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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? |
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Answer» Solution :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 volumeof 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 |
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