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/cm³. (Instructions: Calculate the average sphere density and compare it with the water density. )

Correct answer:

x =  0

Step-by-step explanation:

$$\begin{gathered} d_1=20\text{cm} \\ {d}_2=19\text{cm} \\ \rho =7.8{\text{g/cm}}^3 \\ v=1{\text{g/cm}}^3 \\ {r}_1=d_1/2=20/2=10\text{cm} \\ {r}_2=d_2/2=19/2=\frac{19}{2}=9\frac{1}{2}=9.5\text{cm} \\ {V}_1=\frac{4}{3}\cdot \pi \cdot r_1^3=\frac{4}{3}\cdot 3.1416\cdot 10^3\doteq 4188.7902{\text{cm}}^3 \\ {V}_2=\frac{4}{3}\cdot \pi \cdot r_2^3=\frac{4}{3}\cdot 3.1416\cdot 9.5^3\doteq 3591.364{\text{cm}}^3 \\ m=\rho \cdot (V_1-V_2)=7.8\cdot (4188.7902-3591.364)\doteq 4659.9244\text{g} \\ {\rho }_2=\frac{m}{V_1}=\frac{4659.9244}{4188.7902}\doteq 1.1125{\text{g/cm}}^3 \\ {\rho }_2>v \\ x=0 \end{gathered}$$

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