In the dead sea two-third of an object floating is above the water surface. Assuming that the
density of object is 0.5 g/cm, what is the density of water in the dead sea?
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Answer:
Equilibrium of forces (on the floating body) is expressed as
F
b
=m
body
g⇒ρ
liquid
gV
submerged
=ρ
body
gV
total
which leads to
V
total
V
submerged
=
ρ
liquid
ρ
body
We are told (indirectly) that two-thirds of the body is below the surface, so the fraction above is 2/3. Thus, with ρ
body
=0.98g/cm
3
, we find $$\rho_{liquid} \thickapprox 1.5\hspace{0,05cm} g/cm^3$$
— certainly much more dense than normal seawater (the Dead Sea is about seven times saltier than the ocean due to the high evaporation rate and low rainfall in that region).
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