Question

An iceberg floats in seawater. What fraction of the iceberg lies above the seawater? V0 is the volume of the iceberg and Vf is the volume of the submerged iceberg portion. Density of the ice = 920 kg/m3, and density of the seawater = 1024 kg/m3.

7. An iceberg floats in seawater. What fraction of the iceberg lies above the seawater? Vo is the volume of the iceberg and Vf is the volume of the submerged iceberg portion. Density of the ice Pice 920 kg/m', and density of seawater Peawater1024 kg/m'. (10 pts)

Answer 1

Points to remember:

Density = Mass/Volume

Mass = Density * Volume

Buoyant force = Density of the fluid * Volume of fluid displaced*g

Volume of fluid displaced = Volume of the submerged portion.

Volume of submerged portion = $V_{f}$

Total volume of iceberg=$V_{0}$

Volume of iceberg above water = Total volume - Voulme of submerged portion = $V_{0}-V_{f}$

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Consider the iceberg

The weight of the iceberg is acting downwards which is balanced by the buoyant force.

$W_{ice}=F_{buoyant}$

$m_{ice}*g=\rho _{fluid}*V_{displaced}*g$

$m_{ice}=\rho _{fluid}*V_{submerged}$

$\rho _{ice}*V_{0}=\rho _{water}*V_{f}$

$\frac{\rho _{ice}}{\rho _{water}}=\frac{V_{f}}{V_{0}}$

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Fraction of iceberg above water = $\frac{V_{0}-V_{f}}{V_{0}}$

$=1-\frac{V_{f}}{V_{0}}$

$=1-\frac{\rho _{ice}}{\rho _{water}}$

$=1-\frac{920kg/m^{3}}{1024kg/m^{3}}$

$=1-\frac{920}{1024}$

$=\frac{1024-920}{1024}$

$=\frac{104}{1024}$

ANSWER: ${\color{Red} \frac{13}{128}}$

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