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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.
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Solution 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`
Solution 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 dS = 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
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