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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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$\frac{{TA}^{2}}{2 V}$

$\frac{{TA}^{2}}{V}$

$\frac{2 {TA}^{2}}{V}$

$\frac{4 {TA}^{2}}{V}$

answer is C.

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Detailed Solution

see fig

The excess of pressure p is given by
$p = T (\frac{1}{R} - \frac{1}{\infty}) = \frac{T}{R} ∴ p = \frac{T}{(d / 2)} = \frac{2 T}{d} Volume V = Ad so p = \frac{2 TA}{V} Now force = pressure x area = \frac{2 {TA}^{2}}{V}$

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