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 ≪ 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{d y}{d x}, g is the acceleration due to gravity) [NEET 2025]

Correct Answer is : d2ydx2=ρgSy

\frac{d^2 y}{d x^2}=\sqrt{\frac{\rho g}{S}}

\frac{d y}{d x}=\sqrt{\frac{\rho g}{s}} x

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

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

Class 11 NEET Physics Properties of Matter Questions

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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)=sin‌θ(x)=tan‌θ(x)=‌

,g is the acceleration due to gravity)

[NEET 2025]

‌
d2y
dx2
=√‌
ρg
S
‌
dy
dx
=√‌
ρg
s
x
‌
d2y
dx2
=‌
ρg
S
x
‌
d2y
dx2
=‌
ρg
S
y

Validate

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