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Get Answers to all your Questions
Q.38) Consider a water tank shown in the figure. It has one wall at $x=L$ and can be taken to belvery wide in the $z$ direction. When filled with a liquid of surface tension $S$ and derisity $P$, the liquid surface makes angle $\theta_0\left(\theta_{\mathrm{p}} \ll 1\right)$ with the $x$-axis at $x=L$. If $y(x)$ is the height of the surface then the equation for $y(x)$ is
(take $\theta(x)=\sin \theta(x)=\tan \theta(x)=\frac{d y}{d x}, g$ is the acceleration due to gravity)
A) $\frac{d y}{d x}-\sqrt{\frac{\rho g}{S}} x$
B) $\frac{d^2 y}{d s^2}=\frac{p g}{S} x$
C) $\frac{d^2 y}{d x^2}=\frac{\rho g}{S} y$
D) $\frac{d^2 y}{d x^2}=\sqrt{\frac{p g}{s}}$
Answers (1)
The shape of the liquid surface is governed by a balance between:
- Pressure due to surface tension (related to the curvature, i.e., second derivative $\frac{d^2 y}{d x^2}$ )
- Hydrostatic pressure (due to gravity, $\rho g y(x)$ )
The Young-Laplace equation for pressure difference due to curvature in a 2 D setup is:
$$
P=S \cdot \frac{d^2 y}{d x^2}
$$
And the hydrostatic pressure at height $y$ is:
$$
P=\rho g y(x)
$$
Equating the two:
$$
S \cdot \frac{d^2 y}{d x^2}=\rho g y(x) \Rightarrow \frac{d^2 y}{d x^2}=\frac{\rho g}{S} y(x)
$$
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