Two cones have their heights in the ratio 1:3 and the radii of their bases in the ratio 3:1.

Solution:

Let ratio of the height of the cone be $h_{x}$

height of the $1^{\text {st }}$ cone $=h_{x}$

height of the of the $2^{\text {nd }}$ cone $=3 h_{x}$

Let the ratio of the radius of the of the cone = rx

radius of the $1^{\text {st }}$ cone $=3 r_{x}$

radius of the $2^{\text {nd }}$ cone $=r_{x}$

The ratio of the volume $=\mathrm{v}_{1} / \mathrm{v}_{2}$

Where $v_{1}=$ volume of $1^{\text {st }}$ cone

$v_{2}=$ volume of $2^{\text {nd }}$ cone

$\Rightarrow \frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\frac{\frac{1}{3} \pi * \mathrm{r}_{\mathrm{a}}^{2} \mathrm{~h}_{\mathrm{a}}}{\frac{1}{3} \pi \mathrm{r}_{\mathrm{b}}^{2} \mathrm{~h}_{\mathrm{b}}}$

$=\frac{r_{a}^{2} h_{a}}{r_{b^{2} h_{b}}}$

$=\frac{3 r_{\mathrm{x}}^{2} \mathrm{~h}_{\mathrm{x}}}{r_{\mathrm{x}}^{2} \mathrm{~h}_{\mathrm{x}}}$

$=\frac{9 \mathrm{r}_{\mathrm{x}}^{2} \mathrm{~h}_{\mathrm{x}}}{3 \mathrm{r}_{\mathrm{x}^{2} \mathrm{~h}_{\mathrm{x}}}}$

$=\frac{3}{1}$

$\Rightarrow \frac{v_{1}}{v_{2}}=\frac{3}{1}$