Past exam of the physics B course of the University of Cambridge / 2019 / Paper 2

The units supplied for 130 indicate specific heat capacity, so the block's total heat capacity is $C=mc=0.5(130)=65\mathrm{J}{\mathrm{K}}^{-1}$. Treat it as constant, and the lake as a thermal reservoir at $T_0=283.15\mathrm{K}$. The initial block temperature is $T_i=373.15\mathrm{K}$.

Compute the block's entropy change using a reversible comparison path, even though the actual cooling is irreversible:The lake receives $Q=C(T_i-T_0)=5850\mathrm{J}$ at constant temperature, soThereforeThis is entropy generated by cooling a finite body in a reservoir. The positive total is consistent with the Second law of thermodynamics, although the cooling block's own entropy decreases.

Initially and finally the block and lake are at the same temperature. After the falling block has come to rest, its lost gravitational potential energy has dissipated into internal energy of the lake and surroundings. In the idealization that all $mgh$ is ultimately thermalized in the lake at $T_0=283.15\mathrm{K}$,The block returns to its original thermal state, so its net entropy change is zero. Mechanical potential energy carries no thermal entropy that must be subtracted. ThusUse of $g=9.80\mathrm{m}{\mathrm{s}}^{-2}$ gives the same three-significant-figure answer. Splashing and any temporary heating do not change this final-state balance if their mechanical energy is eventually dissipated in the same reservoir.

Three phase coexistence curves meet at the triple point, where ice, liquid water and water vapour have equal chemical potential. In the ordinary-pressure region, the low-temperature phase is ice, the high-pressure moderate-temperature phase is liquid, and the low-pressure high-temperature phase is vapour. The triple point is approximatelyBelow $p_t$, warming stable ordinary ice crosses the sublimation boundary directly into vapour; there is no intervening stable liquid. Above $p_t$, melting and then boiling can occur. The liquid–vapour curve terminates at the liquid-gas critical point, approximately $647.1\mathrm{K}$ and $22.06\mathrm{MPa}$, beyond which there is no liquid–vapour phase boundary.

The distinctive feature is the negative slope of the ordinary ice–water melting line. The Clapeyron equation givesMelting increases entropy but reduces volume because ordinary ice is less dense than liquid water, so $\Delta s>0$ and $\Delta v<0$. Increasing pressure consequently lowers the melting temperature locally. Sublimation and vaporization both increase volume and entropy, giving positive slopes. The following original schematic shows these signs; it is not a quantitative fit and omits the additional high-pressure ice polymorphs.

Assume an air-filled or evacuated lossless electromagnetic waveguide, and a pulse bandwidth narrow enough to use its carrier's local group velocity. The longitudinal dispersion relation isIt givesHere $f_c/f=20/40=1/2$, soThe phase velocity exceeds $c$, but the pulse envelope travels at the subluminal group velocity; a moving constant-phase surface does not by itself transport information. Their product is $v_{\mathrm{p}}{v}_{\mathrm{g}}=c^2$. A broad pulse can distort because this waveguide is dispersive.

The original PDF calls ${ϵ}_r$ a relative magnetic permeability while also specifying a nonmagnetic dielectric. These descriptions are inconsistent if read literally. The physically consistent interpretation of the notation is relative permittivity ${ϵ}_r=2$ and relative magnetic permeability ${\mu }_r=1$.

Let $a=3\mathrm{mm}$ and let $b=6\mathrm{mm}$ be the inner radius of the outer conductor. For line charge density $\lambda$, Gauss's law gives $E_r=\lambda /(2\pi ϵr)$, henceThe magnetic field in the dielectric is $H_{\phi }=I/(2\pi r)$. Its energy per unit length givesUsing $b/a=2$,These are the lossless coaxial line parameters, assuming negligible field penetration into the conductors. If the quoted radii instead referred to an outer diameter or if ${\mu }_r=2$ were intended, the data would need clarification; the stated dielectric and nonmagnetic assumptions support the interpretation above.

For an ideal electrical transformer, let $n=N_p/N_s$. The reflected secondary load is $R_{\mathrm{in}}=n^2{R}_L$. Matching a $4\Omega$ load to an $8\Omega$ source therefore requiresWith the generator replaced by its internal resistance, its resistance referred to the secondary isThis is the source output resistance relevant to matching and maximum power transfer.

