Liquid mercury (viscosity \(=1.55 \times 10^{-3} \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}^{2}\) ) flows through a horizontal pipe of internal radius \(1.88 \mathrm{~cm}\) and length \(1.26 \mathrm{~m}\). The volume flux is \(5.35 \times 10^{-2} \mathrm{~L} / \mathrm{min}\). \((a)\) Show that the flow is laminar. (b) Calculate the difference in pressure between the two ends of the pipe.

Firstly, the Reynolds number (Re) should be calculated using the following formula: \(Re = \frac{VDρ}{μ}\) where V is the average velocity of the fluid, D is the diameter of the pipe, ρ is the fluid density, and μ is the fluid viscosity. The average velocity can be calculated by dividing the volume flux by the area of the pipe, which in turn is calculated by: \(A=\pi r^{2}\) where r is the radius. The radius of the pipe can be converted to meters, the volume flux to m³/s, then those values can be used to calculate the average velocity. The diameter of the pipe is twice the radius. Using known value of mercury's density (13.6 g/cm⁻³) which should be converted to kg/m⁻³, and given viscosity, now Reynolds number can be calculated. If the Reynolds number is smaller than 2000, this indicates a laminar flow.

Next, the pressure difference between the two ends of the pipe can be calculated using Poiseuille's Law, with the formula: \(ΔP=8μLVQ/πr²\) where μ is the dynamic viscosity, L is the length of the pipe, V is the average velocity calculated before, Q is the volume flow rate, and r is the radius of the pipe. The values given in the problem can be substituted directly into the equation because they are already in proper SI units. The result will be the pressure difference in Pascals (Pa).