How many turns does it take to wind up a $22 \mathrm{~mm}$ diameter empty spool of a tape recorder with a $0,018 \mathrm{~mm}$ thick, $90 \mathrm{~m}$ long tape?
Let $D$ denote the diameter of the reel, and $\delta$ the thickness of the tape. In one revolution, the diameter of the reel increases by $2 \delta$, so for the $k$-th revolution, the length of the tape wound up, $h_{k}$, is:

$$
(D+2(k-1) \delta) \pi \cdot 90000 \mathrm{~mm} > h_{1} + h_{2} + \ldots + h_{n-1}
$$

Therefore,

$$
\sum_{k=1}^{n}(D+2 k \delta) \pi > h_{1} + \ldots + h_{n} > 90000 > h_{1} + \ldots + h_{n-1} > \sum_{k=1}^{n-1}(D+2(k-1) \delta) \pi
$$

The sums of the two arithmetic series in the estimates are calculated as follows:

$$
\frac{(2 D + (2 n + 2) \delta) \pi n}{2} > 90000 > \frac{(2 D + (2 n - 2) \delta) \pi (n-1)}{2}
$$

Thus,

$$
(D + (n + 1) \delta) n > \frac{90000}{\pi} > (D + (n - 1) \delta)(n - 1)
$$

Substituting the values of $\pi$, $D$, and $\delta$, we get the following two inequalities for $n$:

$$
0.018 n^{2} + 22.018 n - 28648.643 > 0 \quad \text{and} \quad 0.018 n^{2} + 21.964 n - 28670.625 < 0
$$

The positive numbers $n$ that satisfy these two inequalities are precisely those for which

$$
790.40867 < n < 791.69061
$$

Therefore, the tape winds up in 791 revolutions.