Mensuration Practice Test 2

Let the radius of the cone = r
Height of the cone = h
According to the question,
New radius = 2 r
New height = $\frac{\mathit{h}}{2}$
Initial Volume = $\frac{1}{3}\mathit{\pi }{\mathit{r}}^2\mathit{h}$
New volume = $\frac{1}{3}\mathit{\pi }{\left(2\mathit{r}\right)}^2\frac{\mathit{h}}{2}$ = $\frac{2}{3}\mathit{\pi }{\mathit{r}}^2\mathit{h}$
Increase in volume = $\frac{2}{3}\mathit{\pi }{\mathit{r}}^2\mathit{h}-\frac{1}{3}\mathit{\pi }{\mathit{r}}^{\mathit{h}}$ = $\mathit{\pi }{\mathit{r}}^2\mathit{h}$
Percentage increase = $\frac{\mathit{\pi }{\mathit{r}}^2\mathit{h}}{/3\mathit{\pi }{\mathit{r}}^2\mathit{h}}\times 100$ = 300 %