Posted by **Physics Problem** on Sunday, December 10, 2006 at 11:16am.

Hi, I need some help with this problem:

The density of ice is 920 kg/m3, and that of sea water is 1030 kg/m3. What fraction of the total volume of an iceberg is exposed?

I took the density of the iceberg divided by the density of the water and multiplied this by 100, but this did not give me the correct answer.

I don't know why you mulitiplied by 100, however...

Consider 1000 kg of ice. It has a volume of 1000/920 m^3.

To meed the bouyance of this, 1000 kg of sea water has to be displaced> a volume of 1000/1030 m^3.

How much is exposed: the difference..

exposed= 1000/920 - 1000/1030

or 1000(1030 - 920)/920*1030

fraction exposed = exposed/total ice volume

= 1000(1030 - 920)/920*1030*(1000/920)

= (1030 - 920) / *1030

check that.

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