Question

A cube made of material having a density of $900{\text{kg m}}^{-3}$ floats between water of density $1000{\text{kgm}}^{-3}$ and a liquid of density $700{\text{kgm}}^{-3}$ which is immiscible in water. What part of cube is inside the water?

• A. 1/2
• B. 2/3
• C. 3/4
• D. 3/7

Answer 1

The correct option is B 2/3

Let $B_1$ and $B_2$ be the buoyancy due to liquid and water of density ${\rho }_1$ and ${\rho }_2$ respectively and $x$ part of cube immersed in water. The density of the body$\left(\sigma \right)$ is 700 $kg/m^3$


Equating the forces acting onthe cube,
$\begin{array}{}{F}_{B_1}+F_{B_2}=Mg& \text{(1)}\end{array}$
where $F_{B_1}$ and $F_{B_2}$ are buoyant forces and $Mg$ is weight of the body.
Since buoyant force, $F_B=\rho Vg$
where $V$ is the liquid of density $\rho$
Equation $\left(1\right)$ can be written as,
${\rho }_1(l-x)l^{2}g+{\rho }_{2}x{l}^{2}g=\sigma l^{3}g$
Substituting the values of ${\rho }_1$, ${\rho }_2$ and $\sigma$, we get
$700l-700x+1000x=900l$
$\Rightarrow x=\frac{2}{3}l$
Hence $\frac{2}{3}$rd part of cube is inside water.