question_answer

Two glass plates are separated by water. If surface tension of water is 75 dynes/cm and area of each plate wetted by water is \[\text{8}\,\text{c}{{\text{m}}^{\text{2}}}\]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 \[=0.12\text{ }mm\] \[=0.012\text{ }cm\] 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\[~=\text{ }A\] Force F required to separate the two plates is given by F = pressure difference \[\times \] area \[=\frac{2T}{d}A\] Putting the given values, we get \[F=\frac{2\times 75\times 8}{0.012}={{10}^{5}}\,\text{dynes}\]