Solution:
$$
h=\frac{2 S \cos \theta}{r \rho g}
$$
Mass of water in the first tube,
$$
\begin{aligned}
&m=\pi r^{2} h \rho=\pi r^{2} \times\left(\frac{2 S \cos \theta}{r \rho g}\right) \times \rho \\
&=\frac{2 \pi r \cos \theta}{g} \\
&\therefore m \propto r . \text { Hence, } \frac{m^{\prime}}{m}=\frac{2 r}{r}=2 \\
&\text { or } m^{\prime}=2 m=2 \times 5 g=10 \mathrm{~g} .
\end{aligned}
$$
