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?

0.125cm^-1

0.250gcm^-1

0.625cm^-1

2.500cm^-1

Correct answer is C

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

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