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ICSE • Class 9 • Science • Ch 5
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Upthrust in Fluids, Archimedes’ Principle and Floatation

In ICSE Class 9 Physics, "Upthrust in Fluids, Archimedes' Principle and Floatation" investigates the physics of buoyancy and stability of floating bodies. When a body is partially or wholly submerged in a fluid, it experiences an upward buoyant force called Upthrust ($F_B$). The fundamental physical origin of upthrust arises from the pressure difference between the lower and upper surfaces of the submerged body: because liquid pressure increases with depth ($P_2 > P_1$), the upward thrust on the bottom face exceeds the downward thrust on the top face ($F_2 - F_1 > 0$), resulting in a net upward force. Archimedes' Principle states: "When a body is immersed partially or completely in a fluid, it experiences an upthrust equal to the weight of the fluid displaced by it": $\mathbf{F_B = V\rho_L g}$ (where $V$ is submerged volume, $\rho_L$ is liquid density, and $g$ is gravity). The apparent weight of a submerged body is $\text{Apparent Weight} = W - F_B$. The Principle of Floatation establishes that a body floats when its total weight equals the upthrust exerted by the displaced liquid: $W = F_B \implies V\rho_S g = v\rho_L g \implies \frac{v}{V} = \frac{\rho_S}{\rho_L}$ (fraction submerged equals relative density of the solid). The chapter covers Relative Density ($\text{RD} = \frac{\rho_{\text{substance}}}{\rho_{\text{water at } 4^\circ\text{C}}} = \frac{W_{\text{air}}}{W_{\text{air}} - W_{\text{water}}}$), Plimsoll lines on ocean cargo ships, submarines (ballast tanks), and hydrometers.

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The Eureka Epiphany: How a Gold Crown and an Overflowing Bathtub Solved the King's Theft Mystery

In the 3rd century BCE, King Hiero II of Syracuse suspected that his royal goldsmith had cheated him. The king had supplied a precise weight of pure gold to make a sacred temple crown, but rumor held that the craftsman had melted in cheap silver and kept a fortune in gold for himself. Hiero challenged the resident genius Archimedes to prove the crime without damaging the crown. Archimedes pondered the riddle day and night without success. Then, stepping into a brimming public bathhouse tub, he watched the water spill over the rim onto the stone floor. In a flash of divine insight, Archimedes realized that his submerged body displaced an exact volume of water equal to his own volume! Since gold is nearly twice as dense as silver, a fraudulent crown diluted with silver would have a larger volume and displace significantly more water than an honest crown of pure gold! Legend tells that Archimedes jumped out of the bath and sprinted naked through the streets of Syracuse crying "EUREKA! EUREKA!" (I have found it!). How does this buoyant upthrust allow a 100,000-ton steel aircraft carrier to float like a feather? Let us master Archimedes' principle!

Why This Chapter Matters

Buoyancy and floatation govern naval architecture (cargo ship loading, stability metacenters), submarine dive control, hot-air balloons, oceanographic buoys, and hydrometric fuel quality testing.

Before You Begin (Prerequisites)

  • Hydrostatic liquid pressure formula $P = h\rho g$ from Chapter 4.
  • Density and weight concepts ($W = mg = V\rho g$).

What You Will Learn (Core Objectives)

  • Define upthrust (buoyancy) and explain its origin in the hydrostatic pressure gradient.
  • State and prove Archimedes' Principle experimentally and mathematically.
  • State the Principle of Floatation and calculate the submerged fraction of a floating body.
  • Define relative density and determine it experimentally using the displacement method.
  • Explain why a heavy iron needle sinks while a massive steel ship floats.
  • Explain the operation of submarine ballast tanks and hot-air balloons.

Chapter Roadmap & Progression

1 1. Upthrust and Its Physical Origin
2 2. Archimedes' Principle & Relative...
3 3. The Principle of Floatation
4 4. Worked ICSE Problem Archetypes

Complete Concept Guide (100% Curriculum Coverage)

1. Upthrust and Its Physical Origin

Buoyant Force Mechanism
A. Definition of Upthrust ($F_B$):

The upward buoyant force exerted on a body by the fluid in which it is submerged is called upthrust or buoyant force.

