To solve this problem it is necessary to apply the concepts related to density, such as the relationship between density and Volume.
The volume of a sphere can be expressed as
[tex]V = \frac{4}{3} \pi r^3[/tex]
Here r is the radius of the sphere and V is the volume of Sphere
Using the expression of the density we know that
[tex]\rho = \frac{m}{V} \rightarrow V = \frac{m}{\rho}[/tex]
The density is given as
[tex]\rho = (19.5g/cm^3)(\frac{10^3kg/m^3}{1g/cm^3})[/tex]
[tex]\rho = 19.5*10^3kg/m^3[/tex]
Now replacing the mass given and the actual density we have that the volume is
[tex]V = \frac{60kg}{19.5*10^3kg/m^3 }[/tex]
[tex]V = 3.0769*10^{-3} m ^3[/tex]
The radius then is,
[tex]V = \frac{4}{3} \pi r^3[/tex]
[tex]r = \sqrt[3]{\frac{3V}{4\pi}}[/tex]
Replacing,
[tex]r = \sqrt[3]{\frac{3(3.0769*10^{-3})}{4\pi}}[/tex]
The radius of a sphere made of this material that has a critical mass is 9.02 cm.
