2026 and 2025 paper questions

An equation is dimensionally correct only if every term has the same dimensions. In option A, $[v_f]=[v_i]=[L{T}^{-1}]$ and $[at]=[L{T}^{-2}][T]=[L{T}^{-1}]$, so every term has dimensions $[L{T}^{-1}]$. So A is correct.

The tempting options look like equations of motion but are wrong. In B the term $2at$ has dimensions $[L{T}^{-1}]$, while $v^2$ has $[L^2{T}^{-2}]$ (the correct equation has $2aS$, with $[L{T}^{-2}][L]=[L^2{T}^{-2}]$). In C the term $\frac{1}{2}at$ has dimensions $[L{T}^{-1}]$ but $s$ has $[L]$ (the correct term is $\frac{1}{2}a{t}^2$). In D, $[m{a}^2]=[M][L{T}^{-2}{]}^2=[M{L}^2{T}^{-4}]$, which is not the dimensions of force $[ML{T}^{-2}]$.

Work is force times displacement, so $[W]=[F][s]=[ML{T}^{-2}][L]=[M{L}^2{T}^{-2}]$. These are the dimensions given, so the quantity is work (energy has the same dimensions). So B is correct.

Option A (power) is work per unit time, $[M{L}^2{T}^{-2}]/[T]=[M{L}^2{T}^{-3}]$, which has an extra $T^{-1}$. Velocity is $[L{T}^{-1}]$ and acceleration is $[L{T}^{-2}]$, and neither contains mass.

The pascal is the SI unit of pressure. Pressure is force per unit area, so $1\mathrm{Pa}=1\mathrm{N}{\mathrm{m}}^{-2}$. In base units, $\mathrm{N}{\mathrm{m}}^{-2}=\mathrm{kg}\mathrm{m}{\mathrm{s}}^{-2}{\mathrm{m}}^{-2}=\mathrm{kg}{\mathrm{m}}^{-1}{\mathrm{s}}^{-2}$. So B is correct.

Option A, N/m, is force per unit length (the unit of surface tension), not force per unit area. Option C squares both force and area, which does not give pressure. Option D, kg/m, has the wrong base units: pressure is $\mathrm{kg}{\mathrm{m}}^{-1}{\mathrm{s}}^{-2}$, so a ${\mathrm{s}}^{-2}$ is missing.

Percentage uncertainty is the absolute uncertainty divided by the measured value, times 100: $\frac{0.01}{2.25}\times 100\%=0.444\%\approx 0.4\%$. An uncertainty is quoted to one significant figure, so C is correct.

Option A ($0.2\%$) would need an absolute uncertainty of about $0.005\mathrm{cm}$, and option D ($0.5\%$) comes from rounding $0.444$ upwards, which is not how $0.444$ rounds to one figure.

By the unit vector rules $\hat{k}\times \hat{ı}=\hat{ȷ}$, so reversing the order gives $\hat{ı}\times \hat{k}=-\hat{ȷ}$. Then $\hat{ȷ}\cdot (\hat{ı}\times \hat{k})=\hat{ȷ}\cdot (-\hat{ȷ})=-(\hat{ȷ}\cdot \hat{ȷ})=-1$. So B is correct.

Option C ($1$) comes from taking $\hat{ı}\times \hat{k}=+\hat{ȷ}$, which is true only for the cyclic order $\hat{ı}\to \hat{ȷ}\to \hat{k}\to \hat{ı}$. Option A ($0$) would hold if the bracket were perpendicular to $\hat{ȷ}$, but $\hat{ı}\times \hat{k}$ lies along $\hat{ȷ}$.

Time of flight is $T=\frac{2v_i\sin \theta }{g}=\frac{2v_{iy}}{g}$, so it is fixed by the vertical component of the launch velocity (the projectile stays up for as long as the vertical motion lasts). So C is correct.

Horizontal velocity (A) only decides how far the projectile travels in that time, not how long it stays in the air. Mass (B) does not appear because every body has the same acceleration $g$. Option D is incomplete: the acceleration $g$ matters, but the time of flight also needs $v_{iy}$.

The acceleration is uniform and the car starts from rest, so the average velocity is $\frac{0+60}{2}=30\mathrm{m}{\mathrm{s}}^{-1}$. Then $S=\bar{v}t=30\times 5=150\mathrm{m}$. Using the equations of motion: $a=\frac{60-0}{5}=12\mathrm{m}{\mathrm{s}}^{-2}$ and $S=\frac{1}{2}a{t}^2=\frac{1}{2}(12)(5{)}^2=150\mathrm{m}$. So A is correct.

Option C ($300\mathrm{m}$) is $60\times 5$, which would be the distance if the car moved at the final speed for the whole 5 s. The car's speed is less than $60\mathrm{m}{\mathrm{s}}^{-1}$ during the whole interval until the last instant.

For motion in a circle the centripetal force is $F_c=\frac{m{v}^2}{r}$. Put $v=r\omega$: $F_c=\frac{m(r\omega {)}^2}{r}=mr{\omega }^2$. Option C ($m{\omega }^2$) has lost the radius, so it has the wrong dimensions; option D ($m{v}^2$) is missing the factor $1/r$.

The blades act like a rotating rigid body whose moment of inertia is $I=\sum m{r}^2$. Longer blades put the mass at larger $r$, so $I$ increases a lot (it goes as $r^2$). The motor gives the same torque for the same voltage and current, and $\alpha =\tau /I$, so the angular acceleration and the speed reached are smaller. The tempting option A is wrong because a bigger fan does not get more torque from the motor; it only gets more rotational inertia (and more air resistance).

Young's modulus is the ratio of the tensile stress to the tensile strain, $Y=\frac{F/A}{\Delta L/L}$ (equation 7.1), so option A is correct.

Option D, $\frac{F\Delta L}{LA}=\frac{F}{A}\times \frac{\Delta L}{L}$, is the stress multiplied by the strain, not divided by it. Options B and C do not have the form stress over strain.

Kinetic energy is $\frac{1}{2}m{v}^2$. For equal masses, $\frac{K_A}{K_B}=\frac{v_A^2}{v_B^2}={\left(\frac{2}{1}\right)}^2=4:1$. Option A (2:1) forgets that the speed is squared.

