Answer: D
Let’s assume the initial radius was \(r\) and now it is \(2r\). The initial height was \(h\) and now it is \(2h\). Let the new volume be \(V\) and the previous volume be \(v\).
\(\therefore\) \(\cfrac{V}{v}=\cfrac{\cfrac{1}{3}\pi (2r)^2.2h}{\cfrac{1}{3}\pi (r)^2.h}=\cfrac{8r^2h}{r^2h}=\cfrac{8}{1}\)
\(\therefore\) \(V=8\times v\)
So, the new volume is 8 times the original volume.
Let’s assume the initial radius was \(r\) and now it is \(2r\). The initial height was \(h\) and now it is \(2h\). Let the new volume be \(V\) and the previous volume be \(v\).
\(\therefore\) \(\cfrac{V}{v}=\cfrac{\cfrac{1}{3}\pi (2r)^2.2h}{\cfrac{1}{3}\pi (r)^2.h}=\cfrac{8r^2h}{r^2h}=\cfrac{8}{1}\)
\(\therefore\) \(V=8\times v\)
So, the new volume is 8 times the original volume.