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.