We fill a cone (with its vertex pointing downwards) with water, so the water stands $10 \mathrm{~cm}$ high inside it. After closing the opening, we place the cone on its base, and now the water stands $2 \mathrm{~cm}$ high. What is the height of the cone?
In the first case, the volume of water $V_{1}$ is equal to the volume of a cone whose height is $10 \mathrm{~cm}$. Let's denote the height of the original cone by $m$ and its volume by $V$.

It is known that $\frac{V_{1}}{V}=\frac{10^{3}}{m^{3}}$, hence $V_{1}=V \frac{10^{3}}{m^{3}}$.

In the second case, the volume of the empty part of the cone is $V_{2}$, and $\frac{V_{2}}{V}=\frac{(m-2)^{3}}{m^{3}}$. Since the volume of water has not changed in between, it is clear that $V_{1}+V_{2}=V$, that is,

$$
V \frac{10^{3}}{m^{3}}+V \frac{(m-2)^{3}}{m^{3}}=V
$$

Simplifying the equation by $V$ (since $V \neq 0$) and rearranging, we get

$$
6 m^{2}-12 m-992=0
$$

The product of the two real roots of the equation is negative, so we must consider the unique positive root. The height of the cone is $13.897 \mathrm{~cm}$.
![](https://cdn.mathpix.com/cropped/2024_05_02_9f828248278c32fca8b8g-1.jpg?height=400&width=822&top_left_y=764&top_left_x=650)