§ 9.2NCERT Class 11 · Physics · Chapter 9

Archimedes' Principle, Buoyancy & The Law of Floatation

Q: What is the physical origin of the buoyant force (upthrust)?

The buoyant force arises directly from the vertical gradient of hydrostatic pressure in a fluid under gravity. Because pressure increases linearly with depth (P = P_0 + rho * g * h), the upward hydrostatic thrust exerted on the bottom surface of an immersed object is strictly greater than the downward hydrostatic thrust on its top surface. The upward vector difference produces the buoyant force F_B.

Q: What is Archimedes' Principle?

Archimedes' Principle states that when a body is completely or partially immersed in a fluid at rest, it experiences an upward buoyant force equal in magnitude to the weight of the fluid displaced by the body: F_B = V_submerged * rho_fluid * g.

Q: What are the three equilibrium conditions for a floating body?

For a body of mass m and volume V immersed in a fluid of density rho_f: (1) If rho_s > rho_f, gravity exceeds maximum upthrust (W > F_B_max) and the body sinks; (2) If rho_s = rho_f, W = F_B_max and the body floats fully submerged in neutral equilibrium at any depth; (3) If rho_s < rho_f, the body floats partially submerged with fractional submerged volume V_in / V = rho_s / rho_f.

Q: Does a piece of ice floating in water cause the water level to rise when it melts?

No, the water level remains strictly unchanged! While floating, the ice displaces a mass of water equal to its own total mass (m_ice = m_displaced). When the ice melts completely, it converts into a liquid water volume V_melt = m_ice / rho_water, which exactly equals the submerged volume V_in previously displaced. Therefore, the melted ice precisely fills the cavity it created, leaving the water level constant.

QCC Notes · Class 11 Physics NCERT-aligned Updated 26 Sep 2026 10 min read

When an object is immersed in a liquid, it appears lighter than in air. This everyday observation is the consequence of buoyancy: the net upward force exerted by a surrounding static fluid on an immersed body. The quantitative foundation was established in the 3rd century BC by the Greek polymath Archimedes of Syracuse.

1. Physical Origin & Derivation of Buoyancy

Consider a solid block of height h and uniform horizontal cross-sectional area A immersed in a liquid of density ρ_f:

       Free Surface (Liquid Level)  ~~~~~~~~~~~~~~~~~~~~~~~
                                    |                     |
                                    |         h₁         |
                                    |                     |
                     Top Face:      +---------------------+  Downward Thrust:
                                    |                     |  F₁ = P₁ * A
                                    |      Solid Body     |  P₁ = P₀ + rho_f * g * h₁
                                    |     Height h        |
                                    |     Volume V        |
                     Bottom Face:   +---------------------+  Upward Thrust:
                                                             F₂ = P₂ * A
                                                             P₂ = P₀ + rho_f * g * h₂
             
         Net Vertical Force (Buoyant Force):
         F_B = F₂ - F₁ = (P₂ - P₁) * A = rho_f * g * (h₂ - h₁) * A = rho_f * V * g

Fig 9.2A: Derivation of buoyant upthrust from vertical hydrostatic pressure differential (P₂ > P₁).

Hydrostatic pressure at top surface: P₁ = P₀ + ρ_f g h₁ ⇒ Downward force F₁ = P₁ A.
Hydrostatic pressure at bottom surface: P₂ = P₀ + ρ_f g h₂ ⇒ Upward force F₂ = P₂ A.
Lateral forces cancel out symmetrically in opposite pairs.
Net vertical upward force (Buoyant Force / Upthrust F_B):
F_B = F₂ - F₁ = (P₂ - P₁)A = ρ_f g (h₂ - h₁)A
Because (h₂ - h₁)A = V_submerged:
F_B = V_sub ρ_f g = m_disp g = Weight of displaced fluid

2. Archimedes' Principle & Apparent Weight

Archimedes' Principle: When a body is wholly or partially immersed in a fluid at rest, it experiences an upward buoyant force whose magnitude equals the weight of the fluid displaced by the body. The buoyant force acts vertically upward through the Center of Buoyancy (the center of gravity of the displaced fluid before displacement).

