A certain hollow iron sphere, weighing 3012 grams, floats in 4-degree water such that $\frac{1}{4}$ of it is above water. What is the thickness of the sphere's wall if we take the specific gravity of iron to be 7.5?
If the outer radius of a sphere is $R$, the inner radius is $r$, the weight is $q$, and the specific weight of iron is $s$, then

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

Since the weight of the sphere is equal to the weight of the water it displaces, we have

$$
\pi R^{3}=q
$$

from which

$$
R^{3}=\frac{q}{\pi}
$$

Substituting this into (1):

$$
r=\sqrt[3]{\frac{q}{\pi}\left(1-\frac{3}{4 s}\right)}
$$

Thus, the thickness of the sphere's wall is:

$$
R-r=\sqrt[3]{\frac{q}{\pi}}\left(1-\sqrt[3]{\frac{4 s-3}{4 s}}\right)=0.34 \text{ cm}
$$

(Károly Burján, Kaposvár.)

The problem was also solved by: Czank K., Demeter J., Faith F., Filkorn S., Grosz K., Kerekes T., Krausz B., Krisztián Gy., Kürth A., Lukhaub Gy., Messik G., Messik V., Perl Gy., Póka Gy., Sasvári G., Sasvári J., Scharff J., Scheuer R., Smodics K., Tézner E.