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 ρ, the liquid surface makes angle θ₀ (θ₀ << 1) with the x-axis at x = L. If y(x) is the height of the surface then the equation for y(x) is:

$\text{(take θ(x) = sinθ(x) = tanθ(x) =}\frac{d}{y}\text{, g is the acceleration due to gravity)}$

Correct Answer:
The correct equation for the surface profile is:

$\frac{d^{2}y}{d{x}^2}=\frac{\rho gy}{S}$

Step-by-Step Explanation:

1. Image Analysis:
From the provided image, we can see the profile of a liquid surface in a tank. The coordinate axes are set such that the horizontal axis represents distance $x$ and the vertical axis represents height $y$. There is a vertical wall located at $x=L$. The liquid surface rises as it approaches the wall, forming a meniscus. At the boundary wall $x=L$, the tangent to the liquid surface forms a small contact angle ${\theta }_0$ with the horizontal line (dashed line parallel to the x-axis).

2. Laplace Pressure across a Curved Interface:
According to the Young-Laplace equation, the difference in pressure across a curved liquid-gas interface is related to the surface tension $S$ and the local radius of curvature $R$ of the interface. The pressure just below the surface is less than the atmospheric pressure $P_0$ above the surface by:

$\Delta P=P_0-P=\frac{S}{R}$

3. Curvature in the Small-Angle Approximation:
For a one-dimensional surface profile defined by the curve $y\left(x\right)$, the exact expression for the curvature is:

$\frac{1}{R}=\frac{\frac{d^{2}y}{d{x}^2}}{{[1+{\left(\frac{dy}{dx}\right)}^2]}^{3/2}}$

Because the slope is very small ($\theta \ll 1$), we can approximate the slope $\frac{dy}{dx}\approx \theta \ll 1$. This allows us to neglect the term ${\left(\frac{dy}{dx}\right)}^2$ in the denominator, simplifying the curvature expression to:

$\frac{1}{R}\approx \frac{d^{2}y}{d{x}^2}$

4. Hydrostatic Balance:
At any point along the interface with height $y$ relative to the flat liquid level far away (where $y=0$ and $P=P_0$), the pressure inside the liquid is governed by hydrostatic balance:

$P=P_0-\rho gy$

This gives the pressure difference across the interface at height $y$:

$P_0-P=\rho gy$

5. Formulating the Differential Equation:
By equating the Laplace pressure difference to the hydrostatic pressure difference:

$S\frac{d^{2}y}{d{x}^2}=\rho gy$

Rearranging this to solve for the second derivative of $y$ with respect to $x$ yields:

$\frac{d^{2}y}{d{x}^2}=\frac{\rho gy}{S}$