Question

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.

Answer 1

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

Answer 2

Answer:

Explanation:

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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