NBPhO 2026

Part (i) — 1.5 / 1.5 pts

The official offers two grading paths. Claude follows the mechanistic “ratio of drag scales” route, so this is graded against Solution 1.

Criterion	Points	Result
$R = F_T/F_L \sim \rho v^{2}/[\mu\,du/dr]$	0.5	✓ "$R = F_T/F_L \sim \rho v^{2}/(\mu v/D)$" set up explicitly as the ratio of the two drag-per-area scales
$du/dr \sim v/D$	0.5	✓ “Across the pipe radius the speed drops from $\sim v$ on the axis to $0$ at the wall over a length of order $D$, so $du/dr \sim v/D$“
$R = \rho v D/\mu$	0.5	✓ Boxed $R = \rho^{1}v^{1}D^{1}\mu^{-1}$, with the powers explicitly read off
Includes numerical factor $> 10$	−0.2	✓ Not applicable — Claude consistently uses $\sim$ (no spurious numerical prefactor introduced)

Part (ii) — 0.5 / 0.5 pts

Criterion	Points	Result
Compute the mean flow speed $v$ from $Q$ via continuity	0.2	✓ $v_{0} = 4Q/(\pi d^{2}) \approx 3.14\,\text{m}\cdot\text{s}^{-1}$
Correctly evaluate $R \approx 37000$	0.2	✓ $R = \rho_{w} v_{0} d/\mu_{w} \approx 3.7\times 10^{4}$
Conclude turbulent flow on the basis of $R \gg 2500$	0.1	✓ ”$R \approx 3.7\times 10^{4} \gg 2500$, the flow in the hose is turbulent”

Part (iii) — 3.0 / 3.0 pts

Criterion	Points	Result
Power balance: gravity term $\rho g h$	0.2	✓ Tabulated as “Lifting water by $h$, $\rho g h$” inside the per-unit-volume budget
Power balance: drag term $\Delta p_\text{drag}$	0.2	✓ “Wall friction in the hose, $\Delta p_{f}$” tabulated as the second budget line
Power balance: exit kinetic-energy term $\rho v_\text{exit}^{2}/2$	0.2	✓ “Kinetic energy of the exiting jet, $\tfrac{1}{2}\rho v_\text{exit}^{2}$” tabulated as the third budget line
Identify $\Delta p_\text{drag} = k v^{2}$ from turbulent-regime scaling	0.3	✓ ”$\Delta p_{f} = (k/2)\rho v_{h}^{2}$, $k$ a dimensionless friction coefficient (constant in $v$)” — equivalent to the official’s $kv^{2}$ form modulo the $\rho/2$ packaging
Calibrate $k \approx 4\times 10^{4}\,\text{Pa}\cdot\text{s}^{2}/\text{m}^{2}$ from the unblocked case	0.3	✓ Claude’s dimensionless $k_{C} \approx 80.97$ converts to the official’s units as $\rho k_{C}/2 = 1000\times 80.97/2 \approx 4.05\times 10^{4}\,\text{Pa}\cdot\text{s}^{2}/\text{m}^{2}$, matching the keyed value
Mass continuity giving $Q_{m} = Q'_{m}$	0.3	✓ “continuity demands $Q_{1} = v_{h}\,A = v_\text{exit}\,fA$“
$v'_\text{exit} = v'/f$	0.2	✓ Boxed $v_\text{exit} = v_{h}/f$
Updated power balance with the kinetic-energy term boosted by $1/f^{2}$	0.5	✓ "$P = Q_{1}[\rho g h + (k+1/f^{2})\tfrac{1}{2}\rho v_{h}^{2}]$" — the $1/f^{2}$ factor on the KE term carried explicitly into the cubic-in-$\alpha$ equation $(\star\star)$
Volumetric flow rate ratio $Q'/Q \approx 0.88$ within $\pm 0.02$	0.4	✓ Boxed $\alpha = Q_{1}/Q_{0} \approx 0.884$, with the cubic residual at $\alpha = 0.884$ down to $\sim 6\,\text{Pa}$ (0.001%)
Formula for excess-pressure ratio $(v'_\text{exit}/v_{0})^{2}$	0.2	✓ "$\Delta p_\text{excess,1}/\Delta p_\text{excess,0} = v_\text{jet,1}^{2}/v_\text{jet,0}^{2} = \alpha^{2}/f^{2}$"
Excess-pressure ratio $\approx 35$ within $\pm 1$	0.5	✓ Boxed $\alpha^{2}/f^{2} \approx 34.7$
Accuracy $> 1\,\%$	−0.1	✓ Not applicable — root accurate to $\sim 0.04\,\%$ on the cubic and $\sim 0.05\,\%$ on the cleaning-pressure ratio

Overall score: 5.0 / 5.0 pts — full marks

Numerical answers match the official key on every part: $R \approx 3.7\times 10^{4}$ (exact match), turbulent classification (match), $\alpha \approx 0.884$ vs official $0.88$ (match within tolerance), cleaning-pressure ratio $\approx 34.7$ vs official $\approx 35$ (match within tolerance).

