一个封闭的圆锥形装饰物，其圆形底面半径为 10 cm，高为 25 cm，内部装有深度为 15 cm 的水银，如下图所示。若将该装饰物倒置，求水银的新深度 $h$ cm。

1. Find the radius of the mercury surface before inversion using similar triangles.

In the cone, radius is proportional to distance from the vertex.

$\frac{r}{15}=\frac{10}{25}$

$r=15\times \frac{10}{25}=6\text{ cm}$

2. Find the original volume of mercury.

The mercury initially forms a smaller cone of radius $6\text{ cm}$ and height $15\text{ cm}$.

$V_{\text{mercury}}=\frac{1}{3}\pi r^2h$

$V_{\text{mercury}}=\frac{1}{3}\pi(6)^2(15)=180\pi\text{ cm}^3$

3. Express the volume of mercury after inversion in terms of $h$.

After inversion, the mercury occupies the bottom part of the cone.

The empty part above the mercury is a smaller cone of height $25-h$.

Its radius is

$\frac{2}{5}(25-h).$

So the volume of mercury is

$\frac{1}{3}\pi(10)^2(25)-\frac{1}{3}\pi\left[\frac{2}{5}(25-h)\right]^2(25-h).$

4. Form and solve the equation using conservation of volume.

Since the volume of mercury is unchanged,

$\frac{1}{3}\pi(10)^2(25)-\frac{1}{3}\pi\left[\frac{2}{5}(25-h)\right]^2(25-h)=180\pi$

$\frac{2500}{3}\pi-\frac{4}{75}\pi(25-h)^3=180\pi$

Divide by $\pi$:

$\frac{2500}{3}-\frac{4}{75}(25-h)^3=180$

$\frac{4}{75}(25-h)^3=\frac{1960}{3}$

$(25-h)^3=12250$

$25-h=\sqrt[3]{12250}=23.052\ldots$

$h=25-23.052\ldots=1.947\ldots$