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?
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Correct Answer: Option C
Explanation:
Height below liquid = 1 - \(\frac{1}{5}\) = \(\frac{4}{5}\)
volume of cube = 103cm3
volume of liquid displaced = \(\frac{4}{5}\) x 1000
= 800cm3
density of liquid displaced;
= \(\frac{500}{800}\)
= 0.625cm-1
Height below liquid = 1 - \(\frac{1}{5}\) = \(\frac{4}{5}\)
volume of cube = 103cm3
volume of liquid displaced = \(\frac{4}{5}\) x 1000
= 800cm3
density of liquid displaced;
= \(\frac{500}{800}\)
= 0.625cm-1