To solve this problem, we need to understand the concept of capillary action.$\\$When a capillary tube is immersed in water, the water rises to a height $h$ due to surface tension.$\\$The mass of the water column in the capillary tube is given by:$\\ m = \rho \cdot V \\$where $\rho$ is the density of water and $V$ is the volume of the water column.$\\$The volume of the water column in a capillary tube of radius $r$ and height $h$ is:$\\ V = \pi r^2 h \\$Thus, the mass of the water in the first capillary tube is:$\\ m_1 = \rho \cdot \pi r^2 h \\$Given that $m_1 = 5$ g, we have:$\\ 5 = \rho \cdot \pi r^2 h \quad$(Equation 1)$\\$Now, consider the second capillary tube with radius $2r. \\$The height of the water column in this tube will be $\frac{h}{2}$ because the height of rise is inversely proportional to the radius.$\\$The volume of the water column in the second tube is:$\\ V_2 = \pi (2r)^2 \cdot \frac{h}{2} \\ = \pi \cdot 4r^2 \cdot \frac{h}{2} \\ = 2 \pi r^2 h \\$The mass of the water in the second capillary tube is:$\\ m_2 = \rho \cdot V_2 \\ = \rho \cdot 2 \pi r^2 h \\$Substituting Equation 1 into this expression, we get:$\\ m_2 = 2 \cdot (\rho \cdot \pi r^2 h) \\ = 2 \cdot 5 \\ = 10$ g$\\$Therefore, the mass of water that will rise in the second tube is $10$ g.$\\$This corresponds to Option 2.