Consider a water tank shown in the figure. It has one wall at and can be taken to be very wide in the -direction. When filled with a liquid of surface tension and density , the liquid surface makes angle with the x-axis at . If is the height of the su

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Solution

Q.
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 $θ_{0}$ $(θ_{0} < < 1)$ with the x-axis at $x = L$ . If $y (x)$ is the height of the surface, then the equation for $y (x)$ is

(Take $θ (x) = \sin θ (x) = \tan θ (x) = \frac{d y}{d x},$ $g$ is the acceleration due to gravity) [2025]

1 $\frac{d^{2} y}{d x^{2}} = \sqrt{\frac{ρ g}{S}}$

2 $\frac{d y}{d x} = \sqrt{\frac{ρ g}{S}} x$

3 $\frac{d^{2} y}{d x^{2}} = \frac{ρ g}{S} x$

4 $\frac{d^{2} y}{d x^{2}} = \frac{ρ g}{S} y$

Ans.
(4)

$P_{A} = P_{B} = P_{O}$

$P_{C} = P_{O} - ρ g y$ ...(i)

$P_{C} = P_{O} - \frac{S}{R}$ ...(ii)

So, $ρ g y = \frac{S}{R}$ ...(iii)

Now, $R = \frac{{1 + {(\frac{d y}{d x})}^{2}}^{3 / 2}}{\frac{d^{2} y}{d x^{2}}}$ (radius of curvature)

As $\frac{d y}{d x}$ is small, so $R = \frac{1}{\frac{d^{2} y}{d x^{2}}}$

Putting in eqn. (iii),

$ρ g y = S \times \frac{d^{2} y}{d x^{2}}$

$\frac{d^{2} y}{d x^{2}} = \frac{ρ g y}{S}$

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