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?
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Correct Answer: Option D
Explanation:
Volume of liquid displaced
 = \(\frac{2}{3}\)(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
Volume of liquid displaced
 = \(\frac{2}{3}\)(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