Question

Two glass plates are separated by water. If surface tension of water is $75\, dyne\, cm^{-1}$ 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

• A. $10^2$ dyne
• B. $10^4$ dyne
• C. $10^5$ dyne
• D. $10^6$ dyne

Answer 1

Answer: (C)

Required force to separate the two glass plates,
$F = \frac{2SA}{t}$
Here, $ S = 75 \, dyn \, cm^{-1} , A = 8 \, cm^2 $
$ t = 0.12 \, mm = 0.012 \, cm$
$\therefore \:\: F = \frac{2 \times 75 \times 8 }{0.012 } = 100000$
$ F = 10^5 \, dyne$