What fraction of an iceberg of density 910 kgm^{-3} will be above the surface of sea water of density 1170 kgm^{-3 }?

#### Solution

Let volume of iceberg = V_{i} = x

Volume of iceberg inside sea water = V_{w}Density of iceberg = ρ_{i} = 910 kmg^{-3}Density of sea waer = ρ_{w} = 1170 kgm^{-3}By law of floatation:

Weight of iceberg = Weight of sea water displaced by the iceberg

`"V"_"i" xx rho_"i" xx "g" = "V"_"w" xx rho_"w" xx "g"`

`"x" xx 910 = "V"_"w" xx 1170`

`"V"_"w" = "x" xx 910/1170`

Volume of iceberg inside seawater = Volume of sea water dispalced by iceberg

`= "V"_"w" = 7/9"x" = 7/9"x"`

Volume of an iceberg above the sea water = `"x" - 7/9`x

`= (1 - 7/9)"x" = (9-7)/9`x

`= 2/9 xx "Volume of iceberg"`

Fraction of iceberg above sea water = `2/9"x" xx 1/"x" = 2/9` part

`=> 2/9`th part of iceberg is above the sea water.