Water rises in a capillary tube of radius r u | vedclass

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(A) The height of water in a capillary tube is given by $h = \frac{2T \cos 	heta}{r ho g}$.Since $T, 	heta, ho,$ and $g$ are constant,$h \propto

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$\frac{m}{4}$

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(A) The height of water in a capillary tube is given by $h = \frac{2T \cos 	heta}{r ho g}$.Since $T, 	heta, ho,$ and $g$ are constant,$h \propto \frac{1}{r}$,which implies $hr = 	ext{constant}$.For a capillary of radius $r_1 = r$,the height is $h_1 = h$. For a capillary of radius $r_2 = \frac{r}{4}$,the new height $h_2$ is given by $h_1 r_1 = h_2 r_2$.$h 	imes r = h_2 	imes \frac{r}{4} \implies h_2 = 4h$.The mass of water in a capillary is given by $m = 	ext{Volume} 	imes 	ext{density} = (\pi r^2 h) ho$.For the new capillary,the mass $m'$ is $m' = \pi (r_2)^2 h_2 ho$.Substituting $r_2 = \frac{r}{4}$ and $h_2 = 4h$:$m' = \pi \left(\frac{r}{4}ight)^2 (4h) ho = \pi \left(\frac{r^2}{16}ight) (4h) ho = \frac{1}{4} (\pi r^2 h ho) = \frac{m}{4}$.

Water rises in a capillary tube of radius $r$ up to a height $h$. The mass of water in the capillary is $m$. What will be the mass of water that rises in a capillary tube of radius $\frac{r}{4}$?

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Two capillary tubes of same diameter are put vertically one each in two liquids whose relative densities are $0.8$ and $0.6$ and surface tensions are $60$ and $50 \text{ dyne/cm}$ respectively. The ratio of the heights of the liquids in the two tubes $\frac{h_1}{h_2}$ is:

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Water rises in a capillary tube of radius $r$ up to a height $h$. The mass of water in the capillary is $m$. The mass of water that will rise in a capillary of radius $r/3$ will be:

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