Q.A capillary tube of internal radius $0.50\ \text{mm}$ is dipped vertically into a trough of mercury. Taking the surface tension of mercury as $0.465\ \text{N/m}$, its angle of contact with glass as $135°$, and its density as $13{,}600\ \text{kg/m}^3$, calculate the depression (fall) of the mercury level inside the capillary tube below the free mercury level in the trough. (Take $g = 9.8\ \text{m/s}^2$.)

Given: $T = 0.465\ \text{N/m}$, $\theta = 135°$, $\rho = 13{,}600\ \text{kg/m}^3$, $r = 0.50\ \text{mm} = 5.0\times10^{-4}\ \text{m}$, $g = 9.8\ \text{m/s}^2$.

Since $\theta = 135°$ is obtuse, $\cos135° = -\cos45° \approx -0.7071$.

Numerator: $2\times0.465\times0.7071 \approx -0.6576$ (magnitude $0.6576$, negative sign from $\cos\theta$).

Denominator: $13{,}600\times9.8\times5.0\times10^{-4} = 66.64$.

$h \approx \frac{-0.6576}{66.64} \approx -9.87\times10^{-3}\ \text{m}$ …

Substitute the given surface tension, angle of contact (using $\cos135°<0$), mercury density, and tube radius into $h = 2T\cos\theta/(\rho g r)$; interpre …

• Treating the obtuse angle of contact as if it were acute and using $+\cos45°$ instead of $\cos135° = -\cos45°$, which would wrongly predict a rise instead of a depression. …