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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
answered-by-expert 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 =   begin mathsize 12px style fraction numerator M over denominator 0.5 cross times open parentheses rho subscript w space minus space rho subscript i close parentheses end fraction space equals space fraction numerator 100 over denominator 0.5 cross times left parenthesis 1000 minus 917 right parenthesis end fraction space equals space 2.41 space m squared end style
Answered by Thiyagarajan K | 24 Apr, 2019, 10:10: PM

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