Pascal’s Law, Upthrust and Archimedes’ Principle Notes with Diagram
🌊 Physics · Pressure & Fluids · Complete Study Notes
Pascal’s Law,
Upthrust &
Archimedes’ Principle
Notes · Diagrams · Definitions · Applications · Numericals · Q&A — All in One
📘 Short Notes 📐 Diagrams 🔢 Numericals ❓ Q&A 🏭 Applications 📝 Practice MCQ
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Key Terms & Definitions
Master all vocabulary before the concepts
Pressure
Force applied per unit area. A scalar quantity measured in Pascals (Pa).
P = F / AFluid
Any substance that can flow and take the shape of its container. Includes both liquids and gases.
Thrust
Total force exerted by a fluid on a surface. A vector quantity measured in Newtons (N).
Thrust = P × AUpthrust (Buoyancy)
The upward force exerted by a fluid on a body submerged or partially submerged in it. Also called buoyant force.
U = V·ρ·gDensity
Mass per unit volume of a substance. Symbol: ρ (rho). Unit: kg/m³ or g/cm³.
ρ = m / VRelative Density (RD)
Ratio of density of a substance to density of water. Dimensionless (no unit). Also called Specific Gravity.
RD = ρ_substance / ρ_waterPascal (Pa)
SI unit of pressure. 1 Pascal = 1 Newton per square metre (1 Pa = 1 N/m²).
1 Pa = 1 N/m²Hydrostatic Pressure
Pressure at a point in a liquid due to the weight of the liquid column above it. Depends on depth, density, and g.
P = h·ρ·gCentre of Buoyancy
The point through which the upthrust (buoyant force) acts. It coincides with the centre of gravity of the displaced fluid.
Metacentre
The point through which the line of action of buoyant force passes when a floating body is slightly tilted. Key concept in ship stability.
Floatation
A body floats when the upthrust equals its weight. The body displaces fluid equal to its own weight.
Weight = UpthrustPascal’s Law
Pressure applied to an enclosed fluid is transmitted equally and undiminished in all directions throughout the fluid.
⭐ Key Units to Remember
Pressure: Pa = N/m² | bar | atm
1 atm = 101,325 Pa ≈ 10⁵ Pa
1 atm = 101,325 Pa ≈ 10⁵ Pa
Density of Water: 1000 kg/m³ = 1 g/cm³
Density of Mercury: 13,600 kg/m³
Density of Mercury: 13,600 kg/m³
g (acceleration due to gravity):
= 9.8 m/s² ≈ 10 m/s² (for calculations)
= 9.8 m/s² ≈ 10 m/s² (for calculations)
Pascal’s Law
Pressure transmission in enclosed fluids
Pascal’s Law — Statement
“Pressure applied to an enclosed fluid is transmitted equally and undiminished in all directions throughout the fluid and acts at right angles to every surface.”
— Blaise Pascal, 1653 | The fluid must be incompressible and at rest
Mathematical Form: If pressure P is applied at one point, every other point in the fluid experiences the same additional pressure P.
For a hydraulic system:
P₁ = P₂ → F₁/A₁ = F₂/A₂
For a hydraulic system:
P₁ = P₂ → F₁/A₁ = F₂/A₂
Hydraulic Principle
F₁/A₁ = F₂/A₂
F₁ = Input force | A₁ = Input area
F₂ = Output force | A₂ = Output area
F₂ = Output force | A₂ = Output area
📌 Key Features of Pascal’s Law
- Applies to enclosed (confined) fluids only
- Fluid must be incompressible (liquids, not gases in practice)
- Pressure is transmitted equally in ALL directions
- Acts perpendicular (at right angles) to any surface
- The law is the basis of hydraulic machines
- A small force on a small area can produce a large force on a large area
Diagram: Hydraulic Press (Pascal’s Law)
F₁ F₂ Pressure transmitted equally → Small Area A₁ Large Area A₂ Piston 1 Piston 2 P₁ = P₂ → F₁/A₁ = F₂/A₂ Incompressible Fluid (e.g. Oil)Fig 1: Hydraulic Press demonstrating Pascal’s Law. A small force F₁ on small piston (area A₁) creates a large force F₂ on large piston (area A₂).
