The correct option is:
$\frac{d^2}{d{x}^2}y=\frac{\rho g}{S}y$

Step-by-Step Explanation:

1. Analyze the System and Image:
As shown in the figure, a liquid of density _ρ and surface tension S is contained in a wide tank. A vertical wall is located at x = L. Near the wall, the liquid surface rises to form a meniscus. The height of the meniscus at any position x is represented by y(x), and the surface makes an angle θ₀ with the horizontal (the x-axis) at x = L.

2. Hydrostatic Pressure:
Let the flat, undisturbed surface of the liquid far from the wall be at y = 0, where the pressure is atmospheric pressure P₀.
At any point x along the meniscus, the height of the liquid is y. Due to hydrostatic equilibrium, the pressure in the liquid just below the curved surface at height y is lower than the atmospheric pressure P₀ by the weight of the raised column of liquid:
$P_{\text{liquid}}=P_0-\rho gy$

3. Excess Pressure due to Surface Tension (Laplace's Law):
The curved boundary of the liquid surface creates a pressure difference across it due to surface tension. According to Laplace's formula for a surface curved in one dimension (since the tank is very wide in the z direction):
$P_0-P_{\text{liquid}}=\frac{S}{R}$
where R is the radius of curvature of the liquid surface profile y(x).

4. Curvature and Small Angle Approximation:
The curvature 1/R of the profile y(x) is mathematically defined as:
$\frac{1}{R}=\frac{\frac{d^{2}y}{d{x}^2}}{{[1+{\left(\frac{dy}{dx}\right)}^2]}^{3/2}}$
We are given that the angle θ(x) of the surface with the horizontal is very small (θ(x) << 1). Therefore, the slope of the surface is also very small:
$\frac{dy}{dx}\approx \theta \ll 1$
Thus, the term ${(dy/dx)}^2$ in the denominator is negligible compared to 1, giving the approximation:
$\frac{1}{R}\approx \frac{d^{2}y}{d{x}^2}$

5. Formulating the Differential Equation:
Now we substitute the pressure relations and the curvature approximation into Laplace's equation:
$P_0-(P_0-\rho gy)=S\left(\frac{d^{2}y}{d{x}^2}\right)$
Simplifying the left-hand side:
$\rho gy=S\frac{d^{2}y}{d{x}^2}$
Rearranging for the second derivative of y with respect to x yields:
$\frac{d^2}{d{x}^2}y=\frac{\rho g}{S}y$