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Consider a liquid drop of radius R arid surface tension T. Due to surface tension the molecules on the surface film experience the net force in inward direction normal to the surface.

Therefore there is more pressure inside than outside. Let p_i be the pressure inside the liquid drop and p_o be the pressure outside the drop. Therefore excess of pressure inside the liquid drop is,

p = p₁– p₀

Due to excess of pressure inside the liquid drop the free surface of the drop will experience the net force in outward direction due to which the drop will expand. Let the free surface displace by dR under isothermal conditions. Therefore excess of pressure does the work in displacing the surface and that work will be stored in the form of potential energy.

The work done by excess of pressure in displacing the surface is, 

dW  = Force x displacement
       = (Excess of pressure x surface area) x displacement of surface
       
       
Increase in the  potential energy is,
dU = surface tension x increase in area of the free surface

  
     

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 Let a capillary tube of uniform bore be dipped vertically in a wet liquid. Since liquid is wet, therefore, the meniscus is concave. Let r be the radius of capillary tube, R be the radius of meniscus and θ the angle of contact.



In figure (i) X is in atmosphere, Z is at the interface of mercury and atmosphere and Y is on convex side of meniscus. Therefore pressure at X and Z is equal and equal to atmospheric pressure but pressure at Y is less than that at X and Z by

, Therefore, the liquid is not in equilibrium.

To attain the equilibrium, the liquid will rise in tube. Let at equilibrium, liquid rise to height h as shown in figure (ii) . Now at equilibrium,



From (1), (2) and (3), we have

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Atmosphere is a gaseous envelope surrounding the earth and the pressure exerted by the atmosphere is called atmospheric pressure.

The atmosphere exerts a huge pressure. The cause of atmospheric pressure is motion of air molecule. The air molecules in continuous motion strike the surface of body placed in it and exert a huge force.

To measure the atmospheric pressure, Torricelli took a meter long graduated tube and he filled it with clean and dry mercury. By closing the tube with thumb, he inverted the tube in a cistern ( a tub filled with mercury) as shown in the figure. He observed that the level of mercury first fell down and finally stayed with a column BC = 76cm in the tube above the free surface of mercury in the cistern leaving behind vacuum above C.

In the tube, above C there is vacuum, therefore pressure at point C is zero. Point B in the tube is 76cm below C, therefore pressure at B is,

P_B =P_C + ρ8^h = ρ8^h

where ρ is density of mercury and h be the height column of mercury in the tube above point B.

In the figure, point A is at the interface of air and mercury, therefore it is both in air and as well as in mercury. Thus pressure at point A is equal to atmospheric pressure, i.e.

P_A = Atmospheric pressure As vertical height between A and B is zero, therefore pressure at A and B is same, i.e.

or Atmospheric pressure = pgh
                  = 13600 x 9.8 x 0.76
                  = 1.013 x 10⁵ N/m² 

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Statement When a body is immersed partially or wholly in a liquid, its weight appears to be reduced and loss of weight is equal to the weight of liquid displaced by the body.

Proof: Consider a body of height h and area of cross-section A placed in liquid of density σ at a depth x below the free surface. Let ρ be density of the body.

Now the pressure at the upper face of the bodi is,   where  P₀ is atmospheric pressure.



Pressure at the lower face of the body is, 

                  

Therefore downward thrust on  upper face of the body is,

                  
and upward thrust on lower face of body is,

                     
Since P₂ > P₁  therefore F₂ > F₁ 

Body will experience the net force F₂ - F₁ in upward direction known as upward thrust.



         (Where V is volume of body)
         = Weight of liquid displaced by body
Now, the  body experiences two forces-weight of the body itself and upward thrust.



i.e. Apparent weight is less than the actual weight by amount equal to weight of liquid displaced by body.

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Let and p₂ be the excess of pressure inside the bubbles of radii a and b respectively. Therefore we have



When the two bubbles come together to form a double bubble, then let r be the radius of common interface.



Now the difference of pressure on the two sides at interface is