Density of a sphere that has a cavity

[MatinSAR]

Homework Statement

At least what percentage of an iron sphere must have cavity, so that the sphere floats on water?
$(\rho _{iron}=5 \times 10^3 kg/m^3 , \rho _{water}=10^3 kg/m^3).$

Relevant Equations

$Density=mass/Volume.$

The sphere floats on water so we should have: $F_b=F_g$
The buoyant force is equal to the weight of the displaced fluid, so : $\rho _wV_wg=\rho _sV_sg$
(w: water, s: sphere)
From last equation we have : $V_w=\frac {\rho _s}{\rho _w} V_s \rightarrow V_w=5 V_s$
The volume of displaced water($V_w$) is equal to the apparent volume of the part of sphere that is inside the water( $V_w=$ Apparent volume of the part of sphere that is inside the water).
I also know that $V_s$ is the true volume of sphere($V_s=$ True volume of sphere).

Volume of cavity($V_{cavity}$) = Apparent volume of sphere($V_{apparent}$) $-$ True volume of sphere($V_{s}$) $\rightarrow$
$V_{cavity}=V_{apparent}-V_s$
If we put $V_{apparent}=V_w$ then we have: $V_{cavity}=V_{w}-V_s=\frac 4 5 V_w=\frac 4 5 V_{apparent}$
So the answer is: $80$%

I think $V_{apparent}$ should be equal to $V_w$(Apparent volume of the part of sphere that is inside the water) because the question asks for the minimum cavity volume. And the volume of the cavity is minimized when the sphere is completely immersed in water because the more the volume of the cavity decreases, the more water is moved to balance the forces.

Am I right?!

 

[erobz]

Alternatively:

$$F_b = F_w$$

$$ \rho_{water} \cancel{g} V \llap{-}_s = \rho_{iron} \left( V \llap{-}_s - V \llap{-}_c \right) \cancel{g} $$

$$ \implies \frac{ V \llap{-}_c }{V \llap{-}_s} = \frac{\rho_{iron} - \rho_{water}}{\rho_{iron}} = \frac{4}{5}$$

 

[Frabjous]

FYI, you are not using the commonly accepted density of iron.

 

[Lnewqban]

One cubic meter of iron or steel weights between 7.75 and 8.05 tons, depending on chemical composition.

https://en.wikipedia.org/wiki/Steel#Material_properties

 

[MatinSAR]

@erobz @Frabjous @Lnewqban Thank you for your time.

 

[MatinSAR]

FYI, you are not using the commonly accepted density of iron.

This assumption was stated in the question(not in my answer). And yes, it is not true. Thank you.