We have threaded a steel ball into a V-shaped groove and filled the gap with paraffin. What is the radius of the cross-section of the gap if this ball floats in water. (The specific gravity of steel is 7.8, and that of paraffin is 0.2.)
Let the radius of the sphere be $r$, the radius of the base of the cylinder (the cross-section of the bore) be $\rho$, the specific gravity of iron be $s$, and the specific gravity of the wooden peg be $s_{1}$. The volume of the remaining part of the sphere is calculated by subtracting the volume of the cylinder and the two complementary spherical caps formed by the bore from the volume of the sphere; that is,

$$
v=2 \pi\left[\frac{2 r^{3}}{3}-\rho^{2} \sqrt{r^{2}-\rho^{2}}-\frac{r-\sqrt{r^{2}-\rho^{2}}}{6}\left(3 \rho^{2}+2 r^{2}-2 r \sqrt{r^{2}-\rho^{2}}-\rho^{2}\right)\right]
$$

from which

$$
v=\frac{4 \pi}{3} \sqrt{\left(r^{2}-\rho^{2}\right)^{3}}
$$

Thus, the volume of the wooden peg is:

$$
v_{1}=\frac{4 r^{3} \pi}{3}-\frac{4 \pi}{3} \sqrt{\left(r^{2}-\rho^{2}\right)^{3}}
$$

If the iron ball floats in water, then

$$
\frac{4 \pi s}{3} \sqrt{\left(r^{2}-\rho^{2}\right)^{3}}+s_{1}\left(\frac{4 r^{3} \pi}{3}-\frac{4 \pi}{3} \sqrt{\left(r^{2}-\rho^{2}\right)^{3}}\right)=\frac{4 r^{3} \pi}{3}
$$

or

$$
\begin{gathered}
s \sqrt{\left(r^{2}-\rho^{2}\right)^{3}}+s_{1} r^{3}-s_{1} \sqrt{\left(r^{2}-\rho^{2}\right)^{3}}=r^{3} \\
\sqrt{r^{2}-\rho^{2}}=r \sqrt[3]{\frac{1-s_{1}}{s-s_{1}}} \\
r^{2}-\rho^{2}=r^{2} \sqrt[3]{\left(\frac{1-s_{1}}{s-s_{1}}\right)^{2}} \\
\rho=r \sqrt{1-\sqrt[3]{\left(\frac{1-s_{1}}{s-s_{1}}\right)^{2}}}
\end{gathered}
$$

Substituting the given values: $\rho=0.881 r$.

(Krisztián György.)

The problem was also solved by: Devecis M., Döme B., Freibauer E., Prohászka J., Sasvári G., Spitzer Ö., Weisz J.