Glycerine flows steadily through a horizontal tube of length $1.5 \; m$ and radius $1.0 \; cm$. If the amount of glycerine collected per second at one end is $4.0 \times 10^{-3} \; kg \; s^{-1}$,what is the pressure difference between the two ends of the tube? (Density of glycerine $= 1.3 \times 10^{3} \; kg \; m^{-3}$ and viscosity of glycerine $= 0.83 \; Pa \; s$). [You may also like to check if the assumption of laminar flow in the tube is correct]

(A) Given:
Length of the tube,$l = 1.5 \; m$
Radius of the tube,$r = 1.0 \; cm = 0.01 \; m$
Mass flow rate,$M = 4.0 \times 10^{-3} \; kg \; s^{-1}$
Density of glycerine,$\rho = 1.3 \times 10^{3} \; kg \; m^{-3}$
Viscosity of glycerine,$\eta = 0.83 \; Pa \; s$
Volume flow rate $V = \frac{M}{\rho} = \frac{4.0 \times 10^{-3}}{1.3 \times 10^{3}} \approx 3.077 \times 10^{-6} \; m^{3} \; s^{-1}$.
Using Poiseuille's formula for pressure difference $p$:
$V = \frac{\pi p r^{4}}{8 \eta l} \implies p = \frac{8 \eta l V}{\pi r^{4}}$
$p = \frac{8 \times 0.83 \times 1.5 \times 3.077 \times 10^{-6}}{\pi \times (0.01)^{4}}$
$p = \frac{3.0647 \times 10^{-5}}{\pi \times 10^{-8}} \approx 9.756 \times 10^{2} \; Pa \approx 9.8 \times 10^{2} \; Pa$.
Checking for laminar flow using Reynolds number $R_e = \frac{4 \rho V}{\pi d \eta}$ where $d = 2r = 0.02 \; m$:
$R_e = \frac{4 \times 1.3 \times 10^{3} \times 3.077 \times 10^{-6}}{\pi \times 0.02 \times 0.83} \approx 0.306$.
Since $R_e < 2000$,the flow is laminar.