2021 AMC 12A Spring Problem 10

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

Answer: E

Concepts:cone similarity power scaling of length, area, and volume

Difficulty rating: 1750

Small hint:

The liquid in each cone is itself a cone; with equal starting volumes, the narrow cone (radius

3

) is four times as tall as the wide cone (radius

6

)

Big hint:

Adding the same marble volume scales the height by the same factor

\sqrt[3]{1 + \frac{\Delta V}{V}}

in both cones, so the rises are in the ratio of the original heights

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 + \frac{\Delta V}{V}},

so each rise equals

h\left(\sqrt[3]{1 + \frac{\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.