🔬 Demonstration of Pascal’s Law Experiment
Pascal’s Rubber Ball Experiment:
- Take a hollow rubber ball and make several small holes all over its surface.
- Fill the ball completely with water (no air bubbles).
- Apply pressure by squeezing the ball tightly with both hands.
- Observation: Water shoots out equally from ALL holes, not just from the squeezed side.
- Conclusion: The applied pressure is transmitted equally in all directions — proving Pascal’s Law.
Another Demo — Pascal’s Vases: Connect vessels of different shapes at the bottom. Pour liquid in one. The liquid rises to the SAME LEVEL in all vessels, regardless of their shape or width. This confirms pressure depends only on height (depth), not shape.
Applications of Pascal’s Law in Daily Life
Hydraulic Brakes
In vehicles, a small force on the brake pedal creates a large braking force on all four wheels simultaneously via hydraulic fluid.
Hydraulic Jack / Lift
A small force on a small piston lifts heavy vehicles. Used in garages, construction, and car service stations.
Aircraft Landing Gear
Hydraulic systems extend/retract the landing gear of aircraft using Pascal’s Law for reliable force multiplication.
Hydraulic Press
Used to press/crush materials like cotton bales, metals in industry. Enormous force generated from small input.
Dentist’s Chair
Hydraulic system in dental chairs allows smooth, controlled height adjustment with minimal effort.
Excavators / Bulldozers
Heavy construction machinery uses hydraulic arms based on Pascal’s Law to dig, lift, and push massive loads.
⭐ Exam Key Points — Pascal’s Law
- Pascal’s Law applies to liquids (incompressible fluids), not gases in practice.
- Mechanical advantage = A₂/A₁. If A₂ is 10× A₁, output force is 10× input force.
- Energy is conserved — the small piston moves a larger distance: F₁×d₁ = F₂×d₂
- Pascal’s Law operates because liquids cannot be compressed significantly.
- The law was proposed by French mathematician Blaise Pascal in 1653.
Concept of Upthrust
Buoyancy in liquids and gases
Definition of Upthrust (Buoyant Force)
U = V × ρ_fluid × g
U = Upthrust (N) | V = Volume of fluid displaced (m³)
ρ_fluid = Density of fluid (kg/m³) | g = 9.8 m/s²
ρ_fluid = Density of fluid (kg/m³) | g = 9.8 m/s²
📌 What is Upthrust?
When a body is partially or fully immersed in a fluid (liquid or gas), the fluid exerts an upward force on the body. This upward force is called upthrust or buoyant force.
- It always acts vertically upward
- It is equal to the weight of the fluid displaced
- It acts through the centre of buoyancy
- It opposes the weight of the body
- It causes the apparent loss of weight when a body is in a fluid
Apparent Weight = Actual Weight − Upthrust
Loss in Weight = Weight of fluid displaced = Upthrust
Loss in Weight = Weight of fluid displaced = Upthrust
Diagram: Upthrust on a Submerged Body
Surface Object W U P_left P_right Water / Fluid Net upward force = Upthrust (U)Upthrust acts upward; Weight acts downward. Apparent weight = W − U
🔍 Why Does Upthrust Occur? Reason
Pressure in a fluid increases with depth (P = hρg). Therefore, the pressure at the bottom face of a submerged object is greater than the pressure at the top face.
Pressure at bottom face (P_bottom): = (h + H)ρg
Pressure at top face (P_top): = h·ρ·g
Net upward pressure: = P_bottom − P_top = H·ρ·g
Net upward force (Upthrust): U = A × H × ρ × g = V·ρ·g
Pressure at top face (P_top): = h·ρ·g
Net upward pressure: = P_bottom − P_top = H·ρ·g
Net upward force (Upthrust): U = A × H × ρ × g = V·ρ·g
Where: h = depth of top face, H = height of object, A = cross-sectional area, V = A×H = volume of object.
This net upward force equals exactly the weight of the fluid displaced by the body.
💨 Upthrust in Gas Air / Gas
- Gases also exert upthrust on objects immersed in them
- Upthrust in air is called buoyancy of air
- Density of air = 1.29 kg/m³ (much less than liquid)
- Upthrust in gas is usually very small and often ignored
- But for large light objects (balloons), it is significant
Hot Air Balloon
Hot air inside is less dense than surrounding air. Upthrust (buoyancy) from air exceeds weight → balloon rises.
