A piece of wood of relative density 0.25 floats in a pot containing oil of relative density 0.81. What will be the fraction of volume of the wood above the surface of the oil. (density of water = 1000 kg/$m^3$)

The correct option is C

0.69

If $V_1$ is the volume of the wood in oil and $V_2$ is the volume of the wood in air, then,

$V_1{\rho }_{1}g=V_2{\rho }_{2}g$

Where ${\rho }_1$ and ${\rho }_2$ are the relative densities of oil and wood respectively

$\frac{V_1}{V_2}$ = $\frac{0.25}{0.81}$ = 0.31

Fraction of the volume above the surface of oil = $1-\frac{V_1}{V_2}$ = 0.69