The radius of the base of a right circular cone is $r=50 \mathrm{~cm}$. What is the radius of a sphere whose surface area is equal to the surface area of the cone; and what is the ratio of the volume of this sphere to the volume of the cone?
The surface area of the cone

$$
S=r^{2} \pi+2 r^{2} \pi=3 r^{2} \pi
$$

For the radius of the sphere, denoted by $R$, we have:

$$
\begin{aligned}
4 R^{2} \pi & =3 r^{2} \pi \\
R=\frac{r}{2} \sqrt{3} & =25 \sqrt{3} \text{ cm}
\end{aligned}
$$

The volume of the cone

$$
V=\frac{1}{3} r^{2} \pi h
$$

where

$$
\begin{gathered}
h^{2}=4 r^{2}-r^{2}=3 r^{2} \\
h=r \sqrt{3}
\end{gathered}
$$

Thus, we get

$$
V=\frac{\sqrt{3}}{3} r^{3} \pi
$$

while the volume of the sphere is

$$
V^{\prime}=\frac{4}{3} R^{3} \pi=\frac{4}{3} \frac{r^{3}}{8} 3 \sqrt{3} \pi=\frac{\sqrt{3}}{2} r^{3} \pi
$$

Therefore,

$$
\begin{gathered}
V: V^{\prime}=\frac{\sqrt{3} r^{3} \pi}{3}: \frac{\sqrt{3} r^{3} \pi}{2} \\
V: V^{\prime}=\frac{1}{3}: \frac{1}{2} \\
V: V^{\prime}=2: 3
\end{gathered}
$$

(Mihálovits Alajos, high school teacher, Kis-Kun-Félegyháza).

The problem was also solved by: Holbay Pál, high school VIII. class student, Nyitra.