In a capillary tube of area of cross-section | vedclass

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(C) The height of water rise in a capillary tube is given by the formula $h = \frac{2T \cos 	heta}{r ho g}$,where $T$ is surface tension,$	heta$ i

Vedclass · Accepted Answer

$h/2$

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(C) The height of water rise in a capillary tube is given by the formula $h = \frac{2T \cos 	heta}{r ho g}$,where $T$ is surface tension,$	heta$ is the angle of contact,$r$ is the radius of the tube,$ho$ is the density of the liquid,and $g$ is the acceleration due to gravity.From this formula,we see that $h \propto \frac{1}{r}$.The area of cross-section $a$ is given by $a = \pi r^2$,which implies $r = \sqrt{\frac{a}{\pi}}$,so $r \propto \sqrt{a}$.Therefore,$h \propto \frac{1}{\sqrt{a}}$.Let $h_1 = h$ for area $a_1 = a$,and $h_2$ be the height for area $a_2 = 4a$.Then,$\frac{h_2}{h_1} = \sqrt{\frac{a_1}{a_2}} = \sqrt{\frac{a}{4a}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.Thus,$h_2 = \frac{h}{2}$.

In a capillary tube of area of cross-section $a$,water rises to a height $h$. To what height will water rise in a capillary tube of area of cross-section $4a$?

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