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.**## Expert Answer

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:\[=100−89.32=10.68\%\approx 11\% \]