Past exam of the physics B course of the University of Cambridge / 2016 / Paper 2 / Section A

The Boltzmann entropy isThe thermodynamic definition of temperature givesHolding the other system parameters fixed, $E=(5N/2)k_{B}T$, so its heat capacity is $C=5N{k}_B/2$. The constants ${\Omega }_0$ and $E_0$ set an additive reference for entropy; neither affects this derivative or the resulting heat capacity.

For a lossless transmission line with characteristic impedance $Z_0$ and resistive termination $R_L$, the voltage reflection coefficient isThe reflected pulse is inverted and has half the incident voltage, so $\Gamma =-1/2$. Substituting the termination givesThe echo delay is a round trip from the observation point to the termination and back, not a one-way travel time. Thus the propagation speed isFor a homogeneous, nonmagnetic coaxial cable dielectric, $v=c/\sqrt{{ϵ}_r}$, giving ${ϵ}_r\simeq 2.25$. More generally the measured speed fixes ${ϵ}_r{\mu }_r=(c/v{)}^2$; identifying this with relative permittivity assumes ${\mu }_r\simeq 1$ and negligible dispersion or attenuation of the pulse.

Separate a ball's orbital motion from its spin. For a rolling ball between concentric races, the ball centre lies $2a$ from the axle axis. Let its tangential centre velocity be $v$ in the direction of axle motion and let its signed spin angular velocity be $\Omega$, positive in the same sense as $\omega$. Rolling without slipping at the moving inner contact and stationary outer contact requiresHenceThe spin is opposite to the axle's rotation, while the centre orbits in the same sense. The total angular momentum about the axle is the sum of its orbital angular momentum and its angular momentum about the centre of mass:For a uniform solid spherical ball, $I_{\mathrm{cm}}=2m{a}^2/5$, soThe solid-ball assumption supplies the usual moment of inertia; the preceding expression in terms of $I_{\mathrm{cm}}$ also states the answer for a different internal mass distribution.

In the thin-gap Couette flow approximation, the tangential velocity varies linearly through the gap. At radius $r$, its difference between the two discs is $r\omega$, so a Newtonian fluid exerts shear stressAn annulus has area $2\pi rdr$, and its tangential force acts at lever arm $r$. Integrating its torque givesThe driving torque is in the direction of rotation; the liquid's resisting torque has the opposite sign. This parallel-disc viscometer result assumes laminar shear with no slip, negligible edge effects and negligible secondary circulation. Its fourth-power radius dependence comes from the combined radial dependence of speed, annular area and lever arm.

At fixed temperature, equilibrium blackbody radiation has fixed energy density $u$, with radiation pressure $p=u/3$. Doubling the volume therefore doubles its internal energy, so $\Delta U=U_0$. The work done by the radiation isThe first law of thermodynamics, with positive work done by the radiation, givesThus $1.00\mathrm{mJ}$ is absorbed from the walls. The walls must replace the energy used to do work as well as supply the extra stored energy; using only $\Delta U=750\mu \mathrm{J}$ would omit that work. This is a reversible isothermal expansion of a photon gas, not an adiabatic expansion.