Question

如下图所示，两个顶点朝下的直圆锥中装有相同体积的液体。两个液面的顶部半径分别为 $3$ cm 和 $6$ cm。向每个圆锥中投入一个半径为 $1$ cm 的球形弹珠，弹珠沉到底部且完全浸没，并且没有液体溢出。窄圆锥中液面上升高度与宽圆锥中液面上升高度的比是多少？

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?

$1 : 1$

$47 : 43$

$2 : 1$

$40 : 13$

$4 : 1$

答案：E

知识点：圆锥相似长度、面积与体积的缩放关系

难度评级：1750

解答：

每个圆锥中的液体形成一个与容器相似的小圆锥。设窄圆锥中液体的半径为 $3$ 、高度为 $h_1$ ，宽圆锥中液体的半径为 $6$ 、高度为 $h_2$ 。体积相等给出 $\tfrac13\pi\cdot 9\cdot h_1 = \tfrac13\pi\cdot 36\cdot h_2$ ，所以 $h_1 = 4h_2$ 。

投入弹珠后，每个圆锥中的体积都增加相同的 $\Delta V = \tfrac43\pi$ 且二者起始体积都是 $V$ 。因为圆锥体积随高度的三次方缩放，新高度为 $h\sqrt[3]{1 + \Delta V/V}$ ，所以上升量为 $h\left(\sqrt[3]{1 + \Delta V/V} - 1\right)$ 。这个因子对两个圆锥相同，因此上升量之比为 $h_1 : h_2 = 4 : 1$ 。

因此，正确答案是 E。

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 第 10 题

答案：E

难度评级：1750

解答：

其他年份的第 10 题