By the equation of continuity, for steady flow of an incompressible fluid, $A_1{v}_1=A_2{v}_2$. Here $A_2=\frac{A_1}{3}$, so

The fluid speeds up and its velocity triples. Option A is the tempting mistake of making the velocity change in the same way as the area; the speed is inversely proportional to the area, not directly. Option D would need the area to change by a factor of nine.

In circular motion the centripetal force is the force that keeps the body on its circle. It must be at right angles to the velocity, which is along the tangent, so that it changes the direction of the velocity without changing the speed. It therefore points along the radius towards the centre, that is, radially inward. Option A is the direction of the velocity (a tangential force would change the speed, not the direction). Option D, outward, is the direction of the so-called centrifugal effect, which is not a real force acting on the body. Option B is true in the sense that the force is perpendicular to the velocity, but it does not give the direction, and 'radially inward' is the complete answer.

A falling sphere speeds up until the drag equals its weight; this is the terminal velocity, and the drag cannot grow beyond this in a steady fall. So the maximum drag is $F_d=mg=1\times 9.8=9.8\mathrm{N}$. Option B ($1\mathrm{N}$) is only the mass in kilograms, not a force, and the drag cannot be larger than the weight in a steady fall.

Young's modulus is the ratio of the tensile stress to the tensile strain: $Y=\frac{F/A}{\Delta L/L}$, which is option D.

Option A, $\frac{\Delta L/L}{F/A}$, is the reciprocal: strain over stress. Options B and C are products of $F$ and $A$ or of $\Delta L$ and $L$, not a ratio of stress to strain.

The efficiency of a Carnot engine is $\eta =1-\frac{T_2}{T_1}$. Take the cold reservoir at $T_2=300\mathrm{K}$ and the source at $T_1=600\mathrm{K}$: $\eta =1-\frac{300}{600}=0.50$.

Doubling the source temperature to $1200\mathrm{K}$ gives ${\eta }^{\prime }=1-\frac{300}{1200}=0.75$, so $\frac{{\eta }^{\prime }}{\eta }=\frac{0.75}{0.50}=1.5$ and the answer is A. The efficiency does not double (B): the term $T_2/T_1$ is halved, not the efficiency, and it can never exceed 1.

In an adiabatic process no heat or mass transfers between the system and its surroundings, so the mass of the gas stays constant (option D). The heat exchanged is zero, $Q=0$, but heat is not one of the options.

The other quantities all change. The first law gives $W=-\Delta U$, so the work done by or on the gas changes the internal energy (A) and with it the temperature (C), and the gas expands or is compressed, so the volume (B) changes. The constant-volume process is isochoric and the constant-temperature process is isothermal.

In a stationary wave an antinode is a point of maximum amplitude, so its displacement is the largest (option B). It is where the two component waves are always in phase.

Zero displacement (A) describes a node, where the waves are always out of phase. An antinode is not a point of zero velocity: it vibrates most rapidly.

Sonar uses the Doppler effect: sound reflected from a moving submarine has a changed frequency, which gives its speed and direction (option A).

Compton's effect (X-ray scattering), Faraday's law (induced emf) and Bernoulli's principle (fluid flow) are not used for detecting submarines.

Resolve the velocity into a component $v\cos 45^{\circ }$ along $\vec{B}$ and a component $v\sin 45^{\circ }$ perpendicular to it. The magnetic force is zero along $\vec{B}$, so the parallel component stays constant and carries the electron forward. The perpendicular component makes it move in a circle. The two together give a spiral path, a helix.

Circular motion (B) occurs only when the velocity is exactly perpendicular to the field ($90^{\circ }$), and straight-line motion (A) only when it is parallel or antiparallel ($0^{\circ }$ or $180^{\circ }$). A magnetic force never gives a parabolic path (C).

The force on a charge is $\vec{F}=q\vec{E}$. The electron is negative, so the force is opposite to $\vec{E}$. The field points from south to north, so the force, and the acceleration, point towards the south. The answer is B.

Option C (north) is the direction in which a positive charge would move. East and west are perpendicular to the field and the force has no component along them.

A potentiometer is a null device. It balances an unknown potential difference (or emf) against a known one, so it is used to compare potential differences, for example the emfs of two cells, using ${\epsilon }_2/{\epsilon }_1={\ell }_2/{\ell }_1$.

Option A is wrong because the comparison of resistances is done with a Wheatstone bridge. Current and power are not compared by a potentiometer.

The electric field strength at a point is the force per unit positive test charge placed there, $E=\frac{F}{q}$ (unit $\mathrm{N}{\mathrm{C}}^{-1}$), so C is correct.

Option A (work done) is wrong: work per unit charge is the potential difference, $V=W/q$, not the field strength. Option D inverts the definition (it would be $q/F$).

A longer slide wire spreads the same potential difference over a greater length, so the potential gradient is smaller. The balance point can then be located more precisely, which gives higher accuracy. The answer is A.

A short wire (B) gives a larger gradient and a coarser balance point. Thickness (C) does not improve the sensitivity of the balance point, and there is no best 'medium length' (D): the longer the wire, the better.

Each wire lies in the field of the other. Between two parallel wires carrying currents in opposite directions the two fields point the same way, so the field between them is strengthened and the wires are pushed apart: they repel. The force per unit length is $\frac{{\mu }_0{I}_1{I}_2}{2\pi r}$.

Option B (attract) is the result for currents in the same direction. Option C is wrong because the force is not zero, and D is wrong because the forces lie along the line joining the wires and give no turning.

The postulates of special relativity are about all the laws of physics. The first says that the laws of physics are the same in every inertial frame, and the second adds that the speed of light is the same for all observers. Option A (only light) is the tempting one: the second postulate is about light, but the first covers every law, so the theory is not limited to light. Options B and D are likewise too narrow.

Magnetic flux is $\Phi =BA$, and $B=\frac{F}{IL}$ has the unit $\mathrm{N}{\mathrm{A}}^{-1}{\mathrm{m}}^{-1}$. Multiplying by an area in ${\mathrm{m}}^2$ gives $1\mathrm{Wb}=1\mathrm{N}{\mathrm{A}}^{-1}{\mathrm{m}}^{-1}\times {\mathrm{m}}^2=1\mathrm{N}\mathrm{m}{\mathrm{A}}^{-1}$, option B.

