$100 \mathrm{~kg} 1.5 \mathrm{~mm}$ diameter copper wire is insulated with $0.5 \mathrm{~mm}$ thick rubber. How many $\mathrm{kg}$ of rubber is needed, if a $1 \mathrm{~m}$ long cord, whose cross-section is a square with a side length of $3 \mathrm{~mm}$, weighs 10 g when made from the same rubber. The specific weight of copper is 8.95.
Let the length of the copper wire be $h \, \text{cm}$, then using the relationship "volume \cdot specific weight = weight"

$$
0.075^{2} \cdot \pi \cdot h \cdot 8.95 = 100000 = 10^{5}
$$

from which

$$
h = \frac{10^{5}}{8.95 \cdot 0.075^{2} \cdot \pi}
$$

The specific weight of rubber, based on $\frac{g \, \text{weight}}{\text{cm}^{3} \, \text{volume}}$, is

$$
\frac{10}{0.3^{2} \cdot 100} = \frac{10}{9}
$$

The volume of the rubber sleeve is: $\left(0.125^{2} - 0.075^{2}\right) h \pi$, and thus the required weight of the rubber (volume \cdot specific weight)

$$
\begin{gathered}
Q = \left(0.125^{2} - 0.075^{2}\right) h \pi \cdot \frac{10}{9} = \frac{0.2 \cdot 0.05 \cdot 10^{5} \cdot 10}{8.95 \cdot 0.075^{2} \cdot 9} = \frac{0.01 \cdot 10^{6}}{8.95 \cdot 0.075^{2} \cdot 9} = \\
= \frac{10^{4}}{8.95 \cdot 0.075^{2} \cdot 9} \approx 22070 \, \text{gr} = 22.07 \, \text{kg}
\end{gathered}
$$

Lajos Almási (Bp., II., Rákóczi g. III. o. t.)