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NEET2008PhysicsMechanical Properties of FluidsActual

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 ~cm ^2 and the distance between the plates is 0.12 mm , then the force applied to separate the two plates is

Options

  1. A10^2 dyne
  2. B10^4 dyne
  3. C10^5 dyne
  4. D10^6 dyne

Correct answer

C. 10^5 dyne

Step-by-step 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 = d 2 Pressure difference across the surface = T R = 2 T d Area of each plate wetted by water =A Force F required to separate the two plates is given by F= pressure difference area = 2 T d A Putting the given values, we get F= 2 75 8 0.012 =10^5 dyne

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