Two vertical parallel glass plates are partially submerged in water. The distance between the plates is $d=0.10mm$, and their width is $l=12cm$. Assuming that the water between the plates does not reach the upper edges of the plates and that the wetting is complete, find the force of their mutual attraction.

The correct option is B $13N$
$Given:$ $d=0.10mm$ ; $l=12cm$

$Solution:$ The energy of the liquid between the

plates is

$E=ldh\rho g\frac{h}{2}-2\alpha lh$

$=\frac{1}{2}\rho gld{h}^2-2\alpha lh$

$=\frac{1}{2}\rho gld{(h-\frac{2\alpha }{\rho gd})}^2-\frac{2{\alpha }^{2}l}{\rho gd}$

This energy is minimum when,

$h=\frac{2\alpha }{\rho gd}\text{and}$

the minimum potential energy is then $E_{min}$

$=-\frac{2{\alpha }^{2}l}{\rho gd}$

The force of attraction between the

plates can be obtained from this as

$F=\frac{-\partial E_{min}}{\partial d}=-\frac{2{\alpha }^{2}l}{\rho g{d}^2}$

$

(minus sign means the force is attractive.)

Thus

$F=-\frac{\alpha lh}{d}=13N$

$So,the$ $correct$ $option:B$