Answer 1

a)

Radius of the artery, r = 2 × 10^–3 m

Diameter of the artery, d = 2 × 2 × 10^–3 m = 4 × 10^{– 3} m

Viscosity of blood, eta = `2.084 xx 10^(-3)` Pas

Density of blood, ρ = 1.06 × 10³ kg/m³

Reynolds’ number for laminar flow, N_R = 2000

The largest average velocity of blood is given by the relation:

`V_"arg" = (N_Reta)/(rhod)`

= `(2000xx2.084xx10^(-3))/(1.06xx10^3xx4xx10^(-3))`

= 0.983 m/s

Therefore, the largest average velocity of blood is 0.983 m/s.

b) Flow rate is given by the relation:

`R = pir^2V_"arg"`

`= 3.14 xx (2xx10^(-3))^2 xx 0.983`

`= 1.235 xx 10^(-5) m^3 s^(-1)`

Therefore, the corresponding flow rate is `1.235 xx 10^(-5) m^3 s^(-1)`

Answer 2

Here `r = 2 xx 10^(-3) m; D = 2r = 2xx2xx10^(-3) =  4xx 10^(-3) m`

`eta = 2.084 xx 10^(-3) Pa-s; rho = 1.06 xx 10^3 kg m^(-3)`

For flow to be laminar , `N_R = 2000`

a) Now `v_c = `(N_Reta)/rho_D` = (2000xx(2.084xx10^(-3)))/((1.06xx10^3)xx(4xx10^(-3))) = 0.98 "m/s"`

b)Volume flowering per second  =`pir^2v_c = 22/7 xx (2xx10^(-3))^2 xx 0.98 = 1.23 xx 10^(-5) m^3s^(-1)`