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:

\[=100âˆ’89.32=10.68\%\approx 11\% \]