The ratio of surface tensions of mercury and water is given to be $7.5$, while the ratio of their densities is $13.6$. Their contact angles, with glass, are close to $135^\circ$ and $0^\circ$, respectively. If it is observed that mercury gets depressed by an amount $h$ in a capillary tube of radius $r_1$, while water rises by the same amount $h$ in a capillary tube of radius $r_2$, then the ratio $\dfrac{r_1}{r_2}$ is close to
Answer: (D) $\dfrac25$
Capillary rise $h = \dfrac{2T\cos\theta}{r\rho g}$. Equating the magnitudes:
$$\frac{2T_m|\cos135^\circ|}{r_1\rho_mg} = \frac{2T_w\cos0^\circ}{r_2\rho_wg}$$
$$\frac{r_1}{r_2} = \frac{T_m}{T_w}\cdot\frac{\rho_w}{\rho_m}\cdot\frac{1}{\sqrt2} = \frac{7.5}{13.6\times1.414} \approx 0.39 \approx \frac25$$
Solution by Sreeraj P, M.Sc Physics