Upthrust in Fluids, Archimedes' Principle and Floatation — Question 5

Given,

Density of ice = 0.9 g cm^-3

Density of sea water = 1.1 g cm^-3

By the principle of floatation,

$\dfrac{\text {Volume of immersed part}}{\text{Total volume}} = \dfrac{\text {Density of ice}}{\text {Density of sea water}} \\[1em] \therefore \dfrac{\text {Volume of immersed part}}{\text{Total volume}} = \dfrac{0.9}{1.1} \\[1 em] = \dfrac{9}{11} = 0.818$

Hence, the volume of iceberg that will remain below the surface of water in the sea = $\dfrac{9}{11}$th of total volume (or 0.818th) part.