Question

Two parallel glass plates are dipped partly in a liquid of density ${}^{\prime }{d}^{\prime }$ keeping them vertical. If the distance between the plates is ${}^{\prime }{x}^{\prime }$, surface tension for the liquid is $T$ and angle of contact is $\theta$, then rise of liquid between the plates due to capillary action will be:

• A. $\frac{T\cos \theta }{xdg}$
• B. $\frac{2T}{xdg\cos \theta }$
• C. $\frac{2T\cos \theta }{xdg}$
• D. $\frac{T\cos \theta }{xd}$

Answer 1

The correct option is C $\frac{2T\cos \theta }{xdg}$

Let the width of each plate be $b$ and due to capillary forces (force due to surface tension), the liquid will rise upto height $h$.


Across the length $b$, surface tension $\left(T\right)$ will be acting from both sides, and its component along the surface of the vertical plates on account of the contact angle $\left(\theta \right)$ made by concave meniscus will give rise to upward force due to surface tension:



Upward force due to surface tension is,
$F=2Tb\cos \theta ...\left(1\right)$
Weight of the liquid that rises in between the plates is,
$W=mg$
$W=Vdg=(b\times x\times h)dg...\left(ii\right)$ where $V\to$volume of liquid risen, $d\to$density of liquid

For the equilibrium of liquid column along vertical direction, the force due to surface tension must balance the weight of risen liquid.
$\Rightarrow 2Tb\cos \theta =(b\times x\times h)dg$
$\therefore h=\frac{2T\cos \theta }{xdg}$