Water is filled up to a height $h$ in a beaker of radius $R$ as shown in the figure. The density of water is $\rho$,the surface tension of water is $T$ and the atmospheric pressure is $P_0$. Consider a vertical section $ABCD$ of the water column through a diameter of the beaker. The force on water on one side of this section by water on the other side of this section has magnitude

  • A
    $\left|2 P_0 Rh+\pi R^2 \rho gh-2 RT\right|$
  • B
    $\left|2 P_0 Rh+R \rho gh^2-2 RT\right|$
  • C
    $\left|P_0 \pi R^2+R \rho g h^2-2 RT\right|$
  • D
    $\left|P_0 \pi R^2+R \rho g h^2+2 RT\right|$

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

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.)

Why does the free surface of a liquid tend to contract?

When water is filled carefully in a glass,one can fill it to a height $h$ above the rim of the glass due to the surface tension of water. To calculate $h$ just before water starts flowing,model the shape of the water above the rim as a disc of thickness $h$ having semicircular edges,as shown schematically in the figure. When the pressure of water at the bottom of this disc exceeds what can be withstood due to the surface tension,the water surface breaks near the rim and water starts flowing from there. If the density of water,its surface tension and the acceleration due to gravity are $10^3 \ kg \ m^{-3}$,$0.07 \ N \ m^{-1}$ and $10 \ m \ s^{-2}$,respectively,the value of $h$ (in $mm$) is:

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