Question

In English the phrase "tip of the Iceberg" means a small visible fraction of something that is mostly hidden. For a real iceberg what is this fraction, if the density of seawater is $1.03\text{g/cc}$ and that of ice is $0.92\text{g/cc}$ ?

• A. $8.4\%$
• B. $10.6\%$
• C. $20\%$
• D. $85\%$

Answer 1

The correct option is B $10.6\%$

Given that,
Density of seawater, ${\rho }_{sw}=1.03\text{g/cc}$
Density of ice, ${\rho }_i=0.92\text{g/cc}$

Let the volume of the iceberg is $V$ and volume of iceberg inside the sea of $V_{in}$
Volume of iceberg outside the seawater, $V_O=V-V_{in}$
In case of flotation, weight = upthrust

i.e. $(mg{)}_{iceberg}=V_{in}{\rho }_{sw}g$
$\left(V{\rho }_i\right)=V_{in}{\rho }_{sw}$
$V_{in}=V\frac{{\rho }_i}{{\rho }_{sw}}$
Volume of iceberg outside the seawater, $V_O=V-V_{in}$
$V_O=V(1-\frac{{\rho }_i}{{\rho }_{sw}})$
Fraction of volume outside the water,
$f_{out}=\frac{V_O}{V}=(1-\frac{{\rho }_i}{{\rho }_{sw}})$
$=(1-\frac{0.92}{1.03})=\frac{0.11}{1.03}=0.106$
=$10.6\%$

$\begin{array}{c}\text{Caution:-}\\ \text{In upthrust expression only the}\\ \text{submerged volume portion is considered.}\end{array}$