Answer:
Their difference will be 7 times the original volume.
Step-by-step explanation:
Let the original radius of the sphere be R
Then, the original volume say V is:
[tex]V = \frac{4}{3} \pi R^{3}[/tex]
Let the new radius be R',
Then, the new volume say V' is:
[tex]V' = \frac{4}{3} \pi R' ^{3}[/tex]
Now, R' = 2R (given)
So,
[tex]V' = \frac{4}{3} \pi (2R) ^{3} = \frac{4}{3} \pi 8R ^{3}[/tex]
Now,
[tex]V' - V = \frac{4}{3} \pi 8R ^{3} - \frac{4}{3} \pi R ^{3}\\ \\V' - V = \frac{4}{3} \pi R ^{3} (8 - 1)\\ \\V' - V = 7 (\frac{4}{3} \pi R ^{3} )\\ \\[/tex]
But [tex]V = \frac{4}{3} \pi R^{3}[/tex]
So V' - V = 7V