Liquid in a cone

liquid in a cone
Solution
The volume of a cone is given by $V = \frac{1}{3}\pi r^2h$. If we half fill the cone, we have a mathematically similar cone of liquid, with volume $$ V_{liquid} = \frac{1}{3}(\alpha r)^2(\alpha h) = \frac{1}{2}V $$ It follows that $\alpha^3 = \frac{1}{2}$, so $\alpha = \sqrt[3]{\frac{1}{2}}$ The height of the liquid is therefore $H_{liquid} = \sqrt[3]{\frac{1}{2}}h$, or more helpfully, approximately 79% of the height of the original cone.