# 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. )

Result

x =  0

#### Solution:

$d_{ 1 } = 20 \ cm \ \\ d_{ 2 } = 19 \ cm \ \\ ρ = 7.8 \ g/cm^3 \ \\ v = 1 \ g/cm^3 \ \\ \ \\ r_{ 1 } = d_{ 1 }/2 = 20/2 = 10 \ cm \ \\ r_{ 2 } = d_{ 2 }/2 = 19/2 = \dfrac{ 19 }{ 2 } = 9.5 \ cm \ \\ \ \\ V_{ 1 } = \dfrac{ 4 }{ 3 } \cdot \ \pi \cdot \ r_{ 1 }^3 = \dfrac{ 4 }{ 3 } \cdot \ 3.1416 \cdot \ 10^3 \doteq 4188.7902 \ cm^3 \ \\ V_{ 2 } = \dfrac{ 4 }{ 3 } \cdot \ \pi \cdot \ r_{ 2 }^3 = \dfrac{ 4 }{ 3 } \cdot \ 3.1416 \cdot \ 9.5^3 \doteq 3591.364 \ cm^3 \ \\ \ \\ \ \\ m = ρ \cdot \ (V_{ 1 }-V_{ 2 }) = 7.8 \cdot \ (4188.7902-3591.364) \doteq 4659.9244 \ g \ \\ \ \\ ρ_{2} = \dfrac{ m }{ V_{ 1 } } = \dfrac{ 4659.9244 }{ 4188.7902 } \doteq 1.1125 \ g/cm^3 \ \\ \ \\ ρ_{2} > v \ \\ \ \\ x = 0$

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