A hollow glass sphere with uniform wall thickness, which is empty inside, has an outer diameter of $16 \mathrm{~cm}$ and floats in water such that $\frac{3}{8}$ of its surface remains dry. What is the wall thickness if the specific gravity of the glass is $s=2.523$?
Since on the same sphere, the surface area of any spherical cap is always proportional to its height, according to the problem, the submerged part of the $16 \mathrm{~cm}$ diameter sphere is a spherical cap whose height is $m=\frac{5}{8} \cdot 16=10 \mathrm{~cm}$.

If the radius of the inner, air-filled sphere is $x$ (i.e., the wall thickness is $8-x \mathrm{~cm}$), then the volume of the glass sphere in $\mathrm{cm}^{3}$ is

$$
K_{1}=\frac{4 \cdot 8^{3} \pi}{3}-\frac{4 x^{3} \pi}{3}=\frac{4 \pi}{3}\left(8^{3}-x^{3}\right)
$$

The volume of the submerged spherical cap is

$$
K_{2}=\frac{m^{2} \pi}{3}(3 r-m)=\frac{10^{2} \pi}{3}(3 \cdot 8-10)=\frac{1400 \pi}{3} \mathrm{~cm}^{3}
$$

According to Archimedes' principle,

$$
\frac{4 \pi}{3}\left(512-x^{3}\right) s=\frac{4 \pi}{3} \cdot 350
$$

which simplifies to

$$
\left(512-x^{3}\right) \cdot 2.523=350
$$

from which

$$
x=\sqrt[3]{512-\frac{350}{2.523}} \approx 7.2 \mathrm{~cm}
$$

and thus the desired wall thickness is $8-7.2=0.8 \mathrm{~cm}$.

Miklós Csapody (Bp. VIII., Piarist g. I. o. t.)