  • SI Unit: Newton ($\text{N}$).
  • Apparent Weight: $\mathbf{W_{\text{apparent}} = W_{\text{actual}} - F_B}$. (A body appears lighter in water).
B. Hydrostatic Origin of Upthrust:

Consider a rectangular block of height $h$ and cross-sectional area $A$ submerged in a liquid of density $\rho_L$. Let its top face be at depth $h_1$ and bottom face at depth $h_2$ ($h_2 - h_1 = h$):

  1. Downward thrust on top face: $F_1 = P_1 \times A = (h_1 \rho_L g) A$
  2. Upward thrust on bottom face: $F_2 = P_2 \times A = (h_2 \rho_L g) A$
  3. The horizontal forces on side walls cancel out due to symmetry.
  4. $$\mathbf{F_B = F_2 - F_1 = (h_2 - h_1) A \rho_L g = h A \rho_L g = V \rho_L g}$$

Since $V \rho_L = \text{Mass of displaced liquid}$ ($m_L$), $V \rho_L g = \text{Weight of displaced liquid}$.

$$\therefore \mathbf{F_B = \text{Weight of Displaced Liquid}} \quad \blacksquare$$

2. Archimedes' Principle & Relative Density

Archimedes' Principle
Statement:

"When a body is immersed wholly or partially in a fluid, it experiences an upthrust which is equal to the weight of the fluid displaced by it."

$$\mathbf{\text{Upthrust } (F_B) = \text{Loss in weight} = \text{Weight of displaced liquid} = V_{\text{submerged}} \times \rho_L \times g}$$
Relative Density (RD):

The ratio of the density of a substance to the density of pure water at $4^\circ\text{C}$ (where water reaches maximum density $\rho_{\text{water}} = 1000\text{ kg/m}^3 = 1\text{ g/cm}^3$):

$$\mathbf{\text{RD} = \frac{\text{Density of Substance}}{\text{Density of Water at } 4^\circ\text{C}} = \frac{\text{Weight in Air}}{\text{Loss of Weight in Water}} = \frac{W_{\text{air}}}{W_{\text{air}} - W_{\text{water}}}}$$

Relative density is a pure dimensionless number (has no units).

3. The Principle of Floatation

Laws of Floatation
Statement:

"A floating body displaces a volume of liquid whose weight is equal to the total weight of the floating body."

$$\mathbf{\text{Weight of Body } (W) = \text{Upthrust } (F_B)}$$
Fractional Submersion Formula:

Let a solid of total volume $V$ and density $\rho_S$ float with submerged volume $v$ in a liquid of density $\rho_L$:

$$W = V \rho_S g \qquad F_B = v \rho_L g$$ $$V \rho_S g = v \rho_L g \implies \mathbf{\frac{v}{V} = \frac{\rho_S}{\rho_L} = \text{Relative Density of Solid (in water)}}$$
  • If $\rho_S < \rho_L$: The body floats partially submerged (e.g., ice $\rho = 0.92\text{ g/cm}^3$ in water $\rho = 1.0\text{ g/cm}^3 \implies 92\%$ submerged).
  • If $\rho_S = \rho_L$: The body floats completely submerged just below the surface.
  • If $\rho_S > \rho_L$: The body sinks to the bottom ($W > F_B$).

4. Worked ICSE Problem Archetypes

Exemplary Solutions
Problem 1: A metal piece weighs $210\text{ gf}$ in air and $180\text{ gf}$ when completely immersed in water. Find: (i) The volume of the metal piece, (ii) Its relative density.

Solution:

Weight in air $W_{\text{air}} = 210\text{ gf}$, Weight in water $W_{\text{water}} = 180\text{ gf}$.

1. Loss of weight in water (Upthrust):

$$F_B = W_{\text{air}} - W_{\text{water}} = 210 - 180 = 30\text{ gf}$$

By Archimedes' principle, $F_B =$ Weight of displaced water $= 30\text{ gf}$.