Apparent Weight of a Submerged Body

When a solid of true mass m, volume V, and density ρ_s is suspended by a spring balance inside a liquid of density ρ_f (ρ_s > ρ_f):

True weight in air/vacuum: W = mg = V ρ_s g.
Upward buoyant force: F_B = V ρ_f g.
Apparent Weight (W_app):
W_app = W - F_B = V ρ_s g - V ρ_f g = V ρ_s g (1 - ρ_f/ρ_s)

W_app = W ( 1 - ρ_f/ρ_s )
Fractional loss in weight:
(Δ W)/W = (W - W_app)/W = ρ_f/ρ_s
Relative Density (Specific Gravity) of Solid:
RD_s = ρ_s/ρ_w = (Weight in Air)/(Loss of Weight in Water) = W_air/(W_air - W_water)

3. The Law of Floatation

A body will float in a liquid if the upward buoyant force can balance the downward gravitational weight of the body (F_B = W):

Case Condition	Density Comparison	Force Relationship	Equilibrium State
Case 1: Sinking	ρ_s > ρ_f	W > F_B,max	Body accelerates downward with a = g(1 - ρ_f/ρ_s)until it hits bottom. Normal reaction from bottomN = W - F_B.
Case 2: Neutral Floatation	ρ_s = ρ_f	W = F_B,max	Body remains in equilibrium completely submerged at any depth inside the liquid (Neutral equilibrium).
Case 3: Partial Floatation	ρ_s < ρ_f	W = F_B with V_in < V	Body floats partially submerged with fraction V_in/V = ρ_s/ρ_f.

Submerged Fraction Formula

For a floating body with total volume V and submerged volume V_in:

W = F_B ⇒ V ρ_s g = V_in ρ_f g ⇒ V_in/V = ρ_s/ρ_f

Fraction of body volume exposed outside the liquid:

V_out/V = 1 - V_in/V = 1 - ρ_s/ρ_f = (ρ_f - ρ_s)/ρ_f

Why Ice Floats on Water: The density of pure ice at 0^°C is ρ_ice ≈ 917 kg/m³, while liquid water at 0^°C is ρ_w ≈ 1000 kg/m³.

V_in/V = 917/1000 ≈ 91.7%

Roughly 92% of an iceberg is hidden below the waterline, with only ≈ 8% visible above the sea surface!

4. Advanced Floatation: Layered Liquids & Melting Ice

Body Floating at the Interface of Two Immiscible Liquids

Consider a solid cylinder of cross-sectional area A and length L floating vertically at the boundary between two immiscible liquids of densities ρ₁ and ρ₂ (ρ₁ < ρ_s < ρ₂), with length L₁ in the upper liquid and length L₂ in the lower liquid (L = L₁ + L₂):

W = F_B1 + F_B2 ⇒ A L ρ_s g = A L₁ ρ₁ g + A L₂ ρ₂ g

L ρ_s = L₁ ρ₁ + L₂ ρ₂

What Happens to Water Level When Floating Ice Melts?

Case A: Pure Ice in Pure Water: The water level remains unchanged. A floating ice cube of mass m displaces a volume of water V_disp = m / ρ_w. When it melts, the resulting liquid water has mass m and volume V_melt = m / ρ_w = V_disp. The liquid water exactly fills the volume previously displaced.
Case B: Ice Containing an Embedded Air Bubble: The water level remains unchanged, because the mass of air is practically zero, so V_disp is determined solely by m_ice.
Case C: Ice Containing a Dense Lead Pellet / Stone (ρ_lead > ρ_w): The water level falls! Initially, the heavy lead stone floats (held by ice buoyancy) and displaces an equivalent weight of water: V₁ = m_lead / ρ_w. When the ice melts, the lead stone sinks to the bottom and displaces only its volume: V₂ = m_lead / ρ_lead. Since ρ_lead > ρ_w, V₂ < V₁, so the total water level drops.
Case D: Ice Containing a Cork Piece (ρ_cork < ρ_w): The water level remains unchanged, as cork continues to float after ice melts.

