NEET 2025 — Physics solution

NEET2025Physics

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 $x = L$ . If $y (x)$ is the height of the surface then the equation for $y (x)$ is:

(take $θ (x) = sin θ (x) = tan θ (x) = \frac{d y}{d x}$ ; $g$ is the acceleration due to gravity)

A
$\frac{d ^{2} y}{d x ^{2}} = \frac{ρ g}{S} x$
B
$\frac{d ^{2} y}{d x ^{2}} = \frac{ρ g}{S} y$
C
$\frac{d ^{2} y}{d x ^{2}} = \frac{ρ g}{S}$
D
$\frac{d y}{d x} = \frac{ρ g}{S} x$

Solution

Consider a thin slice of the liquid surface of length $d$ along the $z$ direction. The surface tension force pulling the element upward differs at its two ends because the surface inclines from $θ$ to $θ + d θ$ across $d x$ .

Net upward force on the element $= [S sin (θ + d θ) - S sin θ] d$ , which equals the weight of the liquid column of height $y$ and width $d x$ under it: $ρ g y d x d$ .

For small $θ$ , $sin θ \approx θ$ , so $S d θ d = ρ g y d x d$ , giving $\frac{d θ}{d x} = \frac{ρ g y}{S}$ .

Since $tan θ = \frac{d y}{d x}$ and $θ$ is small, $θ \approx \frac{d y}{d x}$ , so $\frac{d θ}{d x} = \frac{d ^{2} y}{d x ^{2}}$ .

$∴ \frac{d ^{2} y}{d x ^{2}} = \frac{ρ g}{S} y$ .

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