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A cube made of material having a density of 0.9 x 100 kg/m floats between water and a liquid of density 0.7 x 10 kg/m", which is immiscible with water. What part of the cube is immersed in water? |
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Answer» tion:In the end, the cube floats only. So according to the law of floatation, its WEIGHT = total upthrust Total upthrust = upthrust provided by liquid + upthrust provided by WATER = weight of liquid displaced + weight of water displaced => VDG = Vd(liquid)g + volume of cube submerged x d(water)g ( V is the total volume of SOLID) ATQ;900Vg = 70Vg + 1000*volume of cube submerged*g=> 0.83Vg = volume of solid submerged*g => volume of solid submerged / V = 0.83 or 83/100=> Volume of solid submerged = 0.83 of total V Hence, 0.83 of the total volume is immersed.If we were to CALCULATE how much % of volume is submerged,= volume of solid submerged / volume of solid x 100 = 0.83 x 100 = 83% Hence |
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