A piece of wood of relative density 0.25 floats in oil of relative density 0.81. What percentage of volume the wood is above the surface of oil?

Let volume of wood be $v$ and ${}^{\prime }{a}^{\prime }$ volume of wood in inside, as density of wood $=p_1=0.25$ [Relative density = density]

density of oil $=p_2=0.81$

From Buoyancy principle,

Buoyant force $=a{\rho }_{2}g$ (weight of liquid displace)

weight of wood $=v{\rho }_{1}g$

in equilibrium

$a{\rho }_{2}g=v{\rho }_{1}g$

$\Rightarrow a=\frac{v{\rho }_1}{{\rho }_2}=\frac{25}{81}v$

$\therefore$ volume of wood outside water

$v^{\prime }=v-a=v-\frac{25}{81}v$

$\Rightarrow v^{\prime }=\frac{5b}{81}\Rightarrow \frac{v^{\prime }}{v}=\frac{56}{81}=69.14\%$