Question - Determining the Height of the Water Level in an Inverted Cone

Let the original height of the cone be \( h \) and the radius be \( r \).

When the cone is half-filled with water, the volume \( V \) of the water is \( \frac{1}{3}\pi\left(\frac{r}{2}\right)^2\left(\frac{h}{2}\right) \) because the height of the water is \( \frac{h}{2} \).

When the cone is inverted, let \( x \) be the height of the water level in terms of \( h \), and \( r' \) be the new radius of the water level.

By similar triangles, \( \frac{r'}{x} = \frac{r}{h} \rightarrow r' = \frac{rx}{h} \).

The volume of the water remains constant, so \( V = \frac{1}{3}\pi r'^2 x = \frac{1}{3}\pi \left(\frac{rx}{h}\right)^2 x \).

Equating the two expressions for \( V \), we get \( \frac{1}{3}\pi\left(\frac{r}{2}\right)^2\left(\frac{h}{2}\right) = \frac{1}{3}\pi \left(\frac{rx}{h}\right)^2 x \).

Simplify to find \( x \):

\( \left(\frac{r}{2}\right)^2\left(\frac{h}{2}\right) = \left(\frac{rx}{h}\right)^2 x \).

\( \frac{r^2 h}{4} = \frac{r^2 x^3}{h^2} \).

Multiply both sides by \( h^2 \):

\( h^3 = 4x^3 \).

Take the cube root of both sides to solve for \( x \):

\( x = \frac{h}{\sqrt[3]{4}} \).

Therefore, the height of the water level \( x \) in terms of the original height \( h \) when the cone is inverted is \( x = \frac{h}{\sqrt[3]{4}} \).