We use the formula,
[tex]density=\frac{mass}{volume}[/tex]
Given, [tex]mass=1.90\times 10^{27}\ kg = 1.90\times 10^{30}\ g[/tex] and [tex]density =1.34\ g/cm^3[/tex].
Substituting these values, we get
[tex]volume = \frac{1.90\times 10^{30}\ g}{1.34\ g/cm^3} =1.4179\times 10^{30}\ cm^3[/tex].
As Jupiter were a perfect sphere, therefore the volume of sphere is given by
[tex]volume=\frac{4}{3} \pi r^3[/tex]
Here, r is the radius of sphere.
Substituting the value of volume we get
[tex]1.4179\times 10^{30}\ cm^3=\frac{4}{3}\times 3.14\times r^3 \\\\ r^3=0.339\times 10^{30} \\\\r= 0.697\times 10^{10}\ cm[/tex].
The diameter is twice of radius, thus the diameter of Jupiter would be
[tex]2r=2\times 0.697\times 10^{10}\ cm=1.394\times 10^{10}\ cm[/tex]
