Question

A solid of density 5000`(kg)/(m^3)` weghs 0.5 kg f in air. It is completely immersed in water of density 1000`(kg)/(m^3)`.
(a) Calculate the apparent weight of the solid in water.
(b) what will be its apparent weight if water is replaced by a liquid of density 8000 `(kg)/(m^3)`?

Answer 1

(a) `V=(M)/(d)=(0.5kg)/(5000(kg)/(m^3))=1xx10^-4m^3`
`F_B=`weigth of water displaced`(W)=Vdg`
`=(1xx10^-4m^3)(1000(kg)/(m^3))(10(m)/(s^2)`
`=1N=0.1kgf`
apparent weight `=`true weigth`-F_B=0.5kgf-0.1kgf=0.4kgf`
(b) `F_B=Vdg(1xx10^-4m^3)(8000(kg)/(m^3))(10(m)/(s^2))`
`=8N=0.8kgf`
Since in this case `F_Bgt0.5kgf`, the body floats and its apparent weight is zero.