Physics - Fluid Question with Solution | TestHub

The height of water in a capillary tube is given by $h=\frac{2T\cos \theta }{r\rho g}$.

The mass of water in the capillary tube is $m=\pi r^{2}h\rho$.

Substituting the value of $h$, we get:

$m=\pi r^2\left(\frac{2T\cos \theta }{r\rho g}\right)\rho =\frac{2\pi rT\cos \theta }{g}$

This implies that $m\propto r$.

For a capillary tube of radius $r^{\prime }=\frac{r}{3}$, the new mass $m^{\prime }$ is:

$m^{\prime }=\frac{m}{3}$

Answer: $\frac{\mathrm{m}}{3}$