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ID: past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/section-a/1/solution

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Past exam of the physics course of the year 1 of the natural sciences course of the University of Cambridge / 2022 / Section A / 1 / Solution by

Codex  0 2026-10-18

By dimensional analysis, power has dimensions

M L^{2} T^{- 3}

, mass density has dimensions

M L^{- 3}

, and angular velocity has dimensions

T^{- 1}

; radians are dimensionless. Thus

M L^{2} T^{- 3} = M^{α} L^{γ - 3 α} T^{- β} .

Matching the three independent dimensions gives

α = 1

,

β = 3

and

γ - 3 α = 2

, hence

α = 1, β = 3, γ = 5, P \propto ρ ω^{3} D^{5} .

With unchanged mass density, angular velocity and the same performance coefficient (a dimensionless quantity), the pump power affinity law gives

\frac{P _{new}}{P _{old}} = (\frac{D _{new}}{D _{old}})^{5} = 1. 1^{5} = 1.61051.

The required power therefore increases by $61.1%$ . This uses the stipulated dependence on only these three quantities, with comparable pump geometry and performance; dimensional analysis alone does not guarantee that unrelated pumps have the same performance coefficient.

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