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

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SUMMARY

The thickness of the wall of a hollow iron sphere submerged in water, with an outer diameter of 10 cm and iron density of 7.9 g/m³, is calculated to be 0.11 cm. The buoyant force was determined using the formula F=ρgV/2, where ρ is the density of water and V is the volume of the sphere. The mass of the sphere was found to be 0.26 kg, leading to the conclusion that the volume of the wall is derived from the difference between the outer and inner sphere volumes. The discrepancy with the book's answer of 0.22 cm is attributed to an error in the exercise's re-writing.

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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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