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<mdoxml version="1.0"><p>A right circular cone is filled with liquid to a depth of half its vertical height.</p>
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The cone is inverted. How high up the vertical height of the cone will the liquid rise?</mdoxml>
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<p class="editorial">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.</p>
<p class="editorial">The quick method is to look at the scale
factor.</p>
<p>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 $2^3=8$</p>
<p>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.</p>
<p>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: $$
(\frac{h}{x})^3 =\frac{7}{8}$$</p>
<p>and so $$h = { \frac{\sqrt[3] 7x}{2}}$$</p>
<p>That is, the height of the liquid in the upturned cone is $\frac
{\sqrt[3] 7} {2}$ or 0.9565 of the original height</p>
<p><span class="editorial">Other method is to equate the 2 results
for the volume of the liquid.</span></p>
<p>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.</p>
<p>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 $$ \pi/3 \times (\frac
{rh}{x})^2 \times h (**) $$</p>
<p>(*) and (**) were thus equated and the height of the liquid in
the upturned cone was found, by cancelling, to be:$$h = {
\frac{\sqrt[3] 7x}{2}}$$</p>
<p>or approximately $0.9565$ of the original height.</p>
<p class="editorial">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</p>
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<mdoxml version="1.0">This problem stimulates students to use their understanding of the
relationship between ratio for lengths and ratio for volumes.<br></br></mdoxml>
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<mdoxml version="1.0">What fraction of liquid is present compared to a completely full
cone? <br></br></mdoxml>
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Conical Bottle
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?
Transformations and their Properties
Scale factors
Fractions, Decimals, Percentages, Ratio and Proportion
Calculating with ratio & proportion
3D Geometry, Shape and Space
Cones
Measures and Mensuration
Volume and capacity