Option A ($\mathrm{N}{\mathrm{A}}^{-1}$) has lost the metre, so it is not a flux unit. Options C and D do not follow from $\Phi =BA$: C has $\mathrm{A}$ in the numerator and D has ${\mathrm{m}}^{-1}$, where a flux needs ${\mathrm{m}}^{+1}$. D is close to the unit of the tesla, $\mathrm{N}{\mathrm{A}}^{-1}{\mathrm{m}}^{-1}$, which is a flux density, not a flux.

Einstein's mass-energy relation is $E=m{c}^2$: the energy equivalent of a mass $m$ is the mass times the square of the speed of light. Options C ($mg$) and D ($gh$) look like weight and potential energy per unit mass, not relativity; option A ($mv$) is momentum, not energy.

By length contraction $L=L_0\sqrt{1-v^2/c^2}$, an object moving at very high speed is measured shorter along its direction of motion. Option B is wrong because lengths never expand in this effect (the proper length $L_0$ is the largest). Option D would need $v=c$, which a body with mass cannot reach.

The proton consists of two up quarks and one down quark ($uud$). The charge check: $+\frac{2}{3}+\frac{2}{3}-\frac{1}{3}=+1$. Option A ($udd$, with two down quarks) is the neutron, with charge $+\frac{2}{3}-\frac{1}{3}-\frac{1}{3}=0$. Options C and D have too many quarks, since a baryon has three.

The moment of inertia of a point mass is I=mr^{2}, and for a rigid body I=\sum m_ir_i^{2}. If the distance r from the axis increases, I increases, and it increases as the square of the distance. For example, a ring (MR^{2}) is harder to spin than a disc of the same mass and radius (\tfrac12MR^{2}) because its mass is farther from the axis. Option A is the tempting wrong one, because we often think that a larger distance gives a larger lever arm and so an easier turn, but that is the torque; the rotational inertia grows with distance.

From m=m_0/\sqrt{1-v^{2}/c^{2}} the root is less than 1 for any moving body, so the relativistic mass m is greater than the rest mass m_0 (it rises without limit as v\to c). Option C (equal) is the tempting one: it holds only at v=0, or when the speed is so small that the effect cannot be noticed.

The relation between torque and angular acceleration is \tau=I\alpha (torque is the rotational equivalent of F=ma). Of the four printed options only C has this form, \tau equal to a moment-of-inertia term multiplied by \alpha. Option A has \alpha in the denominator, option B is the inverse relation (\alpha=\tau/I would be the correct one for that arrangement, not \tau=\alpha/I), and option D uses the mass m instead of the moment of inertia, so it has the wrong dimensions (kg s^{-2} instead of N m).

Hadrons are made of quarks. A baryon has three quarks (proton, neutron) and a meson has a quark and an anti-quark (pion, kaon). So a hadron is a baryon or a meson. Option A (lepton) is the tempting wrong one: leptons are elementary particles that do not take part in the strong interaction, so they are not hadrons. Gluons and photons are force carriers.

Coulomb's law is F=k\dfrac{q_1q_2}{r^{2}}. Make k the subject, then put the dimensions of each quantity on the right (1 mark).

As a check, the unit of k is \mathrm{N\,m^{2}\,C^{-2}}=\mathrm{kg\,m^{3}\,s^{-4}\,A^{-2}}, which has the same dimensions.

Precision (1 mark). Precision is the closeness of repeated measurements of the same quantity to each other. It depends on the least count of the instrument and is indicated by the absolute uncertainty: the smaller the least count, the greater the precision.

Accuracy (1 mark). Accuracy is the closeness of a measured value to the standard or true value. It is indicated by the fractional or percentage uncertainty: the smaller it is, the more accurate the measurement.

Diagram (1 mark). Shots at a target show the idea. The centre is the true value, and each dot is a reading.

In (a) the readings agree with each other and lie on the true value (precise and accurate). In (b) they agree with each other but are all shifted from the true value (precise, not accurate). In (c) they scatter but average out around the true value (accurate, not precise). In (d) they scatter and miss (neither).

The cross product of two vectors is \vec A\times\vec B=AB\sin\theta\,\hat n, where \theta is the angle between them and \hat n is a unit vector perpendicular to their plane (right hand rule). Three properties (1 mark each):

• It is anticommutative. \vec A\times\vec B=-\,\vec B\times\vec A. The magnitude AB\sin\theta is the same, but reversing the order reverses the direction. For example \hat\imath\times\hat\jmath=\hat k but \hat\jmath\times\hat\imath=-\hat k.
• It is zero for parallel or antiparallel vectors. For \theta=0^\circ or 180^\circ, \sin\theta=0, so \vec A\times\vec B=\vec 0. In particular \vec A\times\vec A=\vec 0 and \hat\imath\times\hat\imath=\hat\jmath\times\hat\jmath=\hat k\times\hat k=\vec 0.
• It is largest for perpendicular vectors. For \theta=90^\circ, \sin\theta=1, so |\vec A\times\vec B|=AB. For the unit vectors, \hat\imath\times\hat\jmath=\hat k, \hat\jmath\times\hat k=\hat\imath and \hat k\times\hat\imath=\hat\jmath.

Other properties are that the product is a vector (not a scalar) and that, in components, \vec A\times\vec B=\begin{vmatrix}\hat\imath&\hat\jmath&\hat k\\A_x&A_y&A_z\\B_x&B_y&B_z\end{vmatrix}.

Ordinary light is unpolarised: it is made of waves vibrating in all directions perpendicular to the line of propagation. A Polaroid filter blocks all the planes of vibration except one and transmits only that plane.

1. Pass the ordinary (unpolarised) light through the first polaroid (the polariser). It emerges plane polarised, with its vibrations in a single plane, and with its intensity reduced to half.
2. Place a second polaroid (the analyser) behind it. Turning it changes the transmitted intensity as I=I_0\cos^{2}\theta (Malus's law): maximum when the axes are parallel and zero when they are crossed at 90^\circ.

Photon as a force carrier (3 marks). The photon is the exchange particle (force carrier) of the electromagnetic force.

