Mensuration Practice Test 3

Let the radius of the 1st cone = 2r
Radius of the 2nd cone = r
Height of the 1st cone and the 2nd cone = h and H
$\frac{\mathrm{Volume}\mathrm{of}\mathrm{them}1\mathrm{st}\mathrm{cone}}{\mathrm{Volume}\mathrm{of}\mathrm{the}2\mathrm{nd}\mathrm{cone}}$ = $\frac{\frac{1}{3}\mathit{\pi }{\left(2\mathit{r}\right)}^2\mathit{h}}{\frac{1}{3}\mathit{\pi }\left({\mathit{r}}^3\right)\mathit{H}}$
$\frac{4}{3}$ = $\frac{4{\mathit{r}}^2\mathit{h}}{{\mathit{r}}^2\mathit{H}}$
$\frac{\mathit{h}}{\mathit{H}}$ = $\frac{1}{3}$
So, the required ratio of heights = 1 : 3