Question

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?

Answer 1

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

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

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