In a mercury capillary, the mercury level is 8 mm **below** the reservoir level in a tube of radius $5 \times 10^{-4}$ m. Given $T_{Hg} = 0.465$ N/m, $\rho = 13.6 \times 10^3$ kg/m³, $g = 9.8$ m/s², the angle of contact is approximately:

$h = -8\times10^{-3}$ m (depression). $h = 2T\cos\theta/(r\rho g)$ ⇒ $\cos\theta = h r \rho g/(2T)$ $= (-8\times10^{-3})(5\times10^{-4})(13.6\times10^3)(9.8)/(2 \times 0.465) \approx -0.573$ ⇒ $\theta = \arccos(-0.573) \approx 125°$. Mercury–glass contact angle is famously obtuse, ~140°.