If a piece of wood having total volume of 1.6m^3 and density of 800kg/m^3 floats on water of density 1000kg/m^3.Claculate what part of wood is outside water?Pressure & Density
Let the part of wooden volume inside water as x.
When wood floation on water upthrust act on the wooden part that is inside the water. Since the wood is floating the mass of the wood is balanced by the upthrust of the wooden part inside water.
Mass of wood = density*volume = 800*1.6
Upthrust = x*1000
So for equilibrium;
800*1.6 = x*1000
x = 1.28
Part of wooden volume out side water = 1.6-1.28 = 0.32 m^3/s
As a percentage this is equal to 0.32/1.6*100% = 20%
So 20% of the wood volume is outside the water.
A part of substance floating on the liquid
= (Density of substance)/(Sp.Gravity of liquid * Density of liquid)
= (800kg/m^3)/(1 * 1000kg/m^3)
Therefore 4/5 part of the wood floating on the water.
So the part of the wood wihich is outside water is 1-4/5 = 1/5.