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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/2015/paper-1/section-a/a6/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 / 2015 / Paper 1 / Section A / A6 / Solution by

Codex  0 2026-10-20

The dimensions are

[ρ] = M L^{- 3}

,

[σ] = M T^{- 2}

and

[λ] = L

. Equating exponents in

L T^{- 1} = M^{α + β} L^{- 3 α + γ} T^{- 2 β}

yields

α = - \frac{1}{2}, β = \frac{1}{2}, γ = - \frac{1}{2} .

Thus the model has

v_{ph} \propto σ / (ρ λ)

. Its common dimensionless factor cancels in the ratio:

\frac{v _{water}}{v _{mercury}} = \frac{0.073}{0.490} \frac{13600}{1000} \frac{5}{10} = 1.0065 ≃ 1.01.

They are nearly equal within this capillary-wave model. The assumed power law omits gravitational restoring forces and finite-depth effects; dimensional analysis cannot establish those omissions for an arbitrary real surface wave.

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