Commentary

Where this solution goes beyond the grading scheme. The “Overview” front-loads three structural ideas — Reynolds number as the ratio of two drag scales, turbulent friction as a $v^{2}$ law calibratable from one operating point, and the nozzle as an “accelerator” that buys cleaning pressure at the cost of flow rate — and quotes three dimensionless numbers $\rho gh/(P/Q_{0}) \approx 1/3$, $\tfrac{1}{2}\rho v_{0}^{2}/(P/Q_{0}) \approx 0.008$, $1/f^{2} \approx 44$ that already tell the reader (a) friction dominates the unblocked budget, (b) the bare KE term is negligible, and (c) the nozzle promotes KE to the leading channel. Part (i) closes with an “Educational remark” deriving the same answer from pure dimensional analysis, then noting that $\delta \sim D/R = \mu/(\rho v)$ is a consequence of $R$ rather than an independent input. Part (ii) supplies the boundary-layer-thickness number $\delta \sim d/R \sim 0.3\,\mu\text{m}$ as an a posteriori check that the $\delta \ll d$ assumption underpinning the $\rho v^{2}$ wall-stress scaling is comfortably satisfied, and uses this to justify treating the kinetic-energy density as $\tfrac{1}{2}\rho v^{2}$ in part (iii) without a kinetic-energy correction factor. Part (iii) carries the most extensive commentary: an explicit 33%/67%/<1% gravity/friction/KE breakdown of the unblocked pump budget (framing the gardener’s hose as “energetically, a heater”); a small-$f$ scaling argument that the cleaning pressure rises only as $f^{-2/3}$ rather than $f^{-2}$ once the kinetic-energy term saturates the budget (so there are diminishing returns to ever-narrower nozzles); a Reynolds-number sanity check at the new operating point ($R_{1} \approx 3.3\times 10^{4}$, plus a nozzle-exit Reynolds number $\sim 9\times 10^{4}$) confirming the $\Delta p_{f} \propto v^{2}$ law remains valid throughout; an explicit justification for ignoring nozzle friction on the order-of-magnitude grounds $\Delta p_{f}^{\text{nozzle}} \sim \rho v_\text{exit}^{2}\cdot O(1)$ given an unspecified short nozzle length; a discussion of why the constant-power pump idealisation is the natural minimal model for the data given; and a final consistency check that re-allocates the new operating-point budget to gravity (29%), friction (46%), exit KE (25%) and verifies the sum closes to $P/Q_{1} \approx 6.79\times 10^{5}\,\text{Pa}$ — quantifying the headline observation that the KE term has gone from $<1\,\%$ to a quarter of the total.

Where the official solution is sharper. Three places. (1) The official’s part-(i) “Solution 2” supplies an independent dimensional-analysis derivation that explicitly sets up the system $\{r+m=0,\ d-3r-m+u=0,\ -m-u=0\}$ and resolves the residual one-parameter family by appealing to $\mu$ entering the laminar drag with power $-1$. Claude mentions the dimensional-analysis path in a one-sentence aside but never works through this constraint argument; if a student attempted only Solution 2, the constraint step is exactly the subtle point the grading scheme explicitly weights at 0.3 pts. (2) The cubic in part (iii) is recast in dimensionless form $x[a + (b + c/f^{2})x^{2}] = 1$ with $a$, $b$, $c$ identified as the unblocked gravity/friction/KE fractions of the pump budget — and from there $b/c \approx 81$ but $c/f^{2} \approx b$ makes it visually obvious why the friction and boosted-KE contributions are now comparable. Claude has the same fractions (33%/67%/<1%) but only in dimensional form, and his cubic remains in pressure units throughout. The official’s three-fraction normalisation is genuinely slicker pedagogy. (3) The official solves the cubic by the fixed-point iteration $x = (0.972 - 0.318 x)^{1/3}$, displaying the four-step convergence trajectory $1 \to 0.868 \to 0.886 \to 0.884 \to 0.884$ — a self-contained recipe a student can reproduce on a calculator. Claude uses Newton’s method and tabulates four function values without writing out the derivative recomputation per step; the algebra is correct but the official’s iteration is more transparent for a hand-computed problem.