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.

A drop of water of volume `0.05 cm^(3)` is pressed between two glass p

1.	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.
Answer» Correct Answer - D We have discussed above, `F=(2AT)/(d) = (2A^(2)T)/(Ad)` But, `Ad =` volume `:. F = (2A^(2)T)/(V)` Substituting the values we get, `F =(2xx(40xx10^(-4))^(2)xx(70xx10^(-3)))/(0.05 xx 10^(-6))` `= 45 N`

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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.

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