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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 100kg, the area of A should be (specefic gravity of ice 0.917 and the density of water = 1000 kg/m^3)
Asked by jmorakhia | 24 Apr, 2019, 07:08: PM
Expert Answer
weight of man and weight of ice is balanced by upward thrust
Let M be the mass of man, A is the required area of ice block of thickness d, ρi and ρw are the densities of ice and water respectively.
Then we have, Mg + (A×0.5)ρi g = (A×0.5)ρw g or A =
Answered by Thiyagarajan K | 24 Apr, 2019, 10:10: PM
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