Mass of water in the firsr tube,

\(m=\pi r^{2} h \rho=\pi r^{2} \times\left(\frac{2 T \cos \theta}{r \rho g}\right) \times \rho\)

\(=\frac{2 \pi r T \cos \theta}{g}\)

\(m \propto r\)

\(\frac{ m _{2}}{ m _{1}}=\frac{ r _{2}}{ r _{1}}\)

\(\frac{ m _{2}}{5}=\frac{2 r }{ r }\)

\(m _{2}=10 g\)