Neglecting the small mass in the meniscus, for full rise,

\(2 \pi r T=\pi r^2 h \rho g\)

\(\rho=\frac{2 T}{r \rho g}=\frac{2 \times 0.07}{0.25 \times 10^{-3} \times 1000 \times 9.8}=0.057 m =5.7 cm\)

But there the tbe is only \(2 cm\) above the water and so water will rise by \(2 cm\) and meet the tube at an angle such that

\(2 \pi r T \cos 0^{\circ}=\pi r^2 h^{\prime} \rho g\)

\(\Rightarrow 2 T \cos \theta=h^{\prime} r \rho g\)

\(\Rightarrow \cos \theta=\frac{h^{\prime} r \rho g}{2 T}\)

\(\therefore \cos \theta=\frac{2 \times 10^{-2} \times 0.25 \times 10^{-3} \times 1000 \times 9.8}{2 \times 0.07}\)

\(\Rightarrow \theta=70^{\circ}\)