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：圆锥相似体积

难度评级：1660

视频讲解：

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文字解答：

设窄、宽圆锥的初始液面高度分别为 $h_1$ 、 $h_2$ 。

$\frac13\pi(3)^2h_1=\frac13\pi(6)^2h_2,$

由体积相等，所以 $h_1=4h_2$ 。

投入相同弹珠后，两个液面以下的最终体积也相等。若新液面半径分别为 $3x$ 和 $6y$ ，相似性给出新高度 $h_1x$ 和 $h_2y$ 。

$\frac13\pi(3x)^2h_1x=\frac13\pi(6y)^2h_2y.$

利用 $h_1=4h_2$ ，可化简为 $x^3=y^3$ ，所以 $x=y$ 。

$\begin{aligned} h_1(x-1):h_2(y-1) &= h_1:h_2 \\ &= 4:1. \end{aligned}$

所以正确答案是 E。

Let the initial liquid heights in the narrow and wide cones be $h_1$ and $h_2.$ Since the liquid volumes are equal,

$\frac13\pi(3)^2h_1=\frac13\pi(6)^2h_2,$

so $h_1=4h_2.$

After the identical marbles are dropped in, each cone must contain the same final volume below the liquid surface: the original liquid volume plus the volume of one marble. If the new liquid-surface radii are $3x$ and $6y,$ similarity gives new heights $h_1x$ and $h_2y.$ Thus

$\frac13\pi(3x)^2h_1x=\frac13\pi(6y)^2h_2y.$

Using $h_1=4h_2,$ this simplifies to $x^3=y^3,$ so $x=y.$ The rise ratio is therefore

$\begin{aligned} h_1(x-1):h_2(y-1) &= h_1:h_2 \\ &= 4:1. \end{aligned}$

Thus, E is the correct answer.

2021 AMC 10A Spring 第 12 题

答案：E

难度评级：1660

视频讲解：

文字解答：

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