A cube of side 100 mm is placed at the bottom of an empty container on one of its faces. The density of the material of the cube is 800 kg/m³. Liquid of density 1000 kg/m³ is now poured into the container. The minimum height to which the liquid needs to be poured into the container for the cube to just lift up is _________ mm

The correct answer is 80.

Step-by-step Explanation:

Let us analyze the forces acting on the cube to determine the minimum height of the liquid required to lift it.

Let:
- $a$ be the side length of the cube, where $a=100\text{mm}=0.1\text{m}$.
- ${\rho }_c$ be the density of the cube, where ${\rho }_c=800{\text{kg/m}}^3$.
- ${\rho }_l$ be the density of the liquid, where ${\rho }_l=1000{\text{kg/m}}^3$.
- $h$ be the height of the poured liquid in millimeters.
- $g$ be the acceleration due to gravity.

The total volume of the cube is given by:
$V=a^3$

The weight of the cube acting downwards is:
$W={\rho }_c·g·V={\rho }_c·g·a^3$

When the liquid is poured to a height $h$ (where $h\le a$), the volume of the submerged part of the cube is:
$V_{\text{sub}}=a^2·h$

The upward buoyant force ($F_B$) exerted by the displaced liquid is:
$F_B={\rho }_l·g·V_{\text{sub}}={\rho }_l·g·a^2·h$

For the cube to just lift up from the bottom of the container, the buoyant force must balance the weight of the cube:
$F_B=W$

Substituting the expressions for $F_B$ and $W$:
${\rho }_l·g·a^2·h={\rho }_c·g·a^3$

Dividing both sides by $g·a^2$ yields:
${\rho }_l·h={\rho }_c·a$

Solving for $h$:
$h=\frac{{\rho }_c}{{\rho }_l}·a$

Substituting the given values into the equation:
$h=\frac{800}{1000}·100\text{mm}$

Thus, the minimum height to which the liquid needs to be poured is 80 mm.