91Ó°ÊÓ

The primary equation to use is Jurin's Law which states that when capillary action occurs as a result of surface tension, the height \( h \) that the liquid can rise or depressed is given by: \[ h = \frac{2\sigma \cos (\theta)}{\rho g D} \]Here,\( h \) = height of the liquid column,\( \sigma \) = surface tension of the liquid,\( \theta \) = angle of contact,\( \rho \) = density of liquid,\( g \) = acceleration due to gravity, and\( D \) = diameter of tube.The question asks for the height difference \( \Delta h \) which occurs because of two different fluids, so the height difference \( \Delta h \) will be difference of heights of rise or depression of fluid 1 and fluid 2, which can be expressed as:\[ \Delta h = h_1 - h_2 = \frac{2\sigma_1 \cos (\theta_1)}{\rho_1 g D} - \frac{2\sigma_2 \cos (\theta_2)}{\rho_2 g D} \]This is the expression for the height difference.

For water and mercury at the interface, the known values are \( \sigma_1 = 0.073 N/m \) (for water), \( \sigma_2 = 0.465 N/m \) (for mercury), \( \rho_1 = 1000 kg/m^3 \) (for water), \( \rho_2 = 13534 kg/m^3 \) (for mercury), \( \theta_1 = 0^\circ \) (for water), \( \theta_2 = 140^\circ \) (for mercury) along with the given tube diameter of 40 mils which is equivalent to \( D = 40 \times 0.001 in \times 0.0254 m/in = 0.001016 m \). Substituting these known values into the derived formula: \[ \Delta h = \frac{2\sigma_1 \cos (\theta_1)}{\rho_1 g D} - \frac{2\sigma_2 \cos (\theta_2)}{\rho_2 g D} \]we get\[ \Delta h = \frac{2\times0.073\times \cos (0^\circ)}{1000\times9.8\times0.001016} - \frac{2\times0.465\times \cos (140^\circ)}{13534\times9.8\times0.001016} \]

By performing the calculation in the derived formula with the substituted values, we get the height difference \(\Delta h\). Make sure to convert angles to radians if your calculator uses radians for trigonometric functions.