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Q.

A drop of water of volume V is pressed between the two glass plates so as to spread to an area A. If T be the surface tension, the normal force required to separate the glass plates is

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a

TA2V

b

TA22V

c

2TA2V

d

4TA2V

answer is C.

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Detailed Solution

see fig  The excess of pressure p is given byp=T1R−1∞=TR ∴ p=T(d/2)=2Td  Volume V=Ad so p=2TAV Now force = pressure x area = =2TA2V
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