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Sphere floating in liquid

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1. Homework Statement
A hollow spherical shell of external diameter of 1 metre and uniform thickness of 0.2m floats in a liquid with half its volume immersed. If the relative density is 1.5 find t 2 decimal places the relative density of the material of the shell.


2. Homework Equations
Buoyancy = Vρg = Weight


3. The Attempt at a Solution

Relative density of liquid = 1.5
Density of liquid = 1500

Since it floats Buoyancy = Weight

Vρg = Vρg

( [itex]\frac{2}{3}[/itex][itex]\pi[/itex] ) (1500g) = [ [itex]\frac{4}{3}[/itex] [itex]\pi[/itex] - [itex]\frac{4}{3}[/itex] [itex]\pi[/itex] (0.8[itex]^{3}[/itex]) ) ] Xg

1000[itex]\pi[/itex] g = [itex]\frac{244}{375}[/itex] [itex]\pi[/itex] Xg

1000 = [itex]\frac{244}{375}[/itex] X

X = 1536.8

Therefore relative density = 1.5

Answer = 0.96
 

Answers and Replies

ehild
Homework Helper
15,361
1,778
1000[itex]\pi[/itex] g = [itex]\frac{244}{375}[/itex] [itex]\pi[/itex] Xg

1000 = [itex]\frac{244}{375}[/itex] X
4/3 is missing from the right-hand side. It should be 1000=(4/3)*(244/375)*X.

ehild
 

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