Answer
$\frac{P}{P_0} = (\frac{R_0}{R})^4$
If the radius of an artery is reduced to three-fourths of its former value, then the pressure is more than tripled.
Work Step by Step
We can express the rate of flow with the normal values:
$F = \frac{\pi~P_0~R_0^4}{8 ~\eta~ l}$
We can express the rate of flow with the constricted values:
$F = \frac{\pi~P~R^4}{8 ~\eta~ l}$
Since the flow rate is assumed to be the same in both cases, we can equate the two expressions:
$F = \frac{\pi~P~R^4}{8 ~\eta~ l} = \frac{\pi~P_0~R_0^4}{8 ~\eta~ l}$
$P~R^4 = P_0~R_0^4$
$\frac{P}{P_0} = \frac{R_0^4}{R^4}$
$\frac{P}{P_0} = (\frac{R_0}{R})^4$
Suppose that $R = 0.75~R_0$
We can find the value of $P$:
$\frac{P}{P_0} = (\frac{R_0}{R})^4$
$P = (\frac{R_0}{0.75~R_0})^4~P_0$
$P = (\frac{1}{0.75})^4~P_0$
$P = 3.16~P_0$
If the radius of an artery is reduced to three-fourths of its former value, then the pressure is more than tripled.
