Derive an expression for excess of pressure in a liquid drop.

Explanation

Let us consider a water sample of cross sectional area in the form of a cylinder. Let `h_(1)andh_(2)` be the depths from the air-water interface to level 1 and level 2 of the cylinder, respectively as shown in Figure(a). Let `F_(1)` be the force acting downwards on level 1 and `F_(2)` be the force acting upwards on level 2, such that, `F_(1)=P_(1)AandF_(2)=P_(2)A` Let us mass of the sample to be m and under equilibrium condition, the total upward force `(F_2)` is balanced by the total downward force `(F_(1)+mg)`, otherwise, the gravitational force will act downward which is being exactly balanced by the difference between the force `F_(2)=F_(1)`
`F_(2)-F_(1)=mg=F_(G)" "...(1)`
Where m is the mass of the water available in the sample element. Let `rho` be the density of the water then, the mass of water available in the sample element is
`m=rhoV=rhoA(h_(2)-h_(1))`
`V=A(h_(2)-h_(1))`

Hence, gravitational force,
`F_(G)=rhoA(h_(2)-h_(1))g`
On substituting the value of W in equation (1)
`F_(2)=F_(1)+mg`
`rArr" "P_(2)A=P_(1)A+rhoA(h_(2)-h_(1))g`
Cancelling out A on both sides,
`P_(2)=P_(1)+rho(h_(2)-h_(1))g" "...(2)`

If we choose the level 1 at the surface of the liquid (i.e, air-water interface) and the level 2 at a depth 'h' below the surface (as shown in Figure), then the value if `h_(1)` becomes zero `(h_(1)=0)` when `P_(1)` assumes the value of atmospheric pressure (say `P_(a)`). In addition, the pressure `(P_2)` at a depth becomes P. Substituting these values in equation,
`P_(2)=P_(1)+rho(h_(2)-h_(1))g`
we get `P=P_(a)+rhogh`
Which means, the pressure at a depth h is greater than the pressure on the surface of the liquid, where `P_(a)` is the atmospheric pressure `=1.013xx10^(5)P_(a)`. If the atmospheric pressure is neglected then
`P=rhogh`
For a given liquid, `rho` is fixed and g is also constant, then the pressure due to the fluid column is directly proportional to vertical distance or height of the fluid column.