# Sphere floating in liquid

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

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

1000$\pi$ g = $\frac{244}{375}$ $\pi$ Xg

1000 = $\frac{244}{375}$ X

X = 1536.8

Therefore relative density = 1.5

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Homework Helper
1000$\pi$ g = $\frac{244}{375}$ $\pi$ Xg

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

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