What is the thickness of the wall of a hollow iron sphere submerged in water?

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The discussion centers on calculating the thickness of a hollow iron sphere that is half submerged in water, with an outer diameter of 10 cm and an iron density of 7.9 g/m³. The calculated thickness of the wall is 0.11 cm, while a textbook states it should be 0.22 cm. Participants agree that the method used to arrive at 0.11 cm is correct, suggesting a possible error in the textbook. The calculations involve determining the buoyant force and the mass of the sphere based on its volume. Ultimately, the consensus is that the correct thickness is indeed 0.11 cm.
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Homework Statement


Hollow iron sphere is half submerged in water.Sphere has outer diameter of 10 cm. Calculate thickness of the wall , when the iron density is 7.9 g / m3.
I get answer 0.11cm, book says that it is 0.22cm.

Homework Equations


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The Attempt at a Solution


First i found buoyant force F=ρgV/2 where ρ is density of water and V is volume of a sphere.
Then I found mass of a sphere : ρgV/2=mg
m=0.26 kg
I know density of iron ,so: 7900=0.26/v where v is volume of the wall
(v= volume of outside sphere - volume of inside sphere.)
From there I get thikness of the wall to be 0.11cm.
 
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kristjan said:
half submerged in water.
 
kristjan said:
I get answer 0.11cm, book says that it is 0.22cm.
Your method looks correct. I agree with your answer of .11 cm.
 
Someone made my "intuitive" error while re-writing the exercise for the "n-th" edition; agreed, 0.11 cm.
 
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