Question

Water rises to a height of $10cm$ in capillary tube and - mercury falls to a depth of $3.1cm$ 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)

$h=\frac{2\sigma \cos \theta }{r\rho g}\Rightarrow \sigma \propto \frac{h\rho }{\cos \theta }$
$\Rightarrow \frac{{\sigma }_w}{{\sigma }_m}=\frac{h_w{\rho }_w}{\cos {\theta }_w}\times \frac{\cos {\theta }_m}{h_m{\rho }_m}=\frac{10\times 1}{\cos 0^{\circ }}\times \frac{\cos 135^{\circ }}{-3.1\times 13.6}$
$=\frac{10\times (-0.707)}{-3.1\times 13.6}\approx \frac{1}{6}$