Answer: D
Let the initial radius be \( r\), so the new radius is \(2r\) and Let the initial height be \(h\), so the new height is \(2h\) and
The new volume is \(V_1\)
∴\(\cfrac{V_1}{V}=\cfrac{\cfrac{1}{3} π(2r)^2 2h}{\cfrac{1}{3} πr^2 h}= \cfrac{8r^2 h}{r^2 h}=\cfrac{8}{1} \)
\(\therefore V_1=8V\)
Thus, the new volume is \(=8V\)
Let the initial radius be \( r\), so the new radius is \(2r\) and Let the initial height be \(h\), so the new height is \(2h\) and
The new volume is \(V_1\)
∴\(\cfrac{V_1}{V}=\cfrac{\cfrac{1}{3} π(2r)^2 2h}{\cfrac{1}{3} πr^2 h}= \cfrac{8r^2 h}{r^2 h}=\cfrac{8}{1} \)
\(\therefore V_1=8V\)
Thus, the new volume is \(=8V\)