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A block of ice of total area A and thick...

A block of ice of total area A and thicknes `0.5m` is floating in water. In order to just support a man of nass `100kg`, the area A should be (the specific gravity of ice is 0.9)

Text Solution

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For equilibrium `(m_(1)+m_(2)) g=rhoVg`

Here `m_(1)=` mass of man `=100 kg`
`m_(2)=` mass of ice` =0.917xx1000V=917V`
`rho=1000kg//m^(3),h=0.5m`
`:. 100g+917Vg=rhoVg=1000Vg`
`V=(100g)/(1000g-917g)impliesAh=100/83`
`:. A=100/(83xx0.5)=2.41m^(2)`
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