Density of a sphere that has a cavity

In summary, the conversation discusses the relationship between the buoyant force and weight of a sphere floating on water. The equation for this relationship is given, along with the apparent volume of the sphere in the water and its true volume. The minimum cavity volume is determined to be when the sphere is completely immersed in water, resulting in a 80% decrease in the apparent volume. However, it should be noted that the assumption of the density of iron used in the equation is not accurate.
  • #1
MatinSAR
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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?!
 
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  • #2
Alternatively:

$$F_b = F_w$$

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

$$ \implies \frac{ V \llap{-}_c }{V \llap{-}_s} = \frac{\rho_{iron} - \rho_{water}}{\rho_{iron}} = \frac{4}{5}$$
 
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  • #3
FYI, you are not using the commonly accepted density of iron.
 
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  • #6
Frabjous said:
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.
 
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1. What is the formula for calculating the density of a sphere with a cavity?

The formula for calculating the density of a sphere with a cavity is: Density = (3 x Mass of the sphere) / (4 x π x Radius^3).

2. How does the density of a sphere with a cavity compare to that of a solid sphere?

The density of a sphere with a cavity is always less than that of a solid sphere with the same mass and radius. This is because the volume of the cavity reduces the overall volume of the sphere, resulting in a lower density.

3. Can the density of a sphere with a cavity be negative?

No, the density of a sphere with a cavity cannot be negative. Density is a measure of how much mass is contained in a given volume, and since mass cannot be negative, neither can density.

4. How does the size of the cavity affect the density of the sphere?

The size of the cavity has a direct impact on the density of the sphere. The larger the cavity, the lower the density of the sphere will be, and vice versa. This is because a larger cavity will reduce the overall volume of the sphere, resulting in a lower density.

5. Can the density of a sphere with a cavity change?

Yes, the density of a sphere with a cavity can change if the mass or size of the cavity changes. For example, if the mass of the sphere remains the same but the size of the cavity increases, the density of the sphere will decrease. Similarly, if the size of the cavity remains the same but the mass of the sphere increases, the density will also increase.

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