# Iron sphere

Iron sphere has weight 100 kg and density ρ = 7600 kg/m3. Calculate the volume, surface and diameter of the sphere.

Result

V =  13.158 dm3
D =  2.929 dm
S =  26.953 dm2

#### Solution:

$m = 100 \ kg \ \\ h = 7600 \ kg/m^3 \ \\ m = hV \ \\ V_{ 1 } = \dfrac{ m }{ h } = \dfrac{ 100 }{ 7600 } = \dfrac{ 1 }{ 76 } \doteq 0.0132 \ m_{ 3 } \ \\ V = V_{ 1 } \rightarrow dm^3 = V_{ 1 } \cdot \ 1000 \ dm^3 = 13.1578947368 \ dm^3 = 13.158 \ dm^3$
$V = \dfrac{ 4 }{ 3 } \pi r^3 \ \\ r = \sqrt{ V \cdot \ \dfrac{ 3 }{ 4 \pi } } = \sqrt{ 13.1579 \cdot \ \dfrac{ 3 }{ 4 \cdot \ 3.1416 } } \doteq 1.4645 \ dm \ \\ D = 2 \cdot \ r = 2 \cdot \ 1.4645 \doteq 2.9291 = 2.929 \ \text { dm }$
$S = 4 \pi \cdot \ r^2 = 4 \cdot \ 3.1416 \cdot \ 1.4645^2 \doteq 26.9532 = 26.953 \ dm^2$

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