Question

An iceberg floats in sea water of density 1.17 g cm ³, such that 2/9 of its volume is above sea water. Find the density of iceberg.

Answer 1

Density of sea water =ρ_w = 1.17 g cm^-3

Density of solid ice berg =ρ_i = ?

${\displaystyle \frac{2}{9}}$th volume of an iceberg is above the seawater

∴ Volume of iceberg inside water = ${\displaystyle {\text{V}}_{\text{i}}=(1-\frac{2}{9})\text{V}}$

Where V = Total volume of an iceberg

${\displaystyle \Rightarrow {\text{V}}_{\text{i}}=\frac{9-2}{9}\text{V}=\frac{7}{9}\text{V}}$

⇒ Volume of sea water displaced by immersed part of the iceberg

${\displaystyle ={\text{V}}_{\text{w}}=\frac{7}{9}\text{V}}$

By law of floatation:

Weight of iceberg = Weight of sea water displaced by iceberg

${\displaystyle {\text{V}}_{\text{i}}\times {\rho }_{\text{i}}\times \text{g}={\text{V}}_{\text{w}}\times {\rho }_{\text{w}}\times \text{g}}$

${\displaystyle \text{V}\times {\rho }_{\text{i}}=\frac{7}{9}\text{V}\times 1.17}$

${\displaystyle \Rightarrow {\rho }_{\text{i}}=\frac{7}{9}\times 1.17=0.91{\text{gcm}}^{-3}}$

⇒ Density of iceburg = 0.91 g cm^-3