A cube of side 10cm and mass 0.5kg floats in a liquid with only \(\frac{1}{5}\) of its height above the liquid surface. What is the relative density of the liquid?
0.125
0.250
0.625
2.500
Explanation
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Density=Mass/Volume
Volume = 1-1/5=4/5 or 0.8
Mass=0.5kg
Therefore; 0.5/0.8 = 0.625
i don't understand please
but it's relative density we were asked to calculate
please 
can someone explain
I don't understand why we would use 500g for the calculation, The weight of the liquid displaced by the cube is indeed equal to the weight of the portion of the cube that is submerged, not the total weight of the cube. so we must use 4/5 * 500g = 400g to derive the density
=400/800
which should give us 0.5g/cm3.
if you had look for the fraction before calculating the volume, you'll get a wrong answer
ThankGod's explanation has a little fault since the question said 1/5 of the height of the cube was above the liquid meaning 4/5 of the height of the cube was immersed in the liquid. So the volume should be multiplied by 4/5 not 1/5. Myschool is correct. ThankGod's method would have been correct if he used 4/5.
Capitally wrong,,,,,,,the right ansa is D,,,,,let me explain.,,,,,,,,,,,
Total volume of cube=(0,1)^3=0,001m^3
volume of cube immersed=1\5 off total volume of cube=1/5x0,001=2x10^-4m^3
According to d principle of floatation
mass of liquid displace=mass of cube=0.5kg
but mass of liquid displaced=density of liquid x volume of cube immersed
density of liquid=mass of liquid displace/volume of cube immersed=0,5/2x10^-4=2500kgm^-3
sotherefore,,,,relative density=density of the liquid/density of water=2500/1000=2.5
