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Question Number 52377 by Necxx last updated on 07/Jan/19

An iceberg is floating in sea water.  The density of ice is 0.92g/cm^3   and that of sea water is 1.02g/cm^3 .  What % of ice will  be below the  water?

$${An}\:{iceberg}\:{is}\:{floating}\:{in}\:{sea}\:{water}. \\ $$$${The}\:{density}\:{of}\:{ice}\:{is}\:\mathrm{0}.\mathrm{92}{g}/{cm}^{\mathrm{3}} \\ $$$${and}\:{that}\:{of}\:{sea}\:{water}\:{is}\:\mathrm{1}.\mathrm{02}{g}/{cm}^{\mathrm{3}} . \\ $$$${What}\:\%\:{of}\:{ice}\:{will}\:\:{be}\:{below}\:{the} \\ $$$${water}? \\ $$

Commented by mr W last updated on 07/Jan/19

V=volume of iceberg  V_w =part of iceberg below sea water  ρ_i =density of ice  ρ_w =density of sea water  weight of iceberg=Vρ_i g  flotation=V_w ρ_w g  since iceberg floats, ⇒flotation=weight  V_w ρ_w g=Vρ_i g  V_w ρ_w =Vρ_i   ⇒(V_w /V)=(ρ_i /ρ_w )  fraction of iceberg below water  η=(V_w /V)=(ρ_i /ρ_w )=((0.92)/(1.02))≈90%

$${V}={volume}\:{of}\:{iceberg} \\ $$$${V}_{{w}} ={part}\:{of}\:{iceberg}\:{below}\:{sea}\:{water} \\ $$$$\rho_{{i}} ={density}\:{of}\:{ice} \\ $$$$\rho_{{w}} ={density}\:{of}\:{sea}\:{water} \\ $$$${weight}\:{of}\:{iceberg}={V}\rho_{{i}} {g} \\ $$$${flotation}={V}_{{w}} \rho_{{w}} {g} \\ $$$${since}\:{iceberg}\:{floats},\:\Rightarrow{flotation}={weight} \\ $$$${V}_{{w}} \rho_{{w}} {g}={V}\rho_{{i}} {g} \\ $$$${V}_{{w}} \rho_{{w}} ={V}\rho_{{i}} \\ $$$$\Rightarrow\frac{{V}_{{w}} }{{V}}=\frac{\rho_{{i}} }{\rho_{{w}} } \\ $$$${fraction}\:{of}\:{iceberg}\:{below}\:{water} \\ $$$$\eta=\frac{{V}_{{w}} }{{V}}=\frac{\rho_{{i}} }{\rho_{{w}} }=\frac{\mathrm{0}.\mathrm{92}}{\mathrm{1}.\mathrm{02}}\approx\mathrm{90\%} \\ $$

Commented by Necxx last updated on 07/Jan/19

please sir can you give a detailed  explanation. Thanks in advance.

$${please}\:{sir}\:{can}\:{you}\:{give}\:{a}\:{detailed} \\ $$$${explanation}.\:{Thanks}\:{in}\:{advance}. \\ $$

Commented by Necxx last updated on 07/Jan/19

yes.Thanks so much.

$${yes}.{Thanks}\:{so}\:{much}. \\ $$

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