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)

Correct result:

V =  14930.2915 cm³

Solution:

$$\begin{gathered} {\rho }_1=200{\text{kg/m}}^3 \\ {V}_1=0.02{\text{m}}^3 \\ {\rho }_2=789{\text{kg/m}}^3 \\ {m}_1=m_2 \\ V\cdot {\rho }_1=(V_1-V)\cdot ({\rho }_2-{\rho }_1) \\ V\cdot {\rho }_1=V_1\cdot ({\rho }_2-{\rho }_1)-V\cdot ({\rho }_2-{\rho }_1) \\ V\cdot {\rho }_2=V_1\cdot ({\rho }_2-{\rho }_1) \\ v={\displaystyle \frac{V_1\cdot ({\rho }_2-{\rho }_1)}{{\rho }_2}}={\displaystyle \frac{0.02\cdot (789-200)}{789}}\doteq 0.0149{\text{m}}^3 \\ V=v\to {\text{cm}}^3=v\cdot 1000000{\text{cm}}^3=0.0149\cdot 1000000{\text{cm}}^3=14930.292{\text{cm}}^3=14930.2915{\text{cm}}^3=1.493\cdot 10^4{\text{cm}}^3 \end{gathered}$$

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