Physics - Fluid Question with Solution | TestHub

For the given element, (consider length d in Z direction) Net force in upward direction $=$ Weight $(S\sin (\theta +d\theta )-S\sin \theta )d=mg$ Angle is small $\therefore \sin \theta \approx \theta$ $\Rightarrow \frac{d\theta }{ydx}=\frac{\rho g}{S}...(1)$ $\begin{array}{rl}& \tan \theta =\frac{dy}{dx}\Rightarrow \text{ Differentiating wrt }x\\ & {\sec }^2\theta \frac{d\theta }{dx}=\frac{d^{2}y}{d{x}^2}...(2)\end{array}$ Put $d\theta$ from (2) in (1) d take $\cos \theta \approx 1$, we get $\frac{d^{2}y}{d^2}=\frac{\rho gy}{S}$ $\text{ Alternative Solution :}$
$\begin{array}{rl}& P_A=P_B=P_0\\ & P_C=P_0-\rho gy\\ & P_C=P_0-\frac{S}{R}\\ & \rho gy=\frac{S}{R}\\ & \rho gy=S\frac{d^{2}y}{d{x}^2}\\ & \frac{d^{2}y}{d^2}=\frac{\rho gy}{S}\end{array}$ $\mathrm{R}=\frac{{\left\{1+{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2\right\}}^{3/2}}{\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}}$ $\mathrm{dy}/\mathrm{dx}$ is very small $\begin{array}{rl}& \mathrm{R}=\frac{1}{{\mathrm{d}}^2\mathrm{y}/{\mathrm{dx}}^2}\\ & \frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}=\frac{1}{\mathrm{R}}\end{array}$