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 an 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:
(Take $\theta(x) \approx \sin \theta(x) \approx \tan \theta(x) = \frac{dy}{dx}$,where $g$ is the acceleration due to gravity.)

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

Explore More

Similar Questions

$A$ pin or a needle floats on the surface of water,the reason for this is:

If the work done in increasing the size of a soap film from $10\, cm \times 6\, cm$ to $60\, cm \times 11\, cm$ is $2 \times 10^{-4}\, J$,what is the surface tension?

Difficult
View Solution

On what does the value of surface tension depend? Explain.

$A$ needle is $7 \ cm$ long. Assuming that the needle is not wetted by water,what is the weight of the needle,so that it floats on water (in $g \ wt$)? $\left[T = \text{surface tension of water} = 70 \ dyne/cm\right]$ [acceleration due to gravity $= 980 \ cm \ s^{-2}$]

$A$ liquid is coming out from a vertical tube. The relation between the weight of the drop $W$,surface tension of the liquid $T$,and radius of the tube $r$ is given by,if the angle of contact is zero.

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo