Archimedes principle doubt???PLEASE help!!!?
Okay, there is a hollow sphere of relative density 9.Now, my doubt is of the bigger radius of the sphere is R and radius of hollow cavity is r....and if it is completly immersed in water...i my text book....
the eqaation is ...
ok....i understand the LHS....it is the upthrust....what is rhs.???upthrust is equal to the decrease in weight....so...why.....pls just explain to me the whole thing...pls i need it immediately....
- Steve4PhysicsLv 76 years agoBest Answer
The equation in your question is incomplete. And it is not clear what your question is! I hope this helps though.
The external volume of the sphere = (4/3)πR³
When fully immersed, the volume of water displaced is the same = (4/3)πR³
Take the density of water as ρ. The mass of water displaced = (4/3)πR³ρ
So the weight of water displaced = (4/3)πR³ρg and this is the upthrust.
The sphere's external volume is (4/3)πR³ (as above)
The volume of the cavity (4/3)πr³
So the volume of material used to make the hollow sphere is (4/3)πR³ -(4/3)πr³ = (4/3)π(R³ -r³)
The density of the material ios 9ρ. So the mass of the hollow sphere
= (4/3)π(R³ -r³)9ρ = 12π(R³ -r³)ρ
The weight of the hollow sphere = 12π(R³ -r³)ρg
If the sphere is fully immersed, we want neutral buoyancy (upthrust=weight, so no resultant force acts on sphere) then:
(4/3)πR³ρg = 12π(R³ -r³)ρg
which you could simplify if required.
- 6 years ago
The RHS is the volume of the solid part of the sphere, which is the same as the weight of water displaced (water having a unitary density of 1gm per cm^3) which according to Archimedes is the upthrust.