Negligible rail resistance makes all left-hand groove ends equipotential and all right-hand ends equipotential. The $N$ equal <hydraulic resistances> therefore connect in parallel:
$$
\boxed{R_{\rm chip}=\frac{R_g}{N}=\frac{R_g}{50}.}
$$
Here $\Delta p=R_{\rm chip}Q$ uses the total inlet flow. The two blind rail ends carry no flow beyond their final grooves.

Solution

Past exam of the MPhil in micro- and nanotechnology enterprise course of the University of Cambridge / 2019 / Paper 2 / Section A(iii) / 19 / i / Solution

 Solution (source code)