Question

Two right circular cones with vertices facing down as shown in the figure below contain the same amount of liquid. The radii of the tops of the liquid surfaces are $3$ cm and $6$ cm. Into each cone is dropped a spherical marble of radius $1$ cm, which sinks to the bottom and is completely submerged without spilling any liquid. What is the ratio of the rise of the liquid level in the narrow cone to the rise of the liquid level in the wide cone?

$1 : 1$

$47 : 43$

$2 : 1$

$40 : 13$

$4 : 1$

Solution:

The liquid in each cone forms a smaller cone similar to the container. Let the narrow liquid cone have radius $3$ and height $h_1,$ and the wide one radius $6$ and height $h_2.$ Equal volumes give $\tfrac13\pi\cdot 9\cdot h_1 = \tfrac13\pi\cdot 36\cdot h_2,$ so $h_1 = 4h_2.$

Dropping the marble raises the volume by the same amount $\Delta V = \tfrac43\pi$ in each cone, and both start with the same volume $V.$ Because a cone's volume scales as the cube of its height, the new height is $h\sqrt[3]{1 + \Delta V/V},$ so each rise equals $h\left(\sqrt[3]{1 + \Delta V/V} - 1\right).$ This factor is identical for the two cones, so the rises are in the ratio $h_1 : h_2 = 4 : 1.$

Thus, the correct answer is E.

2021 AMC 12A Spring Problem 10

Solution:

Problem 10 in Other Years