To determine the rise in water level when a sphere is completely submerged in a cylindrical vessel, we can use the principle of volume displacement.

Step 1: Calculate the volume of the sphere

The volume $V_{\text{sphere}}$ of a sphere is given by the formula:

$V_{\text{sphere}} = \frac{4}{3} \pi r^3$

where $r$ is the radius of the sphere. Here, $r = 9$ cm.

$V_{\text{sphere}} = \frac{4}{3} \pi (9)^3 = \frac{4}{3} \pi (729) = 972 \pi \, \text{cm}^3$

Step 2: Relating volume displacement to the rise in water level

The volume displaced by the sphere will be equal to the volume of the submerged part of the sphere. This volume will cause the water level in the cylindrical vessel to rise.

The volume $V_{\text{cylinder}}$ of water displaced in a cylindrical vessel is:

$V_{\text{cylinder}} = \pi R^2 h$

where $R$ is the radius of the vessel, and $h$ is the height the water rises by. Here, $R = 12$ cm, and we need to find $h$.

Setting the volume of water displaced equal to the volume of the sphere:

$972 \pi = \pi (12)^2 h$

$972 \pi = \pi (144) h$

Canceling $\pi$ from both sides:

$972 = 144 h$

Step 3: Solve for $h$

$h = \frac{972}{144} = 6.75 \, \text{cm}$

Final Answer:

The surface of the water will rise by 6.75 cm when the sphere is completely submerged.

Would you like any further details or clarifications?

Here are some related questions you might find useful:

1. How does the radius of the vessel affect the rise in water level?
2. What happens if the sphere is only partially submerged?
3. How can we calculate the volume of a hemisphere submerged in water?
4. What if the radius of the sphere changes—how would that affect the rise in water?
5. How do we compute the volume displaced if the vessel isn't cylindrical?

Tip: When solving for volume displacement problems, always equate the displaced volume with the new volume of water added, whether the vessel shape is cylindrical, spherical, or otherwise!