Past exam of the physics B course of the University of Cambridge / 2014 / Paper 2 / Section A

Treat the vapour as an ideal gas, neglect the liquid's molar volume, and take the molar latent heat as constant across this small interval. Integrating the Clausius-Clapeyron equation givesOnly the pressure ratio enters, so conversion from atmospheres to pascals is unnecessary.

The Hohmann transfer orbit has periapsis $r$ and apoapsis $2r$, so its semi-major axis is $3r/2$. Kepler's third law gives a period $(3/2{)}^{3/2}$ times the original orbital period. Only half the ellipse is traversed:The relevant radii are measured from Earth's centre, not its surface.

Use $N$ turns of mean radius $a\simeq 1.05\mathrm{mm}$, including the wire radius, on a light nonmagnetic rod. With wire radius $r_w$, length $l$ and density $\rho$, neglecting helical pitch givesThe last expression is the transverse central moment of inertia of the uniformly distributed turns. The magnetic torque is $-m_B{B}_h\sin \theta$, so the small-angle magnetic dipole oscillation obeys $J\ddot{\theta }+m_B{B}_h\theta =0$. HenceThe number of turns cancels. Taking the turn radius as $1.00\mathrm{mm}$ instead gives $4.04\mathrm{s}$, consistent at the accuracy of the thin-winding model. The rod's inertia, damping and lead-wire torque are neglected.

For slow gyroscopic precession, the gravitational torque turns the spin angular momentum: $Mg\ell \sin \theta =\Omega I_3\omega \sin \theta$, with $I_3=M{R}^2/2$. ThusThe inclination cancels at this order. More precisely, uniform precession of a heavy symmetric top satisfiesIts slow root gives $T_p\simeq 1.57\mathrm{s}$. About $1.6\mathrm{s}$ is the slow-precession answer; arbitrary initial conditions can also produce nutation rather than exact steady precession.

For a waveguide mode in a homogeneous nondispersive filling, ${\omega }^2={\omega }_c^2+v^2{k}_z^2$. Its phase velocity is $v_p=\omega /k_z$ and its group velocity is $v_g=d\omega /d{k}_z=v^2{k}_z/\omega$, giving