A classic Archimedes' Principle problem: A metallic sphere weighing 3 kg in air weighs only 2.5 kg when fully immersed in water. The density of the metallic sphere is 6000 kg/m³ (or 6 g/cm³). This is found using Archimedes' Principle — the buoyant force equals the weight of water displaced, which equals the loss in weight when submerged.
Buoyant force = 3 − 2.5 = 0.5 kg-force = 5 N.
Volume of sphere = 5 × 10⁻⁴ m³ (from Archimedes' principle).
Density of sphere = 3 / (5×10⁻⁴) = 6000 kg/m³ = 6 g/cm³.
Relative density = Wₐ/(Wₐ−Wᵥᵥ) = 3/0.5 = 6.
Archimedes' Principle: buoyant force = weight of fluid displaced.
Loss in weight when immersed = buoyant force.
Density of water = 1000 kg/m³ = 1 g/cm³.
Given: • Weight of sphere in air: Wₐᵢᵣ = 3 kg-force • Apparent weight in water: Wᵥᵥₐₜₑᵣ = 2.5 kg-force • Density of water: ρᵥᵥₐₜₑᵣ = 1000 kg/m³
Step 1: Find the Buoyant Force Buoyant Force (Fb) = Wₐᵢᵣ − Wᵥᵥₐₜₑᵣ Fb = 3 − 2.5 = 0.5 kg-force = 0.5 × 10 = 5 N (using g = 10 m/s²)
Step 2: Find the Volume of the Sphere By Archimedes' Principle: Fb = ρᵥᵥₐₜₑᵣ × V × g 5 = 1000 × V × 10 V = 5 / (1000 × 10) = 5 / 10000 = 5 × 10⁻⁴ m³
Step 3: Find the Density of the Sphere Mass of sphere = 3 kg Density = Mass / Volume Density = 3 / (5 × 10⁻⁴) Density = 3 / 0.0005 = 6000 kg/m³
Answer: Density = 6000 kg/m³ = 6 g/cm³
Verification using Relative Density: Relative Density = Wₐᵢᵣ / (Wₐᵢᵣ − Wᵥᵥₐₜₑᵣ) = 3 / (3 − 2.5) = 3 / 0.5 = 6 Density = 6 × density of water = 6 × 1000 = 6000 kg/m³ ✓
Archimedes' Principle states: 'When a body is partially or fully immersed in a fluid, it experiences a buoyant force (upthrust) equal to the weight of the fluid displaced by it.'
Key equations: Buoyant Force = ρ_fluid × V_submerged × g Loss in weight = Weight in air − Apparent weight in fluid = Buoyant Force
Relative Density (Specific Gravity): Relative Density = Density of substance / Density of water = Weight in air / Loss in weight (when water is the fluid) = Wₐ / (Wₐ − Wᵥᵥ)
For this problem: Relative Density = 3 / 0.5 = 6 → Density = 6 g/cm³ = 6000 kg/m³
Physical meaning: The sphere is 6 times denser than water. This is close to the density of: • Nickel: ~8.9 g/cm³ • Steel: ~7.8 g/cm³ • Cast iron: ~7.2 g/cm³ • Titanium: ~4.5 g/cm³ • Granite: ~2.7 g/cm³ • Aluminium: ~2.7 g/cm³ 6 g/cm³ is consistent with a metal like a chromium-iron alloy or a similar metallic material.
Summary of formulae used:
Buoyant Force: Fb = Weight in air − Apparent weight in fluid Fb = Wₐ − Wₗ
Volume from buoyancy: V = Fb / (ρ_fluid × g)
Density of solid: Density = Mass / Volume = m / V
Relative Density / Specific Gravity: RD = Wₐ / (Wₐ − Wₗ) Density of solid = RD × density of water
Things to remember: • 1 kg-force = 9.8 N ≈ 10 N (for calculation purposes) • Density of water = 1000 kg/m³ = 1 g/cm³ • Apparent weight < true weight (because buoyancy acts upward) • Object floats if density < fluid density; sinks if density > fluid density
Applications of Archimedes' Principle: • Ships float despite being made of dense steel (hull design traps air) • Hydrometer measures liquid density • Submarines use ballast tanks to control buoyancy • Hot air balloons — hot air less dense than surrounding air
Buoyant force = 3 − 2.5 = 0.5 kg-force. Volume = 0.5g/(1000×g) = 5×10⁻⁴ m³. Density = mass/volume = 3/(5×10⁻⁴) = 6000 kg/m³ = 6 g/cm³. Using relative density: 3/(3−2.5) = 6 → density = 6×1000 = 6000 kg/m³.
Archimedes' Principle states that when a body is partially or fully immersed in a fluid, it experiences an upward buoyant force equal to the weight of the fluid it displaces. Mathematically: Buoyant Force = ρ_fluid × V × g. The loss in apparent weight = buoyant force.
Relative Density (Specific Gravity) = Density of substance / Density of water. For experiments using Archimedes' principle: Relative Density = Weight in air / (Weight in air − Apparent weight in water). Example: 3/(3−2.5) = 3/0.5 = 6.
A body weighs less in water because of the upward buoyant force acting on it. This force is equal to the weight of water displaced by the body (Archimedes' Principle). The apparent weight = True weight − Buoyant force. The greater the volume of the body, the greater the buoyant force and the greater the loss in weight.
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