Question

liquid a is known to have a higher surface tension and higher viscosity than liquid b. use these facts to predict the result of each experiment in the table below, if you can.

experiment	predicted outcome
identical wire loops are dipped into liquid a and liquid b, so that a film of liquid forms across the loops (like the bubble solution on a childs bubble blowing wand). the width of each loop is increased slowly and the forces ( f_a ) and ( f_b ) needed to make the loops 5% wider are measured.	( \bullet ) ( f_a ) will be greater than ( f_b ) <br> ( circ ) ( f_a ) will be less than ( f_b ) <br> ( circ ) ( f_a ) will be equal to ( f_b ) <br> ( circ ) its impossible to predict whether ( f_a ) or ( f_b ) will be greater without more information.

Explanation:

First experiment

• Step1: Recall the Hagen - Poiseuille law

The Hagen - Poiseuille law for volumetric flow rate \(Q=\frac{\pi r^{4}\Delta P}{8\eta L}\), where \(Q\) is the volumetric flow rate, \(r\) is the radius of the tube, \(\Delta P\) is the pressure difference, \(\eta\) is the viscosity, and \(L\) is the length of the tube.
We are given that \(Q = 1.9\space mL/s\), \(r=\frac{40.0}{2}=20.0\space mm\) (radius of the tube), and \(L\) (assuming the length of the tube is the same for both liquids). Rearranging for \(\Delta P\) gives \(\Delta P=\frac{8\eta LQ}{\pi r^{4}}\).
Since we are comparing \(P_A\) and \(P_B\) (where \(P\) is related to \(\Delta P\) in the context of the problem, assuming the same driving force and tube length), and we know that \(Q\), \(r\), and \(L\) are the same for both liquids. But we are not given any information about the relationship between the viscosities \(\eta_A\) and \(\eta_B\) in terms of their numerical values relative to the tube radius and length.

Second experiment

• Step1: Recall the formula for surface - tension force

The force due to surface tension \(F = 2\gamma l\) (for a loop, considering two surfaces), where \(\gamma\) is the surface tension and \(l\) is the length (perimeter related to the expansion of the loop).
We are told that liquid \(A\) has a higher surface tension (\(\gamma_A>\gamma_B\)). The wire loops are identical (\(l\) is the same for both loops as we are increasing the width of the loop by the same percentage \(5\%\)). Using the formula \(F = 2\gamma l\), when \(l\) is the same for both loops (since the loops are identical and we are making the same percentage change in width), and \(\gamma_A>\gamma_B\), then \(F_A = 2\gamma_A l\) and \(F_B=2\gamma_B l\).

Answer:

• For the first experiment: It's impossible to predict whether \(P_A\) or \(P_B\) will be greater without more information.
• For the second experiment: \(F_A\) will be greater than \(F_B\)