 Solution

ID: past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/section-b/7/a/solution

The mass and surface area of the spherical droplet are

m = (4 π /3) ρ r^{3}

and

A = 4 π r^{2}

β

\overset{m}{˙} = 4 π r^{2} β = 4 π ρ r^{2} \overset{r}{˙} .

Therefore

\overset{r}{˙} = β / ρ = μ

, and the initial radius gives

r (t) = r_{0} + μ t, μ = \frac{β}{ρ} .

Here

[β] = kg m^{- 2} s^{- 1}

[ρ] = kg m^{- 3}

, and

[μ] = m s^{- 1}

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