# An iceberg (specific gravity 0.917) floats in the ocean (specific gravity 1.025 ). What percent of the volume of the iceberg is under water?

This problem works to familiarize us with the weight of objects underwater. The concept required to solve this problem includes the force density formula and the Buoyant force. Buoyancy, more commonly known as upthrust, is an upward force exercised by a fluid that resists the weight of a partly or fully submerged object. In simple words, it is the ability of an object to float in a fluid either wholly or partly immersed. The formula to calculate buoyancy force is: $F_b = – \rho g V$ Where $F_b$ is the buoyancy force, $\rho$ is the fluid density, $g$ is the acceleration due to gravity, and $V$ is the volume of the fluid.

We will start by calculating the force that the iceberg exerts on the fluid in the form of its weight. Using the buoyancy force formula: $F_w = S_{g_{ice}}\times \rho_{water}\times V$ Where $F_w$ is the force due to the weight of the iceberg, $S_{g_{ice}}$ is the specific gravity of the iceberg = $0.917$, $\rho_{water}$ is the density of water = $62.4 lb/ft^3$, and $V$ is the volume of the water. Substituting the values in the above equation gives us: $F_w = 0.917\times 62.4\times V$ $F_w =57.22V\space lb$ This is the force that the iceberg exerts on the water. Now to find the buoyancy force that acts upwards and cancels out the weight of the iceberg: $F_b = S_{g_{sea water}}\times \rho_{water}\times V’$ Here, $S_{g_{sea water}}$ is the specific gravity of the seawater= $1.025$, $V’$ is the volume of the immersed portion of the iceberg and everything else remains the same. Substituting the values in the above equation gives us: $F_b = 1.025\times 62.4\times V’$ $F_b = 63.96V’\space lb$ Now, we are left with $2$ unknowns here, $V$ and $V’$, but we are to find the percent volume of the iceberg that is immersed in water. To find the volume, we will equate the two forces to relate the formula for the percentage: $F_b=F_w$ $57.22V\space lb=63.96V’\space lb$ $\dfrac{V’}{V} = \dfrac{57.22}{63.96}$ $= 0.895$ Since the ratio is $0.895$, the percentage of the immersed volume to the total volume is: $\bigtriangleup V’ (\%) = 89.5\%$

## Numerical Result

$89.5\%$ of the volume of the iceberg is underwater.

## Example

An iceberg is sailing somewhat immersed in seawater. The density of seawater is $1.03 g/cm^3$ and that of solid ice is $0.92 g/cm^3$. Find the percentage of the total volume of the iceberg above the seawater. Let’s assume $V$ to be the volume of the iceberg, and $x$ to be the volume of the immersed portion of the iceberg. Then with the help of Archimedes’ principle, we can have, $\text{iceberg weight}=\text{weight of the water displaced (weight of x volume of water)}$ $\implies V\times 0.92\times g=x\times 1.03\times g$ $\dfrac{x}{V} = \dfrac{92}{103}$ The percent volume of the iceberg in seawater is =$\dfrac{92}{103}\times 100=89.32\%$. Hence, the percentage of the total volume of the iceberg above the seawater is: $=100−89.32=10.68\%\approx 11\%$