Question

In the figure shown, a rectangular bar of soap having density $800\, kg / m ^{3}$ floats in water of density $1000\, kg / m ^{3}$. Oil of density $300\, kg / m ^{3}$ is slowly added, forming a layer that does not mix with the water. When the top surface of the oil is at the same level as the top surface of the soap, then the ratio of the thickness of oil layer to the soap's thickness is $x / L$. Find $(x+L)$.

Answer 1

Buoyant force on soap,
$F_{B}=A\left[x \rho_{0}+(L-x) \rho_{w}\right] g$
Weight, $W=L A \rho_{s} g$
For equilibrium, $F_{B}=W$
$\Rightarrow L A \rho_{s} g=A\left[x \rho_{0}+(L-x) \rho_{w}\right] g $
$\Rightarrow L \rho_{s}=x \rho_{0}+(L-x) \rho_{w}$
$\Rightarrow \rho_{s}=\frac{x}{L}\left(\rho_{0}-\rho_{w}\right)+\rho_{w} $
$\Rightarrow \frac{x}{L}=\frac{\rho_{w}-\rho_{s}}{\rho_{w}-\rho_{0}}$
$=\frac{1000-800}{1000-300}=\frac{2}{7}$