To find the windings, use the inductance of a toroidal winding. The mean magnetic path is $\ell =2\pi (0.10)\mathrm{m}$, the core area is $A=30\times 10^{-4}=0.003{\mathrm{m}}^2$ and $\mu =500{\mu }_0$. ThusThe ideal secondary turn count is $N_s=500/\sqrt{2}=353.55$, giving the practical nominal designWith integer turns the irrational ratio cannot be exact: $500:354$ reflects the $4\Omega$ load as $7.98\Omega$, an approximately $0.25\%$ resistance mismatch. These turn counts use the stated $0.75\mathrm{H}$ self-inductance; no operating frequency was specified from which to set a separate minimum inductance.

The $200\Omega$ condition must refer to the unloaded magnetizing reactance, not to the matched loaded input resistance. With $L_p=0.75\mathrm{H}$,The secondary self-inductance is $L_s=L_p/n^2=0.375\mathrm{H}$, so its corresponding unloaded coil impedance satisfiesIf output impedance means this transformed coil reactance, $100\Omega$ is its minimum magnitude. It is distinct from the $4\Omega$ transformed source resistance. This is the magnetizing reactance and reflected resistance in a transformer distinction. A finite magnetizing inductance adds a parallel branch, $Z_{\mathrm{in}}=(8\Omega )\parallel (i\omega L_p)$, so exact purely resistive matching is an ideal large-reactance approximation; interpreting the loaded input itself as both $8\Omega$ and at least $200\Omega$ would be impossible.

In a homogeneous linear medium, write $\mathbf{D}=ϵ\mathbf{E}$ and $\mathbf{B}=\mu \mathbf{H}$. Without free charge or current, Maxwell's equations becomeTaking the curl of Faraday's law, using $\nabla \times (\nabla \times \mathbf{E})=\nabla (\nabla \cdot \mathbf{E})-{\nabla }^2\mathbf{E}$, givesTheir plane-wave solutions have ${\omega }^2=k^2/(\mu ϵ)$ and speed $v=1/\sqrt{\mu ϵ}$; in vacuum this is $c=1/\sqrt{{\mu }_0{ϵ}_0}$. For a wave $\mathrm{Re}[{\mathbf{E}}_0{e}^{i(\mathbf{k}\cdot \mathbf{r}-\omega t)}]$, the divergence equations give $\mathbf{k}\cdot {\mathbf{E}}_0=\mathbf{k}\cdot {\mathbf{B}}_0=0$, while Faraday's law gives ${\mathbf{B}}_0=\mathbf{k}\times {\mathbf{E}}_0/\omega$. Thus the fields and propagation direction form an orthogonal triad, with $B_0=E_0/v$.

The two independent transverse electric components allow linear, circular and elliptical polarization of an electromagnetic wave. Their relative amplitudes and phase determine the polarization state. In a lossless nondispersive medium the energy density and energy flux areA travelling plane wave has equal electric and magnetic energy contributions and time-averaged intensity $⟨S⟩=E_0^2/(2Z)$, where $Z=\sqrt{\mu /ϵ}$ is its wave impedance and $E_0$ denotes peak electric-field amplitude. Reflection and refraction follow by imposing the electromagnetic boundary conditions. Frequency-dependent constitutive response introduces dispersion; conductivity introduces attenuation, so the simple real constant-speed result has a specified domain of validity.

A dielectric responds to an electric field mainly by displacing or orienting bound charges, rather than by sustaining a steady conduction current. The polarization density $\mathbf{P}$ is electric dipole moment per unit volume. It creates bound chargeSeparating these charges from free charge motivatesFor an isotropic linear dielectric, $\mathbf{P}={ϵ}_0{\chi }_e\mathbf{E}$, soThe electric susceptibility ${\chi }_e$ describes material polarization and is not the magnetic susceptibility. More generally it is a tensor and can depend on frequency, field strength and temperature.

Electronic polarization displaces an Electron cloud relative to its nucleus; ionic polarization displaces positive and negative ions; orientational polarization partially aligns permanent molecular dipoles against thermal disorder. For independent permanent moments of magnitude $p$, weak-field orientational response gives ${\chi }_e\simeq n{p}^2/(3{ϵ}_0{k}_{B}T)$ when local-field corrections and interactions are negligible. This explains a common inverse-temperature contribution without asserting that every dielectric follows that law.

