Answer
Fraction of iceberg visible $=0.10 $
Work Step by Step
Let $V'$ be the volume of iceberg below water and $V$ be the total volume of iceberg.
Since the iceberg is floating, its weight is equal to the weight of salt water that it displaces (Archimedes Principle).
This can be written as:
$m_{(ice)} g = V'\rho_{w} g $ . . . . . . . . . . . . . . . .(1)
But mass of iceberg can be written in terms of its volume and density :
$mass=volume\times density$
$m_{(ice)}=V \times\rho_{(ice)}$
Thus replacing $m_{(ice)}$ by $V \rho_{(ice)} $ in equation (1) gives :
$V \rho_{(ice)} g = V'\rho_{w} g $
Substituting the known values and solving gives:
$V\times917$ $kg/m^{3}\times g=V'\times1024$ $kg/m^{3}\times g$
$\frac{V'}{V}=\frac{917}{1024}=0.895$
This is the fraction of ice berg inside salt water. To find the fraction outside water (visible part) subtract this fraction from $whole$ :
Fraction of iceberg visible $=1-0.895= 0.105\approx0.10$
