To solve the question regarding the density of the object, we can use the formula for density:

$\text{Density} = \frac{\text{Mass}}{\text{Volume displaced}}$

From the image, we know:

• The mass of the object is 100 g.
• The water level initially is at 50 cm³.
• After the object is immersed, the water level rises to 90 cm³.

So, the volume displaced by the object is:

$\text{Volume displaced} = 90 \, \text{cm}^3 - 50 \, \text{cm}^3 = 40 \, \text{cm}^3$

Now, applying the density formula:

$\text{Density} = \frac{100 \, \text{g}}{40 \, \text{cm}^3} = 2.5 \, \text{g/cm}^3$

Thus, the correct answer is 2.5 g/cm³, as indicated.

Would you like a further breakdown of this solution or details on a related concept?

Here are 5 related questions you might find interesting:

1. How would the density change if the mass were halved but the volume stayed the same?
2. How does the concept of buoyancy relate to the object's density compared to water?
3. Can an object with the same volume but a higher mass have a lower density?
4. What happens to the water level if an object less dense than water is placed in it?
5. How is the Archimedes' principle applied in this scenario?

Tip: To confirm whether an object will float or sink in water, compare its density to that of water (1 g/cm³). If it's less than 1, the object will float; otherwise, it will sink.