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Derive an expression for excess of pressure in a liquid drop. |
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Answer» Solution :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. |
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