📊 Factors Affecting Upthrust Key Factors
| Factor | Effect on Upthrust | Reason |
|---|---|---|
| Volume of object submerged ↑ | Upthrust ↑ | More fluid displaced → U = Vρg |
| Density of fluid ↑ | Upthrust ↑ | More mass displaced per unit volume |
| g increases | Upthrust ↑ | Both weight of fluid and upthrust increase |
| Depth of immersion | No effect | Only volume submerged matters, not depth |
| Shape of object | No effect | Only volume submerged matters |
| Density of object | No effect | Upthrust depends on fluid, not object density |
Diagram: Conditions for Floating, Sinking, and Hovering
HEAVY SINKS ρ_obj > ρ_fluid Weight > Upthrust NEUTRAL HOVERS ρ_obj = ρ_fluid Weight = Upthrust LIGHT FLOATS ρ_obj < ρ_fluid Weight < Upthrust (partially submerged)Fig: Body sinks when ρ_obj > ρ_fluid, hovers when equal, floats when ρ_obj < ρ_fluid
Applications of Upthrust in Daily Life
Ships & Boats
Made of steel but shaped to displace large volume of water → upthrust > weight → floats.
Balloons
Filled with gas (H₂ or He) less dense than air. Upthrust from air makes balloon rise.
Swimming
Water exerts upthrust on human body. Body feels lighter in water (apparent weight reduced).
Fish Swim Bladder
Fish adjust volume of swim bladder to control their density and float at different depths.
⭐ Exam Notes — Upthrust
- Upthrust = Weight of fluid displaced (this IS Archimedes’ Principle)
- Upthrust does NOT depend on depth of immersion — only on volume submerged
- A body fully submerged in denser fluid experiences more upthrust than in less dense fluid
- A body in space (zero-g) has no upthrust (U = Vρg = 0 when g = 0)
- Unit of upthrust = Newton (N), same as force
Archimedes’ Principle
Demonstration, formula & real-world applications
Archimedes’ Principle — Statement
“When a body is wholly or partially immersed in a fluid, it experiences an upthrust equal to the weight of the fluid displaced by it.”
Upthrust (U) = Weight of fluid displaced = m_fluid × g = V_displaced × ρ_fluid × g
🧮 Experimental Verification
Step 1: Weigh the object in air → W₁ (true weight)
Step 2: Weigh the object while submerged in fluid → W₂ (apparent weight)
Step 3: Measure the weight of fluid displaced → W₃
Result: W₁ − W₂ = W₃
i.e., Upthrust = Loss in weight = Weight of fluid displaced
i.e., Upthrust = Loss in weight = Weight of fluid displaced
Key Equations:
Upthrust = W_air − W_liquid
U = V × ρ_fluid × g
Relative Density = W_air / (W_air − W_liquid)
= W_air / Loss in weight in water
Upthrust = W_air − W_liquid
U = V × ρ_fluid × g
Relative Density = W_air / (W_air − W_liquid)
= W_air / Loss in weight in water
Diagram: Demonstrating Archimedes’ Principle (Eureka Can Method)
Step 1: Weight in Air W₁ Object → Step 2: Immersed in Liquid W₂ (submerged) Displaced Fluid W₃ W₁ – W₂ = W₃ Upthrust = Wt. of displaced fluid Eureka (Overflow) Can method — used to collect exactly the displaced fluid W₁ = True WeightFig 2: Archimedes’ Principle verified using Eureka Can. W₁ − W₂ = W₃ (loss in weight = weight of displaced fluid)
🔬 Demonstration of Archimedes’ Principle Experiment
- Take an Eureka can (overflow can) and fill it with water up to the spout level. Place an empty beaker under the spout.
- Tie a stone/metal block to a spring balance. Note its weight in air = W₁.
- Slowly lower the stone into the Eureka can while keeping it attached to the spring balance. Collect all overflowing water in the beaker.
- Note the reading of spring balance now = W₂ (apparent weight in water).
- Weigh the collected overflow water = W₃.
- Observation: W₁ − W₂ = W₃ (within experimental error).
