More than meets the eye

The ocean exerts a pressure on the iceberg equal to the force it would exert on an equal volume of water, in other words, the submerged volume of the iceberg. Therefore, we have:

$\rho_{\text{ocean water}}V_{\text{submerged}}g=\rho_{\text{ice}}V_{\text{total}}g$ $\Rightarrow \dfrac{V_{\text{submerged}}}{V_{\text{total}}}=\dfrac{0.92}{1.02}≈0.9$

This shows that about 90% of our iceberg will remain out of view deep below the water's surface.

The volume of the iceberg = X

The volume of the hidden part = Y

0.92 X = 1.02 Y

Y/X = 0.92 / 1.02 = 0.90196

Iceberg displacement

Let Ρsea = 1.02 density of seawater

Let Ρice = 0.92 density of ice

Let Vbelow = volume in % of iceberg below water

Let Vabove = volume in % of iceberg above water.

By definition

$V_{below} + V_{above} = 1.0$

The iceberg sinks until the volume of ice under water displaces the higher density seawater in a sufficient amount to equal the total weight of the iceberg

$p_{ice}* (V_{below}+V_{above} )=p_{sea}* V_{below}$

Solve for Vbelow; recognizing that

$(V_{below}+V_{above} )=1.0$

$V_{below}=\frac{p_{ice}}{p_{sea} }$

$V_{below}=0.902$

At equilibrium, the weight of the iceberg equals the buoyant force from the water:

$F_B = m_ig$ $\rho_wV_{sub}g = \rho_iV_{tot}g$

where $\rho_w$ and $\rho_i$ are the densities of the water and the iceberg, respectively, and $V_{sub}$ and $V_{tot}$ are the volume of the submerged part of the iceberg and the total volume of the iceberg, respectively. Thus, the fraction of the iceberg's volume that is hidden below the ocean surface is:

$\frac{V_{sub}}{V_{tot}} = \frac{\rho_i}{\rho_w} = \frac{0.92 \text{ g/mL}}{1.02 \text{ g/mL}} = \boxed{0.90}$

volume above water = A ;
Underwater portion's volume = U ;
Buoyant force = B = U * 1.02 (Archimedes principle) ;
Weight = W = (A+U) *0.92 ;
Upward and downward forces must be equal, so W = B ;
then U * 1.02 = (A+U) *0.92 ;
Hidden fraction = U / (A + U) = 0.92/1.02= 46/51~90.2%

If the ice-berg is in equilibrium then its weight is exactly equal to buoyant force of the liquid. Suppose the volume of iceberg is equal to V. The weight of iceberg is then equal to 0.92 V. Buoyant force is equal to 1.02 V', V' being the volume inside water. now 0.92V = 1.02V'. From her V' can be found and hence is the answer.

Density is Mass Divided by Volume i.e, $\frac{M}{v}$

Let Volume Of Iceberg Out Side (Seen Part On The Surface ) = V1

And Volume Of Iceberg Hidden (Inside Ocean) = V2

.92 * V1 = 1.02 * V2

V2/ V1 = .92/1.02 = .90196 (Approx)

I'm curious about the degree of significance being used in the official answer and comments. The problem uses two degrees while the answers use three and more. Is that proper?