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?
\(\frac{1}{6}\)
\(\frac{1}{5}\)
\(\frac{1}{3}\)
\(\frac{4}{5}\)
\(\frac{5}{6}\)
Correct answer is E
Height inside H2O = \(\frac{2}{3}\)
R.d = \(\frac{\text{height inside}H_2O}{\text{height inside liquid}}\)
0.8 = \(\frac{\frac{2}{3}}{X}\)