If both the radius and height of a cone are doubled, the volume of the cone will increase by 8 times the original volume.
Let the initial radius be \(r\), now \(2r\), and the initial height be \(h\), now \(2h\). Let the new volume be \(V\) and the original volume be \(v\).
\(\therefore\) \(\cfrac{V}{v} = \cfrac{\cfrac{1}{3}\pi (2r)^2 \cdot 2h}{\cfrac{1}{3}\pi (r)^2 \cdot h} = \cfrac{8r^2h}{r^2h} = \cfrac{8}{1}\)
\(\therefore\) \(V = 8 \times v\)
[Proved]