a price of ice weights 1kgf to float 1/11th part of the volume above the surface of water.
calculate
I)the density of ice peice
ii)the total volume of ice submerged in water in si unit.
Answers
Answer:
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Given : A piece of ice weights 1 kgf to float 1/11th part of the volume above the surface of water.
To Find : I)the density of ice piece
ii)the total volume of ice submerged in water in SI unit.
Solution:
V = Volume of Ice
Volume of Ice above surface = V/11
Volume of Ice inside water V' = V - V/11 = 10V/11
V'/V = density of Ice/ Density of Water
=> ( 10V/11 )/ V = density of Ice/ Density of Water
=> density of Ice = (10/11)Density of Water
Density of Water = 1000 kg/m³
density of Ice = = 10 * 1000/11 = 909.1 kg/m³
or 0.909 g/cm³
Density = Mass / Volume
=> 10000/11 kg/m³ = 1 / Volume
=> Volume = 11/10000 m³
Volume of Ice inside water = (10/11) ( 11/10000)
= 1/1000 m³
= 10⁻³ m³
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