Answer:
Choice d. Approximately [tex]89\%[/tex] of the volume of this iceberg would be submerged.
Explanation:
Let [tex]V_\text{ice}[/tex] denote the total volume of this iceberg. Let [tex]V_\text{submerged}[/tex] denote the volume of the portion that is under the liquid.
The mass of that iceberg would be [tex]\rho_\text{ice} \cdot V_\text{ice}[/tex]. Let [tex]g[/tex] denote the gravitational field strength ([tex]g \approx 9.81\; \rm N \cdot kg^{-1}[/tex] near the surface of the earth.) The weight of that iceberg would be: [tex]\rho_\text{ice} \cdot V_\text{ice} \cdot g[/tex].
If the iceberg is going to be lifted out of the sea, it would take water with volume [tex]V_\text{submerged}[/tex] to fill the space that the iceberg has previously taken. The mass of that much sea water would be [tex]\rho_\text{sea} \cdot V_\text{submerged}[/tex].
Archimedes' Principle suggests that the weight of that much water will be exactly equal to the buoyancy on the iceberg. By Archimedes' Principle:
[tex]\text{buoyancy} = \rho_\text{sea} \cdot V_\text{submerged} \cdot g[/tex].
The buoyancy on the iceberg should balance the weight of this iceberg. In other words:
[tex]\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}[/tex].
Rearrange this equation to find the ratio between [tex]V_\text{submerged}[/tex] and [tex]V_\text{ice}[/tex]:
[tex]\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}[/tex].
In other words, [tex]89\%[/tex] of the volume of this iceberg would have been submerged for buoyancy to balance the weight of this iceberg.
