Past exam of the physics B course of the University of Cambridge / 2010 / Paper 2B

Original paper: Public original PDF, physical pages 273–278.

Sections A–C coincide with those in paper 2A. Each answer below gives its result and links to the shared derivation, including its assumptions and source qualifications.

The reversible isothermal ideal-gas work is $Q=R{T}_0\ln 2$; the heat would not be fixed by endpoints alone without the process assumption.

Full derivation: solution.

Pairing adjacent contributions from the Biot-Savart law gives magnitude ${\mu }_{0}IN(N+1)/(4\pi a)$, directed opposite to the first wire's field at the observation point.

Full derivation: solution.

The lowest transverse electric mode has cutoff frequency $c/(2a)\simeq 15.0\mathrm{GHz}$; propagation requires frequency strictly above cutoff.

Full derivation: solution.

Two distinct reversible adiabats would form a one-reservoir cycle with an isotherm, contradicting the Kelvin-Planck statement of the second law.

Full derivation: solution.

The trihydrogen cation has one breathing normal mode and two degenerate in-plane distortion normal modes. Rigid translations and rotations do not count as vibrations.

Full derivation: solution.

For steady fields in two leaky dielectric layers, set $\mathcal{D}={\sigma }_2{d}_1+{\sigma }_1{d}_2$. Then $E_1={\sigma }_{2}V/\mathcal{D}$, $E_2={\sigma }_{1}V/\mathcal{D}$, $J={\sigma }_1{\sigma }_{2}V/\mathcal{D}$. The interfacial free and total charges are $({ϵ}_2{\sigma }_1-{ϵ}_1{\sigma }_2)V/\mathcal{D}$ and ${ϵ}_0({\sigma }_1-{\sigma }_2)V/\mathcal{D}$, respectively.

Full derivation: solution.

During partial entry, resistive magnetic braking gives $v=g\tau (1-e^{-t/\tau })$ with $\tau =mR/(B_{0}a{)}^2$. The lossless inductive limit gives $x=g(1-\cos \omega t)/{\omega }^2$, $v=g\sin \omega t/\omega$, with $\omega =B_{0}a/\sqrt{mL}$. Complete immersion changes the equations; the solution below distinguishes reversal from full entry.

Full derivation: solution.

The parallel-plate transmission line has per-length $C=ϵb/a$ and $L=\mu a/b$. The lossless telegrapher's equations imply wave speed $1/\sqrt{LC}=1/\sqrt{\mu ϵ}$, independent of the much slower electron drift velocity.

Full derivation: solution.

The image charge for a grounded conducting sphere is $-qa/d$ at radius $a^2/d$. With symmetric real charges, their exact force-balance equation has a solution close to $d/a=8$. Holding potential $V$ adds force $qaV/d^2$ to the grounded result, approximately $qV/(64a)$ at that separation.

Full derivation: solution.

Bernoulli's theorem and conservation of mass give $t=r_1^2/(2r)$. Ignoring weight, the stated stagnant-rear-pressure model gives zero resultant at $r_2=\sqrt{2}{r}_1$. Including the leading film-weight estimate gives a cancellation radius near $2g{r}_1^2/v^2$. The printed inequality does not establish negligible force at every larger radius; the shared derivation explains this source qualification.

Full derivation: solution.

Rolling without slipping gives ${\omega }^2=g/[(L-r)(1+2R^2/(5r^2))]$. With the supplied dimensions, $T\simeq 80.5\mathrm{s}$ and the maximum centre-of-mass speed for a $10\mathrm{mm}$ rise is $0.0349\mathrm{m}{\mathrm{s}}^{-1}$.

Full derivation: solution.

The torsional wave equation in a circular shaft has speed $\sqrt{G/\rho }$. The given shaft has torque $100\mathrm{N}\mathrm{m}$, twist $9.95\times 10^{-3}\mathrm{rad}$, strain energy $0.497\mathrm{J}$ and kinetic energy $181\mathrm{J}$. The respective transit-time estimates are $786\mathrm{W}$ and $286\mathrm{kW}$; these must not be confused with independent bounds on the actual $30\mathrm{kW}$ stress-work flux.

Full derivation: solution.

Stable circular power-law orbits require $nB>0$ and $n>-2$, excluding $n=0$. Bertrand's theorem selects $n=-1,2$ when every bound orbit must close. For $n=2+ϵ$, the near-circular ellipse orientation precesses at $-ϵ\Omega /8$ to first order.

Full derivation: solution.