An empty silver sphere, whose outer diameter is $2 R=1 \mathrm{dm}$, is exactly half submerged in water. What is the thickness of the sphere wall if the specific gravity of silver is $s=10.5$?
The weight of the entire body is equal to the weight of the displaced water:

$$
\frac{4 \pi s\left(R^{3}-r^{3}\right)}{3}=\frac{2 \pi R^{3}}{3}
$$

and from this

$$
r=\sqrt[3]{\frac{R^{3}(2 s-1)}{2 s}}=0.492 \mathrm{dm}
$$

The thickness of the spherical shell is equal to the difference between the two radii:

$$
d=R-r=0.5-0.492=0.008 \mathrm{dm}
$$

The problem was solved by: Bálint Béla, Berger Hugó, Freund Antal, Friedmann Bernát, Goldstein Zsigmond, Hofbauer Ervin, Kántor Nándor, Riesz Frigyes.