## “20000 Leagues Under the Sea” Part2 Ch13

Question 1: Captain Nemo speaking of how much of an iceberg is submerged and how much is above water says:

“…for one foot of iceberg above the sea there are three below it…”

While he speaks in feet, without knowing the specific shape and orientation of an iceberg, the physics of buoyancy only allow us to calculate what percentage of the iceberg’s volume is unsubmerged. What percentage is this?

Answer 1: The weight W_{i} of an iceberg of total volume V_{i} is:

W_{i} = ρ_{i}V_{i}g

where ρ_{i} is the density of ice = 917 kg/m^{3} and g is the gravitational constant assumed to be 9.8 m/s^{2} on earth’s average surface height.

The weight of the displaced seawater, which is the buoyancy force F_{b}, is

W_{w} = F_{b} = ρ_{w}V_{w}g

where ρ_{w} = 1024 kg/m^{3} is the density of sea water and V_{w} is the volume of the displaced water, that is the submerged volume of the iceberg. For the floating iceberg, these two forces are equal, or

ρ_{i}V_{i}g = ρ_{w}V_{w}g

Since the volume of ice unsubmerged is V_{i} – V_{w}, the fraction of unsubmerged to total iceberg volume is

(V_{i} – V_{w}) / V_{i} = 1 – (V_{w}/V_{i}) = 1 – (ρ_{i}/ρ_{w})

= 1 – (917 kg/m^{3} / 1024 kg/m^{3})

which equals about 0.1044 or roughly 10%

See pg 372 (Chapter 16 Fluids) of Fundamentals of Physics Third Edition Extended by David Halliday & Robert Resnick with the assistance of John Merrill, Copyright 1988, published by John Wiley & Sons, Inc.

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