Two glass plates are separated by water. If surface tension of water is 75 dyne/cm and the area of each plate wetted by water is 8 cm2 and the distance between the plates is 0.12 mm, then the force applied to separate the two plates is:

Engineering

Physics

Surface Tension

Question

Two glass plates are separated by water. If surface tension of water is 75 dyne/cm and the area of each plate wetted by water is 8 cm² and the distance between the plates is 0.12 mm, then the force applied to separate the two plates is:

10⁴ dyne

10⁶ dyne

10² dyne

10⁵ dyne

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$F = \frac{2 Aσ}{t} = \frac{2 \times 8 \times 75 \times 10}{012} dyne = 10^{5} dyne$

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