Question bank/NEET UG/Physics

NEET UGPhysicsModerateNEET 2025

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:
[IMAGE]

(take θ(x) = sinθ(x) = tanθ(x) = dydx\frac{dy}{dx}, g is the acceleration due to gravity)

Show answer and solution
Sol.
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ROCROC = Radius of curvature at point A
Curvature = 1ROC=∣d2ydx2∣(1+(dydx)2)32=∣d2ydx2∣(1+0)32=d2ydx2[∵dydx=tan⁡θ≈0]\frac{1}{ROC} = \frac{\left|\frac{d^2y}{dx^2}\right|}{\left(1 + \left(\frac{dy}{dx}\right)^2\right)^{\frac{3}{2}}} = \frac{\left|\frac{d^2y}{dx^2}\right|}{(1+0)^{\frac{3}{2}}} = \frac{d^2y}{dx^2} \quad \left[\because \frac{dy}{dx} = \tan\theta \approx 0\right]

ΔP=S×\Delta P = S \times curvature

⇒ρgy=Sd2ydx2\Rightarrow \rho g y = S \frac{d^2y}{dx^2}

∴d2ydx2=ρgyS\therefore \frac{d^2y}{dx^2} = \frac{\rho g y}{S}

Source: geekyrakshit/indian-exam-questions / neet2025-code45

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