At a static interface, $\mathbf{n}\cdot ({\mathbf{D}}_2-{\mathbf{D}}_1)={\sigma }_f$ and the tangential electric field is continuous. Bound surface charge is already included through $\mathbf{P}$. For a homogeneous filled capacitor, $C=ϵA/d$, and a reversible linear dielectric stores energy density $u=\mathbf{E}\cdot \mathbf{D}/2$. At fixed voltage, insertion raises the stored field energy but the voltage source also supplies energy; force arguments must include that source. At fixed charge the capacitor energy instead decreases on insertion.

At frequencies too high for a polarization mechanism to follow the field, its contribution weakens. A complex permittivity represents dissipative dielectric loss as well as phase lag; the sign of its imaginary part depends on the time convention. Strong fields can cause nonlinear response or dielectric breakdown, beyond the reversible linear model.

In a homogeneous Ohmic conductor, $\mathbf{J}=\sigma \mathbf{E}$. Combining this with Maxwell's equations givesFor a good conductor, $\sigma \gg \omega ϵ$, so conduction dominates displacement current. With time dependence $e^{-i\omega t}$ and depth $z>0$, take $E\propto e^{ikz}$. ThenThe field is consequently proportional to $e^{-z/\delta }{e}^{iz/\delta }$: it both attenuates and changes phase with depth. The skin depth is the distance for amplitude to fall by $e^{-1}$; power density falls as $e^{-2z/\delta }$, so its attenuation length is $\delta /2$.

Physically, changing fields induce currents whose magnetic field opposes penetration. Equivalently, after neglecting displacement current, magnetic diffusion obeys ${\partial }_t\mathbf{B}=(\mu \sigma {)}^{-1}{\nabla }^2\mathbf{B}$. Faster changes have less time to diffuse into the metal. Increasing frequency, conductivity or magnetic permeability reduces $\delta$, provided the relevant material parameters remain applicable.

For a thick conductor the tangential surface electrical impedance is $Z_s=(1-i)/(\sigma \delta )$ with the stated time convention and consistent field orientation. Its real part $1/(\sigma \delta )$ represents dissipative resistance per square. Current crowding near the surface raises the AC resistance of a wire when $\delta$ is small compared with its radius; thin strands and laminated magnetic cores reduce suitable eddy-current losses. When $\sigma /\gg \omega ϵ$, the full complex propagation constant must replace the good-conductor approximation. In a magnetic metal the magnetic permeability can itself be nonlinear and frequency dependent.

A transmission line must be treated as a distributed circuit when its propagation time is significant on the signal timescale. Let $R^{\prime },L^{\prime },G^{\prime },C^{\prime }$ denote resistance, inductance, leakage conductance and capacitance per unit length. Applying Kirchhoff's voltage law and Kirchhoff's current law to a short length $dx$ gives the telegrapher's equationsFor a lossless line, $R^{\prime }=G^{\prime }=0$. Differentiating again gives wave equations withThe travelling solutions can be written $V=V_{+}(t-x/v)+V_{-}(t+x/v)$ and $I=[V_{+}(t-x/v)-V_{-}(t+x/v)]/Z_0$. The minus sign for the backward wave is essential to power transport.

At a load $Z_L$, continuity of voltage and current gives the voltage reflection coefficientA matched termination $Z_L=Z_0$ absorbs the incident signal without reflection. An open circuit has ${\Gamma }_L=+1$ and a short circuit has ${\Gamma }_L=-1$. Superposed incident and reflected waves produce standing waves. A sinusoidal lossless line of length $\ell$ transforms the termination intofor the $e^{i\omega t}$ convention. At a quarter wavelength, $Z_{\mathrm{in}}=Z_0^2/Z_L$, allowing impedance transformation. A coaxial line supports a transverse electromagnetic mode whose ideal $L^{\prime },C^{\prime }$ are calculated in A4. Its energy travels in the fields between conductors; the characteristic impedance is not the wire's DC resistance. Losses introduce attenuation and make the characteristic impedance and propagation constant complex.

