amma \cos \theta}{\rho g r}$$

where:

From the above formula, it is evident that the height '$$h$$' is inversely proportional to the radius '$$r$$'.

If the radius of the capillary tube is reduced to $$\frac{r}{3}$$, the new height '$$h'$$' of water rise will be:

$$h' = \frac{2 \gamma \cos \theta}{\rho g \left(\frac{r}{3}\right)} = 3 \times \frac{2 \gamma \cos \theta}{\rho g r} = 3h$$

The mass of water in a capillary tube of radius '$$r$$' up to height '$$h$$' is:

$$\mathrm{m} = \rho \times (\text{Volume of water}) = \rho \times (\pi r^2 h)$$

Substituting height $$h'$$ and radius $$\frac{r}{3}$$ into the new volume, the new volume $$V'$$ will be:

$$V' = \pi \left(\frac{r}{3}\right)^2 \times 3h = \pi \frac{r^2}{9} \times 3h = \pi \frac{r^2 h}{3}$$

Thus, the new mass $$m'$$ will be:

$$\mathrm{m'} = \rho \times V' = \rho \times \pi \frac{r^2 h}{3} = \frac{\mathrm{m}}{3}$$

Therefore, the mass of water that will rise in a capillary tube of radius $$\frac{r}{3}$$ is:

Option B: $$\frac{\mathrm{m}}{3}$$

Fluid Mechanics

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