"When a body is partially or completely immerged in a liquid, the buoyant force acting on it is equal to the weight of the displaced liquid and it acts in upward direction at the centre of gravity of the displaced liquid."

Suppose, a solid cube at a depth $x$ from the surface of liquid is as shown in figure.

A solid has height $h$ and cross section area $A$.

Density of liquid is $\rho$.

The forces from body at left and right side are equal and opposite so their effect cancel each

other.

The pressure on the upper surface of body $\mathrm{P}_{1}=x \rho g$

The pressure on below surface of body $\mathrm{P}_{2}=(x+h) \rho g$

The force on upper surface $\mathrm{F}_{1}=\mathrm{P}_{1} \mathrm{~A}=x \rho g \mathrm{~A}$

The force on below surface $\mathrm{F}_{2}=\mathrm{P}_{2} \mathrm{~A}=(x+h) \rho g \mathrm{~A}$

The buoyant (resultant) force acting on the body

$\mathrm{F}_{\mathrm{b}}=\mathrm{F}_{2}-\mathrm{F}_{1}$

$\mathrm{~F}_{\mathrm{b}}=(x+h) \rho g A-x \rho g A$

$\mathrm{~F}_{\mathrm{b}}=h \rho g A$

but $\mathrm{A} h=$ volume of body $\mathrm{V}=$ Mass of displaced liquid

$\therefore \mathrm{F}_{\mathrm{b}}=\mathrm{V} \rho g$

$\therefore \mathrm{F}_{\mathrm{b}}=m g \quad(\because \rho=m / \mathrm{V} \therefore m=\mathrm{V} \rho)$

This force is exerted on upward direction. Hence $\mathrm{m}$ is the mass of displaced liquid. Hence buoyant force = weight of displaced liquid.

It shows Archimedes Principle.