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 \[8\,\,c{{m}^{2}}\] 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\[\times \]area                 \[=\frac{2T}{d}A\] Putting the given values, we get                 \[F=\frac{2\times 75\times 8}{0.012}={{10}^{5}}dynes\]