Question

What fraction of an iceberg is submerged? (ρice = 917 kg/m3, ,ρsea = 1030 kg/m3.)

Answer 1

Answer:

Choice d. Approximately $89\%$ of the volume of this iceberg would be submerged.

Explanation:

Let $V_\text{ice}$ denote the total volume of this iceberg. Let $V_\text{submerged}$ denote the volume of the portion that is under the liquid.

The mass of that iceberg would be $\rho_\text{ice} \cdot V_\text{ice}$. Let $g$ denote the gravitational field strength ($g \approx 9.81\; \rm N \cdot kg^{-1}$ near the surface of the earth.) The weight of that iceberg would be: $\rho_\text{ice} \cdot V_\text{ice} \cdot g$.

If the iceberg is going to be lifted out of the sea, it would take water with volume $V_\text{submerged}$ to fill the space that the iceberg has previously taken. The mass of that much sea water would be $\rho_\text{sea} \cdot V_\text{submerged}$.

Archimedes' Principle suggests that the weight of that much water will be exactly equal to the buoyancy on the iceberg. By Archimedes' Principle:

$\text{buoyancy} = \rho_\text{sea} \cdot V_\text{submerged} \cdot g$.

The buoyancy on the iceberg should balance the weight of this iceberg. In other words:

$\underbrace{\rho_\text{ice} \cdot V_\text{ice} \cdot g}_\text{weight of iceberg} = \underbrace{\rho_\text{sea} \cdot V_\text{submerged} \cdot g}_\text{buoyancy on iceberg}$.

Rearrange this equation to find the ratio between $V_\text{submerged}$ and $V_\text{ice}$:

$\begin{aligned} &\frac{V_\text{submerged}}{V_\text{ice}} \\&= \frac{\rho_\text{ice} \cdot g}{\rho_\text{sea} \cdot g}\\ &= \frac{\rho_\text{ice}}{\rho_\text{sea}}\ = \frac{917\; \rm kg \cdot m^{-3}}{1030\; \rm kg \cdot m^{-3}} \approx 0.89 \end{aligned}$.

In other words, $89\%$ of the volume of this iceberg would have been submerged for buoyancy to balance the weight of this iceberg.