A steel shaft with a circular cross-section floats parallel to the water surface, so that it is submerged to half its radius. What is the specific weight of the material?
The relative density $f$ of a test material with respect to water, based on Archimedes' principle, is given by the ratio of the volume of the submerged circular segment-based cylinder to the volume of the entire cylinder, or, due to the common height, the ratio of the base areas (cross-sections) $t_{1}: t_{2}$. From the half-radius submersion, we can directly see that the cross-section of the submerged part is a circular segment that is cut off from the full circular cross-section of the cylinder by the sides of the inscribed equilateral triangle. Thus, its area is one-third of the difference between the area of the circle and the area of this triangle. If the radius is $r$, then the side of the triangle is $\sqrt{3} r$, and its area is $3 \sqrt{3} r^{2} / 4$, and thus

$$
t_{1}=\frac{1}{3}\left(\pi r^{2}-\frac{3 \sqrt{3} r^{2}}{4}\right) \quad \text { and } t_{2}=\pi r^{2} \text {, from which }
$$

rounded to four decimal places

$$
f=\frac{t_{1}}{t_{2}}=\frac{1}{3}-\frac{\sqrt{3}}{4 \pi}=0.1955
$$

(For the purpose of calculating the subtrahend to four decimal places, four significant figures, the approximate values of $\sqrt{3}$ and $\pi$ used were 1.7321 and 3.1416, respectively, with five significant figures.)

András Jahn (Győr, Czuczor G. Gymnasium III. grade)