A certain hollow steel ball, whose thickness is $1 \mathrm{~cm}$, floats in water; the specific gravity of steel is 7.5. What is the diameter of the ball?
Let the outer radius of the sphere be $R$, the inner radius be $r$, the thickness be $v$, and the specific gravity of iron be $s$.

According to the condition,

$$
\frac{4 s \pi}{3}\left(R^{3}-r^{3}\right)=\frac{4 \pi R^{3}}{3}
$$

from which

$$
r=R \sqrt[3]{\frac{s-1}{s}}
$$

and since

$$
R-r=v
$$

therefore

$$
R-R \sqrt[3]{\frac{s-1}{s}}=v
$$

from which

$$
R=\frac{v \sqrt[3]{s}}{\sqrt[3]{s}-\sqrt[3]{s-1}}
$$

Substituting the appropriate values:

$$
R=21.5 \mathrm{~cm} \mathrm{~s} \text { so } 2 R=43 \mathrm{~cm}
$$

(Kiss Albert.)

The problem was also solved by: Breuer M., Csete F. A., Czank K., Dolowschiák M., Faith F., Filkorn J., Kornis Ö., Krausz B., Krisztián Gy., Lukhaub Gy., Neumann J., Perl Gy., Rehberger Z., Sasvári G., Sasvári J., Szabó J., Weisz J.