Magnetization and field response. A magnetic material contains microscopic magnetic moments associated with Electron spin and orbital motion. The magnetization $\mathbf{M}$ is magnetic dipole moment per unit volume. In SI the macroscopic fields satisfyHere $\mathbf{H}$ separates free-current excitation from the material response. A uniform isotropic linear material has $\mathbf{M}={\chi }_m\mathbf{H}$, givingThis constant scalar magnetic permeability model is useful only within its linear range. Crystals can have a susceptibility tensor; strongly magnetic materials depend on field and magnetic history. For a finite specimen the internal $\mathbf{H}$ also includes its demagnetizing field. In a uniformly magnetized ellipsoid, $H_{\mathrm{int}}=H_{\mathrm{applied}}-N_{d}M$ along a principal axis, so sample shape affects the apparent susceptibility.

Diamagnetic and paramagnetic response. Diamagnetism is a weak negative susceptibility caused by induced orbital response opposing the applied field. It occurs in all matter but can be masked by larger positive contributions. Closed-shell materials such as bismuth or many organic compounds show it clearly. Ordinary diamagnetism is often only weakly temperature dependent; superconducting flux exclusion is a separate collective phenomenon and is not described by an arbitrarily small negative susceptibility.

Paramagnetism arises when available magnetic moments partially align with a field. Thermal disorder prevents complete alignment at weak fields. For independent localized moments of magnitude $m$, the classical weak-field result iswhere the second expression uses $B\simeq {\mu }_{0}H$ in a weakly magnetic sample. Thus localized moments obey a Curie law in this regime. Quantum angular momentum changes the coefficient to the appropriate $J(J+1)$ expression and gives saturation at strong field. Conduction Electrons instead exhibit Pauli paramagnetism: only states near the Fermi surface can change spin populations, so a degenerate gas has a nearly temperature-independent response. Its orbital diamagnetism must also be included for a full metallic susceptibility.

Collective magnetic order. Exchange interactions can favor parallel alignment, producing ferromagnetism below the Curie temperature. Iron, cobalt and nickel are familiar examples. A material may contain regions of spontaneous magnetization, or magnetic domains, while having little net external moment because the domains point in different directions. Applied fields move domain walls and rotate moments; at large fields the magnetization approaches saturation. Pinning and energy barriers make this process hysteretic. Remanence is the magnetization left after removing the driving field, and coercivity measures the reverse field needed to bring the chosen response back to zero.

The area of a closed $B$–$H$ loop is the energy dissipated per unit volume per quasistatic cycle, $\oint HdB$, with positive dissipated magnitude. Consequently the magnetic permeability is neither one universal constant nor enough information to describe a ferromagnet: differential magnetic permeability $dB/dH$, saturation and hysteresis all matter. Above the ordering temperature the spontaneous moment disappears, and a Curie–Weiss law is often an appropriate approximate paramagnetic description away from the critical region.

In antiferromagnetism, ordered opposing sublattices have equal moments and cancel macroscopically in the ideal compensated case. The order disappears above the Néel temperature. In ferrimagnetism, opposing sublattices have unequal moments and leave a net magnetization. Ferrites combine ferrimagnetic response with relatively high electrical resistivity, making them valuable when eddy-current losses would be substantial in a metallic core.

Electromagnets and engineering choice. A coil supplies free-current magnetomotive force $NI$. For a simple uniform magnetic path, Ampère's law gives $H\ell \simeq NI$ and flux $\Phi \simeq \mu ANI/\ell$. A high-magnetic permeability soft magnetic core concentrates flux and reduces the current required for a given field. With an air gap $g$,Even a short gap can dominate the magnetic reluctance. The approximate attractive force across one uniform gap is $F\simeq B^{2}A/(2{\mu }_0)$, obtained from the field energy, with fringing neglected.

Soft magnetic materials have small coercivity and narrow hysteresis loops; they suit relays, electrical transformer cores and controllable electromagnets. Hard magnetic materials retain their magnetization and suit permanent magnets. Laminating metallic cores breaks large eddy-current loops; ferrites are useful at higher frequencies because of their resistivity. Increasing current cannot increase $B$ indefinitely through the linear formula: core saturation, coil heating and leakage limit performance. A good design selects material for the field amplitude, switching frequency and acceptable loss, rather than simply choosing the largest quoted magnetic permeability.

Let the rotating frame have origin acceleration ${\mathbf{A}}_0$ and angular velocity $\mathbf{\Omega }(t)$ relative to an inertial frame. For a vector referred to the moving axes, the transport rule isApply it twice to position. With rotating-frame velocity $\mathbf{v}$ and acceleration ${\mathbf{a}}_R$,Thus Newton's equation expressed in the rotating frame isThe added terms are translational, Coriolis, Euler and centrifugal fictitious forces. They reflect the frame's acceleration, rather than additional interactions with nearby objects.

