Force balance gives $\tau_w=\Delta pD/(4L)$,so $D=4(.6)(12.5)/300=.10$ m. Under the Newtonian laminar-flow relation,$Q=\pi D^4\Delta p/(128\mu L)$. Thus $\mu=\pi(.1)^4(300)/[128(.25)(.6)]=.00490874$ Pa·s, so **n≈4.909** (3.14 for π gives 4.906). The official4.902–4.908 range reflects that π convention. Laminar behavior is assumed by this viscosity inversion; no density is supplied to check Reynolds number independently.
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