First, we calculate the volumes of the two cylindrical containers:

$V_{\text {large }}=\pi(6)^{2}(20)=720 \pi \mathrm{cm}^{3}$

$V_{\text {small }}=\pi(5)^{2}(18)=450 \pi \mathrm{cm}^{3}$

!

Figure 3

The volume of water initially contained in the large cylinder is

!

Figure 4

$$V_{\text {water, initial }}=\pi(6)^{2}(17)=612 \pi \mathrm{cm}^{3}$$

The easiest way to determine the final depth of water in the small cylinder is as follows. Imagine putting a lid on the smaller container and lowering it all the way to the bottom of the larger container, as shown in Figure 3. So there will be water beside and above the smaller container. Note that the larger container will be filled to the brim (since the combined volume of the small container and the initial water is greater than the volume of the large container) and some water will have spilled out of the larger container.

Now if the lid on the small container is removed, all of the water in the large container above the level of the brim of the small container will spill into the small container, as shown in Figure 4. This water occupies a cylindrical region of radius $6 \mathrm{~cm}$ and height $2 \mathrm{~cm}$, and so has a volume of $\pi(6)^{2}(2)=72 \pi \mathrm{cm}^{3}$. This is the volume of water that is finally in the small container. Since the radius of the small container is $5 \mathrm{~cm}$, then the depth of water is Depth $=\frac{72 \pi \mathrm{cm}^{3}}{\pi(5 \mathrm{~cm})^{2}}=\frac{72}{25} \mathrm{~cm}=2.88 \mathrm{~cm}$