An Iceberg is floating partially immersed in seawater. The density of seawater is 1.03 $\dfrac{g}{cm^3 }$ and that of ice is 0.92 $\dfrac{g}{cm^3 }$ , the fraction of the total volume of the Iceberg above this level of seawater is :
A. 8%
B. 11%
C. 34%
D. 89%

Hint: Use the Archimedes principle. The weight of the Iceberg will be equal to the upward thrust force which can be calculated by calculating the weight of the displaced liquid by the Iceberg. Calculate the volume of Iceberg above the sea level and take the ratio of volume above sea level and the total volume.

Complete step-by-step answer:
Let the volume of the Iceberg inside the sea level be $V_{in}$
Let the Total volume of the Iceberg be: V
The density of the Iceberg is given as : 0.92 $\dfrac{g}{cm^3 }$
The density of water is given as: 1.03 $\dfrac{g}{cm^3 }$
Now we know that the upward thrust is equal to the weight of the Iceberg.
We can calculate the upward thrust as below:
Weight is given by $V \rho g$ where $\rho$ is the density , V is the volume
Upward thrust = weight of the displaced liquid = $V_{in} \rho_w g$
The weight of the iceberg can be calculated as below
Weight = volume $\times$ density $\times$ g = $V \rho_i g$
So we get by equating the weight of the Iceberg and upward thrust as below
$V\rho_i g = V_{in} \rho_w g$
$\dfrac{V_{in}}{V} = \dfrac{\rho_i}{\rho_w}$
$1 - \dfrac{V_{in}}{V} = 1 - \dfrac{\rho_i}{\rho_w}$
$\dfrac{V_{in} - V}{V} = \dfrac{V_{above sea level}}{V_{total}} = \dfrac{\rho_w - \rho_i}{\rho_w} = \dfrac{1.03-0.92}{1.03} = 0.1067$
So the percentage is : 0.1067 *100 = 10.67 % which is approximately = 11 %
Using the Archimedes principle we have found the ratio of the volume of the iceberg above the sea level and the total volume as: 11%

Note: One of the points where we find difficulty is finding the volume of the displaced liquid. We need to understand that the volume of the displaced liquid is nothing but the volume of the Iceberg inside the sea level.