Worked examples

2.4 Worked examples

  • Example 1 (lake). Find the gauge pressure and the total pressure at a depth of 12 m in fresh water. (ρ = 1000 kg/m3, g = 9.8 m/s2, Pa = 1.013 × 105 N/m2)
  • ρgh = 1000 × 9.8 × 12 = 1.176 × 105 N/m2; total P = 1.013 × 105 + 1.176 × 105 = 2.189 × 105 N/m2 (about 2.2 atm).
  • Example 2 (force on a tank bottom). An oil tank (ρ = 850 kg/m3) is filled to a depth of 2.5 m. Its base area is 4 m2. (g = 10 m/s2, Pa = 1.013 × 105 N/m2)
  • ρgh = 850 × 10 × 2.5 = 21 250 N/m2.
  • Total pressure P = 1.013 × 105 + 0.2125 × 105 = 1.2255 × 105 N/m2.
  • Total force on the base = P × A = 1.2255 × 105 × 4 = 4.9 × 105 N. (Force due to the oil alone: 21 250 × 4 = 85 000 N.)
Comparison aspectMain ideaKeywords
Example 3 (diver).
  • A diver is at 30 m in sea water (ρ = 1030 kg/m3). ρgh = 1030 × 9.8 × 30 = 3.03 × 105 N/m2
  • total = 4.04 × 105 N/m2 ≈ 4 atm
water1030total
Example 4 (depth from pressure).At what depth in water is the total pressure 3 atm if the surface pressure is 1 atm? The extra 2 atm = 2 × 1.013 × 105 = ρgh → h = 2.026 × 105/(1000 × 9.8) = 20.7 mwhatwatertotalsurface
Chapter 5: Properties of Hydrostatics · 8 / 15
2.5 Balance of two immiscible liquids in a U-tube

2.5 Balance of two immiscible liquids in a U-tube

  • Pour some water into a U-shaped tube.
  • Then pour oil (which does not mix with water) into one arm until the oil/water separating surface is at A in that arm.
  • Let the oil column above the separating level have height ho and the water column above the same horizontal level in the other arm have height hw.
  • The points at the same level in the same liquid have the same pressure.
  • Applying this to the two points at the level of the separating surface:

Pa + ρo g ho = Pa + ρw g hw

→ ρo ho = ρw hw or ρo / ρw = hw / ho

  • So by measuring ho and hw we can find the relative density of the oil (ρo/ρw), and its density if the density of water is known.
Comparison aspectMain ideaKeywords
Example 1.The oil column is 10 cm and the water column is 8 cm: relative density of the oil = 8/10 = 0.8; ρo = 800 kg/m3.columnwaterrelativedensity
Example 2 (unequal arms, book-style).A U-tube has a narrow arm of cross-section 1 cm2 and a wide arm of 2 cm2. Water is in the tube, and 6 cm of oil of density 900 kg/m3 is poured into the narrow arm. Height of the water above the separating level: hw = ρo ho/ρw = 900 × 6/1000 = 5.4 cm. (The cross-sectional areas of the arms do not matter, only heights.)tubenarrowcrosssection
Example 3 (water and mercury).Mercury (13 600 kg/m3) forms a column of 5 cm in one arm; the water column above the mercury level in the other arm: hw = 13 600 × 5/1000 = 68 cm.formscolumnabovelevel
Chapter 5: Properties of Hydrostatics · 9 / 15
Lesson 3: Atmospheric Pressure and the Barometer

Lesson 3: Atmospheric Pressure and the Barometer

3.1 Atmospheric pressure

  • Atmospheric pressure
  • is the pressure produced by the weight of the column of air above a unit area of the Earth’s surface.
  • Going higher, the column of air above you gets shorter, so the atmospheric pressure decreases with height.
  • Torricelli took a glass tube about 1 m long, closed at one end, completely filled it with mercury, and turned it upside down with its open end under the surface of mercury in a dish.
  • The mercury in the tube dropped until the vertical height of its surface above the surface of mercury in the dish was about 0.76 m.

  • The space above the mercury is a vacuum (the Torricelli vacuum), apart from a little mercury vapour whose pressure is negligible.