• According to quantum field theory, charged particles exert the electromagnetic force on each other by exchanging photons. One charged particle emits a photon and another absorbs it, so the photon carries the force between them; hence photons are the mediators of the electromagnetic force.
• A photon is the smallest quantum of electromagnetic radiation and the basic unit of light. It travels through vacuum at 3\times10^{8}\,\mathrm{m\,s^{-1}} and has energy E=hf.
• It has zero rest mass and zero charge, it is stable, and it has spin 1, so it is a boson (a gauge boson).

In short, the electromagnetic interaction between two charges is the exchange of photons between them.

Both terms describe how a solid responds to a force. The difference is whether the body recovers its shape when the force is removed.

Sign of the entropy change (1 mark). The sign is negative because heat is removed from the body. Entropy change is positive when heat is added to a system and negative when heat is removed from it. The entropy of the body decreases, while the entropy of the surroundings that receive the heat increases by at least this amount.

Write each vector in terms of its rectangular components along the unit vectors \hat\imath, \hat\jmath, \hat k, and multiply them with the dot product (1 mark).

Conservative force (1.5 marks). A force is conservative if the work it does in moving a body between two points is independent of the path followed, or equivalently if the work done around a closed path is zero. The field of such a force is a conservative field. Examples: gravitational force and electric force.

Non-conservative force (1.5 marks). A force is non-conservative if the work done depends on the path followed, or if the work done around a closed path is not zero. Examples: friction, viscous drag and air resistance.

Magnus effect (1.5 marks). When a spinning object moves through a fluid, its path is deflected sideways, in a way that does not happen when it is not spinning. The spinning ball drags some of the surrounding air with it, so the air moves faster on one side and slower on the other. By Bernoulli's principle the faster side has lower pressure, and the ball is pushed towards the low-pressure side. The deflection depends on the rotational speed of the object.

Example (1.5 marks). A spinning cricket ball or golf ball. As the ball spins, the air moves faster over the side where the surface moves with the air stream, giving a lower pressure there. The pressure difference gives a sideways force, so the ball curves (swings) away from its straight path in flight, or a golf ball gets extra lift.

By Faraday's law the magnitude of the induced emf is \varepsilon=N\dfrac{\Delta\phi}{\Delta t}, where N is a pure number. So \varepsilon has the same units as \dfrac{\Delta\phi}{\Delta t}, and we show that the units of \Delta\phi/\Delta t reduce to the volt.

Here 1\,\mathrm{Wb}=1\,\mathrm{N\,m\,A^{-1}}, 1\,\mathrm{A\,s}=1\,\mathrm{C} and 1\,\mathrm{J\,C^{-1}}=1\,\mathrm{V}. Both \varepsilon and \Delta\phi/\Delta t are measured in volts, so they have the same units.

Conservative force (1.5 marks). A force is conservative if the work it does in moving a body between two points is independent of the path followed, or equivalently if the work done around a closed path is zero. The field of such a force is a conservative field. Examples: gravitational force and electric force.

Non-conservative force (1.5 marks). A force is non-conservative if the work done depends on the path followed, or if the work done around a closed path is not zero. Examples: friction, viscous drag and air resistance.

Solids are crystalline or amorphous according to how orderly their particles are arranged.

Streamline (steady or laminar) flow is smooth flow in which every particle passing a point follows the same path as the particles before it. Turbulent (non-laminar) flow is irregular, with small whirlpool-like regions, and the flow lines become disordered and mixed.

Radar (radio detection and ranging) uses the Doppler effect to find the speed and direction of aeroplanes.

1. The radar sends out radio waves of fixed frequency towards the aeroplane.
2. The waves are reflected by the moving aeroplane. Because of its motion the reflected waves have a changed frequency.
3. If the frequency of the reflected waves increases, the aeroplane is moving towards the radar; if it decreases, the aeroplane is moving away.
4. The size of the change in frequency is used to calculate the speed. The same principle is used in radar speed traps to measure the speed of cars.

Let \theta be the angle between \vec A and \vec B. Square the magnitude of each product.

Hence the relation is verified. It holds for every angle, because the dot product picks out the part of one vector along the other and the cross product the part perpendicular to it.

The elastic potential energy of a stretched wire equals the work done in stretching it. The work is found from the force-extension graph (the area of the triangle), and Young's modulus is then used to write the force in terms of the material.

Reversible process (1 mark). A process which can be retraced in exactly the reverse order without causing any change in the surroundings. If heat is absorbed in the direct process it is given out in the reverse process, and if work is done by the system in the direct process, it is done on the system in the reverse process. Examples: slow melting, freezing, boiling and condensing (water and ice).

Irreversible process (1 mark). A process which cannot be retraced in exactly the reverse order without causing a change in the surroundings. Examples: any process involving friction or dissipation of energy, and conduction, convection and radiation.

Cyclic process (1 mark). A succession of events which bring the system back to its initial condition. Event A takes the system from condition 1 to condition 2, and event B brings it back from 2 to 1. For a complete cycle the change in internal energy is zero, \Delta U=0. Heat engines work in cycles.

The density of an object increases as its speed approaches the speed of light. Density is \rho=m/V, and both the mass and the volume change with speed.

Since 1-v^{2}/c^{2} is less than 1, \rho>\rho_0. At 0.8c the factor is 1/0.36\approx2.8, so the density is about 2.8 times its rest value, and it grows without limit as v\to c.

The mass of the shell is not needed because the equations of motion describe the motion only.

An LDR works on photoconductivity: the conductance of its semiconductor material increases when light falls on it. Light of greater intensity delivers more energy to the material, so more electrons are freed from their bonds and become available to carry current (this is explained by the energy band theory).

With more charge carriers the conductance rises, and since R=1/G the resistance falls as the intensity of the light increases. In the dark there are few free carriers, so the resistance is high, of the order of megohms.

An atomizer is a device that emits a liquid as a fine spray. It works on Bernoulli's principle: where a fluid moves fast its pressure is low, and where it moves slowly its pressure is high.

The same arrangement is used to spray paint and in the carburettor of an engine.

Statement (2 marks). The average emf induced in a conducting coil of N turns is directly proportional to the rate of change of magnetic flux through the coil.

Mathematical form (1 mark).

Here \varepsilon is the magnitude of the induced emf, N the number of turns and \Delta\Phi/\Delta t the rate of change of flux. With the sign, \varepsilon=-N\dfrac{\Delta\Phi}{\Delta t}, the minus sign shows that the direction of the induced current is given by Lenz's law.

