Water rises in a capillary tube of radius ' $r$ ' upto a height ' $h$ '. The mass of water in a capillary is…
Water rises in a capillary tube of radius ' $r$ ' upto a height ' $h$ '. The mass of water in a capillary is ' $m$ '. The mass of water that will rise in a capillary tube of radius $\frac{\text { ' } r \text { ' }}{3}$ will be
$3 \mathrm{~m}$
$\frac{\mathrm{m}}{3}$
$\mathrm{m}$
$\frac{2 \mathrm{~m}}{3}$
Solution
$\begin{array}{ll} & \mathrm{h} \propto \frac{1}{\mathrm{r}} \\ & \text { Mass of water in a capillary, } \mathrm{m}=\pi \mathrm{r}^2 \mathrm{~h} \rho \\ \therefore \quad & \mathrm{m} \propto \mathrm{r}^2 \mathrm{~h} \\ \therefore \quad & \mathrm{m} \propto \mathrm{r}^2 / \mathrm{r} \\ \therefore \quad & \frac{\mathrm{m}_2}{\mathrm{~m}}=\frac{\frac{\mathrm{r}}{3}}{\mathrm{r}} \\ \therefore \quad & \mathrm{m}_2=\frac{\mathrm{m}}{3}\end{array}$