Finding the Relative Density of a Floating Sphere in Liquid

[Darth Frodo]

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.

Homework Equations

Buoyancy = Vρg = Weight

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

Answer = 0.96

 

[ehild]

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.

ehild