[Geometry]
Problem: A hollow cylindrical tube has a radius of $10 \mathrm{~mm}$ and a height of $100 \mathrm{~mm}$. The tube sits flat on one of its circular faces on a horizontal table. The tube is filled with water to a depth of $h \mathrm{~mm}$. A solid cylindrical rod has a radius of $2.5 \mathrm{~mm}$ and a height of $150 \mathrm{~mm}$. The rod is inserted into the tube so that one of its circular faces sits flat on the bottom of the tube. The height of the water in the tube is now $64 \mathrm{~mm}$. Determine the value of $h$.

Solution: Initially, the water in the hollow tube forms a cylinder with radius $10 \mathrm{~mm}$ and height $h \mathrm{~mm}$. Thus, the volume of the water is $\pi(10 \mathrm{~mm})^{2}(h \mathrm{~mm})=100 \pi h \mathrm{~mm}^{3}$.

After the rod is inserted, the level of the water rises to $64 \mathrm{~mm}$. Note that this does not overflow the tube, since the tube's height is $100 \mathrm{~mm}$.

Up to the height of the water, the tube is a cylinder with radius $10 \mathrm{~mm}$ and height 64 mm.

Thus, the volume of the tube up to the height of the water is

$$
\pi(10 \mathrm{~mm})^{2}(64 \mathrm{~mm})=6400 \pi \mathrm{mm}^{3}
$$

This volume consists of the water that is in the tube (whose volume, which has not changed, is $100 \pi h \mathrm{~mm}^{3}$ ) and the rod up to a height of $64 \mathrm{~mm}$.
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Since the radius of the rod is $2.5 \mathrm{~mm}$, the volume of the rod up to a height of $64 \mathrm{~mm}$ is $\pi(2.5 \mathrm{~mm})^{2}(64 \mathrm{~mm})=400 \pi \mathrm{mm}^{3}$.

Comparing volumes, $6400 \pi \mathrm{mm}^{3}=100 \pi h \mathrm{~mm}^{3}+400 \pi \mathrm{mm}^{3}$ and so $100 h=6000$ which gives $h=60$.

Final answer: 60