  • The vertical height h does not change if the tube is tilted or if the tube is wider or narrower — only the length of mercury in the tube changes when it is tilted.

  • Take two points A (outside the tube, at the mercury surface in the dish) and B (inside the tube, at the same horizontal level). The pressure at B = the pressure at A = atmospheric pressure. The pressure at B is produced by the mercury column above it. So

    Pa = ρHg g h

  • Standard atmospheric pressure is the pressure that supports a column of mercury 0.76 m high at 0 °C (cross-section 1 m2) at sea level; it is the pressure at STP (standard temperature and pressure: 0.76 m Hg and 0 °C), a term often used with gas laws.
  1. Using ρHg = 13 595 kg/m3 at 0 °C and g = 9.8 m/s2:
  2. Pa = 1 atm = 0.76 × 13 595 × 9.8 = 1.013 × 105 N/m2
Unit Equivalent
pascal (Pa) 1 N/m2
bar 105 Pa
mm Hg = torr 1 torr = 1 mm Hg
atmosphere 1 atm = 1.013 × 105 Pa = 1.013 bar = 760 mm Hg = 76 cm Hg = 0.76 m Hg = 760 torr
  • The height of the mercury column does not depend on the cross-section of the tube, nor on tilting it, nor on the amount of mercury in the dish.
  • It depends on the atmospheric pressure, the density of the liquid used and g.
  • (A water barometer would need a column 13.6 times taller: about 10.3 m.)

3.5 Worked examples

Comparison aspectMain ideaKeywords
Example 1.The barometer height on a hilltop is 0.72 m Hg. Pa = 0.72 × 13 600 × 9.8 = 9.60 × 104 N/m2.barometerheighthilltop
Example 2.Water barometer height: Pa = ρw g h → h = 1.013 × 105/(1000 × 9.8) = 10.3 m.Waterbarometerheight1000
Example 3.
  • Convert 1.5 atm: in cm Hg = 1.5 × 76 = 114 cm Hg
  • in Pa = 1.5 × 1.013 × 105 = 1.52 × 105 Pa
Convert
Example 4 (change of height).A mercury barometer reads 76 cm Hg at the bottom of a tower and 74 cm Hg at the top. The pressure difference = 2 cm Hg = 0.02 × 13 600 × 9.8 = 2666 N/m2. If the air density is about 1.2 kg/m3 then ρair g H = 2666 → H = 2666/(1.2 × 9.8) ≈ 227 m.mercurybarometerreadsbottom

3.6 Enrichment: the ear at heights

  • The pressure on the two sides of the eardrum is normally equal.
  • When we rise quickly (in a lift, aeroplane or climbing a mountain), the outside pressure decreases, and the higher pressure inside the middle ear pushes the eardrum outward, producing a feeling of tension.
  • Swallowing or chewing opens the Eustachian tube and equalises the pressure.
  • Why don’t we feel the huge atmospheric pressure (more than 10 tonnes per square metre)?
  • Because the pressure of the fluids inside our bodies equals the outside pressure, so the forces on the surfaces of the body are balanced.
Chapter 5: Properties of Hydrostatics · 10 / 15
Lesson 4: The Manometer and Applications of Pressure

Lesson 4: The Manometer and Applications of Pressure

4.1 The manometer

  • A manometer measures the difference between the pressure of a gas in a container and the atmospheric pressure.
  • It is a U-shaped tube containing a liquid of known density (usually mercury, or water/oil for small pressures).
  • One arm is connected to the gas container; the other is open to the air.
  • Gas pressure greater than atmospheric the liquid is lower in the arm connected to the gas. Using two points at the same level in the same liquid: P = Pa + ρ g h
  • Gas pressure less than atmospheric the liquid in the open arm is lower: P = Pa − ρ g h
  • The pressure difference (gauge pressure) is ΔP = P − Pa = ρ g h (positive or negative).
  • Choice of manometer liquid.
  • Mercury for large pressures (dense, so the height difference is small); a light liquid (water, oil) for small pressure differences, which gives a large and readable height difference.
  • The liquid must not react with or dissolve the gas and should have a low vapour pressure.

