(a) What is the pressure difference required to make blood flow through an artery of inner radius \(2.0 \mathrm{mm}\) and length \(0.20 \mathrm{m}\) at a speed of \(6.0 \mathrm{cm} / \mathrm{s} ?\) (b) What is the pressure difference required to make blood flow at \(0.60 \mathrm{mm} / \mathrm{s}\) through a capillary of radius \(3.0 \mu \mathrm{m}\) and length \(1.0 \mathrm{mm} ?\) (c) Compare both answers to your average blood pressure, about 100 torr.

The formula for Poiseuille's Law is: \(\Delta P = \dfrac{8 \eta L Q}{\pi r^4}\), where \(\Delta P\) is the pressure difference, \(\eta\) is the dynamic viscosity of the blood (assumed constant), \(L\) is the length of the vessel, \(Q\) is the flow rate, and \(r\) is the inner radius of the vessel.

We know that flow rate, \(Q = A \times v\), where \(A\) is the cross-sectional area of the vessel and \(v\) is the flow velocity. So, \(Q = \pi r^2 v\). (a) For the artery, \(r = 2.0 \mathrm{mm}\) and \(v = 6.0 \mathrm{cm} / \mathrm{s}\). First, convert the given values to SI units: \(r = 2.0 \times 10^{-3} \mathrm{m}\), and \(v = 0.06 \mathrm{m} / \mathrm{s}\). Now, calculate the flow rate for the artery: \(Q_{artery} = \pi (2.0 \times 10^{-3})^2 (0.06) = 7.54 \times 10^{-7} \mathrm{m^3/s}\). (b) For the capillary, \(r = 3.0 \mu \mathrm{m}\) and \(v = 0.60 \mathrm{mm} / \mathrm{s}\). Convert the given values to SI units: \(r = 3.0 \times 10^{-6} \mathrm{m}\), and \(v = 6.0 \times 10^{-4} \mathrm{m} / \mathrm{s}\). Now, calculate the flow rate for the capillary: \(Q_{capillary} = \pi (3.0 \times 10^{-6})^2 (6.0 \times 10^{-4}) = 1.69 \times 10^{-11} \mathrm{m^3/s}\).

Assuming the viscosity of blood to be constant, \(\eta = 4 \times 10^{-3} \mathrm{Pa\cdot s}\) (at 37°C): (a) Substitute the given values for the artery and the calculated flow rate into Poiseuille's Law formula: \(\Delta P_{artery} = \dfrac{8(4\times 10^{-3})(0.20)(7.54\times 10^{-7})}{\pi (2.0\times 10^{-3})^4} = 603.53 \mathrm{Pa}\). (b) Substitute the given values for the capillary and the calculated flow rate into Poiseuille's Law formula: \(\Delta P_{capillary} = \dfrac{8(4\times 10^{-3})(1.0\times 10^{-3})(1.69\times 10^{-11})}{\pi (3.0\times 10^{-6})^4} = 5.65 \times 10^4 \mathrm{Pa}\).

The average blood pressure is given as 100 torr. We will convert it to pascals: \(100 \mathrm{torr} \times \dfrac{101325 \mathrm{Pa}}{760 \mathrm{torr}} \approx 13332 \mathrm{Pa}\). (a) For the artery, the pressure difference required is \(603.53 \mathrm{Pa}\), which is less than the average blood pressure of \(13332 \mathrm{Pa}\). (b) For the capillary, the pressure difference required is \(5.65 \times 10^4 \mathrm{Pa}\), which is about 4 times greater than the average blood pressure of \(13332 \mathrm{Pa}\). The pressure difference required in the capillary is much greater than that in the artery and significantly higher than the average blood pressure. This suggests that there may be other factors affecting blood flow in capillaries that are not accounted for in Poiseuille's Law or our simplifying assumptions.