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 ...

(4/3)(pi)(R^3)(1)(g)=(4/3)(pi)(R^3-r^3)(9)(g)

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....

Relevance

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.

• 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.