Question 299: Properties of Matter

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 $ho$, the liquid surface makes angle $	heta_0\left(	heta_0 \ll 1ight)$ with the $x$-axis at $x=L$. If $y(x)$ is the height of the surface then the equation for $y(x)$ is:

(take $	heta(x)=\sin 	heta(x)=	an 	heta(x)=\frac{d y}{d x}, g$ is the acceleration due to gravity)

Accepted Answer

$$\begin{aligned} & R O C=	ext { Radius of curvature at point } A \ & 	ext { Curvature }=\frac{1}{R O C}=\frac{\left|\frac{d^2 y}{d x^2}ight|}{\left(1+\left(\frac{d y}{d x}ight)^2ight)^{\frac{3}{2}}}=\frac{\left|\frac{d^2 y}{d x^2}ight|}{(1+0)^{\frac{3}{2}}}=\frac{d^2 y}{d x^2} \quad\left[\because \frac{d y}{d x}=	an 	heta=0ight] \ & \Delta P=S 	imes 	ext { curvature } \ & \Rightarrow ho g y=S \frac{d^2 y}{d x^2} \ & 	herefore \frac{d^2 y}{d x^2}=\frac{ho g y}{S} \end{aligned}$$
Alternate Solution :
For the given element, (consider length d in Z direction) Net force in upward direction $=$ Weight $(\mathrm{S} \sin (	heta+\mathrm{d} 	heta)-\mathrm{S} \sin 	heta) \mathrm{d}=\mathrm{mg}$
Angle is small $	herefore \sin 	heta \approx 	heta$
$$ \begin{aligned} & \Rightarrow \frac{d 	heta}{y d x}=\frac{ho g}{S} \ & 	an 	heta=\frac{d y}{d x} \Rightarrow 	ext { Differentiating wrt } x \ & \sec ^2 	heta \frac{d 	heta}{d x}=\frac{d^2 y}{d x^2} \end{aligned} $$
Put $d 	heta$ from (2) in (1) $d$ take $\cos 	heta \approx 1$, we get
$$ \frac{d^2 y}{d x^2}=\frac{ho g y}{S} $$

Answer

$\frac{d^2 y}{d x^2}=\sqrt{\frac{\rho g}{s}}$

Answer

$\frac{d y}{d x}=\sqrt{\frac{\rho g}{s}} x$

Answer

$\frac{d^2 y}{d x^2}=\frac{\rho g}{s} x$

Answer

$\frac{d^2 y}{d x^2}=\frac{\rho g}{s} y$

Properties of Matter

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