CBSE Class 11-science Answered
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Asked by Saravanan | 09 Feb, 2015, 12:38: AM
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 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 ... (1)
Increase in surface area of the drop = Final surface area - Initial surface area
Increase in potential energy = Increase in surface area x Surface tension
= ... (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,
or
or
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