Question

ice cube has density of 900kg/m3.what part of the ice will float above the water surface if it is immersed in water at room temperature​

Answer 1

Step-by-step explanation:

Let the volume of the cube be V and the fraction of cube outside water be f.

Thus, volume of ice inside water = volume of water displaced =(1−f)V

Under equilibrium,

Weight of ice = Weight of water displaced

⟹Vρiceg=(1−f)Vρwaterg

⟹(1−f)=109

⟹f=101

Thus, the percentage of volume outside water =10%

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Answer 2

Step-by-step explanation:

let the volume of the cube be v And the fraction of outside water be f the volume of ice Inside the water = the volume of water displeased = (1-f)v p water g. (1-f) =9/10. = f=1/10. Thus the %of the volume outside water is=10%