A circular paper disk of radius $5$ is cut along two radii into two sectors with central angles $144^\circ$ and $216^\circ$. Each sector is rolled up (joining its two straight edges) to form the lateral surface of a cone. What is the ratio of the volume of the smaller cone to the volume of the larger cone?

1. The arc of each sector becomes the base circumference. The base radii are $r_1=\frac{144}{360}\cdot5=2$ and $r_2=\frac{216}{360}\cdot5=3$; both cones have slant height $5$.
2. The heights are $h_1=\sqrt{25-4}=\sqrt{21}$ and $h_2=\sqrt{25-9}=4$.
3. $\dfrac{V_1}{V_2}=\dfrac{r_1^2h_1}{r_2^2h_2}=\dfrac{4\sqrt{21}}{9\cdot4}=\boxed{\dfrac{\sqrt{21}}9}$.