## Task 17 - V00917

At the Berlin Metal Huts and Semi-Tools VEB, copper pipes (outer diameter $32 \mathrm{~mm}$, inner diameter $29 \mathrm{~mm}$) with a length of $3 \mathrm{~m}$ are drawn into pipes with an outer diameter of $27 \mathrm{~mm}$ and an inner diameter of $25 \mathrm{~mm}$.

How long are the drawn pipes?
Let $d_{a}=32 \mathrm{~mm}, d_{i}=29 \mathrm{~mm}, l=3 \mathrm{~m}$ be the dimensions of the original pipes, and $d_{a}^{\prime}=27 \mathrm{~mm}, d_{i}^{\prime}=25$ $\mathrm{mm}, l^{\prime}=x \mathrm{~m}$ be the dimensions of the drawn pipes. Since the volume of the pipes remains constant, we have

$$
\begin{aligned}
\left(\frac{\pi}{4} d_{a}^{2}-\frac{\pi}{4} d_{i}^{2}\right) \cdot l & =V=\left(\frac{\pi}{4}{d^{\prime}}_{a}^{2}-\frac{\pi}{4}{d^{\prime}}_{i}^{2}\right) \cdot x \\
\left(d_{a}^{2}-d_{i}^{2}\right) \cdot l & =\left({d^{\prime}}_{a}^{2}-{d^{\prime}}_{i}^{2}\right) \cdot x \\
x & =\frac{d_{a}^{2}-d_{i}^{2}}{{d^{\prime}}_{a}^{2}-{d^{\prime}}_{i}^{2}} \cdot l
\end{aligned}
$$

Substituting the values gives $x \approx 5.28 \mathrm{~m}$.