question_answer

Two glass plates are separated by water. If surface tension of water is 75 dynes/cm and 1 area of each plate wetted by water is \[8\text{ }c{{m}^{2}}\]and) the distance between the plates is 0.12 mm, then the force applied to separate the two plates is:

A) 102 dyne             

B) 104 dyne

C) 105 dyne             

D) 106 dyne

Correct Answer: C

Solution :

The shape of water layer between the two plates is shown in the figure. Thickness d of the film = 0.12 mm                                         = 0.012 cm Radius R of the cylindrical face \[=\frac{d}{2}\] Pressure difference across the surface                        \[=\frac{T}{R}=\frac{2T}{d}\] Area of each plate wetted by water = A Force F required to separate the two plates is given by F = pressure difference x area\[=\frac{2T}{d}A\] Putting the given values, we get\[F=\frac{2\times 75\times 8}{0.012}={{10}^{5}}\,dynes\]