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?
A
\(\frac{1}{6}\)
B
\(\frac{1}{5}\)
C
\(\frac{1}{3}\)
D
\(\frac{4}{5}\)
E
\(\frac{5}{6}\)
Verified Correct Answer: A.
Physical Principles & Numerical Solution
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–\(\frac{5}{6}\)) = \(\frac{1}{6}\).
Source: Myschool Past Questions Archive (Record #28574)