How Tall is the Cone?

Let the distance from the cone vertex to the water surface be $x$ . Then the volume of the cone of height $x$ is directly proportional to $x^3$ . That is $V \propto x^3$ or $V = kv^3$ , where $k$ is a constant. Let the height of the conical bottle be $h$ .

Then, the volume of water on flat base is $V_w = kh^3 - k(8^3)$ , and that when the bottle is turned upside down $V_w = k(h-2)^3$ . Therefore,

$\begin{aligned} kh^3 - k(8^3) & = k(h-2)^3 & \small \color{#3D99F6} \text{Divide both sides by }k \\ h^3 - 512 & = h^3 - 6h^2 + 12h - 8 & \small \color{#3D99F6} \text{Rearrange} \\ 6h^2 - 12h - 504 & = 0 & \small \color{#3D99F6} \text{Divide both sides by }6 \\ h^2 - 2h - 84 & = 0 & \small \color{#3D99F6} \text{Solving the quadratic for }h \\ h & = \frac {2+\sqrt{340}}2 & \small \color{#3D99F6} \text{Note that }h >0 \\ & \approx \boxed{10.2195} \end{aligned}$

Let the conical bottle have a radius of $r$ and a height of $h$ . The volume of a cone is $V = \frac{1}{3}\pi r^2h$ .

The volume of the water in the right side up bottle is the difference of the volume of the whole cone and the volume of the cone of air (with a radius of $r_a$ and a height of $8$ ), so $V_w = \frac{1}{3}\pi r^2 h - \frac{1}{3}\pi r_a^2 8$ . Since these two cones are similar, $\frac{r_a}{r} = \frac{8}{h}$ , and substituting this gives $V_w = \frac{1}{3}\pi r^2 h - \frac{1}{3}\pi (\frac{8r}{h})^2 8$ .

The volume of the water in the upside down bottle is a cone (with a radius of $r_b$ and a height of $h - 2$ ), so $V_w = \frac{1}{3}\pi r_b^2 (h - 2)$ . Since this cone is similar to the conical bottle, $\frac{r_b}{r} = \frac{h - 2}{h}$ , and substituting this gives $V_w = \frac{1}{3}\pi (\frac{(h - 2)r}{h})^2 (h - 2)$ .

Equating the two equations gives $V_w = \frac{1}{3}\pi r^2 h - \frac{1}{3}\pi (\frac{8r}{h})^2 8 = \frac{1}{3}\pi (\frac{(h - 2)r}{h})^2 (h - 2)$ , which simplifies to $h^3 - 512 = (h - 2)^3$ , and solves to a positive solution of $h = 1 + \sqrt{85} \approx \boxed{10.2195}$ .

Since a cone's volume is proportional to its height cubed,
$\begin{aligned} h^3-8^3&=(h-2)^3 \\ h&=1+\sqrt{85} \\ &\approx10.2195 \text{ cm} \end{aligned}$

Key is that radius is propotional to height from the top (vertex). Setup the equation

h^3 - 8^3 = (h-2)^3 and solve.

Answer is h = 1+√85=10.219

Just take the volume of each cone and make them equal each other as they are the same cone. Then you get

$h^3 - 8^3 = (h-2)^3$

Solving for $h$ you will get approximately 10.22

Let the volume of cone is v, and the volume of water w, and the height of the cone is h.Using similarity : (8/h)^3 = (v – w)/v, or (8/h)^3 = 1 – w/v, ....(1), for the left figure,and [(h – 2)/h]^3 = w/v .......(2), for right figure. Substitute eqs.(2) to eqs.(1), then we get, (8/h)^3 = 1 – [(h – 2)/h]^3, 8^3+(h-2)^3/h^3 = 1, h^2 – 3 h – 84 = 0, solving the quadratic equation, we get h=10.2195

1+sqrt(85)

I just added 8+2 and other people come up with these complicated equations xD

Total cone volume=1/3 pi r^2(2+h)-----(1) Total cone volume=1/3 pi r^2(8+h1)-------(2) ratio=1 divide 1 &2 then h+2-h1-8=0 @ h=8 then h1=2 total height = 2+h=2+8=10 cm.