Question

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 \(g\) as the acceleration due to gravity) 

• A. \( \frac{d^2 y}{dx^2} = \frac{\rho g}{S} y \)
• B. \( \frac{d^2 y}{dx^2} = \sqrt{\frac{\rho g}{S}} y \)
• C. \( \frac{d^2 y}{dx^2} = \sqrt{\frac{S}{\rho g}} y \)
• D. \( \frac{d^2 y}{dx^2} = \frac{S}{\rho g} y \)

Hint:

The shape of the liquid meniscus near a wall is determined by the balance between surface tension forces (related to the curvature) and gravitational forces (related to hydrostatic pressure). The Young-Laplace equation provides the fundamental relationship, which can be approximated for small slopes.

Answer 1

The Correct Option is A

To determine the equation for \(y(x)\), we must analyze the forces acting on the liquid surface, specifically surface tension and gravity, and their influence on surface curvature.

Surface tension \(S\) acts to minimize liquid surface area, inducing curvature. This curvature creates a pressure difference governed by the Young-Laplace equation:

\(\Delta P = S \left(\frac{d^2 y}{dx^2}\right)\)

where \(\frac{d^2 y}{dx^2}\) represents the surface curvature. For small angles \(\theta_0\), the governing equations for the liquid surface simplify to linear terms.

The pressure difference due to the liquid's weight at height \(y(x)\) is given by:

\(\Delta P = \rho g y(x)\)

where \(\rho\) is the liquid density and \(g\) is the acceleration due to gravity.

By equating the pressure differences from surface tension and gravity, we obtain:

\(S\left(\frac{d^2 y}{dx^2}\right) = \rho g y(x)\)

Rearranging these terms yields the differential equation for \(y(x)\):

\(\frac{d^2 y}{dx^2} = \frac{\rho g}{S} y(x)\)

Consequently, the equation describing the liquid surface height in the tank is:

\(\frac{d^2 y}{dx^2} = \frac{\rho g}{S} y\)