A capillary tube has its lower end immersed in mercury, and the...

The height of a liquid in a capillary tube is given by Jurin's law:

$h = \dfrac{2T \cos \theta}{r \rho g}$

For mercury, the capillary fall is:

$h_{Hg} = \dfrac{2T_{Hg} \cos \theta_{Hg}}{r \rho_{Hg} g} = -2 \text{ cm}$

For water, the capillary rise is:

$h_w = \dfrac{2T_w \cos \theta_w}{r \rho_w g}$

Taking the ratio of the two heights to eliminate the radius $r and acceleration due to gravity$g$:<br><br>$\dfrac{h_w}{h_{Hg}} = \left( \dfrac{T_w}{T_{Hg}} \right) \times \left( \dfrac{\rho_{Hg}}{\rho_w} \right) \times \left( \dfrac{\cos \theta_w}{\cos \theta_{Hg}} \right)<br><br>Using the standard physical constants for water and mercury:<br>Surface tension of water,T_w \approx 0.075 \text{ N/m}$<br>Surface tension of mercury,$T_{Hg} \approx 0.465 \text{ N/m}$<br>Density of water,$\rho_w = 1000 \text{ kg/m}^3$<br>Density of mercury,$\rho_{Hg} = 13600 \text{ kg/m}^3$<br>Angle of contact for water,$\theta_w = 0^{\circ} \Rightarrow \cos 0^{\circ} = 1$<br>Angle of contact for mercury,$\theta_{Hg} \approx 140^{\circ} \Rightarrow \cos 140^{\circ} \approx -0.766$<br><br>Substituting these values into the ratio equation:<br><br>\dfrac{h_w}{-2} = \left( \dfrac{0.075}{0.465} \right) \times \left( \dfrac{13600}{1000} \right) \times \left( \dfrac{1}{-0.766} \right)$

$h_w = -2 \times \dfrac{0.075}{0.465} \times 13.6 \times \dfrac{1}{-0.766}$

$h_w = \dfrac{2.04}{0.35619} \approx 5.727 \text{ cm}$

Rounding to two decimal places, the water will rise to a height of $5.73 \text{ cm}$.

Answer: $5.73 \text{ cm}$