# Iron ball

The iron ball has a weight of 100 kilograms. Calculate the volume, radius, and surface if the iron's density is h = 7.6g/cm3.

Correct result:

V =  13.158 dm3
r =  1.465 dm
S =  26.97 dm2

#### Solution:

$m=100 \ \text{kg} \ \\ h=7600 \ \text{kg/m}^3 \ \\ \ \\ m=h V \ \\ \ \\ V_{1}=m/h=100/7600 \doteq \dfrac{ 1 }{ 76 } \doteq 0.0132 \ \text{m}^3 \ \\ \ \\ V=V_{1} \rightarrow dm^3=V_{1} \cdot \ 1000 \ dm^3=0.0131578947368 \cdot \ 1000 \ dm^3=13.158 \ dm^3=13.158 \ \text{dm}^3$
$V=\dfrac{ 4 }{ 3 } \cdot \ \pi \cdot \ r^3 \ \\ \ \\ r=\sqrt[3]{ \dfrac{ V \cdot \ 3 }{ 4 \pi } }=\sqrt[3]{ \dfrac{ 13.1579 \cdot \ 3 }{ 4 \cdot \ 3.1416 } }=1.465 \ \text{dm}$
$S=4 \pi \cdot \ r^2=4 \cdot \ 3.1416 \cdot \ 1.4645^2=26.97 \ \text{dm}^2$

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I think it’s wrong

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