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
The correct answer is: D
Explanation
Volume of liquid displaced
= \(\frac{2}{3}\)(0.5)\(^3\) = 0.0833cm3
Mass of liquid displaced = mass of floating cube = 75kg
Density of liquid = \(\frac{mass}{volume}\)
= \(\frac{75}{0.0833}\)
= 900kgm\(^{-3}\)
R.D of liquid = \(\frac{(900 )}{(1000)}\)
= 0.9
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