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Physics · PYQ NEET UG 2025

हिन्दी

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)

Question diagram
  1. A
    \( \frac{d^2y}{dx^2} = \frac{\rho g}{S} x \)
  2. B
    \( \frac{d^2y}{dx^2} = \frac{\rho g}{S} y \)
  3. C
    \( \frac{d^2y}{dx^2} = \sqrt{\frac{\rho g}{S}} \)
  4. D
    \( \frac{dy}{dx} = \sqrt{\frac{\rho g}{S}} x \)
View Answer & Solution

Correct answer · Option B

\( \frac{d^2y}{dx^2} = \frac{\rho g}{S} y \)

Explanation

Equating excess pressure due to curvature with hydrostatic pressure: \( S \frac{d^2y}{dx^2} = \rho g y \), giving \( \frac{d^2y}{dx^2} = \frac{\rho g}{S} y \).
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