(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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. Relevant 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

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# Homework Help: Sphere floating in liquid

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