**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: \[=100âˆ’89.32=10.68\%\approx 11\% \]

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