A capillary tube of radius r is immersed in water and water rises...

The relation between mass and radius of the capillary tube is given by,$$\begin{array}{ll}m ∝ r\end{array}$$

$$\begin{array}{ll}⇒ \dfrac{{m}_{2}}{{m}_{1}} = \dfrac{{r}_{2}}{{r}_{1}}\end{array}$$

$⇒ \dfrac{{m}_{2}}{5} = \dfrac{2r}{r}$

$⇒ {m}_{2} = 10g$