If the radius of a sphere is doubled, what is the ratio of the volume of the first sphere to that of the second sphere?

Volume of first sphere = 4×r³×π / 3 = V₁
Volume of second sphere = 4×(2r)³×π / 3 = 4×8r³×π / 3 = V₂
Ratio = V₁ / V₂
        = (4×r³×π / 3) / (4×8r³×π / 3)
        = 4×r³×π / 4×8r³×π
        = r³ / 8r³
        = 1 / 8
        = 1:8
As it turns out, if the radius is increased by n (2 in this case), the volume of the sphere is increased by n³ (2³=8)

 

Volume of a sphere = (\(\frac{4}{3}\))πr³ 

Let the radius of first sphere be 'r’. 

Radius of 2^nd sphere = 2r 

Ratio of the volume of the first sphere to that of the second sphere = \(\frac{\frac{4}{3}\pi r^3}{\frac{4}{3}\pi(2r)^3}\) = 1 : 8