Precision (1 mark). Precision is the closeness of repeated measurements of the same quantity to each other. It depends on the least count (resolution) of the instrument and is indicated by the absolute uncertainty: the smaller the uncertainty, the greater the precision.

Importance of multiple readings (2 marks).

• Every reading contains some random error, so single readings fall on either side of the true value. Taking many readings and averaging them lets these random errors partly cancel, so the mean is closer to the true value and more reliable.
• The spread of the readings shows how well they agree. The mean deviation of the readings from their average gives the uncertainty in the mean, so more readings give a better estimate of the precision.
• A reading that lies far from the others (a mistake) is easy to spot among many readings and can be checked or rejected.

Induced emf (1.5 marks). No. By Faraday's law \varepsilon=N\dfrac{\Delta\Phi}{\Delta t}, so the emf depends only on the number of turns and the rate of change of flux. The resistance of the circuit does not appear.

Induced current (1.5 marks). Yes. The induced current in the circuit is found from Ohm's law:

For the same induced emf, a larger resistance gives a smaller current. For example, if the circuit is open (R\to\infty) an emf is still induced but no current flows.

The area of the parallelogram whose adjacent sides are the vectors \vec A and \vec B is the magnitude of their cross product, \text{Area}=|\vec A\times\vec B|=AB\sin\theta (base A times height B\sin\theta) (1 mark).

LDR (2 marks). A light-dependent resistor (photo-resistor) is a resistor whose resistance depends on the intensity of the light falling on it. It works on the principle of photoconductivity: when the light intensity increases the resistance decreases, and when the light is reduced the resistance increases. In the dark its resistance is of the order of megohms, and in bright light it falls to a few hundred ohms.

Application (1 mark). Street-light control: an LDR in a potential divider switches the lamp on automatically at sunset and off at sunrise. (Other uses are security systems based on light detection and light meters.)

The magnetic force on a moving charge is always perpendicular to its velocity, \vec F=q(\vec v\times\vec B). A force perpendicular to the displacement does no work, because W=\vec F\cdot\vec d=0.

Since no work is done, the change in kinetic energy is zero, \Delta K=W=0, so \tfrac12mv^{2} stays constant and the speed does not change. Only the direction of the velocity changes, which is why the particle moves in a circle.

In each alpha decay the nucleus emits an alpha particle {}^{4}_{2}\mathrm{He}: the mass number falls by 4 and the atomic number falls by 2. Charge and nucleon number are conserved in each step.

The resulting isotope is radium-225, {}^{225}_{88}\mathrm{Ra}.

Both are effects of the superposition of waves from two coherent sources. They differ in the phase of the waves when they meet and in the result.

Faraday cage (1 mark). A Faraday cage is a cage made of a conducting material. It is a hollow conductor in which any charge stays on the outside surface, and the electric field inside it is zero.

Working (2 marks).

Torque on a point mass

The same result holds for a rigid body, because the torques on its parts add and every part has the same angular acceleration: \tau_{\text{net}}=\sum m_ir_i^{2}\,\alpha=I\alpha.

Gluons and photons are both force carriers (gauge bosons), but they belong to different forces.

Both are massless, travel at the speed of light and have spin 1. The main difference is that the gluon carries colour charge and acts on quarks, while the photon is colourless and acts on electric charge.

First law of thermodynamics (1.5 marks). When heat Q is added to a system, it increases the internal energy \Delta U of the system and also does work W by the system on the environment. It is the law of conservation of energy applied to thermodynamic processes:

Here Q is positive when heat is added, W is positive when the gas expands (work done by the system) and \Delta U is positive when the temperature rises.

Internal energy (1.5 marks). The internal energy of a system is the sum of all the forms of molecular energies, kinetic and potential, of its substance. For an ideal gas there are no intermolecular forces, so it is only the translational kinetic energy of the molecules, with average value \tfrac32kT per molecule. It is therefore directly proportional to the absolute temperature. Internal energy is increased by heating and by doing mechanical work, and it is a function of state: its change \Delta U=U_2-U_1 depends only on the initial and final states, not on the path.

Meaning (1 mark). Annihilation is the process in which a particle and its anti-particle meet and destroy each other. Their mass is converted into energy, in accordance with E=mc^{2}, usually as gamma-ray photons. A particle and its anti-particle cannot exist in the same place.

Reaction (1 mark). An electron and a positron that come close annihilate and give two gamma-ray photons:

Two photons are needed to conserve momentum. The pair at rest has zero total momentum, and two photons moving in opposite directions have momenta that cancel. Each photon carries 0.511\,\mathrm{MeV}, the rest energy of an electron.

Second case (1 mark). The released energy may instead form a new particle and anti-particle pair, as when an up quark and an anti-up quark annihilate and produce a top quark and an anti-top quark.

The amplitude must be increased by a factor of about 3.16 (it becomes 3.16 times the original).

Potential gradient (1 mark). The potential gradient is the rate of change of electric potential \Delta V with distance \Delta r, that is \dfrac{\Delta V}{\Delta r}. It is the slope of the potential along the direction in which the potential changes.

Formula (2 marks). Take a uniform field E between two parallel plates and move a test charge q a small distance \Delta r from A to B along the field.

So the field strength equals the potential gradient. The minus sign shows that the potential falls in the direction of the field; ignoring it, E=\Delta V/\Delta r, with the unit volt per metre.

Relation (2 marks). For a fixed amount of gas at constant temperature, the ideal gas equation PV=nRT has a constant right-hand side, so

This is Boyle's law: the volume of a given amount of gas at constant temperature is inversely proportional to the pressure applied. If the pressure is increased the volume decreases, and vice versa. For two states of the same gas at the same temperature, P_1V_1=P_2V_2.

Graph (1 mark). The graph of P against V at constant temperature is a curve (a hyperbola, the isotherm). It falls as the volume increases and never touches either axis.

The negative charge spreads over the ring, so every part of the ring carries negative charge. Like charges repel: for two charges of the same sign q_1q_2 is positive and Coulomb's law gives a repulsive force.

The parts of the ring therefore push one another apart, and the radius of the ring increases. As the charges move farther apart the repulsion between them falls. The ring behaves the same way as for a positive charge, since only the sign of both charges is changed.

