A drop of water volume 0.05 $c m^{3}$ is pressed between two glass-plates, as a consequence of which, it spreads between the plates. The area of contact with each plate is 40 $c m^{2}$ . If the surface tension of water is 70 dyne/cm, the minimum normal force required to separate out the two glass plates in newton is approximately ______ $N$ .
(assumed angle of contact is zero)

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Solution

Volume $= 0.05 c m^{3}$
Thickness $= \frac{0.05}{40} c m$
Since contact angle is $0 °$ , $d = 2 r$
Excess pressure inside the liquid $= T (\frac{1}{r_{1}} + \frac{1}{r_{2}})$
where, $r_{1}$ and $r_{2}$ are radius in two perpendicular directions
Here,
$F = P A = T (\frac{2}{d} + \frac{1}{\infty}) A$ [Another surface is flat]
$F = \frac{2 T A}{d} F = \frac{2 \times 70 \times 40 \times 40}{0.05 \times 10^{- 2}} \times 10^{- 7} N F = 44.8 N F = 45 N$

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Volume=0.05cm3Thickness=0.0540cmSince contact angle is 0°, d=2rExcess pressure inside the liquid=T(1r1+1r2)where, r1 and r2 are radius in two perpendicular directionsHere, F=PA=T(2d+1∞)A [Another surface is flat] F=2TAdF=2×70×40×400.05×10−2×10−7NF=44.8NF=45N