A water drop of `0.05 cm^(3)` is squeezed between two glass plates and spreads into area of `40 cm^(2)`. If the surface tension of water is 70 dyne `cm^(-1)`, then the normal force required to separate the glass plates from each other will be

Options

A. 22.5 N
B. 45 N
C. 90 N
D. 450 N

Answer

B

Explanation

`Ft = 2SA rArr F = (2SA)/(t) = (2SA^(2))/(At) = (2SA^(2))/(V)`
Force `F = (2 xx 70 xx (40)^(2))/(0.05) = 44.8 xx 10^(5)` dyne
`= 44.8 N ~~ 45N`