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 $\theta_0$ ($\theta_0 << 1$) with the $x$-axis at $x = L$. If $y(x)$ is the height of the surface then the equation for $y(x)$ is :
(take $\theta(x) = \sin\theta(x) = \tan\theta(x) = \dfrac{dy}{dx}$, $g$ is the acceleration due to gravity)
Answer: (B) $\dfrac{d^2y}{dx^2} = \dfrac{\rho g}{S}y$
Concept. Balance the vertical forces on a thin strip of the surface: the net surface-tension pull against the pressure deficit $\rho gy$ below a surface at height $y$.
Step 1. Surface tension. Net vertical component on a strip $dx$:
$$S\sin\theta(x + dx) - S\sin\theta(x) = S\frac{d^2y}{dx^2}dx$$
Step 2. Pressure. Just below the surface at height $y$ the liquid is $\rho gy$ below atmospheric, so the net downward push is $\rho gy\,dx$.
Step 3. Equilibrium.
$$S\frac{d^2y}{dx^2} = \rho gy \;\Rightarrow\; \frac{d^2y}{dx^2} = \frac{\rho g}{S}y$$
Answer. $\dfrac{d^2y}{dx^2} = \dfrac{\rho g}{S}y$ (Option B).
Solution by Sreeraj P, M.Sc Physics