To solve this problem, we need to calculate how much the water level rises when the cube is placed inside the cubical container.

Step 1: Find the volume of the cube.

The cube has an edge length of 3 cm. The volume of a cube is given by:

$V_{\text{cube}} = \text{edge}^3 = 3^3 = 27 \, \text{cm}^3$

Step 2: Calculate the rise in the water level.

Since the cube is fully submerged, its volume displaces an equivalent volume of water. We now determine how much this displaced volume increases the water level in the container.

Given:

• The base area of the container is 36 cm².
• Volume of water displaced = Volume of the cube = 27 cm³.
• Rise in water level = $\frac{\text{Volume of the cube}}{\text{Base area of the container}}$

$\text{Rise in water level} = \frac{27 \, \text{cm}^3}{36 \, \text{cm}^2} = 0.75 \, \text{cm}$

Conclusion:

The water level rises by 0.75 cm when the cube is placed inside the container.

Would you like more details, or do you have any other questions?

---

Here are 5 related questions:

1. How would the water level change if the cube had an edge of 4 cm instead?
2. What would happen if the container's base area was different but the same volume of cube was placed in it?
3. How does the water level change if the container is not cubical but rectangular?
4. If only a quarter of the container was filled, how would the water level be affected?
5. What would happen if the cube was only partially submerged?

Tip: The displacement of water by an object is equal to the volume of the object submerged in the water.