Since the density of water is $1\text{ g/cm}^3$, mass of displaced water $= 30\text{ g}$, so:

$$\mathbf{\text{Volume of metal piece} = 30\text{ cm}^3}$$

2. Relative Density (RD):

$$\text{RD} = \frac{W_{\text{air}}}{W_{\text{air}} - W_{\text{water}}} = \frac{210}{30} = \mathbf{7.0}$$
Problem 2: A block of wood of volume $250\text{ cm}^3$ floats in water with $200\text{ cm}^3$ of its volume submerged. Calculate the density of the wood.

Solution:

Total volume $V = 250\text{ cm}^3$, Submerged volume $v = 200\text{ cm}^3$, Density of water $\rho_w = 1.0\text{ g/cm}^3$.

By the Principle of Floatation:

$$\frac{v}{V} = \frac{\rho_{\text{wood}}}{\rho_{\text{water}}} \implies \frac{200}{250} = \frac{\rho_{\text{wood}}}{1.0}$$ $$\rho_{\text{wood}} = \frac{4}{5} = \mathbf{0.8\text{ g/cm}^3} = \mathbf{800\text{ kg/m}^3}$$

Key Formulas, Reactions & Definitions

Archimedes Upthrust
$$F_B = V_{\text{submerged}} \rho_L g$$
Equal to weight of displaced liquid.
Apparent Submerged Weight
$$W_{\text{apparent}} = W_{\text{actual}} - F_B$$
Loss in weight in fluid.
Relative Density
$$\text{RD} = \frac{\rho}{\rho_{\text{water}}} = \frac{W_{\text{air}}}{W_{\text{air}} - W_{\text{water}}}$$
Unitless density ratio.
Floatation Ratio
$$\frac{v_{\text{submerged}}}{V_{\text{total}}} = \frac{\rho_{\text{solid}}}{\rho_{\text{liquid}}}$$
Fraction of floating solid submerged.

Physics: Upthrust Hydrostatic Pressure Mechanism & Floatation Balance

Buoyancy & Floatation: Hydrostatic Origin of Upthrust & Archimedes' Principle Origin of Upthrust: F_B = F₂ - F₁ Block (V = Ah) F₁ = h₁ρgA F₂ = h₂ρgA (Larger!) Net Upthrust F_B = F₂ - F₁ = (h₂ - h₁)Aρg F_B = Vρ_L g = Weight of Displaced Fluid Principle of Floatation: W = F_B Water (ρ_L) Above (V - v) Submerged v W = Vρ_S g F_B = vρ_L g At Equilibrium: W = F_B ⇒ v / V = ρ_S / ρ_L Fraction submerged equals relative density of solid

Chapter Summary & 10 Key Takeaways

Takeaway 1
Upthrust (buoyant force) is the net upward force exerted by a fluid on a submerged body.
Takeaway 2
Upthrust arises because hydrostatic liquid pressure increases with depth (P2 > P1).
Takeaway 3
Archimedes' Principle: The upthrust equals the weight of the fluid displaced by the body: F_B = V * ρ_L * g.
Takeaway 4
Apparent weight of a body submerged in a fluid is W_apparent = W_actual - F_B.
Takeaway 5
Relative density (RD) is the ratio of substance density to water density at 4°C: RD = W_air / (W_air - W_water).
Takeaway 6
Relative density is a pure dimensionless number without units.
Takeaway 7
Principle of Floatation: A floating body displaces liquid equal to its own total weight: W = F_B.
Takeaway 8
The fraction of a floating body submerged is v / V = ρ_solid / ρ_liquid.
Takeaway 9
A steel ship floats because its hollow shape encloses air, making its average density less than water.
Takeaway 10
Submarines dive by taking water into ballast tanks and surface by expelling water using compressed air.

Check Your Understanding (Diagnostic Practice Questions)

Diagnostic questions testing core conceptual clarity. Answers are hidden initially — solve each problem first, then click to reveal the step-by-step verified solution.