A superfluid is a liquid with zero viscosity. It flows without friction through any surface and so loses no kinetic energy. When stirred it forms vortices that keep spinning indefinitely, and it can creep over the walls of a container to 'empty' itself.

Superfluidity is observed in liquid helium-4 below 2.17 K (helium-3 becomes superfluid below 0.0025 K). To make it, helium gas cooled to a few degrees above absolute zero is compressed and then expelled through a small nozzle. As the gas expands rapidly it cools into the superfluid state (Fig. 6.17).

An inertial frame of reference is at rest or moves with a uniform velocity along a straight line, so it does not accelerate. Newton's laws hold in it: a body at rest stays at rest, and a body in motion keeps a constant velocity unless an external force acts. The Earth (for many experiments) and the inside of a car moving at constant velocity are inertial frames.

A non-inertial frame of reference is accelerating. Bodies in it appear to accelerate even when no force acts on them, because the frame itself accelerates, so Newton's laws do not hold in it. An accelerating car is an example.

By the principle of homogeneity of dimensions, an equation can be correct only if every term has the same dimensions. Here both sides have the dimensions [ML^{-1}T^{-2}], so the equation P=\rho gh is dimensionally homogeneous. (This shows the dimensions are consistent; it cannot fix a numerical constant, which here is 1.)

The second law of thermodynamics can be stated in three ways. Two are the classical statements and the third is in terms of entropy.

1. Kelvin's statement: it is impossible for a heat engine to convert heat taken from a single reservoir entirely into work without leaving any change in the working system.
2. Clausius's statement: it is impossible to cause heat to flow from a cold body to a hot body without the expenditure of work.
3. In terms of entropy: if a system undergoes a natural process, it proceeds in the direction that causes the entropy of the system plus the environment to increase. For all reversible processes the entropy stays constant, and for an irreversible process it increases, so the entropy of the universe always increases.

Torque on a point mass

The same result holds for a rigid body, because the torques on its parts add and every part has the same angular acceleration: \tau_{\text{net}}=\sum m_ir_i^{2}\,\alpha=I\alpha.

Interference (1 mark). Interference is the effect produced by the superposition of waves from two coherent sources (waves of the same frequency and a constant phase difference). When they meet in phase they reinforce each other (constructive interference); when they meet 180^\circ out of phase they cancel (destructive interference).

Conditions (2 marks).

1. The waves must travel in the same direction.
2. They must arrive at the same place at the same time.
3. They must be in phase for constructive interference and out of phase (180^\circ) for destructive interference.
4. The path difference must be a whole number of wavelengths, d=m\lambda, for constructive interference, and an odd multiple of half a wavelength, d=\left(m+\tfrac12\right)\lambda, for destructive interference (m=0,1,2,\dots).

Vector product: statement and explanation (3 marks)

When the product of two vectors gives a vector, it is called the vector product or cross product, written \vec A\times\vec B. For two vectors of magnitudes A and B with angle \theta between them it is defined as

The magnitude is the product of the magnitudes and the sine of the angle between them. The unit vector \hat n gives the direction: it is perpendicular to the plane containing \vec A and \vec B, and it is found by the right hand rule. Curl the fingers of the right hand from the first vector towards the second through the smaller angle; the erect thumb gives the direction of \hat n.

Example: torque. The torque of a force \vec F acting at position vector \vec r is the cross product \vec\tau=\vec r\times\vec F=rF\sin\theta\,\hat n. Its magnitude is largest when \vec F is perpendicular to \vec r (\sin90^\circ=1) and zero when \vec F is along \vec r. Another example is angular momentum, \vec L=\vec r\times\vec p.

Showing that A × B is not equal to B × A (4 marks)

Compare the two products in magnitude and in direction.

Check with unit vectors. \hat\imath\times\hat\jmath=\hat k, but reversing the order gives \hat\jmath\times\hat\imath=-\hat k. The two products are equal only when each is the null vector, that is when \vec A and \vec B are parallel or antiparallel (\sin\theta=0).

A projectile is launched from level ground with speed v_i at angle \theta to the horizontal. Air resistance is neglected, so horizontally a_x=0 (the horizontal velocity stays constant) and vertically a_y=-g. The initial components are

(a) Range (3 marks)

(b) Height (3 marks)

Projectile (1 mark)

A projectile is an object thrown into the air that then moves under the action of gravity only. Its motion is two-dimensional, along a curved path called its trajectory. Air resistance and the rotation of the Earth are neglected. The horizontal and vertical motions are independent: horizontally a_x=0, so the horizontal velocity stays constant, and vertically a_y=-g. Examples are a football that is kicked, a cricket ball and a long jumper.

Let the projectile be launched from level ground with speed v_i at angle \theta to the horizontal. The initial components are

Time of flight (2 marks)

Range and maximum range (4 marks)

(a) Postulates of special relativity (2 marks)

Einstein's special theory of relativity (1905) rests on two postulates.

1. Principle of relativity: the laws of physics are the same for all observers in any inertial frame of reference (a frame at rest or moving with constant velocity). No experiment can tell whether such a frame is at rest or moving uniformly. For example, a ball falls vertically for a person in a stationary train and for a person in a train moving at constant velocity.
2. Constancy of the speed of light: the speed of light in vacuum, c=299\,792\,458\,\mathrm{m\,s^{-1}}, is the same for all observers, whatever the relative velocity of the source and the observer. An observer moving at 0.5c along a light beam still measures its speed as c, not 0.5c.

(b) Time dilation and length contraction (2 marks)

Time dilation. A moving clock runs slower than a clock at rest relative to the observer. If t_0 is the proper time (the time between two events measured by one clock at rest at the events) and t is the time measured by an observer who sees that clock moving with speed v, then

Since \sqrt{1-v^{2}/c^{2}}<1, t>t_0: the moving clock ticks more slowly. For example, a pendulum of period 5.0\,\mathrm{s} in its own frame has a period of 10.0\,\mathrm{s} for an observer moving at 0.866c, since 1/\sqrt{1-0.75}=2. The effect is noticeable only at speeds close to c.

Length contraction. An object is measured to be shorter in the direction of its motion relative to the observer. If L_0 is the proper length (measured in the rest frame of the object) and L the length measured by an observer for whom the object moves with speed v, then

Since the root is less than 1, L<L_0. The contraction is only along the direction of motion; the width and height are unchanged. At v=0.8c the factor is 0.6, so a 10\,\mathrm{m} rod is measured as 6\,\mathrm{m}.

