Question Number 52377 by Necxx last updated on 07/Jan/19
$${An}\:{iceberg}\:{is}\:{floating}\:{in}\:{sea}\:{water}. \\ $$$${The}\:{density}\:{of}\:{ice}\:{is}\:\mathrm{0}.\mathrm{92}{g}/{cm}^{\mathrm{3}} \\ $$$${and}\:{that}\:{of}\:{sea}\:{water}\:{is}\:\mathrm{1}.\mathrm{02}{g}/{cm}^{\mathrm{3}} . \\ $$$${What}\:\%\:{of}\:{ice}\:{will}\:\:{be}\:{below}\:{the} \\ $$$${water}? \\ $$
Commented by mr W last updated on 07/Jan/19
$${V}={volume}\:{of}\:{iceberg} \\ $$$${V}_{{w}} ={part}\:{of}\:{iceberg}\:{below}\:{sea}\:{water} \\ $$$$\rho_{{i}} ={density}\:{of}\:{ice} \\ $$$$\rho_{{w}} ={density}\:{of}\:{sea}\:{water} \\ $$$${weight}\:{of}\:{iceberg}={V}\rho_{{i}} {g} \\ $$$${flotation}={V}_{{w}} \rho_{{w}} {g} \\ $$$${since}\:{iceberg}\:{floats},\:\Rightarrow{flotation}={weight} \\ $$$${V}_{{w}} \rho_{{w}} {g}={V}\rho_{{i}} {g} \\ $$$${V}_{{w}} \rho_{{w}} ={V}\rho_{{i}} \\ $$$$\Rightarrow\frac{{V}_{{w}} }{{V}}=\frac{\rho_{{i}} }{\rho_{{w}} } \\ $$$${fraction}\:{of}\:{iceberg}\:{below}\:{water} \\ $$$$\eta=\frac{{V}_{{w}} }{{V}}=\frac{\rho_{{i}} }{\rho_{{w}} }=\frac{\mathrm{0}.\mathrm{92}}{\mathrm{1}.\mathrm{02}}\approx\mathrm{90\%} \\ $$
Commented by Necxx last updated on 07/Jan/19
$${please}\:{sir}\:{can}\:{you}\:{give}\:{a}\:{detailed} \\ $$$${explanation}.\:{Thanks}\:{in}\:{advance}. \\ $$
Commented by Necxx last updated on 07/Jan/19
$${yes}.{Thanks}\:{so}\:{much}. \\ $$