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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 $135^o$, the approximate ratio of surface tensions of water and mercury is
$1 : 0.15$
$1 : 3$
$1 : 6$
$1.5 : 1$
Solution
$\mathrm{h}=\frac{2 \sigma \cos \theta}{\mathrm{r} \rho \mathrm{g}} \Rightarrow \sigma \propto \frac{\mathrm{h} \rho}{\cos \theta}$
$\Rightarrow \frac{\sigma_{\mathrm{w}}}{\sigma_{\mathrm{m}}}=\frac{\mathrm{h}_{\mathrm{w}} \rho_{\mathrm{w}}}{\cos \theta_{\mathrm{w}}} \times \frac{\cos \theta_{\mathrm{m}}}{\mathrm{h}_{\mathrm{m}} \rho_{\mathrm{m}}}$
$=\frac{10 \times 1}{\cos 0^{\circ}} \times \frac{\cos 135^{\circ}}{-3.1 \times 13.6}$
$=\frac{10 \times(-0.707)}{-3.1 \times 13.6} \approx \frac{1}{6}$
