A capillary tube of radius 1 mm is held vertical with the lower end...

The height of water raised in a capillary tube is given by Jurin's law:

$h = \dfrac{2T \cos \theta}{\rho g r}$

For water, assuming standard values $T = 0.075 \text{ N/m}$, $\rho = 1000 \text{ kg/m}^3$, $g = 10 \text{ m/s}^2$, and angle of contact \theta = 0^{}^{{}^{\circ}}:

h = \dfrac{2 \times 0.075 \times \cos 0^{}^{{}^{\circ}}}{1000 \times 10 \times 10^{-3}} = 0.015 \text{ m} = 1.5 \text{ cm}

When the capillary tube is of insufficient length $h'$, the water rises to the top and the meniscus adjusts its angle of contact to $\theta'$ such that the radius of curvature $R$ changes, keeping $h R$ constant. Since $R = \dfrac{r}{\cos \theta}$, we have:

$\dfrac{h}{\cos \theta} = \dfrac{h'}{\cos \theta'}$

Given that the new length is half of the calculated height, $h' = \dfrac{h}{2}$:

$\dfrac{h}{1} = \dfrac{h/2}{\cos \theta'}$

$\cos \theta' = \dfrac{1}{2}$

\theta' = 60^{}^{{}^{\circ}}