Past exam of the physics B course of the University of Cambridge / 2012 / Paper 1 / Section A

Use absolute thermodynamic temperatures and assume a constant-parameter Arrhenius equation, with composition held fixed so that reaction rate is proportional to the rate constant. Write $R(T)=A{e}^{-\beta /T}$, where $\beta =E_a/\mathcal{R}$. two-temperature Arrhenius extrapolation eliminates the unknown Arrhenius prefactor and givesConsequentlyA tenfold increase for one $20\mathrm{K}$ interval does not imply another tenfold increase for the next interval: the Arrhenius plot is linear in inverse temperature. Two measured rates alone do not specify a unique extrapolation without a rate law for its temperature dependence; the result above states that modelling assumption explicitly.

Choose the plate polarity so that the gap electric field points along $+z$ when $V>0$. Away from the plate edges, the quasistatic electromagnetic approximation gives $E_z=V_0\sin \omega t/d$. A circular Ampere-Maxwell law contour of radius $r<a$ then givesIt is displacement current, rather than a conduction current across the air, that supplies the enclosed current term. The direction follows the right-hand rule for increasing positive electric flux; it reverses every half cycle. In the ideal extension that sets the gap electric field to zero outside the plate footprint, $B_{\varphi }={\mu }_0{ϵ}_0\omega V_0{a}^2\cos \omega t/(2dr)$ for $r>a$. That extension is an edge approximation, not an exact exterior solution for finite plates. Neglect fringing field of a capacitor and retardation, requiring $d\ll a$ and $\omega a/c\ll 1$.

For a hollow, air-filled rectangular electromagnetic waveguide with conducting walls, transverse confinement givesA transverse electric mode of a rectangular waveguide permits one zero index, but not two; a transverse magnetic mode of a rectangular waveguide requires both indices positive. Take $w=20\mathrm{mm}$ and $h=10\mathrm{mm}$. The dominant mode is ${\mathrm{TE}}_{10}$, withThe waveguide cutoff frequency is the limiting value: positive axial group velocity and propagating power require $f>f_c$. There is no transverse electromagnetic mode in this single hollow conducting pipe; otherwise a zero cutoff would incorrectly follow.

The Stefan–Boltzmann law gives solar luminosity $L_{\odot }=4\pi R_{\odot }^2\sigma T_{\odot }^4$. At orbital distance $r$, the incident irradiance is $\sigma T_{\odot }^4{R}_{\odot }^2/r^2$. A spherical absorber intercepts a disc of area $\pi R_M^2$ but radiates over $4\pi R_M^2$. Thus the uniform-temperature radiative equilibrium estimate isThe planetary radius cancels. This is a black-body radiation estimate with zero reflected fraction and complete redistribution of absorbed heat. More generally the balance fixes $⟨T^4{⟩}^{1/4}$, rather than the arithmetic mean of a nonuniform surface temperature. It excludes atmospheric heating and thermal lag; no measured surface temperature is inferred.

For the inertia tensor of a uniform cube, reflection symmetry makes every off-diagonal element of the central inertia tensor vanish. Permutation of the three axes makes its diagonal elements equal. Since $M=8\rho a^3$ and the mean of each squared coordinate in the cube is $a^2/3$,Every central axis is therefore a principal axis, including the body diagonal. The angular momentum is parallel to the angular velocity, and its magnitude is