Question

The potential energy of the liquid of surface tension and density that rises into the capillary tube is (angle of contact is 0°)

Answer 1

Answer:

The potential energy of the liquid of the surface tension S and density rho is given as

$U = \frac{2\pi S^2}{\rho g}$

Explanation:

let the radius of the capillary is "r"

so capillary rise of the liquid into the tube is given as

$h = \frac{2Scos\theta}{\rho g r}$

here we know that

$\theta = 0$

so we will have

$h = \frac{2S}{\rho g r}$

Now we know that total mass of the liquid in the capillary is

$m = \rho(\pi r^2 h)$

$m = \rho(\pi r^2)(\frac{2S}{\rho g r})$

$m = \frac{2 \pi r S}{g}$

now potential energy of the liquid is given as

$U = mgH$

here H = height of center of mass

so we have

$U = (\frac{2 \pi r S}{g})g(\frac{S}{\rho g r})$

$U = \frac{2\pi S^2}{\rho g}$

#Learn

Topic : capillary Rise

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