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 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:

\(\left(\right.\)take \(\theta(x)=\sin \theta(x)=\tan \theta(x)=\frac{d y}{d x}, g\) is the acceleration due to gravity\()\)

For the given element, (consider length \(d\) in \(Z\) direction) Net force in upward direction \(=\) Weight \((S \sin (\theta+d \theta)-S \sin \theta) d=m g\)
Angle is small \(\therefore \sin \theta \approx \theta\)
\(\Rightarrow \frac{d \theta}{y d x}=\frac{\rho g}{S} \ldots\text{(1)}\)
\(\tan \theta=\frac{d y}{d x} \Rightarrow\) Differentiating wrt \(x\)
\(\sec ^2 \theta \frac{d \theta}{d x}=\frac{d^2 y}{d x^2} \ldots\text{(2)}\)
Put \(d \theta\) from (2) in (1) d take \(\cos \theta \approx 1\), we get \(\frac{d^2 y}{d^2}=\frac{\rho g y}{S}\)
Alternative Solution :

\(P_A=P_B=P_0 \)
\( P_C=P_0-\rho g y \)
\( P_C=P_0-\frac{S}{R} \)
\( \rho g y=\frac{S}{R} \)
\( \rho g y=S \frac{d^2 y}{d x^2} \)
\( \frac{d^2 y}{d^2}=\frac{\rho g y}{S}\)
\(R=\frac{\left\{1+\left(\frac{d y}{d x}\right)^2\right\}^{3 / 2}}{\frac{d^2 y}{d x^2}}\)
\(dy / dx\) is very small
\(R =\frac{1}{d^2 y / dx ^2}\)
\(\frac{d^2 y }{ dx ^2}=\frac{1}{ R}\)