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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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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