4.2 Worked examples

Comparison aspectMain ideaKeywords
Example 1.A mercury manometer connected to a gas container shows the mercury in the open arm 24 cm higher than in the arm connected to the container. The gas pressure P = 76 + 24 = 100 cm Hg = 100/76 = 1.316 atm = 1.316 × 1.013 × 105 = 1.333 × 105 N/m2mercurymanometerconnectedcontainer
Example 2.
  • If the mercury in the open arm were 18 cm lower
  • then P = 76 − 18 = 58 cm Hg (below atmospheric: a partial vacuum)
mercuryopenwerelower
Comparison aspectMain ideaKeywords
Example 3 (water manometer).A water manometer shows a height difference of 40 cm. ΔP = ρgh = 1000 × 9.8 × 0.4 = 3920 N/m2showsheightdifference1000
Example 4 (mercury, SI).
  • In a manometer h = 25 cm of mercury
  • ρ = 13 600
  • g = 9.8: ΔP = 13 600 × 9.8 × 0.25 = 3.332 × 104 N/m2
  • the absolute gas pressure = 1.013 × 105 + 0.333 × 105 = 1.346 × 105 N/m2
manometerabsolutepressure
  1. Blood pressure. Blood is a viscous fluid pumped by the heart through arteries and veins.
  2. Blood flow is normally steady.
  3. If it becomes turbulent it produces noise, detected by a stethoscope on the artery while measuring blood pressure.
  4. Two values are recorded:
Comparison aspectMain ideaKeywords
Systolic pressurethe maximum pressure in the artery when the heart muscle contracts and pushes blood from the left ventricle into the aorta. Normal: about 120 torrmaximumarteryheartmuscle
Diastolic pressure
  • the minimum pressure when the heart muscle relaxes. Normal: about 80 torr. (The cuff is placed on the upper arm at heart level so that the height of the blood column does not add or subtract ρgh to the reading. Long sitting increases the risk of leg blood clots because blood flow in the legs is slow
  • moving and exercising the legs helps.)
minimumheartmusclerelaxes
  1. Car-tyre pressure. A tyre filled with air at high pressure has a small contact area with the road.
  2. A tyre with too low a pressure has a large contact area, so more friction and more heat, and wear.
  3. Pressure is measured by a tyre-pressure gauge (a piston pushed against a spring with a graduated scale).
4.4 More worked examples
Example 1 (tyre).A tyre gauge shows 2.5 × 105 N/m2 above the atmosphere. Absolute pressure = 2.5 × 105 + 1.013 × 105 = 3.51 × 105 N/m2 (3.5 atm).
Example 2.Blood pressure of 120 torr = 120 mm Hg = 0.12 × 13 600 × 9.8 = 1.6 × 104 N/m2.
Chapter 5: Properties of Hydrostatics · 11 / 15
Lesson 5: Pascal’s Principle and the Hydraulic Press

Lesson 5: Pascal’s Principle and the Hydraulic Press

5.1 Pascal’s principle

  • Consider a liquid in a container fitted with a piston.
  • At a point A at depth h below the piston the pressure is P = P₁ + ρgh, where P₁ is the pressure just under the piston.
  • If we add a weight on the piston, the pressure at the piston surface rises by ΔP.
  • The piston does not move because the liquid is incompressible, but the pressure at every point increases by the same ΔP.
  • Pascal’s principle (law) When pressure is applied to a liquid enclosed in a container, the pressure is transmitted in full to every part of the liquid and to the walls of the container.
  • (If the pressure is increased too much, the container may burst.)

