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
$2 \pi r T \cos 0^{\circ}=\pi r^2 h^{\prime} \rho g$
$\Rightarrow 2 T \cos \theta=h^{\prime} r \rho g$
$\Rightarrow \cos \theta=\frac{h^{\prime} r \rho g}{2 T}$
$\therefore \cos \theta=\frac{2 \times 10^{-2} \times 0.25 \times 10^{-3} \times 1000 \times 9.8}{2 \times 0.07}$
$\Rightarrow \theta=70^{\circ}$
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$\left[ g =10 m / s ^{2} ; \sin 60^{\circ}=\frac{\sqrt{3}}{2} ; \cos 60^{\circ}=\frac{1}{2}\right]$