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

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(B) The height $h$ of a water column in a capillary tube is given by the formula: $h = \frac{2T \cos 	heta}{r ho g}$.Since $T, 	heta, ho,$ and $

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$\frac{m}{4}$

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(B) The height $h$ of a water column in a capillary tube is given by the formula: $h = \frac{2T \cos 	heta}{r ho g}$.Since $T, 	heta, ho,$ and $g$ are constant,we have $h \propto \frac{1}{r}$.The mass $m$ of the liquid in the capillary is given by $m = V ho = (\pi r^2 h) ho$.Substituting $h \propto \frac{1}{r}$ into the mass equation,we get $m \propto r^2 	imes \frac{1}{r}$,which simplifies to $m \propto r$.If the radius changes from $r$ to $r' = \frac{r}{4}$,the new mass $m'$ will be $m' = m 	imes \frac{r'}{r} = m 	imes \frac{r/4}{r} = \frac{m}{4}$.

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 of radius $\frac{r}{4}$?

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