A $20 \mathrm{~cm}$ high, $10 \mathrm{~cm}$ diameter cylinder has "sagged". Its height and bases have not changed, but its slant height has stretched by $1 \mathrm{~mm}$, thus forming two congruent frustums. By what percentage has the volume of the cylinder increased?
![](https://cdn.mathpix.com/cropped/2024_05_02_1089fbf3bf18fa132895g-1.jpg?height=302&width=550&top_left_y=294&top_left_x=797)
The base circle radius of the cylinder is $r=5 \mathrm{~cm}$, the height is $M=20 \mathrm{~cm}$, and the volume is: $V_{\mathrm{h}}=5^{2} \pi \cdot 20$.

The "wetted" cylinder consists of two truncated cones, with base circle radius $r=5 \mathrm{~cm}$, height $m=10 \mathrm{~cm}$, and slant height $a=10.05 \mathrm{~cm}$. The radius $R$ of the top circle can be found from the right triangle shown in the diagram:

$$
\begin{gathered}
(R-r)^{2}=a^{2}-m^{2}=10.05^{2}-10^{2}=20.05 \cdot 0.05=1.0025 \\
R-r=\sqrt{1.0025} \approx 1.0012, \quad \text { hence } \quad R=6.0012
\end{gathered}
$$

The volume of the truncated cone is $V_{\mathrm{csk}}=\frac{m \pi}{3}\left(R^{2}+R r+r^{2}\right)$. The combined volume of the two truncated cones is:

$$
V \approx \frac{1}{3} 20 \pi \cdot 91.0204
$$

The ratio of the two volumes is:

$$
\frac{\frac{1}{3} 20 \pi \cdot 91.0204}{25 \cdot \pi \cdot 20} \approx \frac{91.0204}{75} \approx 1.2136
$$

Thus, the volume of the cylinder has increased by approximately $21.36 \%$.
![](https://cdn.mathpix.com/cropped/2024_05_02_61c84da93a67ba68c38dg-1.jpg?height=412&width=806&top_left_y=838&top_left_x=667)