Question

An iceberg of density $900\,kg{m^{ - 3}}$ is floating in water of density $1000\,kg{m^{ - 3}}$. The percentage of volume of ice-cube outside the water is
A. $20\% $
B. $35\% $
C. $10\% $
D. $25\% $

Answer 1

Hint:Here we have to use the concept of buoyancy.The pressure exerted by the fluid in which an object is submerged causes the buoyancy force. As the friction of a fluid increases with depth, the buoyancy force often points upwards. Buoyancy is directly proportional to the mass of the submerged fluid, or the buoyant force.We have to put the weight of ice and water displaced under equilibrium condition to get the answer.

Complete step by step answer:
The theory of Archimedes states that a buoyant force equal in magnitude to the force of gravity on the displaced fluid is encountered by an object submerged in a fluid.
The theory of Archimedes is very useful for measuring an object's volume that does not have a normal form. It is possible to submerge the strangely shaped object, and the volume of the fluid displaced is equal to the object's volume. It may also be used in the measurement of an object's mass or specific gravity. For e.g., the material should be weighted in air and then weighed when immersed in water for an object denser than water.

Let $V$ be the volume of the cube and $f$ be the fraction of the exterior water cube. Therefore, within water ice volume$ = $ volume of water displaced$ = \left( {1 - f} \right) \times V$
Under equilibrium condition, Weight of ice $ = $ weight of water displaced
$
\Rightarrow V{\rho _{ice}}g = \left( {1 - f} \right)V{\rho _{water}}g \\
\Rightarrow \left( {1 - f} \right) = \dfrac{9}{{10}} \\
\therefore f = \dfrac{1}{{10}} \\ $
Hence, the percentage of volume outside water is $10\% $ .

Thus, option C is correct.

Note:Here we have to subtract one to the fraction of water to get the answer. Otherwise if we only divide the densities then the answer would be wrong. When the object is submerged, it weighs less because of the buoyant force pushing upward. The object's specific gravity is then the object's weight in air divided by how much weight the object loses when placed in water.