Water rises to a height of 10cm in a capillary tube and mercury falls to a depth of 3.42cm in the same capillary tube. If the density of mercury is 13.6gm/cc and the angle of contact is 135⁰, ratio of surface tension for water and mercury is 

In a capillary tube with radius r, the height h through which a liquid will rise is determined by the equation 

$\mathrm{h}=\frac{2\mathrm{Scos\theta }}{\mathrm{r\rho g}}$
In the case of two liquids (water "1" and mercury "2") in similar capillary tubes

$\frac{{\mathrm{h}}_1}{{\mathrm{h}}_2}=\frac{2{\mathrm{S}}_1{\mathrm{cos\theta }}_1}{{\mathrm{r\rho }}_1\mathrm{g}}\times \frac{{\mathrm{r\rho }}_2\mathrm{g}}{2{\mathrm{S}}_2{\mathrm{cos\theta }}_2}$

$\Rightarrow \frac{{\mathrm{h}}_1}{{\mathrm{h}}_2}=\frac{{\mathrm{S}}_1{\mathrm{cos\theta }}_1}{{\mathrm{\rho }}_1}\times \frac{{\mathrm{\rho }}_2}{2{\mathrm{S}}_2{\mathrm{cos\theta }}_2}$

$\Rightarrow \frac{{\mathrm{S}}_1}{{\mathrm{S}}_2}=\frac{{\mathrm{h}}_1}{{\mathrm{h}}_2}\times \frac{{\mathrm{cos\theta }}_2}{{\mathrm{cos\theta }}_1}\times \frac{{\mathrm{\rho }}_1}{{\mathrm{\rho }}_2}=\frac{10}{-3.42}\times \frac{\cos {135}^{°}}{\cos 0^{°}}\times \frac{1}{13.6}$

$\Rightarrow \frac{{\mathrm{S}}_1}{{\mathrm{S}}_2}=\frac{10}{-3.42}\times \frac{-0.707}{1}\times \frac{1}{13.6}=0.152\approx \frac{1}{6.5}$

Hence the correct answer is 1:6.5.