Question

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 $5g$. Another capillary tube of radius $2r$ is immersed in water. The mass of water in grams that will rise in this tube is

Answer 1

Let $h$ be the height through which the liquid rises in the capillary tube of radius $r$
$h=\frac{2Scos\theta }{r\rho g}$
Mass of the water in the first tube is
$m=\pi r^{2}h\rho =\pi r^2\times \left(\frac{2Scos\theta }{r\rho g}\right)\times \rho =\frac{2\pi r.Scos\theta }{g}$
Hence, $m\propto r$
Let mass of water in capillary tube of radius $2r$ be $m^{\prime }$, then
$\frac{m^{\prime }}{m}=\frac{r^{\prime }}{r}=\frac{2r}{r}=2$
$m^{\prime }=2m=2\times 5g=10g$