Each daughter has the planar Poiseuille relation
$$
Q_i=\frac{2h^3}{3\mu}G_i,
\qquad
G_i=-\frac{dp_i}{dx}.
$$
Hence
$$
\boxed{
G_i=\frac{3\mu Q_i}{2h^3}
=O\left(\frac{\mu Q_i}{h^3}\right)}.
$$
The higher-flow daughter therefore requires the larger pressure-gradient magnitude:
$$
Q_2>Q_1\quad\Longrightarrow\quad G_2>G_1.
$$


Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 334 / 1 / c / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-334/1/c/ii/solution