## 96. Math Puzzle 5/73

A copper wire that is $l_{1}=50 \mathrm{~m}$ long and $d_{1}=1 \mathrm{~mm}$ thick is drawn to a length of $l_{2}=1800 \mathrm{~m}$. What is the new diameter $d_{2}$?
$$
\frac{d_{1}^{2} \pi l_{1}}{4}=\frac{d_{2}^{2} \pi l_{2}}{4} \quad \rightarrow \quad d_{2}=d_{1} \sqrt{\frac{l_{1}}{l_{2}}}=0.17 \text{ mm}
$$