According to the sources, the height $h$ to which a liquid rises in a capillary tube of radius $a$ is given by the formula $h = \frac{2S \cos \theta}{\rho g a}$ . The mass ($m$) of the liquid column is the product of its density ($\rho$) and volume ($V$). Assuming the liquid column is a cylinder of height $h$, the volume is $V = \pi a^2 h$. Substituting the expression for $h$ into the mass formula: $m = \rho \times \pi a^2 \times \left( \frac{2S \cos \theta}{\rho g a} \right)$. Simplifying this gives $m = \frac{2 \pi a S \cos \theta}{g}$. This derivation shows that the mass of the liquid in the capillary is directly proportional to the radius of the tube ($m \propto a$) . Given that the first tube has radius $r$ and contains 5 g of water, a second tube with radius $2r$ (double the radius) will contain double the mass of water. Therefore, $m_2 = 2 \times 5$ g $= 10.0$ g.