A $350 \mathrm{~cm}^{3}$ volume brass cone, whose height is equal to the radius of its base, floats in mercury with its tip downward. How deep does this cone sink into the mercury, given that the specific gravity of the former is 7.2 and that of the latter is $13,6 \mathrm{gr}$.
The submerged part of the iron cone is again a cone, whose height is equal to the radius of its base. Its volume is $v=\frac{\pi x^{3}}{3}$, where $x$ represents the unknown depth to which the iron cone is submerged. According to Archimedes' principle,

$$
\frac{13.6 \pi x^{3}}{3}=7.2 \times 350
$$

and from this,

$$
\begin{aligned}
& x^{3}=\frac{3 \times 7.2 \times 350}{13.6 \times 3.14259} \\
& 3 \log x=\log 3+\log 7.2+\log 350-\log 13.6-\log 3.14259= \\
& =0.47712+ \\
& +0.85733+ \\
& +2.54407- \\
& -1.13353- \\
& -0.49715= \\
& =2.24783 \text {. } \\
& \log x=0.74928 \\
& x=5.6141 \text{ cm} .
\end{aligned}
$$

Sándor Pollák, 7th grade student, Győr. György Jankovich, Losoncz; Ferencz Kintzig, Arad; Sándor Kugel, Losoncz; Mihály Seidner, Losoncz.