Question

When you float an ice cube in water, you notice that 90% of it is submerged beneath the surface. Now suppose you put the same ice cube in a glass of some liquid whose density is less than that of water. 1)How much of the ice cube will be submerged below the surface of this liquid?

Answer 1

Answer:

The variation is inversely proportional to the decrease density

Explanation:

This is an exercise where we will use the Archimedes principle that states that the thrust is equal to the weight of the dislodged liquid

         B = m g

         ρf = m / Vf

         m = ρf Vf

         B = ρf Vf g

If we use Newton's second law for equilibrium

          B - W = 0

          B = m g

          ρ Vf g = ρb Vb g

          ρb / ρf = Vf / Vb               (1)

Let's apply this expression to our case

In water

           Vf / Vg = 0.90

replace in equation 1

           ρb / ρf = 0.90

           ρb = ρf 0.90

           ρb = 1000 0.90

           ρb = 900 kg / m3

Now we change the liquid to one with lower density, let's calculate the volume ratio

           Vf / Vg = ρb / ρf

The density of the body (ρb) remains constant if the density of the fluid decreases, as in the denominator the volume fraction increases, whereby the submerged part decreases

The variation is inversely proportional to the decrease density