The centrifugal force points outward from the rotation axis and, for constant $\mathbf{\Omega }$, has potential $-m|\mathbf{\Omega }\times \mathbf{r}{|}^2/2$. The Coriolis force is perpendicular to velocity, so does no work on the particle measured in that rotating frame; it changes direction rather than speed directly. The Euler force is present when the frame's rotation rate or direction changes.

On Earth, gravity and centrifugal acceleration combine into effective gravity. For slow horizontal motion at latitude $\lambda$, the vertical component of Earth's rotation gives Coriolis parameter $f=2\Omega \sin \lambda$: an initially northward motion deflects eastward in the northern hemisphere. This supplies a useful local explanation of rotating flows and Foucault precession. A bucket of rotating liquid illustrates the centrifugal potential: its steady free surface is an equipotential of $gz-{\Omega }^2{r}^2/2$, giving $z={\Omega }^2{r}^2/(2g)+\mathrm{constant}$. These equations require a declared frame; adding a centrifugal force while retaining inertial-frame acceleration would count the same kinematic effect twice.

For a slender elastic beam undergoing small bending, cross-sections remain approximately plane and perpendicular to the centreline. At distance $z$ from the neutral axis, longitudinal strain is ${ϵ}_{xx}=-z\kappa$, where $\kappa$ is curvature. Linear elasticity gives ${\sigma }_{xx}=E{ϵ}_{xx}$, so the bending moment has magnitudeHere $I$ is the second moment of cross-sectional area, not a mass moment of inertia. For width $b$ and depth $h$ in the bending direction, $I=b{h}^3/12$. The cubic depth dependence explains why a deep beam resists bending more effectively than the same material spread sideways.

For a uniform beam with transverse displacement $y(x)$, $\kappa \simeq y^{\prime \prime }$ and bending energy is $\frac{1}{2}\int EI(y^{\prime \prime }{)}^{2}dx$. Balancing transverse load $q(x)$ yields the Euler-Bernoulli beam equation $EI{y}^{\prime \prime \prime \prime }=q$ under a consistent sign convention. Supports specify values of deflection, slope, moment or shear; the equation alone does not fix the response.

Now apply a compressive end load $P>0$. A small deflection shortens the axial projection by $\frac{1}{2}\int (y^{\prime }{)}^{2}dx$, so the total potential energy relative to the straight configuration isIts stationary equation isFor pinned ends, $y(0)=y(\ell )=0$ and the end moments vanish, so $y^{\prime \prime }(0)=y^{\prime \prime }(\ell )=0$. The candidate shapes $y=A\sin (n\pi x/\ell )$ require $P=EI(n\pi /\ell {)}^2$. The first loss of stability is consequentlyAt smaller $P$, the quadratic energy is positive; beyond this threshold the straight configuration no longer minimizes it. End restraint changes the threshold: clamped–clamped gives $4{\pi }^{2}EI/{\ell }^2$ and clamped–free gives ${\pi }^{2}EI/(4{\ell }^2)$, equivalently an effective-length correction. A column buckles in the direction of smaller flexural rigidity. Euler buckling is a geometric instability within elastic response; yielding or crushing can occur first for a short thick strut. Large deflections, shear deformation and imperfections require refinements of the ideal slender-beam model.

Take a fixed sphere of radius $R$ in an unbounded incompressible inviscid fluid, with uniform velocity $U{\mathbf{e}}_z$ at infinity. Assume irrotational flow, so $\mathbf{v}=\nabla \varphi$ and incompressibility gives ${\nabla }^2\varphi =0$. Axisymmetry and the dipolar angular dependence of the far field suggestMatching the far field gives $A=U$. The no-penetration condition $v_r(R)=0$ then gives $B=U{R}^3/2$. ThusThe surface has stagnation points at its front and rear poles, while the equatorial speed is $3U/2$. Fluid slips tangentially past the sphere; imposing a no-slip condition would contradict this inviscid potential-flow model.