- Conclusion: Upthrust (W₁ − W₂) equals exactly the weight of fluid displaced (W₃) — Archimedes’ Principle is verified.
📐 Relative Density Using Archimedes’ Principle Important
For Solids
RD = W_air / (W_air − W_water)
= Weight in air / Upthrust in water
= Weight in air / Loss in weight in water
= Weight in air / Loss in weight in water
For Liquids
RD_liquid = Loss in liquid / Loss in water
Using same object weighed in liquid and water
RD = (W_air − W_liquid)/(W_air − W_water)
RD = (W_air − W_liquid)/(W_air − W_water)
Note: Relative Density has NO unit (dimensionless). If RD > 1 → substance sinks in water. If RD < 1 → substance floats in water. Water has RD = 1 exactly.
⛵ Law of Floatation Extension of Archimedes
Law of Floatation
“A floating body displaces fluid equal to its own weight.”
- Weight of floating body = Weight of displaced fluid
- W_body = m_displaced × g = V_displaced × ρ_fluid × g
- A floating body is always partially submerged (unless ρ_body = ρ_fluid)
- Fraction submerged = ρ_body / ρ_fluid
Example: An iceberg floats with only about 1/9th of its volume above water.
ρ_ice/ρ_water = 900/1000 = 0.9 → 90% below water, 10% above.
ρ_ice/ρ_water = 900/1000 = 0.9 → 90% below water, 10% above.
Applications of Archimedes’ Principle in Daily Life
Designing Ships
Ship hulls designed so displaced water weight equals ship’s weight. Steel ship floats because its average density is less than water.
Hydrometer
Measures relative density of liquids using Archimedes’ Principle. Used to test battery acid, milk purity, alcohol content.
Submarine
Ballast tanks filled with/emptied of water to control average density. Dive → fill tanks; Surface → blow tanks with air.
Life Jackets
Filled with low-density material. Increases volume without much mass → average density < water → person floats.
Finding Density of Irregular Solids
Measure loss in weight in water → calculate volume → density = mass/volume. Used in labs for irregular-shaped objects.
Lactometer / Alcohol Meter
Hydrometers calibrated specifically for milk (lactometer) or alcohol solutions to check purity and concentration.
⭐ Exam Key Points — Archimedes’ Principle
- Archimedes shouted “Eureka!” (“I have found it!”) when he discovered this principle in his bathtub.
- The principle applies to both liquids and gases.
- A body of RD exactly 1 will be completely submerged but hover (neutral buoyancy).
- Upthrust ≠ weight of object. Upthrust = weight of fluid displaced by object.
- Steel is denser than water (RD ≈ 8) but a steel ship floats because it displaces water equal to its weight.
- The principle is used to find density and relative density of both solids and liquids.
Solved Numericals
Step-by-step solutions with key formulas
NUM 01Pascal’s Law — Hydraulic Press Force
A hydraulic press has pistons of areas 5 cm² and 200 cm². If a force of 20 N is applied on the smaller piston, find the force produced on the larger piston.
Given: A₁ = 5 cm² | A₂ = 200 cm² | F₁ = 20 N | Find: F₂ = ?
Formula
By Pascal’s Law: F₁/A₁ = F₂/A₂Rearrange
F₂ = F₁ × (A₂/A₁) = 20 × (200/5)Calculate
F₂ = 20 × 40 = 800 N✔ Answer: F₂ = 800 N (Mechanical Advantage = 40)
NUM 02Hydrostatic Pressure at Depth
Find the pressure at a depth of 50 m below the surface of sea water. (Density of sea water = 1025 kg/m³, g = 9.8 m/s²)
Given: h = 50 m | ρ = 1025 kg/m³ | g = 9.8 m/s²
Formula
P = h × ρ × gSubstitute
P = 50 × 1025 × 9.8Calculate
P = 50 × 10,045 = 502,250 Pa✔ Answer: P ≈ 5.02 × 10⁵ Pa ≈ 5.02 × 10⁵ N/m² (about 4.96 atm)
NUM 03Upthrust Calculation
A metal block of volume 500 cm³ is fully submerged in water. Calculate the upthrust acting on it. (ρ_water = 1000 kg/m³, g = 10 m/s²)
Given: V = 500 cm³ = 500 × 10⁻⁶ m³ | ρ = 1000 kg/m³ | g = 10 m/s²
Formula
U = V × ρ_fluid × gSubstitute
U = (500 × 10⁻⁶) × 1000 × 10Calculate
U = 500 × 10⁻⁶ × 10,000 = 5 N✔ Answer: Upthrust U = 5 N
NUM 04Relative Density of a Solid
A stone weighs 80 N in air and 65 N when fully submerged in water. Find: (a) Upthrust (b) Relative Density of stone.
