## College Physics (4th Edition)

The total rate of flow in the two pipes is $\frac{1}{8}$ of the original flow rate.
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.