A cylindrical cone of semi-angle α, height H and mass M is submerged in a fluid of density ρ. Givecn that the volume of a cone is 1/3 area of base x height, find the depth to which it is submerged.

Any ideas?

There are 2 things that can happen: the cone can sink, or float, depending on whether it is less dense or more dense than the fluid...

So (1/3)(pie.r^2)(H)>p

Now thats a test, lets see if it can be proved...

archimedies principle applies here I think..

well change density to pressure

(1/3)(pie. r^2) H = row.g.H

cancelt he Hs

(1/3)(pie.r^2)=row.9.8

(pie.r^2)/row=9.8/1.333
ditto            =7.351

I lost myself now, but it goes something liek that Im sure. Hopefully my notes might be of some use. Ill come back later and try.

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Punctured bicycle on a hillside, desolate,
Could nature make a man of me yet?