Two right circular cones with vertices facing down contain the same amount of liquid. The radii of the tops of the liquid surfaces are $4\text{ cm}$ and $12\text{ cm}$. Into each cone is dropped a spherical marble of radius $1\text{ cm}$, which sinks to the bottom, fully submerged, without spilling any liquid. What is the ratio of the rise of the liquid level in the narrow cone to that in the wide cone?

1. As in the problem, both depths grow by the same factor, so the ratio of the rises equals the ratio $h_1:h_2$ of the initial depths, independent of the marble's radius.
2. Equal volumes give $4^2h_1=12^2h_2$, so $h_1:h_2=9:1$, and the ratio of the rises is $9:1$.