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I have a question that I have been puzzling over for a couple of hours now, but I can't seem to understand.

"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 would the liquid rise?"

That's it. No other numbers specified.

Any help would be appreciated. Thanks.

FruitBangSkoda
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  • The volume of the liquid is $\frac{1}{8}$ of the volume of the cone. For which $x$ we have $(1-x)^3=\frac{7}{8}$? – Jack D'Aurizio Feb 16 '15 at 19:35
  • Ah, I didn't think of it in that way - the whole cone being an enlargement of the smaller cone. Thanks very much indeed. Much appreciated. – FruitBangSkoda Feb 16 '15 at 20:07

2 Answers2

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Let h be the height of the right circular cone and R be the radius.

Let r(x) be the radius at a cross-section of the cone where we assume x=0 is at the bottom. So, $r(0)=0, r(h)=R$. Thus, $r(x) = \frac Rh x$.

Then, the volume of the whole cone is $$V = \frac 13 \pi R^2 h.$$ Initially, the cone is only filled up to half its vertical height, so the volume of the liquid is $$V_0 = \frac 13 \pi (r(\frac h2))^2 h = \frac 13 \pi (\frac R2)^2 h$$ because it is only filled half.

Now, when you invert the cone, you want to find that height x for which $$V-\frac 13 \pi (r(x))^2 h = V_0.$$

That leads to $1-(\frac xh)^2 = \frac 14$ and $x=\frac{\sqrt 3}2 h$.

Remember $x=0$ is the bottom of the original cone. Thus, your answer will be that the inverted cone is filled up to a height of $1-\frac{\sqrt 3}2$ of the total vertical height.

user3117090
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Wrong! When radius $r(\frac h2)=\frac R2$, the height is not $h$! $$\text{Volume }= \frac13 \pi \left(\frac R2\right) ^2 \cdot \left(\frac h2\right) = \frac V8 \qquad\text{or}\qquad \frac V{2^3}.$$

Tianlalu
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corneliusgoh
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