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 $8 \mathrm{~cm}^2$ 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
• Correct answer
  C

  $10^5$ dyne
• D

  $10^6$ dyne

Solution

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

Hence, the correct answer is option: 'C'.