The resultant density of the sphere when its radius is halved is eight times the initial density of the sphere.
Let the radius of the sphere = r
The volume of the sphere = [tex]\frac{4}{3} \pi r^{3}[/tex]
Mass of the sphere = M
Now, The density of the sphere = D = Mass/Volume
So, The density will be,
[tex]D = \frac{M}{\frac{4}{3} \pi r^{3} } = \frac{3M}{4\pi r^{3} }[/tex]
It is given that the radius of the sphere is halved, then
[tex]D' = \frac{M}{\frac{4}{3} \pi (\frac{r}{2} )^{3} } = \frac{24M}{4\pi r^{3} } = \frac{6M}{\pi r^{3} }[/tex]
Now, Dividing both the densities, D and D', we get
[tex]\frac{D}{D'} = \frac{1}{8}\\ \\D' = 8D[/tex]
Hence, The resultant density of the sphere when its radius is halved is eight times the initial density of the sphere.
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