Two glass plates are separated by water. If the surface tension of water is 75 dyne/cm and the area of each plate wetted by water is $8 c m^{2}$ and the distance between the plates is 1.2 cm, then the force applied to separate the two plates is

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Solution

Excess pressure inside a liquid is
$P = T (\frac{1}{r_{1}} + \frac{1}{r_{2}})$
where, $r_{1}$ and $r_{2}$ is radius in perpendicular directions
Here,
$P = T [\frac{2}{d} + 0] [r_{2} = \infty] F = P A = \frac{2 T}{d} A F = \frac{2 \times 75 \times 8}{1.2} d y n e F = 1000 d y n e$

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Two glass plates are separated by water. If the surface tension of water is 75 dyne/cm and the area of each plate wetted by water is 8cm2 and the distance between the plates is 1.2 cm, then the force applied to separate the two plates is

Excess pressure inside a liquid is P=T(1r1+1r2)where, r1 and r2 is radius in perpendicular directionsHere, P=T[2d+0][r2=∞]F=PA=2TdAF=2×75×81.2dyneF=1000dyne

Two glass plates are separated by water. If the surface tension of water is 75 dyne/cm and the area of each plate wetted by water is $8 c m^{2}$ and the distance between the plates is 1.2 cm, then the force applied to separate the two plates is

Excess pressure inside a liquid is
$P = T (\frac{1}{r_{1}} + \frac{1}{r_{2}})$
where, $r_{1}$ and $r_{2}$ is radius in perpendicular directions
Here,
$P = T [\frac{2}{d} + 0] [r_{2} = \infty] F = P A = \frac{2 T}{d} A F = \frac{2 \times 75 \times 8}{1.2} d y n e F = 1000 d y n e$