11.103. A metal sphere of radius $R$ is recast into a cone, the lateral surface area of which is three times the area of the base. Calculate the height of the cone.
Solution.

Since the radius of the sphere is $R$, the volumes of the bodies are $\frac{4}{3} \pi R^{3}$. Let $r$ be the radius of the base of the cone; then, by the condition, $S_{\text {side }}=3 \pi r^{2}$. On the other hand, $S_{\text {side }}=\pi r l$, where $l$ is the slant height of the cone. Therefore, $l=3 r$, from which $h=\sqrt{l^{2}-r^{2}}=2 r \sqrt{2}$, i.e., $r=\frac{\sqrt{2}}{4} h$. We find $V=\frac{1}{3} \pi r^{2} h=\frac{1}{3} \pi \cdot \frac{2}{16} h^{2} \cdot h=\frac{1}{24} \pi h^{3}$. From the equality $\frac{4}{3} \pi R^{3}=\frac{1}{24} \pi h^{3}$, we get $h=2 R \sqrt[3]{4}$.

Answer: $2 R \sqrt[3]{4}$.