\section*{Problem 3 - V01103}

\(500 \mathrm{~m}\) of paper with a thickness of \(0.1 \mathrm{~mm}\) is to be wound onto a roll with a diameter of \(15 \mathrm{~cm}\).

a) How many layers of paper will be on the roll at the end, and

b) what will be the diameter of the roll when all the paper has been wound?
}

The first layer of paper has a length of \(\pi d\) with \(d=150 \mathrm{~mm}\). With each layer, the diameter increases by \(0.2 \mathrm{~mm}\). Therefore, for the total length of the paper to be \(500000 \mathrm{~mm}\) with \(n\) layers of paper,

\[
\begin{aligned}
500000 & =\pi d+\pi(d+0.2)+\pi(d+2 \cdot 0.2)+\ldots+\pi(d+(n-1) \cdot 0.2) \\
& =\pi(n \cdot d+0.2+0.4+\ldots(n-1) \cdot 0.2) \\
& =\pi\left(n \cdot d+0.2 \cdot \frac{n(n-1)}{2}\right)
\end{aligned}
\]

This quadratic equation has the solutions \(n_{1} \approx 717.91\) and \(n_{2} \approx -2216.91\), where \(n_{2}\), being less than 0, is discarded.

At the end, there are 718 layers of paper on the roll. The roll then has a diameter of \(150+717 \cdot 0.2=293.4 \mathrm{~mm}\), or \(29.34 \mathrm{~cm}\).