Given: W_air = 80 N | W_water = 65 N
Part (a) — Upthrust
U = W_air − W_water = 80 − 65 = 15 NPart (b) — Relative Density
RD = W_air / (W_air − W_water) = 80 / (80 − 65) = 80 / 15Calculate
RD = 80 / 15 = 5.33✔ Upthrust = 15 N | Relative Density = 5.33 (no unit)
NUM 05Relative Density of a Liquid
A solid weighs 50 N in air, 44 N in water, and 45.5 N in a liquid. Find the relative density of the liquid.
Given: W_air = 50 N | W_water = 44 N | W_liquid = 45.5 N
Loss in weight in water
= 50 − 44 = 6 NLoss in weight in liquid
= 50 − 45.5 = 4.5 NFormula: RD of liquid
= Loss in liquid / Loss in water = 4.5 / 6✔ Relative Density of liquid = 0.75 (liquid is less dense than water)
NUM 06Apparent Weight in a Liquid
A body has mass 200 g and volume 80 cm³. Find its apparent weight when fully immersed in a liquid of density 1.2 g/cm³. (g = 10 m/s²)
Given: mass = 200 g = 0.2 kg | V = 80 cm³ = 80×10⁻⁶ m³ | ρ_liquid = 1200 kg/m³ | g = 10 m/s²
True Weight
W = mg = 0.2 × 10 = 2 NUpthrust
U = V × ρ_liquid × g = 80×10⁻⁶ × 1200 × 10 = 0.96 NApparent Weight
W_app = W − U = 2 − 0.96 = 1.04 N✔ Apparent weight = 1.04 N (body still sinks as W > U)
NUM 07Fraction of Iceberg Above Water
The density of ice is 917 kg/m³ and density of sea water is 1025 kg/m³. What fraction of an iceberg is above the surface?
Given: ρ_ice = 917 kg/m³ | ρ_seawater = 1025 kg/m³
For floating body
V_submerged / V_total = ρ_body / ρ_fluid = 917/1025 = 0.895Fraction submerged
= 0.895 = 89.5%Fraction above water
= 1 − 0.895 = 0.105 = 10.5%✔ About 10.5% of the iceberg is above water (89.5% is submerged)
NUM 08Hydraulic Brake — Pressure & Force
In a car’s hydraulic brake system, the brake pedal creates a force of 150 N on a master cylinder of area 3 cm². If the wheel cylinder has area 12 cm², find the force on each wheel and the pressure in the system.
Given: F₁ = 150 N | A₁ = 3 cm² = 3×10⁻⁴ m² | A₂ = 12 cm² = 12×10⁻⁴ m²
Pressure in system
P = F₁/A₁ = 150 / (3×10⁻⁴) = 500,000 Pa = 5×10⁵ PaForce on each wheel
F₂ = P × A₂ = 5×10⁵ × 12×10⁻⁴ = 600 NMechanical Advantage
MA = F₂/F₁ = 600/150 = 4✔ System pressure = 5×10⁵ Pa | Force on each wheel = 600 N | MA = 4
Practice Questions & MCQ
Test your understanding — click to check answers
Short Answer Questions
Q01 State Pascal’s Law in your own words.▼
Pascal’s Law states that when pressure is applied to any point of an enclosed (confined) incompressible fluid (liquid), it is transmitted equally and without reduction in all directions throughout the fluid, and acts perpendicularly on every surface it contacts. This forms the scientific basis of hydraulic machines.
Q02 What is upthrust? On what factors does it depend?▼
Upthrust (buoyant force) is the upward force exerted by a fluid on a body partially or fully immersed in it. It acts vertically upward and equals the weight of fluid displaced.
