JAMB - Physics (1985 - No. 15)
A uniform cylindrical block of wood floats in water with one-third of its height above the water level, in a liquid of relative density 0.8, what fraction of its height will be above the liquid level?
\(\frac{1}{6}\)
\(\frac{1}{5}\)
\(\frac{1}{3}\)
\(\frac{4}{5}\)
\(\frac{5}{6}\)
Explanation
A floating body displaces its own weight of the liquid in which it floats.
i.e mass of block = mass of water = mass of liquid
let the fraction of the volume of the block in liquid be q
\(\frac{2}{3}\)V x 1 = Vq X 0.8
q = \(\frac{2V}{2.4V}\) = \(\frac{5}{6}\)
Therefore, the fraction of the height above the liquid level is (1–\(\frac{5}{6}\)) = \(\frac{1}{6}\).
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