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Question: Iceberg floats in water with part of it submerged. What is the fraction of the volume of the iceberg...

Iceberg floats in water with part of it submerged. What is the fraction of the volume of the iceberg submerged if the density of ice is ρi=0.917gmcm3{\rho _i} = 0.917\,gm\,c{m^{ - 3}} ?

Explanation

Solution

Archimedes principle states that any body partially or fully submerged in a liquid or gas at the state of rest is acted upon by an upward buoyant force the magnitude of which is equal to the fluid displaced by the body. In this question we shall equate the weight of the iceberg with the weight of the displaced liquid.

Complete step by step solution:
Since the iceberg is floating on water with some part submerged under the sea, it should follow the Archimedes principle.
The weight of the iceberg must be equal to the weight of the liquid displaced from the sea.
We know that weight is given by W=mgW = mg where m is the mass and g is the acceleration due to gravity.
Also, we know that the density of a material is given as ρ=mV\rho = \dfrac{m}{V} where V is the volume of the object.
And so, the mass can be given as m=ρVm = \rho V .
For iceberg,
Let V denote the volume and ρi{\rho _i} denote the density of the ice. The weight of the iceberg is given by Wiceberg=micebergg{W_{iceberg}} = {m_{iceberg}}g
But we know that m=ρVm = \rho V . So, the weight can be expressed as Wiceberg=ρiVg{W_{iceberg}} = {\rho _i}Vg
For the displaced liquid,
Let Vwater{V_{water}} denote the volume and ρw{\rho _w} denote the density of water. The weight of the displaced liquid is given by Wwater=mwaterg{W_{water}} = {m_{water}}g
But we know that m=ρVm = \rho V . So, the weight can be expressed as Wwater=ρwVwaterg{W_{water}} = {\rho _w}{V_{water}}g
We know that the weight of the iceberg must be equal to the weight of the liquid displaced from the sea.
Hence, Wiceberg=Wwater{W_{iceberg}} = {W_{water}}
This can be rewritten as ρwVwaterg=ρiVg{\rho _w}{V_{water}}g = {\rho _i}Vg
The above expression simplifies to VwaterV=ρiρw\dfrac{{{V_{water}}}}{V} = \dfrac{{{\rho _i}}}{{{\rho _w}}}
We are given that ρi=0.917gmcm3{\rho _i} = 0.917\,gm\,c{m^{ - 3}} . So, VwaterV=ρiρw=0.917\dfrac{{{V_{water}}}}{V} = \dfrac{{{\rho _i}}}{{{\rho _w}}} = 0.917
Hence the fraction of volume submerged in water is 0.9170.917 .

Note: Very often weight is misunderstood as mass of the object. These are two different terms and must be noted carefully. Mass is the inherent property of an object which tells how much matter it contains while weight is the normal reaction acting on the body. Mass is constant while the weight may vary.