How high must I fill?

Extend the sides of the frustrum as follows. Then the larger cone and the smaller cone are similar by $AA$ . By similar triangles, $\frac{2}{h} = \frac{4}{h+10} \Rightarrow h = 10$ .

Now, $V \propto L^3$ , so the volume is proportional to any of the lengths cubed. Let the volume of the entire cone be $V$ . Then the volume of the frustrum can be expressed as (large cone - small cone), and the volume of the small cone can be represented as $\left(\frac{\text{small cone radius}}{\text{large cone radius}}\right)^3 V$ : a proportion of the large cone's volume. Thus the frustrum's volume is $V - \left(\frac{2}{4}\right)^3 V = \frac{7}{8}V$ , so the volume of the water is $\frac{7}{16}V$ .

We use similarity again but in terms of the heights. $\left(\frac{10+h}{20}\right)^3 V - \left(\frac{10}{20}\right)^3 V = \frac{7}{16}V$ . Cancelling the $V$ s and rearranging gives $\left(\frac{10+h}{20}\right)^3 = \frac{7}{16} + \frac{1}{8}$ , and so $10+h = 20 \left( \frac{7}{16} + \frac{1}{8} \right)^{1/3} \Rightarrow h \approx 6.5096$ .

Extend the sides of the frustum to form a triangle at the bottom (and hence a right circular cone). If H is the height of this triangle, similar triangles shows:

2/H = 4 / (H+10)

Solving yields

H = 10.

We then have two cones formed (a "big" one with radius 4 and height 20 and a "small" one with radius 2 and height 10). Utilizing the formula for the volume of a right circular cone:

Volume of frustum = Volume of "big" right cone - volume of "small" cone = (Pi/3) (4^2) 20 - (Pi/3) (2^2) 10

Volume of frustum = 280*Pi/3 (equation 1)

After the water is filled, let the radius be r and the height of the water above the bottom of the frustum be h. Using similar triangles again, we find:

r/(h+10) = 2/10, or

h = 5*r-10 (equation 2)

Again, subtracting volumes yields:

Volume of water = Pi (r^2) (h+10) / 3 - (Pi/3) (2) (2) 10 = (1/2) 280*Pi/3 (the last equation follows from equation 1 and the fact that the volume of water is 1/2 the volume of the frustum).

Now, substitution equation 2 into this and simplifying yields:

5*r^3 = 180

or, r = (36)^(1/3) and h = 5*(36^(1/3)) - 10 = 6.5096 (approximately)

Let the bottom radius, the top radius and the height of the conical frustum be $r, R, H$ respectively, and let the required height be $H'$ .

Then,

$H'=H\times \dfrac {\left (\frac {R^3+r^3}{2}\right )^{\frac 13}-r}{R-r}$

Substituting values we get

$H'=5\times \left (36^{\frac 13}-2\right )\approx \boxed {6.5096}$ .