5.2 The hydraulic press

  • It consists of two cylinders connected by a pipe and filled with a suitable liquid (oil).
  • One cylinder has a small piston of area a; the other has a large piston of area A.
  • A force f on the small piston creates a pressure P = f/a.
  • By Pascal’s principle this pressure is transmitted to the large piston, which is at the same level, so P = F/A also.
  • f / a = F / A → F = f · (A / a)
  • Hence a small force on the small piston produces a large force on the large piston.
  • Mechanical advantage of the press:
  • η = F / f = A / a (a ratio, no unit)
  • Energy conservation. If the small piston moves down a distance y₁ and the large piston moves up y₂, then (ideal press, efficiency 100 %) the work done on the small piston equals the work done by the large piston:

f y₁ = F y₂ → F/f = y₁/y₂ = A/a

  • So η = F/f = A/a = y₁/y₂. The large piston moves a smaller distance than the small one — we gain force but not work.
  • Because the area depends on the square of the radius or diameter, if the ratio of the radii is rlarge : rsmall = n, then η = n2. (E.g. a radius ratio of 9 : 1 between the large and the small piston gives η = 81.)
Comparison aspectMain ideaKeywords
Hydraulic brakes
  • in cars: the pedal pushes a piston in the master cylinder
  • the pressure is transmitted through the brake fluid to the pistons of the wheel cylinders
  • which push brake shoes against the drum (rear drum brakes) or pads against the disc (front disc brakes). A small foot force produces large friction forces at all four wheels
  • and the pressure is the same to all wheels
carspedalpushespiston
Hydraulic lift(car lift in a garage)garage
Hydraulic excavator
  • and other hydraulic machines (jacks
  • presses
  • aircraft control systems)
machinesjackspressesaircraft

5.4 Worked examples

Example 1.→In a hydraulic press the small piston has diameter 4 cm and the large piston has diameter 40 cm. A force of 50 N is applied to the small piston. (g = 10 m/s2)
Comparison aspectMain ideaKeywords
•
  • Area ratio A/a = (40/4)2 = 100
  • so η = 100
Arearatio
•
  • F = 50 × 100 = 5000 N
  • the largest mass it can lift = 5000/10 = 500 kg
5000largestmasslift
To raise the large piston by 2 cmy₁ = η y₂ = 100 × 2 = 200 cm
Example 2 (car lift).
  • A car of mass 1200 kg (weight 12 000 N) rests on a piston of radius 20 cm
  • the small piston has a radius of 2 cm. f = F(a/A) = 12 000 × (2/20)2 = 12 000/100 = 120 N
mass1200weightrests
Example 3 (pressure).What pressure is transmitted if 120 N acts on a small piston of area 4 × 10−4 m2? P = 120/4 × 10−4 = 3 × 105 N/m2Whattransmittedactssmall
Example 4 (efficiency idea).
  • A press has η = 40. A force of 250 N acts on the small piston
  • which is pushed 10 cm. F = 10 000 N
  • the large piston rises y₂ = 10/40 = 0.25 cm. Work input = 250 × 0.10 = 25 J
  • work output = 10 000 × 0.0025 = 25 J (equal)
pressforceactssmall
Example 5 (given η and F).→A hydraulic press lifts a 2000 N load on the large piston of area 0.05 m2 with a small-piston area of 2.5 × 10−3 m2. η = 0.05/2.5 × 10−3 = 20, so f = 2000/20 = 100 N.
Chapter 5: Properties of Hydrostatics · 12 / 15
Short answers to typical discussion questions

Short answers to typical discussion questions

  • Which factor does not affect the pressure at the bottom of a liquid-filled container? The area of the base (and the shape of the vessel): pressure depends on depth, density, g and atmospheric pressure.
  • Which factor does not affect the height of mercury in a barometer? The cross-sectional area of the tube.
  • The ratio of the radii of the pistons of a hydraulic press is 2 : 9; what is the ratio between the forces on the two pistons? f/F = a/A = (2/9)2 = 4/81.
  • Prove P = Pa + ρgh see Lesson 2.2 above.
Chapter 5: Properties of Hydrostatics · 13 / 15
Summary

