A solid cube of side 50cm and mass 75kg floats in a liquid with \(\frac{1}{3}\) of its height above the liquid surface. The relative density of the liquid is?
0.33
0.50
0.67
0.90
Correct answer is D
= \(\frac{2}{3}\)(0.5)\(^3\)
Mass of liquid displaced = mass of floating cube = 75kg
Density of liquid = \(\frac{mass}{volume}\)
= \(\frac{75}{(\frac{7}{3(0.5)})}\) × 3
= 0.9 × 103kgm\(^{-3}\)
R.D of liquid = \(\frac{(0.9 )}{(1.0)}\) × 103
= 0.9