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 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) = sin θ(x) = tan θ(x) = dy/dx, g is the acceleration due to gravity) (a) (d²y)/(dx²) = √((ρ g)/(S)) (b) dy/dx = √((ρ g)/(S)) x (c) (d²y)/(dx²) = (ρ g)/(S)x (d) (d²y)/(dx²) = (ρ g)/(S)y

Answer: (d). For small slope, curvature of the surface is 1/R = (d²y)/(dx²), so the pressure jump across it is S/R = S(d²y)/(dx²). Liquid just below the surface at height y (above the flat level) has pressure p₀ - ρ g y. Balance: p₀ - (p₀ - ρ g y) = S(d²y)/(dx²) So the answer is (d) (d²y)/(dx²) = (ρ g)/(S)y.

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Velocity of Efflux (Torricelli)

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)

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Answer: (d)

For small slope, curvature of the surface is $\frac{1}{R} = \frac{d^2y}{dx^2}$ , so the pressure jump across it is $\frac{S}{R} = S\frac{d^2y}{dx^2}$ .
Liquid just below the surface at height $y$ (above the flat level) has pressure $p_0 - \rho g y$ .
Balance: $p_0 - (p_0 - \rho g y) = S\frac{d^2y}{dx^2}$
So the answer is (d) $\frac{d^2y}{dx^2} = \frac{\rho g}{S}y$ .

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