A piece of wood of volume `200cm^(3)` and density `0.84gcm^(-3)` floats in a liquid of density `1.05gcm^(-3)`
(i) What volume of wood will remain above the surface of liquid?
(ii) What force must be exerted on wood to keep it totally submerged?

Explanation

Given `V=200cm^(3),rho_(s)=0.84gcm^(-3),rho_(L)=1.05gcm^(-3)`
(i) Let `V.cm^(3)` be the volume of wood which remains above the surface of liquid. Then submerged volume of wood `v-V-V.=(200-V.)cm^(3)` and by the principle of floatation.
Weight of wood piece
= Upthrust due to submerged part of wood
`200xx0.84xxg=(200-V.)xx1.05xxg`
`:.V.=(200xx(1.05-0.84))/1.05`
`=(200xx0.21)/1.05=40cm^(3)`
(ii) When wood piece is totally submerged then
Upthrust `=Vxxrho_(L)xxg=200xx1.05xxg`
`=210` gf (upwards)
Weight of wood piece `=Vxxrho_(s)xxg=200xx0.84xxg`
`=168` gf(downwards)
`:.` Force to be exerted to keep the wood totally submerged
= Upthrust - Weight of wood piece
`=210-168=42` gf