To solve this problem we can use the concepts related to the change of flow of a fluid within a tube, which is without a rubuleous movement and therefore has a laminar fluid.
It is sometimes called Poiseuille’s law for laminar flow, or simply Poiseuille’s law.
The mathematical equation that expresses this concept is
[tex]\dot{Q} = \frac{\pi r^4 (P_2-P_1)}{8\eta L}[/tex]
Where
P = Pressure at each point
r = Radius
[tex]\eta =[/tex] Viscosity
l = Length
Of all these variables we have so much that the change in pressure and viscosity remains constant so the ratio between the two flows would be
[tex]\frac{\dot{Q_A}}{\dot{Q_B}} = \frac{r_A^4}{r_B^4}\frac{L_B}{L_A}[/tex]
From the problem two terms are given
[tex]R_A = \frac{R_B}{2}[/tex]
[tex]L_A = 2L_B[/tex]
Replacing we have to
[tex]\frac{\dot{Q_A}}{\dot{Q_B}} = \frac{r_A^4}{r_B^4}\frac{L_B}{L_A}[/tex]
[tex]\frac{\dot{Q_A}}{\dot{Q_B}} = \frac{r_B^4}{16*r_B^4}\frac{L_B}{2*L_B}[/tex]
[tex]\frac{\dot{Q_A}}{\dot{Q_B}} = \frac{1}{32}[/tex]
Therefore the ratio of the flow rate through capillary tubes A and B is 1/32
