question_answer

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

A) \[~{{10}^{2}}\] dyne                                      

B) \[{{10}^{4}}\]dyne          

C)        \[{{10}^{5}}\]dyne                          

D) \[{{10}^{6}}\]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 \[F=\frac{2\times 75\times 8}{0.012}={{10}^{5}}dyne\]