IIT JEE , AIEEE Forum - Physics , Mechanics - Surface Tension-Liquid Drop

harsha_27

A drop of water of volume V is pressed between two glass plates , as a consequence of which it spreads and occupies an area A.If the surface tension of water is T,the normal force required to separate the two glass plates is ____________.

a)2TV/A

b)2TV/A²

c)2TA/V²

d)2TA²/V

Solve.Get Rates.

anchitsaini

when water is between two plates excess pressure=T/r
=TA/V
therefor F required=TA*A/V
i don't know why this option is not there, hoping that someone will explain!!

anandghegde

is it 2TA²/V ?

harsha_27

Right.

But explanation needed.

Aatish

hey anchit you are right......but it will become two times for each plate......

anchitsaini

ok perhaps now i understand. when we apply the force the area on which we would apply it =2A henc ans=2A*A/v

anchitsaini

ie 2TA*A/V

anandghegde

download image for clarity
if d is the distance between the plates,

is the angle of contact
from the figure, cos

= d/2r

can be taken as zero for glass -water
cos

= 1
therefore pressure = 2T/d
force = pressure x area
=2TA/d
=2TA²/V

anandghegde

please rate if satisfied

anchitsaini

hey anandghegde
when water is in between plates the radius of curvature of water surface=r while that of the st line is infinity.hence--
excess p=T(1/r + 1/infinity)=T/r

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