Floating of wood - Archimedes law

What will be the volume of the floating part of a wooden (balsa) block with a density of 200 kg/m3 and a volume of 0.02 m3 that floats in alcohol? (alcohol density is 789 kg/m3)

Result

V =  14930.292 cm3

Solution:

ρ1=200 kg/m3 V1=0.02 m3 ρ2=789 kg/m3  m1=m2 V⋅ ρ1=(V1−V)⋅ (ρ2−ρ1) V⋅ ρ1=V1⋅ (ρ2−ρ1)−V⋅ (ρ2−ρ1) V⋅ ρ2=V1⋅ (ρ2−ρ1)  v=V1⋅ (ρ2−ρ1)ρ2=0.02⋅ (789−200)789≐0.0149 m3  V=v→cm3=v⋅ 1000000 cm3=0.0149302915082⋅ 1000000 cm3=14930.29151 cm3=14930.292 cm3ρ_{1}=200 \ \text{kg/m}^3 \ \\ V_{1}=0.02 \ \text{m}^3 \ \\ ρ_{2}=789 \ \text{kg/m}^3 \ \\ \ \\ m_{1}=m_{2} \ \\ V \cdot \ ρ_{1}=(V_{1}-V) \cdot \ (ρ_{2}-ρ_{1}) \ \\ V \cdot \ ρ_{1}=V_{1} \cdot \ (ρ_{2}-ρ_{1})-V \cdot \ (ρ_{2}-ρ_{1}) \ \\ V \cdot \ ρ_{2}=V_{1} \cdot \ (ρ_{2}-ρ_{1}) \ \\ \ \\ v=\dfrac{ V_{1} \cdot \ (ρ_{2}-ρ_{1}) }{ ρ_{2} }=\dfrac{ 0.02 \cdot \ (789-200) }{ 789 } \doteq 0.0149 \ \text{m}^3 \ \\ \ \\ V=v \rightarrow cm^3=v \cdot \ 1000000 \ cm^3=0.0149302915082 \cdot \ 1000000 \ cm^3=14930.29151 \ cm^3=14930.292 \ \text{cm}^3

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