Sphere floating

Will float a hollow iron ball with an outer diameter d1 = 20cm and an inside diameter d2 = 19cm in the water? The iron density is 7.8 g/cm 3. (Instructions: Calculate the average sphere density and compare with the density of the water. )

Correct result:

x =  0

Solution:

$$\begin{gathered} d_1=20\text{ cm } \\ {d}_2=19\text{ cm } \\ \rho =7.8{\text{g/cm}}^3 \\ v=1{\text{g/cm}}^3 \\ {r}_1=d_1\mathrm{/}2=20\mathrm{/}2=10\text{ cm } \\ {r}_2=d_2\mathrm{/}2=19\mathrm{/}2={\displaystyle \frac{19}{2}}=9.5\text{ cm } \\ {V}_1={\displaystyle \frac{4}{3}}\cdot \pi \cdot r_1^3={\displaystyle \frac{4}{3}}\cdot 3.1416\cdot 10^3\doteq 4188.7902{\text{cm}}^3 \\ {V}_2={\displaystyle \frac{4}{3}}\cdot \pi \cdot r_2^3={\displaystyle \frac{4}{3}}\cdot 3.1416\cdot 9.5^3\doteq 3591.364{\text{cm}}^3 \\ m=\rho \cdot (V_1-V_2)=7.8\cdot (4188.7902-3591.364)\doteq 4659.9244\text{ g } \\ {\rho }_2={\displaystyle \frac{m}{V_1}}={\displaystyle \frac{4659.9244}{4188.7902}}\doteq 1.1125{\text{g/cm}}^3 \\ {\rho }_2>v \\ x=0 \end{gathered}$$

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