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 find the correct equation for \(y(x)\), we need to consider the forces acting on the liquid surface and how they relate to surface tension, gravity, and the curvature of the surface.

Surface tension \(S\) acts to minimize the surface area of the liquid, leading to a curved surface. The curvature of the surface, in turn, produces a pressure difference described by the Young-Laplace equation:

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

where \(\frac{d^2 y}{dx^2}\) is the curvature of the surface. Since the angle \(\theta_0\) is small, the equations governing the liquid surface simplify to linear terms.

The pressure difference caused by the liquid's weight at height \(y(x)\) is expressed as:

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

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

Equating the pressure differences due to surface tension and gravity, we have:

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

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

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

Thus, the correct equation describing the height of the liquid surface in the tank is:

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

Answer 2

To derive the differential equation for the height \(y(x)\) of the liquid surface, we need to consider the forces involved due to surface tension and gravity. The surface tension causes a force that depends on the curvature of the surface, while gravity causes a downward force proportional to the height of the liquid column.

The Young-Laplace equation relates the curvature of a surface to the pressure difference across the surface due to surface tension. For a surface described by a function \(y(x)\), the curvature \(\kappa\) can be approximated by the second derivative \(\kappa \approx \frac{d^2y}{dx^2}\) when \(\theta_0 \ll 1\).

The pressure difference \(\Delta P\) across the liquid surface due to surface tension is given by:

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

The hydrostatic pressure due to gravity at height \(y(x)\) is:

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

For equilibrium, set the pressure due to surface tension equal to the hydrostatic pressure:

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

Dividing both sides by \(S\), we obtain the differential equation:

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

This is a linear second-order differential equation describing the shape of the surface of the liquid.