Question

Find the fraction of ice of density 900 kg/m3 inside the water of density 1000 kg/m3.

Answer 1

Answer:

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

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

Under equilibrium,

Weight of ice is gonna be equal to Weight of water displaced

⟹Vρ  

ice

​  

g=(1−f)Vρ  

water

​  

g

⟹(1−f)=  10

9

​  

 

⟹f=  10

1

​  

Thus, the percentage of volume outside water = 10%

i hope it helps you buddy

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have a gud day!!