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

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(B) The height $h$ to which water rises in a capillary tube is given by $h = \frac{2T \cos 	heta}{r ho g}$,which implies $h \propto \frac{1}{r}$.If

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$\frac{m}{3}$

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(B) The height $h$ to which water rises in a capillary tube is given by $h = \frac{2T \cos 	heta}{r ho g}$,which implies $h \propto \frac{1}{r}$.If the radius changes from $r_1 = r$ to $r_2 = \frac{r}{3}$,the new height $h_2$ becomes $h_2 = h_1 	imes \frac{r_1}{r_2} = h 	imes \frac{r}{r/3} = 3h$.The mass of water in the capillary is given by $m = V ho = (\pi r^2 h) ho$.For the first tube,$m_1 = \pi r^2 h ho = m$.For the second tube,$m_2 = \pi (r_2)^2 h_2 ho = \pi (\frac{r}{3})^2 (3h) ho$.$m_2 = \pi (\frac{r^2}{9}) (3h) ho = \frac{1}{3} \pi r^2 h ho$.Therefore,$m_2 = \frac{m}{3}$.

Water rises in a capillary tube of radius $r$ 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 $\frac{r}{3}$?

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