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

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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

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 θ_o(θ_o<< 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) $\displaystyle \frac{{{d}^{2}}y}{d{{x}^{2}}}=\sqrt{\frac{\rho g}{S}}$

(2) $\displaystyle \frac{dy}{dx}=\sqrt{\frac{\rho g}{S}}x$

(3) $\displaystyle \frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{\rho g}{S}x$

(4) $\displaystyle \frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{\rho g}{S}y$

Solution :

For equilibrium,

total force on left wall = total force on the right wall

$\displaystyle Sl=\frac{\rho g{{y}^{2}}l}{2}+Slcos\theta$

$\displaystyle \Rightarrow S(1-cos\theta )=\frac{\rho g{{y}^{2}}}{2}$

$\displaystyle \Rightarrow S\left[ 1-(1-2{{\sin }^{2}}\frac{\theta }{2}) \right]=\frac{\rho g{{y}^{2}}}{2}$

$\displaystyle \Rightarrow S\left[ 2{{\left( \sin \frac{\theta }{2} \right)}^{2}} \right]=\frac{\rho g{{y}^{2}}}{2}$

$\displaystyle \Rightarrow S\left[ 2{{\left( \frac{\theta }{2} \right)}^{2}} \right]=\frac{\rho g{{y}^{2}}}{2}$

$\displaystyle \Rightarrow S{{\theta }^{2}}=\rho g{{y}^{2}}$

$\displaystyle \Rightarrow \theta =\sqrt{\frac{\rho g}{S}}y$

$\displaystyle \Rightarrow \frac{dy}{dx}=\sqrt{\frac{\rho g}{S}}y$

$\displaystyle \Rightarrow \frac{{{d}^{2}}y}{d{{x}^{2}}}=\sqrt{\frac{\rho g}{S}}\frac{dy}{dx}=\sqrt{\frac{\rho g}{S}}\times \sqrt{\frac{\rho g}{S}}y$

$\displaystyle \therefore \frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{\rho g}{S}y$

Hence (4) is the correct option.

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