Answer: D
Let the original radius be \(r\), and the original height be \(h\). Now, the new radius is \(2r\) and the new height is \(2h\). Let the original volume be \(v\) and the new volume be \(V\)
\(\therefore\) \(\cfrac{V}{v}=\cfrac{\cfrac{1}{3}\pi (2r)^2.2h}{\cfrac{1}{3}\pi (r)^2.h}=\cfrac{8r^2h}{r^2h}=\cfrac{8}{1}\)
\(\therefore\) \(V=8\times v\)
Therefore, the volume of the new cone is 8 times the volume of the original cone.
Let the original radius be \(r\), and the original height be \(h\). Now, the new radius is \(2r\) and the new height is \(2h\). Let the original volume be \(v\) and the new volume be \(V\)
\(\therefore\) \(\cfrac{V}{v}=\cfrac{\cfrac{1}{3}\pi (2r)^2.2h}{\cfrac{1}{3}\pi (r)^2.h}=\cfrac{8r^2h}{r^2h}=\cfrac{8}{1}\)
\(\therefore\) \(V=8\times v\)
Therefore, the volume of the new cone is 8 times the volume of the original cone.