Consider a water tank shown in the figure. It has one wall at x=L...

For the given element, (consider length d in Z direction) Net force in upward direction = Weight $(S \sin (\theta+d \theta)-S \sin \theta) d=m g$ Angle is small $\therefore \sin \theta \approx \theta\Rightarrow \frac{d \theta}{y d x}=\frac{\rho g}{S}...(1)$$\begin{aligned} & \tan \theta=\frac{d y}{d x} \Rightarrow \text{ Differentiating wrt } x \\ & \sec ^2 \theta \frac{d \theta}{d x}=\frac{d^2 y}{d x^2}...(2) \end{aligned}$Put $d \theta$ from (2) in (1) d take $\cos \theta \approx 1$, we get $\frac{d^2 y}{d^2}=\frac{\rho g y}{S}\text{ Alternative Solution :}$<img width="250px"src="https://cdn-question-pool.Quantrex.app/pyq/neet/ev5Mymddcd9XeENdy_3ruKhpP_FJFV6x5M6UTMhR3PY.original.fullsize.png"/><br/>\begin{aligned} & P_A=P_B=P_0 \\ & P_C=P_0-\rho g y \\ & P_C=P_0-\frac{S}{R} \\ & \rho g y=\frac{S}{R} \\ & \rho g y=S \frac{d^2 y}{d x^2} \\ & \frac{d^2 y}{d^2}=\frac{\rho g y}{S}\end{aligned}\mathrm{R}=\frac{\left\{1+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^2\right\}^{3 / 2}}{\frac{\mathrm{~d}^2 \mathrm{y}}{\mathrm{dx}^2}} \mathrm{dy} / \mathrm{dx}$ is very small\begin{aligned} & \mathrm{R}=\frac{1}{\mathrm{~d}^2 \mathrm{y} / \mathrm{dx}^2} \\ & \frac{\mathrm{~d}^2 \mathrm{y}}{\mathrm{dx}^2}=\frac{1}{\mathrm{R}} \end{aligned}$$