(c) Speed at which the mass of a proton is tripled (2 marks)

Bernoulli's equation (1 mark)

Statement. For the steady flow of an ideal fluid (non-viscous, incompressible, laminar), the sum of the pressure, the kinetic energy per unit volume and the potential energy per unit volume has the same value at every point along a streamline:

Bernoulli's equation is the law of conservation of energy for a flowing fluid. It shows that where the fluid moves fast the pressure is low.

Velocity of efflux, Torricelli's theorem (3 marks) and diagram (2 marks)

Torricelli's theorem states that the speed of flow of fluid from an orifice at depth h below the top surface of a liquid is equal to the speed gained by the fluid in free fall through the height h.

Torricelli's theorem gives the speed of water leaving a hole at the bottom of a bucket of depth h as \sqrt{2gh} (numerical problem 6.8).

Working of a refrigerator (4 marks)

Definition

A refrigerator is a device that works in the reverse direction to a heat engine. It removes heat from a region at a lower temperature than its surroundings and gives it to a region at a higher temperature, and work must be done on the working substance to make this happen.

Working

• The working substance, a gas called the refrigerant, takes in heat Q_2 from the cold reservoir at T_2 (the inside of the refrigerator).
• A compressor does work W on the refrigerant, raising its pressure and temperature.
• The refrigerant gives out heat Q_1 to the hot reservoir at T_1 (the room), where Q_1=Q_2+W.

Coefficient of performance (2 marks)

For heating, the coefficient of performance is \mathrm{CP}_{\text{heating}}=\dfrac{Q_1}{W}=\dfrac{T_1}{T_1-T_2}. No cyclic device can extract heat from a cold reservoir and deliver it entirely to a hot one without work.

Work-energy theorem (3 marks)

Statement. The net work done on an object is equal to the change in its kinetic energy.

Work-energy theorem in a resistive medium (3 marks)

A resistive force (friction, viscous drag) opposes the motion. When an applied force does work W on an object, part of it increases the object's kinetic energy and the rest is the work W_r done by the object against the resistive force:

The work against the resistive force takes energy away from the object. If the speed is constant, \Delta\text{K.E.}=0 and all the work is spent against resistance, W_r=W. This method works even when the forces are not constant, because only the work and the change in kinetic energy are needed. A parachute, which greatly increases the resistive force, makes a skydiver slow down.

Expression for the force (4 marks)

A charge at rest feels no magnetic force. A charge q moving with velocity \vec v in a uniform field \vec B feels a force perpendicular to both. It is found by treating the moving charge as a current in a conductor.

Direction of the force (2 marks)

The force \vec F=q(\vec v\times\vec B) is perpendicular to the plane containing \vec v and \vec B. Use the right-hand rule: curl the fingers of the right hand from \vec v towards \vec B through the smaller angle, and the thumb points along the force on a positive charge. (Equivalently, use Fleming's left-hand rule with the current replaced by the velocity of the positive charge.) For a negative charge the force is in the opposite direction.

• The force is largest when \theta=90^\circ, F=qvB.
• It is zero when \vec v is parallel or antiparallel to \vec B (\theta=0^\circ or 180^\circ).
• Because \vec F\perp\vec v, the force changes only the direction of the motion, not the speed: the particle moves in a circle for \theta=90^\circ and in a helix for other angles.

Equation of continuity

The equation of continuity is the statement of conservation of mass for the steady flow of an ideal (incompressible, non-viscous) fluid. It states that the product of the cross-sectional area and the speed of the fluid is constant at every point of a pipe: Av=\text{constant}.

Example from daily life

When water falls from a tap, its speed increases as it falls under gravity, so the stream becomes narrower (Fig. 6.8a). Squeezing the end of a rubber pipe reduces the area of the opening and makes the water leave faster (Fig. 6.8b).

Basis (the 'or' part). The equation of continuity is derived on the basis of the law of conservation of mass: for an ideal fluid in steady flow the mass entering a pipe in a given time equals the mass leaving it (\rho_1A_1v_1=\rho_2A_2v_2), and for an incompressible fluid this becomes A_1v_1=A_2v_2.

(a) Working of a refrigerator (3 marks)

Definition

A refrigerator is a device that works in the reverse direction to a heat engine. It removes heat from a region at a lower temperature than its surroundings and gives it to a region at a higher temperature, and work must be done on the working substance to make this happen.

Working

• The working substance, a gas called the refrigerant, takes in heat Q_2 from the cold reservoir at T_2 (the inside of the refrigerator).
• A compressor does work W on the refrigerant, raising its pressure and temperature.
• The refrigerant gives out heat Q_1 to the hot reservoir at T_1 (the room), where Q_1=Q_2+W.

The coefficient of performance for cooling is the ratio of the heat removed from the cold reservoir to the work done:

(b) Increase in entropy means degradation of energy (3 marks)

Entropy S measures the unavailability of a system's thermal energy for conversion into mechanical work, and the degree of disorder. For a reversible process at absolute temperature T, \Delta S=\dfrac{\Delta Q}{T}. In natural (irreversible) processes the entropy of the system plus its surroundings increases. Every time entropy increases, the opportunity to convert some heat into work is lost, so an increase in entropy means the degradation of energy: the energy is still there, but less of it can do useful work.

Example. In a car engine the burning of fuel produces some work, while the remaining heat energy is released into the atmosphere. Before the change all the heat from the fuel was available for work, but afterwards a major part of it is lost to the atmosphere and can no longer be used. The entropy of the universe has increased and the energy has been degraded.

In a reversible process the entropy is constant and no energy is degraded. Natural processes are irreversible, so energy is always being degraded.

Elastic potential energy (1 mark)

The elastic potential energy (strain energy) is the energy stored in a material due to the strain produced in it. For a material deformed within its elastic limit it is equal to the work done in producing the deformation.

Relation for the elastic potential energy (6 marks)

The work done in stretching a wire by an extension x is the area under its force-extension graph. Hooke's law and Young's modulus then give the energy in terms of the material.