Upthrust depends on: (1) Volume of body submerged — more volume = more upthrust, (2) Density of fluid — denser fluid = more upthrust, (3) Acceleration due to gravity (g). It does NOT depend on depth, shape, or density of the object.
Upthrust depends on: (1) Volume of body submerged — more volume = more upthrust, (2) Density of fluid — denser fluid = more upthrust, (3) Acceleration due to gravity (g). It does NOT depend on depth, shape, or density of the object.
Q03 State Archimedes’ Principle and give its mathematical form.▼
“When a body is wholly or partially immersed in a fluid, it experiences an upthrust (buoyant force) equal to the weight of fluid displaced by it.”
Mathematical form: U = V × ρ_fluid × g
Where U = Upthrust, V = Volume of fluid displaced, ρ_fluid = density of fluid, g = acceleration due to gravity.
Mathematical form: U = V × ρ_fluid × g
Where U = Upthrust, V = Volume of fluid displaced, ρ_fluid = density of fluid, g = acceleration due to gravity.
Q04 Why does a steel ship float but a steel nail sinks?▼
A steel nail is solid — its average density (≈7800 kg/m³) is much greater than water (1000 kg/m³), so it sinks. A steel ship is hollow inside — it contains large volumes of air. This makes the ship’s average density (total mass/total volume including air spaces) much less than water. So it displaces water equal to its weight before becoming fully submerged, and floats (Law of Floatation). This is the principle of ship design.
Q05 How does a submarine submerge and surface?▼
A submarine has special ballast tanks. To submerge (dive): the ballast tanks are flooded with sea water → average density of submarine increases beyond that of sea water → submarine sinks. To surface: compressed air is pumped into the ballast tanks, pushing the water out → average density decreases below sea water → upthrust exceeds weight → submarine rises to the surface. This is a perfect application of Archimedes’ Principle.
Q06 Why is it easier to swim in sea water than in fresh water?▼
Sea water is denser (≈1025 kg/m³) than fresh water (1000 kg/m³) because of dissolved salts. Upthrust = V × ρ_fluid × g. Since the density of sea water is greater, the upthrust on a person’s body is greater in sea water than in fresh water. This greater upthrust makes a person float more easily (with a greater fraction of body above the water surface), making swimming easier.
Q07 Differentiate between Thrust and Pressure.▼
Thrust: Total force exerted by a fluid on a surface. It is a vector quantity. Unit = Newton (N). Formula: Thrust = P × A.
Pressure: Force per unit area. It is a scalar quantity. Unit = Pascal (Pa = N/m²). Formula: P = F/A.
Key difference: Thrust depends on area (a large surface gets a large thrust even at low pressure). Pressure is independent of area and describes how concentrated the force is.
Pressure: Force per unit area. It is a scalar quantity. Unit = Pascal (Pa = N/m²). Formula: P = F/A.
Key difference: Thrust depends on area (a large surface gets a large thrust even at low pressure). Pressure is independent of area and describes how concentrated the force is.
Q08 What is a hydrometer? How does it work?▼
A hydrometer is an instrument used to measure the relative density (specific gravity) of liquids. It works on Archimedes’ Principle and the Law of Floatation. It consists of a weighted glass bulb with a graduated stem. When placed in a liquid, it floats. The deeper it sinks, the less dense the liquid (less upthrust needed). The reading on the stem at the liquid surface gives the relative density directly. Examples: lactometer (milk purity), battery hydrometer (acid density), alcohol hydrometer.
Multiple Choice Questions — Click to Answer
MCQ 01 Pressure at depth h in a fluid of density ρ is:
MCQ 02 According to Pascal’s Law, if area A₂ is 5 times A₁, the output force F₂ is:
MCQ 03 Upthrust on a body submerged in water depends on:
MCQ 04 A body sinks in a fluid when:
MCQ 05 Relative Density is also called:
MCQ 06 A stone weighs 60 N in air and 48 N in water. Its upthrust is:
MCQ 07 Which of these is NOT an application of Pascal’s Law?
MCQ 08 Archimedes’ Principle was discovered when Archimedes:
MCQ 09 The unit of upthrust is:
MCQ 10 A floating body displaces fluid equal to:
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