Introduction
Let us define thrust before we define upthrust. Thrust is a force acting on any object perpendicular to its surface. So, UPthrust is a force acting UPWARD on any object perpendicular to its surface.
The direction of application of upthrust is vertically opposite to the direction of application of weight.
We generally denote upthrust by U. Its SI unit is Newton (N).
Upthrust exerted by Air
When an object travels in the air, such as an airplane or balloon, it experiences an upward force due to air. This upward force is called the upthrust exerted by air.
Upthrust exerted by Liquid
When an object is partially or fully submerged in a liquid, it experiences an upward force due to the liquid. This upward force is called the upthrust exerted by liquid.
Origin of Upthrust exerted by Liquid
Consider a situation when an object, say a cuboid box, is submerged fully in a liquid. All of its six surfaces are exposed to the liquid. In such conditions, the liquid molecules that are in direct contact with the surface of the cuboid exert pressure on the surface.
However, the pressure is not uniform at all points. Pressure in a static liquid increases by a quantity $\rm \rho g d$ on every depth d in the liquid. Thus, pressure in a liquid at a shallower part A is less than the pressure at a deeper part B in the liquid.
Let $\rm P_{A}$ be the pressure at surface A and $\rm P_{B}$ be the pressure at surface B, then $\rm P_{A} < P_{B}$. If the perpendicular distance between these surfaces is ‘d’, then we can write $\rm P_{B} = P_{A} + \rho g d$.
The horizontal sides of the cuboid experience equal pressure on each side but in opposite directions. As their surface area is the same, the force exerted by the liquid pressure has zero effect on these sides.
The vertical sides of the cuboid, as described earlier, experience unequal pressure on each side and in opposite directions. As their surface area is the same, the force exerted by the liquid pressure is more at the bottom as compared to the top.
Let $\rm F_{A}$ be the force due to liquid pressure at surface A and $\rm F_{B}$ be the force due to liquid pressure at surface B, then we can write:
$\rm F_{A} = P_{A} \cdot S$ and $\rm F_{B} = P_{B} \cdot S$, where S is the surface area of each side.
These forces act in opposite directions. Let $\rm U$ be the resultant of these forces in the vertically upward direction, then
$\rm U = F_{B} - F{A}$
$\rm or, U = P_{B} \cdot S - P_{A} \cdot S$
$\rm or, U = S ( P_{A} + \rho g d - P_{A}$
$\rm or, U = S \cdot \rho g d$
$\rm or, U = \rho g \cdot ( S \cdot d)$
$\rm \therefore U = \rho g V$, where V = $\rm S \cdot d$ is the volume of the object immersed in the fluid.
This expression is valid for an object partially immersed in liquid, also. However, one must be careful while calculating the volume (V). The volume (V) is the part of the solid that is immersed in the liquid. One must neglect the volume of the solid that is outside the liquid.
IN SIMPLER TERMS
$\rm U = W_{1} - W_{2}$, where $\rm W_{1}$ is the actual weight of the object and $\rm W_{2}$ is the apparent weight of the object.