Question 143 of 192

If two glass plates have water between them and are separated by very small distance (see figure), it is very difficult to pull them apart. It is because the water in between forms cylindrical surface on the side that gives rise to lower pressure in the water in comparison to atmosphere. If the radius of the cylindrical surface is R and surface tension of water is T then the pressure in water between the plates is lower by:

A$\dfrac{T}{R}$B$\dfrac{T}{4R}$C$\dfrac{4T}{R}$D$\dfrac{2T}{R}$

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Answer: (D) $\dfrac{2T}{R}$

$d=2Rcos\theta$

$∴$ Pressure difference across a liquid surface is
$∆P=T\left(\dfrac{1}{{R}_{1}}+\dfrac{1}{{R}_{2}}\right)$

$∵{R}_{1}=Rand{R}_{2}=R$

$∆P=T\left(\dfrac{1}{R}+\dfrac{1}{R}\right)$

$∆P=\dfrac{2T}{R}$

$∴$ Pressure is more in the concave side hence pressure in water between the plates is lower by $\dfrac{2T}{R}$