Question

Water rises to a height of $10 \,cm$ in capillary tube and - mercury falls to a depth of $3.1 \,cm$ in the same capillary tube. If the density of mercury is $13.6$ and the angle of contact for mercury is $135^{\circ}$, the approximate ratio of surface tensions of water and mercury is

• A. $1: 0.15$
• B. $1: 3$
• C. $1: 6$
• D. $1.5: 1$

Answer 1

Answer: (C)

$ h =\frac{2 \sigma \cos \theta}{r \rho g} \Rightarrow \sigma \propto \frac{h \rho}{\cos \theta}$
$\Rightarrow \frac{\sigma_{ w }}{\sigma_{ m }} =\frac{ h _{ w } \rho_{ w }}{\cos \theta_{ w }} \times \frac{\cos \theta_{ m }}{ h _{ m } \rho_{ m }}=\frac{10 \times 1}{\cos 0^{\circ}} \times \frac{\cos 135^{\circ}}{-3.1 \times 13.6} $
$=\frac{10 \times(-0.707)}{-3.1 \times 13.6} \approx \frac{1}{6} $