An ice cube of volume 10 cm³ is floating over a glass of water of 10 cm² cross-section area and 10 cm height. The level of the water is exactly at the brim of the glass. Given that the density of ice is 10% less than that of water, what will be the situation when ice melts completely?

Concept:

The buoyant force is given by F = ρVg, where:

• ρ = Density of water
• V = Volume of water displaced
• g = Acceleration due to gravity

Calculation:

Let the density of water be ρ_w, then the density of ice \(\frac{9ρ_w}{10}\)

For floating, the Weight of the cube = Buoyant force

⇒ ρ_iV_ig = ρ_wVg

⇒  \(\frac{9ρ_w}{10}\) × 10 cm³ = ρ_w × V

⇒ V = 9 cm3 

⇒ 9 cm³ of ice is initially submerged in water.

When ice melts, its volume changes from 10 cm³ to (10 cm³) × \(\frac{9}{10}\) = 9 cm³.

⇒ There is no change in the volume of water.

∴ There is no change in the level of the water.