A compact aluminum sphere has a diameter of $25 \mathrm{~cm}$. What uniform thickness of paraffin should it be coated with so that it just floats in water? (The specific gravity of aluminum is 2.7, and that of paraffin is 0.24. - Only precise numerical calculations will be accepted.)
I. Solution: The aluminum sphere coated with paraffin, according to Archimedes' principle, floats in a certain liquid if the buoyant force acting on the sphere, which is equal to the weight of the displaced liquid, equals the weight of the composite sphere.

If we denote the sought thickness of the paraffin coating by $x$, the radius of the aluminum sphere by $r$, the specific gravity of aluminum by $f_{a}$, the specific gravity of paraffin by $f_{p}$, and the specific gravity of the liquid by $f_{v}$, then

$$
\frac{4 r^{3} \pi}{3} f_{a}+\frac{4 \pi\left[(r+x)^{3}-r^{3}\right]}{3} f_{p}=\frac{4(r+x)^{3} \pi}{3} f_{v}
$$

Dividing by $\frac{4 \pi}{3}$ and rearranging,

$$
(r+x)^{3}\left(f_{p}-f_{v}\right)=r^{3}\left(f_{p}-f_{a}\right)
$$

from which

$$
r+x=r \sqrt[3]{\frac{f_{a}-f_{p}}{f_{v}-f_{p}}}
$$

Substituting the given values $\left(f_{v}=1\right)$

$$
12.5+x=12.5 \sqrt[3]{\frac{2.7-0.24}{1-0.24}}=12.5 \sqrt[3]{\frac{2.46}{0.76}}
$$

from which (working with a 4-digit logarithmic table)

$$
x=5.99 \mathrm{~cm}
$$

László Farkas (Ózd, József Attila g. III. o. t.)

II. Solution: Since the condition for floating in water is that the specific gravity of the submerged material equals the specific gravity of water, we mix aluminum and paraffin to the specific gravity of water: 1, thus we can determine the ratio of the volume of aluminum to the volume of paraffin:

$$
V_{a}: V_{p}=(1-0.24):(2.7-1)=0.76: 1.7
$$

from which

$$
V_{p}=\frac{1.7 V_{a}}{0.76}
$$

and thus we arrive at the equation

$$
\frac{4 \pi r^{3}}{3}+\frac{1.7 \cdot 4 \pi r^{3}}{0.76 \cdot 3}=\frac{4 \pi(r+x)^{3}}{3}
$$

(This is how the problem was solved by the chemical technical students.)

András Szántó (Bp. XIV., Chemical Technical III. o. t.)