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A block of ice of area A and thickness 0...

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 , the area A should be (specific gravity of ice `0.917` and density of water `=1000 kg//m^(2))`

A

`1.24 m^(2)`

B

`4.21 m^(2)`

C

`2.41 m^(2)`

D

`7.23 m^(2)`

Text Solution

Verified by Experts

(c) For equilibrium
`(m_(1)+m_(2))g=rhoVg`
Here, `m_(1)`=mass of man =100 kg
`m_(2)` msss of ice
`=0.917xx1000 V`
`=917 V`
`rho=1000 kg//m^(2)`
h=0.5 m`
`:. 100g+917Vg=rho=1000Vg`

`:. V=(100g)/(1000g-917g)`
`:. Ah=(100)/(83)`
`:. A=(100)/(83xx0.5)`
`=2.41m^(2)`
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