 Solution

ID: past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2015/paper-1/section-a/a5/solution

Δ P = 200

Pa and the central flow estimate is

Q = \frac{π ( 2.3 \times 1 0 ^{- 3} ) ^{4} ( 200 )}{8 ( 8.9 \times 1 0 ^{- 4} ) ( 0.50 )} = 4.939 \times 1 0^{- 6} m^{3} s^{- 1} .

σ_{Δ P} = 1 0^{2} + 1 0^{2} = 14.14

Pa. First-order propagation gives

(\frac{σ _{Q}}{Q})^{2} = (4 \frac{0.1}{2.3})^{2} + (\frac{0.3}{8.9})^{2} + (\frac{0.01}{0.50})^{2} + \frac{1 0 ^{2} + 1 0 ^{2}}{20 0 ^{2}} = 0.03678.

Hence

Q = (4.94 \pm 0.95) \times 1 0^{- 6} m^{3} s^{- 1} = (4.94 \pm 0.95) mL s^{- 1} .

The radius sensitivity of Poiseuille-flow uncertainty comes from the fourth-power dependence on radius. If the quoted errors instead mean independent worst-case bounds, the first-order absolute fractional bound is

4 (0.1/2.3) + 0.3/8.9 + 0.01/0.50 + 20/200 = 0.3276

, giving approximately

\pm 1.62 \times 1 0^{- 6}

m³/s. That conservative bound and the standard uncertainty result are different conventions, not interchangeable errors. The supplied flow model presumes a Newtonian fluid and laminar flow that is fully developed; no density is given for an independent Reynolds-number check.

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