Given that, Surface tension of water, $T = 7.3 \times 10^{-2} \, \mathrm{Nm^{-1}}$ Density of water, $\rho = 1.0 \times 10^{3} \, \mathrm{kgm^{-3}}$ Contact angle, $\theta = 0^\circ$ Acceleration due to gravity, $g = 9.8 \, \mathrm{ms^{-2}}$ Diameter of the first capillary tube, $d_1 = 3.0 \, \mathrm{mm} = 3.0 \times 10^{-3} \, \mathrm{m}$ Radius of the first capillary tube, $r_1 = \frac{d_1}{2} = 1.5 \times 10^{-3} \, \mathrm{m}$ Diameter of the second capillary tube, $d_2 = 6.0 \, \mathrm{mm} = 6.0 \times 10^{-3} \, \mathrm{m}$ Radius of the second capillary tube, $r_2 = \frac{d_2}{2} = 3.0 \times 10^{-3} \, \mathrm{m}$ The height of the water column in a capillary tube is given by the formula: $$ \begin{aligned} h &= \frac{2T \cos \theta}{r \rho g} \end{aligned} $$ Since $\theta = 0^\circ$, $\cos 0^\circ = 1$, so $h = \frac{2T}{r \rho g}$. The difference in heights of the water columns in the two tubes is: $$ \begin{aligned} \Delta h &= h_1 - h_2 \\ &= \frac{2T}{r_1 \rho g} - \frac{2T}{r_2 \rho g} \\ &= \frac{2T}{\rho g} \left( \frac{1}{r_1} - \frac{1}{r_2} \right) \end{aligned} $$ Substituting the values: $$ \begin{aligned} \Delta h &= \frac{2 \times 7.3 \times 10^{-2}}{10^3 \times 9.8} \left( \frac{1}{1.5 \times 10^{-3}} - \frac{1}{3.0 \times 10^{-3}} \right) \\ &= \frac{14.6 \times 10^{-2}}{9800} \left( 666.67 - 333.33 \right) \\ &= 1.4898 \times 10^{-5} \times 333.34 \\ &= 4.966 \times 10^{-3} \, \mathrm{m} \\ &= 4.966 \, \mathrm{mm} \end{aligned} $$ Therefore, the difference in height between the two liquid surfaces is $4.966 \, \mathrm{mm}$.