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=14930.2915082 cm3=14930.292 cm3ρ_{1} = 200 \ kg/m^3 \ \\ V_{ 1 } = 0.02 \ m^3 \ \\ ρ_{2} = 789 \ 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 \ m^3 \ \\ \ \\ V = v \rightarrow cm^3 = v \cdot \ 1000000 \ cm^3 = 14930.2915082 \ cm^3 = 14930.292 \ cm^3

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