Past exam of the physics B course of the University of Cambridge / 2021 / Paper 1 / Section A

For the intended ideal Venturi effect, take steady flow, incompressible flow, negligible viscous losses, and the same height at the two stations shown. The Bernoulli equation along a streamline givesUsing water mass density $\rho \simeq 1000\mathrm{kg}{\mathrm{m}}^{-3}$,Thus the outlet pressure is lower by $16\mathrm{kPa}$. The increased kinetic energy per unit volume comes from the drop in pressure.

The source's assumption of laminar flow alone does not imply negligible viscosity or losses. The numerical result uses the usual lossless Bernoulli equation approximation, with the quoted speeds interpreted as representative streamline speeds. A real tube with significant viscous losses requires additional data; in a uniform-velocity energy balance its pressure drop would include an extra positive loss term.

The principal moments of inertia are the eigenvalues of the symmetric inertia tensor. In units of $\mathrm{kg}{\mathrm{m}}^2$, its characteristic determinant isTherefore the requested algebraic answer isAn orthonormal basis of principal axes is $(1,1,0)/\sqrt{2}$, $(1,-1,0)/\sqrt{2}$, and $(0,0,1)$; the last two span the degenerate eigenspace and may be replaced by any orthonormal basis of it.

There is a physical inconsistency in calling the supplied matrix a rigid body's inertia tensor. The moment-of-inertia triangle inequalities require each principal moment to be no greater than the sum of the other two. Here $3>1+1$. In principal coordinates, $I_2+I_3-I_1=2\int x_1^{2}dm$ cannot be negative for a nonnegative mass distribution. Thus the eigenvalue calculation is well defined, but no ordinary rigid mass distribution realizes this tensor. The positive second-moment matrix test for an inertia tensor makes the same obstruction explicit.

Use local east, north and upward unit vectors at the equator, with Earth's rotation $\mathbf{\Omega }={\Omega }_E\hat{\mathbf{n}}$. The parcel starts at rest relative to the ground. To first order in ${\Omega }_E$, its unperturbed vertical velocity is $\dot{z}=-gt$ and its fall time from height $h$ isThe Coriolis acceleration is $-2\mathbf{\Omega }\times \mathbf{v}$. Since north crossed with downward points west, its negative points east:Integrating with $x_E(0)={\dot{x}}_E(0)=0$ gives the equatorial Coriolis deflection in free fall,For $h=200\mathrm{m}$, $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$ and ${\Omega }_E=2\pi /(86164\mathrm{s})=7.2921\times 10^{-5}{\mathrm{s}}^{-1}$, $t_f=6.39\mathrm{s}$ andThis neglects air resistance, variation of $g$ over the fall, and terms of order ${\Omega }_E^2$. The measured local $g$ already includes the background centrifugal correction. “Stationary” is interpreted as hovering above a fixed ground location, not being stationary in an inertial frame.

Under a uniform compressive pressure $p$, each normal stress is $-p$. The isotropic Hooke's law gives each longitudinal strain asThus $\Delta V/V=-3p(1-2\sigma )/E$. By the definition of bulk modulus, $B=-p/(\Delta V/V)$, so the isotropic elastic modulus relation isPositive finite Young's modulus and bulk modulus requireThe incompressible limit is $\sigma \to 1/2$ with $B\to \infty$ at finite $E$. If positivity of the shear modulus $G_s=E/[2(1+\sigma )]$ is also imposed, the full stable isotropic range is $-1<\sigma <1/2$. Negative Poisson's ratio is not ruled out; the displayed bulk-modulus relation alone supplies the upper bound.

For the intended reversible adiabatic process, an ideal gas obeys $T{V}^{\gamma -1}=\text{constant}$. A monatomic gas has heat capacity ratio $\gamma =5/3$, soThe required extra assumption is that “adiabatic” denotes the usual quasistatic isentropic process. Absence of heat transfer alone would not fix the final temperature: an ideal gas undergoing free expansion into vacuum is also adiabatic but leaves its temperature unchanged.