1
State Archimedes' Principle. A body weighs $300\text{ N}$ in air and $220\text{ N}$ when completely submerged in water. Calculate: (i) Upthrust on the body, (ii) Volume of the body (take density of water $= 1000\text{ kg/m}^3, g = 10\text{ m/s}^2$).
Reveal Answer & Explanation
Answer:

• Archimedes' Principle: When a body is immersed wholly or partially in a fluid, it experiences an upward buoyant force (upthrust) equal to the weight of the fluid displaced by it.
• (i) Upthrust ($F_B$):

$$F_B = \text{Loss in weight} = W_{\text{air}} - W_{\text{water}} = 300\text{ N} - 220\text{ N} = \mathbf{80\text{ Newtons}}$$


• (ii) Volume of the body ($V$):

$$F_B = V \rho_w g \implies 80 = V \times 1000 \times 10$$


$$80 = 10000 V \implies V = \frac{80}{10000} = \mathbf{0.008\text{ m}^3} \quad (\text{or } \mathbf{8,000\text{ cm}^3})$$


Upthrust = 300 - 220 = 80 N. V = F_B / (ρ_w * g) = 80 / 10000 = 0.008 m^3.
2
Explain the physical origin of upthrust inside a liquid in terms of the hydrostatic pressure gradient.
Reveal Answer & Explanation
Answer:

• Hydrostatic liquid pressure increases directly with depth ($P = h\rho g$).
• When a solid body is submerged, its bottom surface is at a greater depth ($h_2$) than its top surface ($h_1$).
• Therefore, the upward hydrostatic pressure pushing up against the bottom face ($P_2 = h_2\rho g$) is strictly greater than the downward pressure pushing down on the top face ($P_1 = h_1\rho g$).
• Multiplying by the surface area $A$, the upward thrust $F_2$ exceeds the downward thrust $F_1$.
• The resulting net upward force ($F_B = F_2 - F_1$) is the Upthrust.


Bottom surface is deeper than top surface, so upward force F2 exceeds downward force F1.
3
A solid block of wood of density $600\text{ kg/m}^3$ floats in water of density $1000\text{ kg/m}^3$. What fraction of the volume of the block remains above the water surface?
Reveal Answer & Explanation
Answer:

• By the Principle of Floatation, the fraction of volume submerged ($v/V$) is:

$$\frac{v}{V} = \frac{\rho_{\text{wood}}}{\rho_{\text{water}}} = \frac{600}{1000} = \frac{3}{5} = 0.60 \quad (60\% \text{ submerged})$$


• The fraction remaining above the water surface is:

$$\text{Fraction above} = 1 - \frac{v}{V} = 1 - \frac{3}{5} = \mathbf{\frac{2}{5}} \quad (\text{or } \mathbf{0.40} / \mathbf{40\%})$$


Fraction submerged = 600/1000 = 3/5. Fraction above = 1 - 3/5 = 2/5 (40%).
4
Why does an iron nail sink in water, but a huge ship made of thousands of tons of iron and steel floats on water?
Reveal Answer & Explanation
Answer:

• Iron Nail: An iron nail is solid with a density of $7.8\text{ g/cm}^3$, which is much greater than the density of water ($1.0\text{ g/cm}^3$). The weight of the tiny volume of water displaced by the nail is much less than the weight of the nail ($W > F_B$). Hence, it sinks.
• Steel Ship: A ship is not a solid block of iron; it is constructed as a hollow vessel containing vast compartments of air. The overall average density of the ship (Total Mass / Total Enclosed Volume) is substantially less than the density of water. Consequently, it displaces a volume of water whose weight easily equals the entire weight of the ship ($W = F_B$), allowing it to float.


