Answer
The total rate of flow in the two pipes is $\frac{1}{8}$ of the original flow rate.
Work Step by Step
We can use Poiseuille's law to find an expression for the original flow rate:
$Q_1 = \frac{\pi~P~r^4}{8~\eta~L}$
$Q$ is the flow rate ($m^3~s^{-1}$)
$P$ is the pressure ($Pa$)
$r$ is the radius ($m$)
$\eta$ is the fluid viscosity
$L$ is the length of the tube ($m$)
We can use Poiseuille's law to find an expression for the total flow rate in the two pipes:
$Q_2 = 2\times \frac{\pi~P~(\frac{r}{2})^4}{8~\eta~L}$
$Q_2 = \frac{1}{8}\times \frac{\pi~P~r^4}{8~\eta~L}$
$Q_2 = \frac{1}{8}\times Q_1$
The total rate of flow in the two pipes is $\frac{1}{8}$ of the original flow rate.
