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(C) The height $h$ to which water rises in a capillary tube of radius $r$ is given by $h = \frac{2T \cos 	heta}{r ho g}$.The mass of water $m$ in t

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$10$

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(C) The height $h$ to which water rises in a capillary tube of radius $r$ is given by $h = \frac{2T \cos 	heta}{r ho g}$.The mass of water $m$ in the capillary tube is given by $m = V ho = (\pi r^2 h) ho$.Substituting the expression for $h$:$m = \pi r^2 \left( \frac{2T \cos 	heta}{r ho g} ight) ho = \frac{2 \pi r T \cos 	heta}{g}$.From this expression,we see that $m \propto r$.For the second capillary tube with radius $r' = 2r$,the mass $m'$ will be:$\frac{m'}{m} = \frac{r'}{r} = \frac{2r}{r} = 2$.Therefore,$m' = 2m = 2 	imes 5 \, g = 10 \, g$.

$A$ capillary tube of radius $r$ is immersed in water and water rises in it to a height $h$. The mass of water in the capillary tube is $5 \, g$. Another capillary tube of radius $2r$ is immersed in water. The mass of water that will rise in this tube is ...... $g$

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