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