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Derive an equation for the total pressure at a depth 'h' below the liquid surface. |
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Answer» Solution :Consider a water sample of cross sectional area in the form of cylinder. Let `h_(1)` and `h_(2)` be the depths from th e air -water interface to level I and the 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)` A and `F_(2) = P_(2)` A the 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)`, i.e., 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)` 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 = rho V = rhoA(h_(2) - h_(1))` `V = A(h_(2) - h_(1))`  Hence, gravitational force `F_(G) = rho A(h_(2) - h_(1))g` On substituting the value of W in equation `F_(2) = F_(1) + mg rArr P_(2)A = P_(1)A + rho A(h_(2) - h_(1))g` Cancelling out A on both sides, `P_(2) = P_(1) + rho(h_(2) - h_(1))g` 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 Figuer (B)), then the value of `h_(1)` becomes ZERO `(h_(1) = 0)`. and `P_(1) = P_(a)` atmospheric pressure (say `P_(a)`). In the pressure `(P_(2))` at a depth becoes P. Substituting these values in equation. `P = P_(a) + rho gh` 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 which is equal to `1.013 xx 10^(5)P_(a)`. If the atmospheric pressure is neglected or ingnored, then `P = rho gh` 
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