Summary

Comparison aspectMain ideaKeywords
•P = F/A (pascal = N/m2)pascal
In a liquid
  • P = Pa + ρgh
  • same level
  • same pressure
  • connecting vessels
  • U-tube of immiscible liquids ρ₁h₁ = ρ₂h₂
samelevelpressureconnecting
Atmospheric pressure
  • Pa = ρHg g h
  • 1 atm = 1.013 × 105 Pa = 760 mm Hg = 1.013 bar
  • STP
ManometerΔP = P − Pa = ρgh
Blood pressure
  • 120/80 torr
  • tyre gauge
torrtyregauge
Pascal’s principle
  • pressure on an enclosed liquid is transmitted in full
  • hydraulic press f/a = F/A
  • η = F/f = A/a = y₁/y₂
  • applications: brakes
  • lifts
  • excavators
pressureenclosedliquidtransmitted
•
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Chapter 5: Properties of Hydrostatics · 14 / 15
Part 15

Part 15

  • 📖

📌 Comprehensive Scientific Terminology & Definitions:

Comparison aspectMain ideaKeywords
Density and Relative DensityDensity (ρ) is mass per unit volume: ρ = m / V (kg/m3). Relative density (specific gravity) is the ratio of substance density to pure water density at 4°C (dimensionless)massunitvolumespecific
Hydrostatic Pressure (P)The perpendicular force exerted by a fluid per unit surface area: P = Patm + ρ g h (where Patm is atmospheric pressure, ρ is liquid density, and h is depth).perpendicularforceexertedfluid
Communicating Vessels Principle
  • In open connected containers of any shape
  • the liquid reaches exactly the same horizontal level because pressure depends only on vertical depth
  • not on container cross-section
openconnectedcontainersshape
Pascal’s Principle
  • Any external pressure applied to an enclosed
  • incompressible fluid is transmitted undiminished and equally in all directions throughout the fluid and to the walls of its container
externalpressureappliedenclosed
Hydraulic Press and Mechanical AdvantagePractical application of Pascal’s principle enabling a small force (f) on a small piston (a) to lift a heavy load (F) on a large piston (A): Mechanical Advantage η = A / a = F / f = y₁ / y₂PracticalapplicationPascalprinciple

⚖️ Governing Scientific Laws, Principles & Mathematical Formulas:

Comparison aspectMain ideaKeywords
Densityρ = m / V (kg/m3)
Total Pressure in LiquidP = Patm + ρ g h | Hydrostatic Pressure Difference: ΔP = ρ g hPatmHydrostaticDifference
U-Tube Manometerρ₁ h₁ = ρ₂ h₂ (immiscible fluids in equilibrium)immisciblefluidsequilibrium
Standard Atmospheric Pressure1 atm = 76 cm-Hg = 1.013 × 105 Pa (N/m2) = 1.013 bar
Mechanical Advantageη = A / a = R2 / r2 = F / f = d₁ / d₂ (always > 1)always

🔍 In-Depth Explanations, Deductions & Core Concepts:

Comparison aspectMain ideaKeywords
Why are dams built thicker at the base than at the top?
  • Because liquid pressure increases directly with vertical depth (P = ρ g h)
  • reaching its maximum at the reservoir bottom and requiring massive structural reinforcement
Becauseliquidpressureincreases
Why is the mechanical efficiency of a hydraulic press never 100%?
  • Due to frictional resistance between the pistons and cylinder walls
  • and the presence of dissolved micro-air bubbles that compress and absorb work
frictionalresistancebetweenpistons
Why does Pascal’s principle not apply to gases?
  • Because gases are highly compressible due to large intermolecular distances
  • applied pressure goes into compressing the gas volume rather than transmitting undiminished
Becausehighlycompressiblelarge
Instrument Underlying Physical Principle Primary Function & Application
Mercury Barometer (Torricelli) Points at the same depth in a static fluid have equal pressure Measuring atmospheric pressure and mountain/building heights
U-Tube Manometer Pressure difference ΔP = ρ g h between fluid columns Measuring gas pressure in tanks and vehicle tire gauges
Hydraulic Press / Car Lift Pascal’s principle of transmitted pressure
  • Force multiplication: hydraulic jacks
  • heavy machinery brakes

💡 High-Yield Exam Traps & Key Takeaways:

  • Pressure is a scalar quantity, acting perpendicularly to any surface immersed in fluid.
  • Barometric height h depends solely on vertical depth, completely unaffected by tube diameter or tilt angle.
Chapter 5: Properties of Hydrostatics · 15 / 15