Consider a water tank shown in the figure. It has one wall at $x=L$ and can be taken to be very wide in the $z$ direction. When filled with a liquid of surface tension $S$ and density $\rho$,the liquid surface makes an angle $\theta_0 \left(\theta_0 \ll 1\right)$ with the $x$-axis at $x=L$. If $y(x)$ is the height of the surface,then the equation for $y(x)$ is:
(Take $\theta(x) \approx \sin \theta(x) \approx \tan \theta(x) = \frac{dy}{dx}$,where $g$ is the acceleration due to gravity.)

Give two practical illustrations of surface tension.

When water is filled carefully in a glass,one can fill it to a height $h$ above the rim of the glass due to the surface tension of water. To calculate $h$ just before water starts flowing,model the shape of the water above the rim as a disc of thickness $h$ having semicircular edges,as shown schematically in the figure. When the pressure of water at the bottom of this disc exceeds what can be withstood due to the surface tension,the water surface breaks near the rim and water starts flowing from there. If the density of water,its surface tension and the acceleration due to gravity are $10^3 \ kg \ m^{-3}$,$0.07 \ N \ m^{-1}$ and $10 \ m \ s^{-2}$,respectively,the value of $h$ (in $mm$) is:

$A$ drop of some liquid of volume $0.04 \ cm^{3}$ is placed on the surface of a glass slide. Then another glass slide is placed on it in such a way that the liquid forms a thin layer of area $20 \ cm^{2}$ between the surfaces of the two slides. To separate the slides a force of $16 \times 10^{5} \ dyne$ has to be applied normal to the surfaces. The surface tension of the liquid is (in $dyne \ cm^{-1}$):

The value of surface tension of a liquid at critical temperature is

The force required to take away a flat circular plate of radius $2 \ cm$ from the surface of water is $[$ Surface tension of water $= 70 \times 10^{-3} \ Nm^{-1}, \pi = \frac{22}{7} ]$