JEE & NEET Solved Practice Problems

Problem 1: A piece of metal weighs 210 g in air, 180 g in water, and 120 g in an unknown liquid. Calculate: (a) the relative density of the metal, and (b) the density of the unknown liquid.

Solution:
(a) Loss of weight in water:

Δ W_w = W_air - W_water = 210 - 180 = 30 g-wt

Because Δ W_w = V ρ_w g = 30 g-wt, the volume of the metal isV = 30 cm³.
Relative density of the metal:

RD_metal = W_air/(Δ W_w) = 210/30 = 7.0

(b) Loss of weight in the unknown liquid:

Δ W_L = W_air - W_L = 210 - 120 = 90 g-wt

Since Δ W_L = V ρ_L g:

ρ_L = (Δ W_L)/V = (90 g)/(30 cm³) = 3.0 g/cm³ = 3000 kg/m³

Problem 2: A wooden cylinder of density 900 kg/m³ and length 1 m floats vertically in a vessel containing water (ρ_w = 1000 kg/m³) and oil (ρ_oil = 800 kg/m³). If the oil layer is sufficiently deep to submerge the upper part of the cylinder completely, find the length of the cylinder immersed in water.

Solution:
Let L = 1 m be the total length of the cylinder, L_w be the length immersed in water, and L_oil = L - L_w be the length immersed in oil.
By equilibrium of floatation:

W = F_B,water + F_B,oil

A L ρ_wood g = A L_w ρ_w g + A (L - L_w) ρ_oil g

Dividing by A g:

(1)(900) = L_w(1000) + (1 − L_w)(800)

900 = 1000 L_w + 800 − 800 L_w = 200 L_w + 800

200 L_w = 900 − 800 = 100 ⇒ L_w = 0.5 m

Half the cylinder (0.5 m) is in water and 0.5 m is in oil. Check: the wood (900 kg/m³) is denser than oil but lighter than water, so it must rest across the interface.

Frequently Asked Questions

Q1. What is the physical origin of the buoyant force (upthrust)?

The buoyant force arises directly from the vertical gradient of hydrostatic pressure in a fluid under gravity. Because pressure increases linearly with depth (P = P₀ + rho * g * h), the upward hydrostatic thrust exerted on the bottom surface of an immersed object is strictly greater than the downward hydrostatic thrust on its top surface. The upward vector difference produces the buoyant force F_B.

Q2. What is Archimedes' Principle?

Archimedes' Principle states that when a body is completely or partially immersed in a fluid at rest, it experiences an upward buoyant force equal in magnitude to the weight of the fluid displaced by the body: F_B = V_submerged * rho_fluid * g.

Q3. What are the three equilibrium conditions for a floating body?

For a body of mass m and volume V immersed in a fluid of density rho_f: (1) If rho_s > rho_f, gravity exceeds maximum upthrust (W > F_B_max) and the body sinks; (2) If rho_s = rho_f, W = F_B_max and the body floats fully submerged in neutral equilibrium at any depth; (3) If rho_s < rho_f, the body floats partially submerged with fractional submerged volume V_in / V = rho_s / rho_f.

Q4. Does a piece of ice floating in water cause the water level to rise when it melts?

No, the water level remains strictly unchanged! While floating, the ice displaces a mass of water equal to its own total mass (m_ice = m_displaced). When the ice melts completely, it converts into a liquid water volume V_melt = m_ice / rho_water, which exactly equals the submerged volume V_in previously displaced. Therefore, the melted ice precisely fills the cavity it created, leaving the water level constant.

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