A block of ice of area A and thickness 0.5 m is floating in the fresh water. In order to just support a man of 100 kg, find the area A. (the specific gravity of ice is 0.917 anddensity of water = 1000 kg/m3).

For equilibrium, (m1+m2)g = ρVgHere, m1 = mass of man = 100 kg         m2 = mass of ice                = 0.917x1000 V = 917 V              ρ = 1000 kg/m3              h = 0.5 m∴ 100 g + 917 Vg = ρVg = 1000 Vg∴ V = 100g1000g-917g∴ Ah = 10083  ∴A = 10083×0.5 = 2.41 m2