With gravity omitted or the hydrostatic pressure subtracted, steady Bernoulli givesAt the surface,The front–rear symmetry makes the integral of pressure force along the flow vanish: terms proportional to $\cos \theta$ integrate to zero over the sphere. This is D'Alembert's paradox, not a prediction that real spheres have no drag. Viscosity creates boundary layers; separation and a wake break the fore–aft symmetry. Potential flow remains valuable for the outer velocity field when viscous regions are thin, but it cannot supply their drag by itself.

A spatially uniform external gravitational acceleration moves a body's centre of mass without stretching it. Tidal forces arise from the variation of that acceleration across the body. Let a distant mass $M$ lie at distance $D$ along unit vector $\mathbf{n}$, and let $|\mathbf{r}|\ll D$ denote position relative to the body's centre. Subtract the centre's acceleration. To first order,The tidal tensor is $T_{ij}=GM(3n_i{n}_j-{\delta }_{ij})/D^3$. It stretches along the line to the source with eigenvalue $2GM/D^3$ and compresses in the two perpendicular directions with eigenvalues $-GM/D^3$. Its trace vanishes outside the source, consistently with Laplace's equation for the gravitational potential.

After removal of the constant and uniform-acceleration terms, the tidal potential isAn ideal equilibrium ocean on a sphere of radius $R$ follows an equipotential of the body's gravity plus this perturbation. For surface gravity $g$, the leading displacement isup to a constant fixing total water volume. The quadrupolar form produces bulges on both the near and far sides; the far-side bulge follows from differential acceleration, without postulating a repulsive gravitational force.

The Moon and Sun both raise terrestrial tides, with relative strength set mainly by $M/D^3$, rather than by the total gravitational pull $M/D^2$. Their contributions reinforce near new and full Moon to produce spring tides, and partly cancel near quadrature to produce neap tides. Earth's rotation relative to the tidal pattern gives a strong semidiurnal component in the simplest geometry. Real tide heights and timing depend on ocean depth, coastlines, basin resonances, Coriolis acceleration, friction and the declination of the forcing bodies. Tidal dissipation produces phase lags and torques, exchanging rotational and orbital angular momentum and gradually altering the spin and orbit. The equilibrium bulge is therefore a useful starting model, not an exact local tide forecast.

Inertia is a tensor. For a rigid body, distances between material points are fixed. Write each position relative to the centre of mass as $\mathbf{r}$, so its rotational velocity is $\mathbf{\omega }\times \mathbf{r}$. Integrating $\mathbf{r}\times (\mathbf{\omega }\times \mathbf{r})$ gives angular momentumThe rotational kinetic energy isThe inertia tensor is real and symmetric, so orthogonal principal axes diagonalize it. Angular momentum is parallel to angular velocity only along a principal axis, or within a subspace of equal principal moments. For translation plus rotation about the centre of mass, $T=M{v}_{\mathrm{CM}}^2/2+T_{\mathrm{rot}}$. Moving the reference to a parallel axis adds $M(d^{2}1-\mathbf{d}{\mathbf{d}}^{\mathsf{T}})$ to the tensor, the tensor form of the parallel-axis theorem.

Euler's equations. The external torque about the centre of mass obeys $\mathbf{\tau }=(d\mathbf{L}/dt{)}_{\mathrm{space}}$. In body axes, the transport rule givesThe principal moments are constant in that frame, henceThese are Euler's equations. With zero torque, spatial angular momentum and kinetic energy are conserved, although the body components generally vary. For distinct moments, small perturbations about rotation around the largest or smallest principal moment oscillate, whereas those about the intermediate moment grow. This is the intermediate-axis instability.

A lamina. Choose the lamina's plane as $z=0$. Then $I_{xz}=I_{yz}=0$, so the normal is a principal axis, andThis perpendicular-axis theorem applies to a planar mass distribution; it is not a general identity for a three-dimensional body. Rotate the in-plane axes to diagonalize the remaining symmetric two-by-two block. For a uniform rectangular lamina with sides $a,b$ and mass $M$, its central principal moments are $I_x=M{b}^2/12$, $I_y=M{a}^2/12$ and $I_z=M(a^2+b^2)/12$. If $a>b$, then $I_x<I_y<I_z$: torque-free rotation about the long in-plane axis or the normal is stable, while that about the short in-plane axis is unstable. Coincident moments in a square require the degenerate, rather than strict intermediate-axis, case. Under a constraint to rotate only about the normal, the motion reduces to the scalar equation ${\tau }_z=I_z{\dot{\omega }}_z$.

