A closed conical ornament has a circular base of radius 10 cm and height 25 cm, and contains mercury to a depth of 15 cm as shown in the figure below. The ornament is then inverted. Find the new depth, $h$ cm, of the mercury. Answer $h$ = ... cm

1. Find the ratio of the volume of mercury to the total volume of the cone.

Let $V_{total}$ be the total volume of the cone and $V_{Hg}$ be the volume of the mercury.
Since the mercury forms a cone similar to the entire cone, the ratio of their volumes is the cube of the ratio of their heights:
$\frac{V_{Hg}}{V_{total}} = \left(\frac{15}{25}\right)^3 = \left(\frac{3}{5}\right)^3 = \frac{27}{125}$

2. Find the ratio of the volume of the empty space to the total volume when the cone is inverted.

When inverted, the empty space at the top forms a cone similar to the entire cone.
Let $V_{empty}$ be the volume of this empty space.
$\frac{V_{empty}}{V_{total}} = 1 - \frac{V_{Hg}}{V_{total}} = 1 - \frac{27}{125} = \frac{98}{125}$

3. Relate the volume ratio of the empty space to its height.

Let $h_{empty}$ be the height of the empty space cone. The height of the empty space is $25 - h$.
$\left(\frac{25 - h}{25}\right)^3 = \frac{98}{125}$

4. Solve for $h$.

$\frac{25 - h}{25} = \sqrt[3]{\frac{98}{125}} = \frac{\sqrt[3]{98}}{5}$
$25 - h = 5\sqrt[3]{98}$
$h = 25 - 5\sqrt[3]{98}$
$h \approx 25 - 5(4.6104) = 25 - 23.052 = 1.948 \text{ cm}$