Solid iron density is 7.8 > 1 (sinks). A ship is hollowed out with air, making average density < 1.
5
An iceberg floats in seawater (density $= 1.03\text{ g/cm}^3$). If the density of ice is $0.92\text{ g/cm}^3$, calculate the percentage of the iceberg that lies submerged beneath the sea.
Reveal Answer & Explanation
Answer: • Using the floatation ratio $\frac{v}{V} = \frac{\rho_{\text{ice}}}{\rho_{\text{seawater}}}$:
$$\frac{v}{V} = \frac{0.92}{1.03} \approx 0.8932$$
• Converting to percentage:
$$\text{Percentage Submerged} = 0.8932 \times 100\% = \mathbf{89.32\%}$$
• *Observation:* Nearly $90\%$ of an iceberg is hidden underwater, explaining why icebergs are so dangerous to maritime navigation ("tip of the iceberg").
Percentage submerged = (0.92 / 1.03) * 100% ≈ 89.3%.
6
A body weighs $20\text{ gf}$ in air and $16\text{ gf}$ in a liquid of density $0.8\text{ g/cm}^3$. Find the volume of the body.
Reveal Answer & Explanation
Answer: • Upthrust in the liquid:
$$F_B = W_{\text{air}} - W_{\text{liquid}} = 20 - 16 = 4\text{ gf}$$
• By Archimedes' principle, $F_B =$ Weight of displaced liquid $= 4\text{ gf} \implies$ mass of displaced liquid $= 4\text{ g}$.
• Volume of displaced liquid (which equals volume of the submerged body):
$$V = \frac{\text{Mass}}{\text{Density}} = \frac{4\text{ g}}{0.8\text{ g/cm}^3} = \mathbf{5\text{ cm}^3}$$
Upthrust = 4 gf -> mass of liquid = 4 g. Volume = mass / density = 4 / 0.8 = 5 cm^3.
7
Explain how a submarine can both dive beneath the ocean surface and surface back up.
Reveal Answer & Explanation
Answer:

• A submarine is equipped with specialized ballast tanks:
1. To Dive: Sea vents are opened, allowing dense seawater to flood into the ballast tanks while venting out air. This increases the average density of the submarine until its total weight exceeds upthrust ($W > F_B$), causing the submarine to submerge.
2. To Surface: High-pressure compressed air is blown into the ballast tanks, forcing the seawater out through exhaust valves. This dramatically reduces the submarine's average density until upthrust exceeds weight ($F_B > W$), causing it to rise to the ocean surface.


Flooding ballast tanks with water increases weight to dive; blowing in compressed air expels water to surface.
8
Why is it easier to swim in seawater than in freshwater in a swimming pool?
Reveal Answer & Explanation
Answer:

• Seawater contains dissolved mineral salts, giving it a higher density ($\approx 1030\text{ kg/m}^3$) than freshwater ($1000\text{ kg/m}^3$).
• Since upthrust is directly proportional to liquid density ($F_B = V\rho_L g$), seawater exerts a greater upward buoyant force on the swimmer's body than freshwater.
• Consequently, less of the swimmer's body needs to be submerged to balance their weight, making floating and swimming significantly easier.


Seawater is denser due to dissolved salts, providing greater upthrust F_B = Vρ_L g.
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All Class 9 Science Chapters

Ch 1: Measurements and Experimentation Ch 2: Motion in One Dimension Ch 3: Laws of Motion Ch 4: Pressure in Fluids and Atmospheric Pressure Ch 5: Upthrust in Fluids, Archimedes’ Principle and Floatation Ch 6: Heat and Energy Ch 7: Reflection of Light Ch 8: Propagation of Sound Waves Ch 9: Current Electricity Ch 10: Magnetism Ch 11: Language of Chemistry Ch 12: Chemical Changes and Reactions Ch 13: Water Ch 14: Atomic Structure and Chemical Bonding Ch 15: The Periodic Table Ch 16: Study of the First Element - Hydrogen Ch 17: Study of Gas Laws Ch 18: Atmospheric Pollution Ch 19: Introducing Biology Ch 20: Cell: The Unit of Life Ch 21: Tissues: Plant and Animal Tissues Ch 22: The Flower Ch 23: Vegetative Propagation and Micropropagation Ch 24: Pollination and Fertilisation Ch 25: Seeds - Structure and Germination Ch 26: Respiration in Plants Ch 27: Five Kingdom Classification Ch 28: Economic Importance of Bacteria and Fungi Ch 29: Nutrition Ch 30: Digestive System Ch 31: Movement and Locomotion Ch 32: Skin - The Jack of All Trades Ch 33: The Respiratory System Ch 34: Hygiene - A Key to Healthy Life Ch 35: Diseases: Cause and Control Ch 36: Aids to Health Ch 37: Health Organisations Ch 38: Waste Generation and Management

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