A symmetric top without torque. Let $I_1=I_2=I_{\perp }$ and $I_3=I_{\parallel }$. Euler's equations reduce toThus ${\omega }_1+i{\omega }_2$ rotates at signed body-frame frequency $\nu$, while the axial component stays fixed. The symmetry axis has constant angle to fixed spatial angular momentum. Indeed $\mathbf{L}=I_{\perp }\mathbf{\omega }+(I_{\parallel }-I_{\perp }){\omega }_3{\mathbf{e}}_3$, soIt precesses in space about $\mathbf{L}$ with angular speed $L/I_{\perp }$. This spatial precession and the body-frame frequency $\nu$ describe different motions and should not be identified.

A heavy top. If a symmetric top is pivoted a distance $\ell$ from its centre of mass, gravity gives torque ${\mathbf{r}}_{\mathrm{CM}}\times M\mathbf{g}$. Here the relevant moments are about the fixed pivot. The axial angular momentum and the vertical angular-momentum component are conserved, because gravity supplies neither an axial torque nor a torque about the vertical. Using tilt $\theta$, precession angle $\varphi$ and spin angle $\psi$, its Lagrangian isEliminating the two conserved cyclic momenta leaves one-dimensional motion in an effective tilt potential; bounded oscillation in tilt is nutation. Steady precession at $\dot{\varphi }=\Omega$ and fixed tilt satisfiesFor a rapidly spinning top on the slow branch, the first term is small andGravity then changes the direction of angular momentum rather than simply tipping the top straight down. The approximation fails near conditions where nutation or the fast-precession branch matters; the full equations retain those possibilities.

Pressure and its physical terms. Solving the molar Van der Waals equation for pressure givesThe excluded-volume parameter $b$ reduces the volume available for translational motion, raising the kinetic pressure relative to $RT/V$. The attraction parameter $a$ lowers the pressure because attractive interactions reduce the momentum transferred to the boundary. This mean-field model treats $a,b$ as temperature-independent material constants. For one mole, $V$ is the sample volume with the molar amount already fixed; $b$ has units of that volume and $a$ of pressure times volume squared.

volume dependence of entropy. The Helmholtz free energy has differential $dF=-SdT-PdV$. Equality of mixed derivatives yields the Maxwell relationFor $a=0$ this derivative is $R/(V-b)$, givingEquivalently, $S=R\ln (V-b)+f(T)$ with a common reference volume understood inside the logarithm. The equation of state fixes this volume dependence, not the otherwise unspecified function of temperature.

Excess entropy at fixed volume and temperature. The ideal gas has $(\partial S_{\mathrm{id}}/\partial V{)}_T=R/V$. ThereforeIntegrate from infinite volume, where the entropy difference is specified to vanish:It is negative because excluding configurations reduces the accessible state count. It tends to zero as $V\to \infty$ and has leading dilute limit $-Rb/V$. This is the entropy excess of a van der Waals gas.

Effect of attraction. Restoring a temperature-independent $a$ changes the pressure by $-a/V^2$, but not its derivative with respect to temperature. With the same infinite-volume entropy reference, it therefore does not change $\Delta S^{\mathrm{ex}}$ at fixed $V,T$. This can also be seen fromThe attraction affects internal energy and the equilibrium volume, so it can affect entropy indirectly when different state points are compared. If $a$ were temperature dependent, its derivative would contribute directly to the entropy; that is outside the stipulated constant-parameter model.

Coexistence isotherm. Below the Van der Waals critical point, the formal homogeneous isotherm has a local minimum and maximum. Its middle branch has $\partial P/\partial V>0$, hence negative isothermal compressibility, and is mechanically unstable. Portions of the negative-slope branches can be metastable rather than globally stable. The equilibrium isotherm replaces the loop between liquid volume $V_l$ and vapour volume $V_g$ by a constant saturation pressure $P_s$. The two volume are fixed byThe Maxwell equal-area construction expresses equality of the two phases' chemical potentials. Intermediate total volume are mixtures: changing the vapour fraction changes total volume at the same pressure, rather than following the unstable homogeneous curve. The spinodals are the turning points of the formal isotherm; they are distinct from the coexistence endpoints.

Dense-phase volume at low temperature. At a general subcritical temperature, the two coexistence equations above must be solved together. For this equation of state the area condition is explicitlyThis supplies an exact implicit determination of the liquid volume even when a simple closed form is unavailable.

