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 θ₀(θ₀ ll 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. d²y/dx²=ρ g/Sx 2. d²y/dx²=ρ g/Sy 3. d²y/dx²=√(ρ g/S) 4. dy/dx=√(ρ g/S)x

Answer: 2. d²y/dx²=ρ g/Sy Key idea: the surface tension pulls the curved surface so that, for each small piece, the net vertical force balances the extra hydrostatic pressure at that height. Take a strip of surface between x and x+dx (width w in the z direction). The tension S w acts along the tangent at both ends, so the net vertical force is S w [θ(x+dx)-θ(x)] = S w dθ/dxdx , where θ=dy/dx . The liquid below the strip at height y exerts a pressure difference ρ g y on it, giving a force ρ g y w dx . Balancing the two: Sd²y/dx²=ρ g y ⇒ d²y/dx²=ρ g/S y Options 1 and 3 do not contain y , and option 4 is first order, so they cannot describe a height profile that depends on y itself.

Topic: Surface energy, surface tension, angle of contact, excess pressure, drops, bubbles, capillary rise· Properties of Solids and Liquids

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_0(θ_0 1) with the x-axis at x = L.…

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 $\rho$, the liquid surface makes angle $\theta_0(\theta_0 \ll 1)$ with the $x$-axis at $x = L$. If $y(x)$ is the height of the surface then the equation for $y(x)$ is :

(take $\theta(x) = \sin\theta(x) = \tan\theta(x) = \frac{dy}{dx}$, $g$ is the acceleration due to gravity)

water tank wall at x=L, liquid surface curve y(x) making angle theta0 with x-axis at wall

1. $\frac{d^2y}{dx^2}=\frac{\rho g}{S}x$

2. $\frac{d^2y}{dx^2}=\frac{\rho g}{S}y$

3. $\frac{d^2y}{dx^2}=\sqrt{\frac{\rho g}{S}}$

4. $\frac{dy}{dx}=\sqrt{\frac{\rho g}{S}}x$

Doubtly Asked 30 Oct 2026 NEET UG 2025 Practice test neet Save WhatsApp Report

1 answer

Best answer

Answer: 2. $\frac{d^2y}{dx^2}=\frac{\rho g}{S}y$

Key idea: the surface tension pulls the curved surface so that, for each small piece, the net vertical force balances the extra hydrostatic pressure at that height.

Take a strip of surface between $x$ and $x+dx$ (width $w$ in the $z$ direction). The tension $S w$ acts along the tangent at both ends, so the net vertical force is $S w\,[\theta(x+dx)-\theta(x)] = S w\,\dfrac{d\theta}{dx}dx$, where $\theta=\dfrac{dy}{dx}$.

The liquid below the strip at height $y$ exerts a pressure difference $\rho g y$ on it, giving a force $\rho g y\,w\,dx$. Balancing the two:

$$S\frac{d^2y}{dx^2}=\rho g\,y \;\Rightarrow\; \frac{d^2y}{dx^2}=\frac{\rho g}{S}\,y$$

Options 1 and 3 do not contain $y$, and option 4 is first order, so they cannot describe a height profile that depends on $y$ itself.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

Was this answer helpful?

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_0(θ_0 1) with the x-axis at x = L.…

1 answer

Your answer