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 fille…

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 ρ, the liquid surface makes angle 0 (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 θ(x)=sinθ(x)=tanθ(x)=dy/dx; g is the acceleration due to gravity):

1. A. d^2y/dx^2=ρ g/S\,x
2. B. d^2y/dx^2=ρ g/S\,y (Correct answer)
3. C. d^2y/dx^2=ρ g/S
4. D. dy/dx=ρ g/S\,x

Correct Answer

Option B — d^2y/dx^2=ρ g/S\,y

Detailed Solution & Explanation

The correct answer is Option B.

Key Points

• At the free surface, the excess pressure due to curvature balances the hydrostatic pressure: $\Delta P = S\times(\text{curvature})$.
• For a nearly flat surface the curvature $\approx \dfrac{d^2y}{dx^2}$ (since $\dfrac{dy}{dx}=\tan\theta\approx 0$).
• Hydrostatic pressure at height $y$ is $\Delta P=\rho g y$.
• Equating: $\rho g y = S\,\dfrac{d^2y}{dx^2}\;\Rightarrow\; \dfrac{d^2y}{dx^2}=\dfrac{\rho g}{S}\,y$.