For the asymptotically low-temperature limit, put $\tau =RTb/a\ll 1$. The liquid is dense, $V_l\simeq b$, while the vapour is dilute, $P_s{V}_g\simeq RT$ and $V_g\gg V_l$. The area condition then givesThus the vapour volume grows exponentially in $1/\tau$, and the saturation pressure is exponentially small. It is consequently consistent to neglect $P_s$ in the liquid equation to algebraic order in $\tau$:Choose the root near $b$, not the large-volume zero-pressure root:Expanding gives the leading requested dense-phase behavior,This is the low-temperature liquid volume of a van der Waals gas. Setting $P_s=0$ is an asymptotic approximation, not an exact coexistence condition at every temperature below $T_c$. The figure illustrates generic subcritical coexistence; its finite-temperature liquid volume is determined by the full equal-area equations. The van der Waals liquid also ceases to be a realistic description at sufficiently low physical temperature, where other phases or quantum effects can matter.

Two equivalent second-law statements. The Clausius statement of the second law forbids a cyclic device whose sole net effect is transferring heat from a colder reservoir to a hotter one. A refrigerator can make that transfer when supplied with work, so the sole-effect qualification matters. The Kelvin-Planck statement of the second law forbids a cyclic device whose sole net effect is drawing heat from a single reservoir and converting all of it into work. A heat engine can produce work when it also rejects heat to a colder reservoir.

The two statements are equivalent by combining a supposed violating device with an ordinary reversible engine or refrigerator: one can cancel one reservoir exchange and obtain a violation of the other statement. The following original schematic shows the forbidden sole effects, with arrows indicating energy flow.

Microscopic fluctuations. The second law is a macroscopic statistical law. A finite fluctuation can transfer a small heat amount $Q$ from cold to hot, with reservoir entropy change $Q/T_h-Q/T_c<0$. Such an individual event does not imply that a cyclic engine can extract systematic useful work from equilibrium. Entropy-decreasing trajectories become exceedingly improbable when the entropy involved is large compared with $k_B$. Under suitable microscopic reversibility conditions, fluctuation relations quantify the relative probabilities of positive and negative entropy production and give nonnegative mean total entropy production. Thus a deterministic no-fluctuation reading has a limit at small scales or short times; the statistical second law remains applicable.

Maximum heating rate. Interpret the quoted $0.5\mathrm{kW}$ as the maximum work-input rate ${\dot{W}}_{\max }$ to an ideal reversible heat pump. This interpretation is needed to obtain a heating rate larger than its electrical input; the wording does not explicitly distinguish input power from delivered heat. A reversible cycle satisfiesTherefore the heating coefficient of performance isWith $T_h=293\mathrm{K}$ and $T_c=273\mathrm{K}$,The additional heat comes from the outdoor reservoir; it is not produced by converting work into more energy than was supplied.

Lowest outdoor temperature. Let $\Delta T=T_h-T_c>0$. The largest available heating rate is ${\dot{W}}_{\max }{T}_h/\Delta T$, while the house loses $\lambda \Delta T$. Equilibrium can be maintained whenAt the limiting outdoor temperature,HenceThis is the indoor-temperature limit of a Carnot heat pump. It assumes the ideal pump retains its stated input-power capability over the operating range. Real heat exchangers require finite temperature differences and incur irreversibility, so the reversible result is an ideal upper performance bound.

Conventional heater. At this limiting outdoor temperature, the required delivered heating power is simply the house's heat loss:This is heat delivered to the house. The fuel-input power would be $3.83/\eta \mathrm{kW}$ for heater efficiency $\eta$. At the earlier outdoor temperature of $273\mathrm{K}$, the required delivered heat would instead be $0.1(293-273)=2.00\mathrm{kW}$; the final question's reference temperature must be kept clear.

If $0.5\mathrm{kW}$ is read literally as a maximum delivered heat rate rather than a work-input rating, the alternative answers are ${\dot{Q}}_{h,\max }=0.5\mathrm{kW}$, $T_{c,\min }=293-0.5/0.1=288\mathrm{K}$, and a conventional heater delivering $0.5\mathrm{kW}$ at that limit. Such a pump could not balance the stated house loss at $273\mathrm{K}$. Both readings follow consistently once the ambiguous power rating is defined; the Carnot calculation above uses the work-input reading.