A glass capillary tube of internal radius r=0.25mm is immersed in...

Water wets glass and so the angle of contact is zero. For full rise, neglecting the small mass in the meniscus $$\begin{aligned}2 \pi \mathrm{T}&=\pi \mathrmr^{2} \mathrm{ h} \rho \mathrm{g} \Rightarrow \mathrm{h}=\frac{2 \mathrm{ T}}{\mathrm{r} \rho \mathrm{g}}\left[\because \text{ water wets glass, } \theta=0^{{}^{\circ}}\right] \\ \text{\text{ and } } =\frac{2 \times 007}{0.25 \times 10^{-3} \times 1000 \times 9.8} \end{aligned}$$ As the tube is only $2 \mathrm{ cm}$ above the water and so, water will rise by $2 \mathrm{ cm}$ and meet the tube at an angle such that, $$\begin{aligned}2 \pi r & \operatorname{T} \cos \theta=\pi r^{2} h^{\prime} \rho g \\ \Rightarrow & 2 T \cos \theta= h^{\prime} r \rho g \\ \Rightarrow \cos \theta&=\frac{h^{\prime} r \rho g}{2 T}\\ \text{\text{ and } } =\frac{2 \times 10^{-2} \times 0.25 \times 10^{-3} \times 1000 \times 9.8}{2 \times 0.07} \end{aligned}$$ The liquid will meet the tube at an angle, $\theta\cong 70^{{}^{\circ}}$