In a capillary tube having area of cross-sect | vedclass

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(B) 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

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$3 h$

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(B) 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 contact angle,$r$ is the radius of the tube,$ho$ is the density of the liquid,and $g$ is the acceleration due to gravity.Since $T, 	heta, ho,$ and $g$ are constant,we have $h \propto \frac{1}{r}$,which means $rh = 	ext{constant}$.The area of cross-section is $A = \pi r^2$,so $r = \sqrt{\frac{A}{\pi}}$,which implies $r \propto \sqrt{A}$.Given the initial area $A_1 = A$ and final area $A_2 = \frac{A}{9}$.The ratio of radii is $\frac{r_2}{r_1} = \sqrt{\frac{A_2}{A_1}} = \sqrt{\frac{A/9}{A}} = \sqrt{\frac{1}{9}} = \frac{1}{3}$.Using the relation $r_1 h_1 = r_2 h_2$,we get $h_2 = h_1 \left( \frac{r_1}{r_2} ight)$.Substituting the values,$h_2 = h 	imes 3 = 3h$.

In a capillary tube having area of cross-section $A$,the water rises to a height $h$. If the cross-sectional area is reduced to $\frac{A}{9}$,the rise of water in the capillary tube is

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