A drop of water volume `0.05 cm^(3)` is pressed between two glass -paltes, as a consequence of which, it spreads and occupies an area of `40 cm^(2)`. If the surface tension of water is `70 dyn e//cm`, find the normal force requried to separte out the two glass plates in newton.

Explanation

Pressure inside the surface `P_("in") = P_(0) - (T)/(r_(C)) = P_(0) - (T)/(t//2) = P_(0) - (2T)/(t)`,
So, net inwards force `= P_(0)A - P_("in") A = (P_(0) - (2T)/(t))A - P_(0)A = (2TA)/(t)`
Here volume between the paltes `V = A xx t rArr t = (V)/(A)` Putting the value of `t`
`F = (2A^(2)T)/(V) = (2 xx (40 xx 10^(-4))^(2) xx (70 xx 10^(-3)))/(0.05 xx 10^(-6)) = 45 N ,` So this much force is required to separate the plates