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CBSE Class 11-science Answered

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Asked by Saravanan | 09 Feb, 2015, 12:38: AM
answered-by-expert Expert Answer
1.
If the liquid surface is curved then the molecule O on the liquid surface acts uopn by the forces due to the surface tension along the tangents to the surface as shown in fig. (a) and (b).


On resolving these forces into horizontal and vertical components, we know that the horizontal components cancel each other whereas vertical components add up.
Thus, a resultant force acts on the curved surface of the liquid which acts towards its centre of curvature. This means that the resultant force is directed inwards in case of convex surface and is directed outwards in case of concave surface. In order that the curved surface of the liquid may be in equilibrium, there must be an excess presuure on its concave side over that on the convex side, so that the forces of excess pressure will balance the resultant forces due to the surface tension. 
Hence, for the curved surface of liquid in equilibrium the pressure on concave side of liuid will be greater than pressure on its convex side.
 
2.
Consider a liquid drop of radius R. The molecules lying on the surface of the liquid drop due to the surface tension will experience a resultant force acting inwards, perpendicular to the surface. As the size of liquid drop cannot be reduced to zero due to force of surface tension, therefore the pressure inside the drop must be greater than the pressure outside.
The liquid drop is in equilibrium.
Let S be the surface tension of liquid drop
Po be the atmospheric pressure
Pi be the total pressure inside the liquid drop
Excess of pressure inside the drop, p=Pi-Po
Let there be an increase in the radius of the drop by a small quantity begin mathsize 14px style increment straight R end style due to excess pressure.
Then, work done by the excess pressure
W = Force x Displacement
    = (excess pressure x area) x increase in radius
    = p x 4 begin mathsize 14px style πR squared space straight x space increment straight R end style                          ... (1)
Increase in surface area of the drop = Final surface area - Initial surface area
                                                   begin mathsize 14px style equals space 4 space straight pi space left parenthesis straight R space plus space increment straight R right parenthesis squared space minus space 4 space straight pi space straight R squared equals space 4 space straight pi space left square bracket straight R squared space plus space 2 space straight R space left parenthesis increment straight R right parenthesis space plus space left parenthesis increment straight R right parenthesis squared space minus space straight R squared right square bracket equals space 8 space straight pi space straight R space increment straight R space space space space space space space space space space space space space space space space space space space space space space left parenthesis because Neglecting space left parenthesis increment straight R right parenthesis squared space as space it space is space very space small right parenthesis end style
begin mathsize 14px style therefore end styleIncrease in potential energy = Increase in surface area x Surface tension
                                             = begin mathsize 14px style 8 space straight pi space straight R space left parenthesis increment straight R right parenthesis space straight x space straight S end style                   ... (2)
As the drop is in equilibrium, the increase in the potential energy is at the cost of work done by the excess pressure, therefore, from (1) and (2), we get,
begin mathsize 14px style straight p space straight x space 4 space straight pi space straight R squared space straight x space increment straight R space equals space 8 space straight pi space straight R space increment straight R space straight x space straight S end style
or
begin mathsize 14px style straight p equals fraction numerator 2 space straight S over denominator straight R end fraction end style
or
begin mathsize 14px style straight P subscript straight i minus straight P subscript straight o equals fraction numerator 2 space straight S over denominator straight R end fraction end style

Answered by Faiza Lambe | 09 Feb, 2015, 12:07: PM
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