Q.Iceberg floats in water with part of it submerged. What is the fraction of the volume of iceberg submerged if the density of ice is $\rho_i = 0.917$ g cm$^{-3}$?

The key idea is Archimedes' principle: the weight of the displaced water equals the weight of the floating iceberg.

Let $V$ be the total volume of the iceberg and $V_{\text{sub}}$ be the submerged volume. The weight of the iceberg is $\rho_i V g$, and the weight of the displaced water is $\rho_w V_{\text{sub}} g$, where $\rho_w = 1.00$ g cm$^{-3}$.

Equating the two:

$\rho_i V g = \rho_w V_{\text{sub}} g$ …

The question is a classic application of Archimedes’ principle and the condition for floating equilibrium. When an object floats, the upward buoyant force exactly balances its weight. The buoyant force equals the weight of the displaced fluid — here, water. So the weight of the iceberg equals the weight of the water it pushes aside.

That immediately tells you: the volume submerged times the density of water equals the total volume times the density of ice. The fraction submerged is therefore just the ratio of densities. No extra forces, no complications — just a clean ratio.

Let’s walk through it step by step.

1. Set up the variables.

   Let $V$ be the total volume of the iceberg. Let $V_{\text{sub}}$ be the volume submerged in water. The density of ice is $\rho_i = 0.917$ g cm$^{-3}$, and the density of water is $\rho_w = 1.000$ g cm$^{-3}$ (fresh water at standard conditions; the problem implies this value).

2. Write the equilibrium condition.

   Weight of iceberg: $W = \rho_i V g$

   Buoyant force: $F_b = \rho_w V_{\text{sub}} g$

   For floating: $W = F_b$

   So $\rho_i V g = \rho_w V_{\text{sub}} g$

3. Cancel $g$ and solve for the fraction.

   $\rho_i V = \rho_w V_{\text{sub}}$

   $\frac{V_{\text{sub}}}{V} = \frac{\rho_i}{\rho_w}$

4. Plug in the numbers.

   $\frac{V_{\text{sub}}}{V} = \frac{0.917}{1.000} = 0.917$ …

Archimedes: rho_iVg=rho_wV_subg => V_sub/V=rho_i/rho_w=0.917/1 …