A solid of density 5000 kg/m^3 weighs 0.5 kgf in air. It is completely immersed in water of density 1000 kg/m^3. Calculate the apparent weight of the solid in water.

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Solution

Density ρ = 5000 kg/m³ Weight = 0.5 kg f in air
mass M = 0.5 kg
Volume V = M / ρ = 0.5 / 5000 = 10⁻⁴ m³

mass of water displaced = V * density of water
= 10⁻⁴ * 1000 = 0.1 kg

Apparent weight = wt in air - wt of water displaced
= 0.5 kg f - 0.1 kg f
= 0.4 kg f

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A solid of density 5000 kg/m^3 weighs 0.5 kgf in air. It is completely immersed in water of density 1000 kg/m^3. Calculate the apparent weight of the solid in water.

Density ρ = 5000 kg/m³ Weight = 0.5 kg f in air mass M = 0.5 kg Volume V = M / ρ = 0.5 / 5000 = 10⁻⁴ m³ mass of water displaced = V * density of water = 10⁻⁴ * 1000 = 0.1 kg Apparent weight = wt in air - wt of water displaced = 0.5 kg f - 0.1 kg f = 0.4 kg f

Density ρ = 5000 kg/m³ Weight = 0.5 kg f in air
mass M = 0.5 kg
Volume V = M / ρ = 0.5 / 5000 = 10⁻⁴ m³

mass of water displaced = V * density of water
= 10⁻⁴ * 1000 = 0.1 kg

Apparent weight = wt in air - wt of water displaced
= 0.5 kg f - 0.1 kg f
= 0.4 kg f