Question

A wooden block floats in water with two-third of its volume submerged.
(a) Calculate the density of wood. 
(b) When the same block is placed in oil, three-quarters of its volume is immersed in oil. Calculate the density of oil. 

Answer 1

Volume of wooden block submerged in water(v) = `2/3 xx "total volume (V)"`

Volume of wooden block submerged in oil (v') = `3/4 xx "total volume (V)"`

Say density of wood = `ρ_"wood"  "gcm"^-3`

Say density of oil = `ρ_"oil"  "gcm"^-3`

According to the law of floatation, 

`"v"/"V" = ρ_"wood"/ρ_"water"`

or , `2/3 = ρ_"wood"/ρ_"water" = ρ_"wood"/1000`

or , `ρ_"wood" = 1000 xx 2/3 = 667  "kgm"^-3`

Again, according to the law of floatation, 

`"v'"/"V" = ρ_"wood"/ρ_"oil"`

or , `3/4 = ρ_"wood"/ρ_"oil"`

or , `3/4 = 667/ρ_"oil"`

or , `ρ_"oil" = 4/3 xx 667 = 889  "kgm"^-3`

Answer 2

(1) Let volume of wood = V

Volume of wood submerged v' = `2/3` V

`"d"_"S"/"d"_"w" = "v'"/"V" = (2/3"V")/"V" = 2/3`

`"d"_"S" = 2/3  "d"_"W" = 2/3 xx 1000 = 667` kg m^-3 

but `"d"_"W" = 1000` kg m^-3 

Density of wood d_S = 667 kg m^-3 

(2) Now `"d"_"S"/"d"_"w" = "v'"/"V"`

`(2000/3)/"d"_"L" = (3/4"V")/"V" = 3/4`

`=> 2000/(3"d"_"L") = 3/4`

`=> "d"_"L" = (2000 xx 4)/(3 xx 3) = 8000/9 = 889` kg m^-3 

∴ Density of oil = 889 kg m^-3