A drop of water of volume `0.05 cm^(3)` is pressed between two glass plates, as a consequence of which, it spreads and occupies an are of `40 cm^(2)`. If the surface tension of water is `70 "dyne"//cm`, find the normal force required to separate out the two glass plates is newton.

Options

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

Answer

B

Explanation

Required force is
`Ft=2TA`
or `F=(2TA)/(t)=(2TA^(2))/(At)=(2TA^(2))/(V)`
`impliesF=(2xx70xx(40)^(2))/(0.05)`
`=44.8xx10^(5)` dyne
`=44.8N~~45N`