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 $\theta_0\left(\theta_0 \ll 1\right)$ with the $x$-axis at $x=L$. If $y(x)$ is the height of the surface then the equation for $y(x)$ is: ![NEET 2025 Physics, Mechanical Properties of Fluids — question figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/616059730283de43c87e3e22/Physics/images/Mechanical_Properties_of_Fluids/682c9a92c2281abe1696a0a1/question_1__q_682c9a92c2281abe1696a0a1__cdn-question-pool.getmarks.app__apNa77tO8zggC1qbUK0Cc-__LAswz6bHpIOciKbseBc.original.fullsiz__2ce9bc34e2_final_ppt_sync.png) $\left(\operatorname{take} \theta(x)=\sin \theta(x)=\tan \theta(x)=\frac{d y}{d x}, g\right.$ is the acceleration due to gravity)

NEET UG 2025 — Physics Mechanics

2025mcqhard

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 $\theta_0\left(\theta_0 \ll 1\right)$ with the $x$ -axis at $x=L$ . If $y(x)$ is the height of the surface then the equation for $y(x)$ is: NEET 2025 Physics, Mechanical Properties of Fluids — question figure

$\left(\operatorname{take} \theta(x)=\sin \theta(x)=\tan \theta(x)=\frac{d y}{d x}, g\right.$ is the acceleration due to gravity)

Official previous-year question

Held on 30 Apr 2025 · Verified 9 Jul 2026.

Options

A
$\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}=\frac{\rho \mathrm{g}}{\mathrm{S}} \mathrm{x}$
B
$\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}=\frac{\rho \mathrm{g}}{\mathrm{S}} y$
C
$\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}=\sqrt{\frac{\rho \mathrm{g}}{\mathrm{S}}}$
D
$\frac{d y}{d x}=\sqrt{\frac{\rho g}{S}} x$

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Solution

NEET 2025 Physics, Mechanical Properties of Fluids — solution figure 1

For the given element, (consider length d in Z direction) Net force in upward direction $=$ Weight $(S \sin (\theta+d \theta)-S \sin \theta) d=m g$ Angle is small $\therefore \sin \theta \approx \theta$ $\Rightarrow \frac{d \theta}{y d x}=\frac{\rho g}{S}...(1)$ $\begin{aligned} & \tan \theta=\frac{d y}{d x} \Rightarrow \text { Differentiating wrt } x \\ & \sec ^2 \theta \frac{d \theta}{d x}=\frac{d^2 y}{d x^2}...(2) \end{aligned}$ Put $d \theta$ from (2) in (1) d take $\cos \theta \approx 1$ , we get $\frac{d^2 y}{d^2}=\frac{\rho g y}{S}$ $\text { Alternative Solution :}$ NEET 2025 Physics, Mechanical Properties of Fluids — solution figure 2

$\begin{aligned} & P_A=P_B=P_0 \\ & P_C=P_0-\rho g y \\ & P_C=P_0-\frac{S}{R} \\ & \rho g y=\frac{S}{R} \\ & \rho g y=S \frac{d^2 y}{d x^2} \\ & \frac{d^2 y}{d^2}=\frac{\rho g y}{S}\end{aligned}$ $\mathrm{R}=\frac{\left\{1+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^2\right\}^{3 / 2}}{\frac{\mathrm{~d}^2 \mathrm{y}}{\mathrm{dx}^2}}$ $\mathrm{dy} / \mathrm{dx}$ is very small $\begin{aligned} & \mathrm{R}=\frac{1}{\mathrm{~d}^2 \mathrm{y} / \mathrm{dx}^2} \\ & \frac{\mathrm{~d}^2 \mathrm{y}}{\mathrm{dx}^2}=\frac{1}{\mathrm{R}} \end{aligned}$

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Options

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

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

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

$\frac{d y}{d x}=\sqrt{\frac{\rho g}{S}} x$

Solution

NEET 2025 Physics, Mechanical Properties of Fluids — solution figure 1

NEET UG 2025 — Physics Mechanics

Options

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Solution

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NEET UG Mechanics in other years

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NEET UG 2025 — Physics Mechanics

Options

Did you get this right?

Solution

Your note

NEET UG Mechanics in other years

More NEET UG Mechanics questions

Other Physics topics for NEET UG

Electromagnetism

Modern Physics

Thermodynamics

Optics

Waves & Oscillations

Keep practicing Mechanics