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…

For the given element, (consider length d in Z direction) Net force in upward direction $=$= Weight $(S \sin (\theta+d \theta)-S \sin \theta) d=m g$(Ssin(θ+dθ)−Ssinθ)d=mg Angle is small $\therefore \sin \theta \approx \theta$∴sinθ≈θ $\Rightarrow \frac{d \theta}{y d x}=\frac{\rho g}{S}...(1)$⇒ydxdθ​=Sρg​...(1) $$\begin{aligned} & \tan \theta=\frac{d y}{d x} \Rightarrow \text { Differentiating wrt } x \\ & \sec ^2 \theta \frac{d \theta}{d x}=\frac{d^2 y}{d x^2}...(2) \end{aligned}$$​tanθ=dxdy​⇒ Differentiating wrt xsec2θdxdθ​=dx2d2y​...(2)​ Put $d \theta$dθ from (2) in (1) d take $\cos \theta \approx 1$cosθ≈1, we get $\frac{d^2 y}{d^2}=\frac{\rho g y}{S}$d2d2y​=Sρgy​ $\text { Alternative Solution :}$ Alternative Solution :

$$\begin{aligned} & P_A=P_B=P_0 \\ & P_C=P_0-\rho g y \\ & P_C=P_0-\frac{S}{R} \\ & \rho g y=\frac{S}{R} \\ & \rho g y=S \frac{d^2 y}{d x^2} \\ & \frac{d^2 y}{d^2}=\frac{\rho g y}{S}\end{aligned}$$​PA​=PB​=P0​PC​=P0​−ρgyPC​=P0​−RS​ρgy=RS​ρgy=Sdx2d2y​d2d2y​=Sρgy​​ $\mathrm{R}=\frac{\left\{1+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^2\right\}^{3 / 2}}{\frac{\mathrm{~d}^2 \mathrm{y}}{\mathrm{dx}^2}}$R=dx2 d2y​{1+(dxdy​)2}3/2​ $\mathrm{dy} / \mathrm{dx}$dy/dx is very small $$\begin{aligned} & \mathrm{R}=\frac{1}{\mathrm{~d}^2 \mathrm{y} / \mathrm{dx}^2} \\ & \frac{\mathrm{~d}^2 \mathrm{y}}{\mathrm{dx}^2}=\frac{1}{\mathrm{R}} \end{aligned}$$​R= d2y/dx21​dx2 d2y​=R1​​