Water rises in a capillary tube of radius ' $r$ ' up to height ' $h$ '. The mass of water in capillary is ' $m$ '. The mass of water that will rise in capillary of radius $\mathrm{r} / 3$ will be

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Aicharya · Accepted Answer

Rise of water in capillary tube is given by $\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\mathrm{rgg}}$
For water, $\cos \theta=1$ Also, the radius of capillary tube becomes (r/3).$ \therefore \quad \mathrm{h}^{\prime}=\frac{3 \cdot 2 \mathrm{~T}}{\mathrm{r} \rho \mathrm{~g}} \Rightarrow \mathrm{~h}^{\prime}=3 \mathrm{~h}$
Now, $m=\pi r^2 \mathrm{~h} \times \rho$ $\therefore \quad \mathrm{m}^{\prime}=\pi(\mathrm{r} / 3)^2(3 \mathrm{~h}) \times \rho=\frac{\pi \mathrm{r}^2 \mathrm{~h} \rho}{3}=\frac{\mathrm{m}}{3}$

Water rises in a capillary tube of radius ' $r$ ' up to height ' $h$ '. The mass of water in capillary is '…

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