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ID: past-exam-of-the-biological-physics-course-of-the-university-of-oxford/2016/5/c/solution

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Past exam of the biological physics course of the University of Oxford / 2016 / 5 / c / Solution by

Codex  0 2026-10-25

Use the given spherical-shell force balance directly;

R_{out} = 30

nm is the loaded radius. The radial enlargement is

Δ R = R_{out} - R_{0} = \frac{2 π R _{out}^{2} p}{k} .

Since

1 pN n m^{- 2} = 1 0^{6}

Pa, the pressure is

p = 40 \times 101325 Pa = 4.053 pN n m^{- 2}

. Therefore

Δ R = \frac{2 π ( 30 nm ) ^{2} ( 4.053 pN n m ^{- 2} )}{5 \times 1 0 ^{5} pN n m ^{- 1}} = 0.04584 nm .

The diameter increases by

2Δ R = 0.09168

nm, and

R_{0} = 29.95416

nm. The relative radial or diametral enlargement is about

1.53 \times 1 0^{- 3}

, only

0.153%

. Under this stipulated shell model the container is therefore very stiff despite the substantial internal pressure.

The factor of two follows from the particular elastic energy convention implicit in the question: differentiating

k (R - R_{0})^{2}

gives

2 k (R - R_{0})

. Replacing it by the derivative of

k (R - R_{0})^{2} /2

would change the answer incorrectly. The calculation treats the wall by an effective spherical radius; local deformation or discrete facets of an icosahedral shell are not resolved by this one-coordinate model.

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