Conical Bottle
Problem
A right circular cone is filled with liquid to a depth of half its vertical height.
The cone is inverted. How high up the vertical height of the cone will the liquid rise?
Getting Started
Student Solutions
Most people start out by calculating the volume of liquid. As with many mathematical tasks some thought in advance may save a lot of work . Failing that, if you review your method you may find a neater and more efficient way to do it. Try to evaluate your own work, think about it and ask yourself questions like: "what is the key issue here?", "does my answer suggest a connection in the problem I did not use?","have I done it the best way?" . Very often the best approach leads to a really pretty bit of maths.
The quick method is to look at the scale factor.
Consider the enlargement of the conical space above the liquid to the whole cone. The scale factor is $2$ (linearly) so the volume scale factor is $2^3=8$
The space above the liquid has an eighth of the volume of the whole cone and the liquid takes up seven eighths of the volume.
When it is inverted the volume of water is still seven eighths of the volume of the cone so we use the fact again that the cube of the linear scale factor gives the volume scale factor to get: $$ (\frac{h}{x})^3 =\frac{7}{8}$$
and so $$h = { \frac{\sqrt[3] 7x}{2}}$$
That is, the height of the liquid in the upturned cone is $\frac {\sqrt[3] 7} {2}$ or 0.9565 of the original height
Other method is to equate the 2 results for the volume of the liquid.
No-one had any trouble in showing that the volume of liquid was $7 \frac{\pi r^2x}{24}$ (*) where $r$ was the base radius and x the height of the cone.
All used the properties of similar triangles to find the radius of the base of liquid in the upturned cone, which is (rh/x). Hence the volume of liquid in the upturned cone is $$ \pi/3 \times (\frac {rh}{x})^2 \times h (**) $$
(*) and (**) were thus equated and the height of the liquid in the upturned cone was found, by cancelling, to be:$$h = { \frac{\sqrt[3] 7x}{2}}$$
or approximately $0.9565$ of the original height.
Easy to follow solutions to this problem were received from: Sam, Jonathan and Kevin, Tom, Euan, Michael and James of Madras College and Moray and Richard of Wellingborough School