Floating Iceberg Displacing 723m³ of Water

[Felafel]

Homework Statement

I think it should be pretty simple, but my result and that of the book are different:
How much water does an iceberg displace (Its emerged part is $V_i=100m^3$)

The Attempt at a Solution

knowing the density of sea water is $d_w=1.03*10^3 kg/m^3$ and that of ice $d_i=0.92*10^3 kg/m^3$ i can calculate the volume of the submerged part of the iceberg:

$\frac{d_i}{d_w}=0.89%$ meaning the total volume of the iceberg is $100:11=x:100$
x=909. The submerged part is then 909-100=809m^3. now, since the iceberg floats, its weight and archimedes' force should be equal, then:

$m \cdot g= d_w \cdot g \cdot V_w$ and so:
$809 \cdot 92=103 \cdot v_w$ $\Rightarrow$ $V_w=723 m^3$
but according to the book it should be 1050m^3
what's wrong in my reasoning?
thank you in advance

 

[SteamKing]

IMO, the book's answer is incorrect.
I don't follow your ratio method.

What I did was to say the x = submerged volume of the iceberg. Then I wrote an equation setting the weight of the iceberg = to the buoyancy of the iceberg. Solve for x.

 

[Felafel]

I don't understand your method. I get two incognitas:

Weight of the iceberg: mass multiplied g, where the mass is volume divided density and x is the submerged volume. Buoyancy= g multiplied the volume of displaced water (which i don't know) multiplied the density of the water

$\frac{x+100}{d_{ice}} \cdot g=V_w \cdot g \cdot d_{w}$
but i don't know both x and V_w

 

[Borek]

How is displaced water different from submerged volume?

 

[haruspex]

$\frac{d_i}{d_w}=0.89%$ meaning the total volume of the iceberg is $100:11=x:100$

Because it gets subtracted from 1, effectively, the rounding error in truncating it to 0.89 becomes significant. You need to use a couple more digits of precision.

x=909. The submerged part is then 909-100=809m^3.

The method looks ok to here, but as Borek points out this should also be the volume of water displaced. I don't understand what you did from here.
Fwiw, I get 836 cu m.

 

[Borek]

Fwiw, I get 836 cu m.

And you are not alone

 

[Felafel]

now everything's clear :) thank you!

 

[SteamKing]

I don't understand your method. I get two incognitas:

Weight of the iceberg: mass multiplied g, where the mass is volume divided density and x is the submerged volume. Buoyancy= g multiplied the volume of displaced water (which i don't know) multiplied the density of the water

$\frac{x+100}{d_{ice}} \cdot g=V_w \cdot g \cdot d_{w}$
but i don't know both x and V_w

Let x = submerged volume of iceberg

Total volume of iceberg = (x + 100) cu. m.

Mass of iceberg = 920 kg/m^3 * (x + 100) m^3

Since the iceberg is floating, the mass of the iceberg = mass of the displaced water
(this is Archimedes Principle),

Therefore,
mass of displaced water = x * 1030

equating displacement of iceberg to mass of iceberg,

1030*x = 920 * (x + 100)

Solve for x