(a) Kirchhoff's laws (3 marks)

Kirchhoff's first law (current law). The currents entering a node are equal to the currents leaving it. Taking entering currents as positive and leaving currents as negative, the algebraic sum of the currents at a node is zero:

It is a consequence of the conservation of charge: the charge that enters a junction in a given time must leave it in the same time.

Kirchhoff's second law (voltage law). The algebraic sum of all the potential changes around any closed loop is zero. A rise in potential is taken as positive and a drop as negative:

It is a consequence of the conservation of energy: the energy per unit charge gained from the sources equals the energy per unit charge lost in the resistances.

(b) Resistance of the coil (3 marks)

Why an astronaut feels weightless (1 mark)

A satellite and everything inside it are in free fall towards the Earth. Gravity supplies the centripetal force that keeps the satellite in its orbit, and the satellite and the astronaut have the same acceleration. The floor therefore does not push on the astronaut: the contact force (apparent weight) is zero, and he floats. Gravity has not disappeared; at the height of a space station it is still about 90\,\% of its surface value.

Artificial gravity (1 mark)

Artificial gravity is produced by rotating the satellite (space station) about its own axis. The outer wall of the rotating rim pushes the occupants inwards and provides the centripetal force that keeps them on a circle. They feel this contact force as weight, so the centripetal acceleration plays the part of g.

Derivation (4 marks)

Kinetic energy (3 marks)

Kinetic energy is the energy a body possesses because of its motion. Take a ball, initially at rest, which is given a horizontal force F so that it moves through a displacement d and reaches a final speed v. The work done on the ball appears as its kinetic energy (Figure 5.7).

Since \mathrm{kg\,m^{2}\,s^{-2}}=\mathrm{N\,m}=\mathrm{J}, the SI unit of kinetic energy is the joule. Although derived for a cricket ball, the formula holds for any moving body.

Work-energy theorem in a resistive medium (3 marks)

Statement. The net work done on an object is equal to the change in its kinetic energy, W_{net}=\Delta\text{K.E.}. In a resistive medium (friction, air resistance, viscous drag) a force opposes the motion, so only part of the work W done by the applied force increases the kinetic energy and the rest, W_r, is the work done by the object against the resistive force:

The work against the resistive force takes energy away from the object. If the speed is constant, \Delta\text{K.E.}=0 and all the work is spent against resistance, W_r=W. This method works even when the forces are not constant, because only the work and the change in kinetic energy are needed. A parachute, which greatly increases the resistive force, makes a skydiver slow down.

In a closed pipe of length L the closed end is a node (the air cannot move) and the open end is an antinode. Each node-to-antinode gap is \lambda/4. A vibrating tuning fork is held at the open end. Fig. 9.10 shows the three lowest modes.

In general the nth mode has \lambda_n=\dfrac{4L}{2n-1} and f_n=(2n-1)\dfrac{v}{4L}. A closed pipe therefore has only odd harmonics, f_1,\,3f_1,\,5f_1.

The cells are taken as ideal (no internal resistance), because none is given. Both cells have their long plate, the positive terminal, at the top. Three branches join the top and bottom wires, so choose a current in each: I_1 upward through the 10\,\mathrm{V} cell, I_2 downward through the 2\,\Omega resistor, and I_3 downward through the 6\,\mathrm{V} cell and the 1\,\Omega resistor.

Current in each resistor: 2\,\Omega resistor, 5\,\mathrm{A} (downward); 1\,\Omega resistor, 4\,\mathrm{A} (in the direction of I_3, towards the junction at the bottom of the middle branch). The 10\,\mathrm{V} cell supplies 9\,\mathrm{A}.

For a product, the percentage uncertainties add, and a power multiplies the percentage uncertainty of that quantity by the power. In LT^{2}: \%\Delta(LT^{2})=\%\Delta L+2\,\%\Delta T=0.1\,\%+2\times2\,\%=4.1\,\%. So A is correct.

Option C (2.1\,\%) is what you get from 0.1\,\%+2\,\%, forgetting that T is squared, so its percentage uncertainty is doubled.

Torque equals the moment of inertia times the angular acceleration, \tau=I\alpha, the rotational form of F=ma (Section 4.9 of the book). Option A squares I without reason, so the dimensions are wrong. Option B divides instead of multiplying, which would make the angular acceleration larger for a larger I; it is the other way round. Option C adds two quantities of different kinds, which is not allowed.

Velocity is displacement divided by time, so [v]=\dfrac{[L]}{[T]}=[LT^{-1}]. So B is correct.

Option C, [LT^{-2}], is the dimensions of acceleration (velocity per unit time). Option A, [M^{0}L^{0}T^{0}], belongs to a dimensionless quantity such as an angle, and D, [L^{2}T^{-1}], is area per unit time.

Put \vec A\cdot\vec B=AB\cos\theta and |\vec A\times\vec B|=AB\sin\theta in the given condition: AB\cos\theta=AB\sin\theta. Cancel AB to get \tan\theta=1, so \theta=\pi/4 (45^\circ). So B is correct.

At \theta=0 or \pi the cross product is zero while the dot product is \pm AB, and at \theta=\pi/2 the dot product is zero while the cross product is AB. So A, C and D cannot satisfy the condition.

A joule is the unit of work, which is force times distance, so \mathrm{J}=\mathrm{N\,m}=(\mathrm{kg\,m\,s^{-2}})\,\mathrm{m}=\mathrm{kg\,m^{2}\,s^{-2}}. So D is correct.

Option B, \mathrm{kg\,m\,s^{-1}}, is the unit of momentum, and C, \mathrm{kg\,m^{2}\,s^{-1}}, is the unit of angular momentum (and of Planck's constant). They differ from the joule by a factor of \mathrm{s^{-1}}. Option A has \mathrm{m^{-2}} instead of \mathrm{m^{2}}, which would come from dividing by a distance instead of multiplying.

For a head-on elastic collision with the second body at rest (u_2=0), the book gives v_1=\dfrac{m_1-m_2}{m_1+m_2}\,u_1. If m_1\gg m_2, then m_2 can be neglected, so v_1\approx u_1 and v_2=\dfrac{2m_1}{m_1+m_2}u_1\approx2u_1. The massive body keeps moving with its original velocity, so V_1'=V_1 and C is correct.

Option A (2V_1) is the velocity of the lighter body after the collision, not of the massive one. Option B would be true only for bodies of equal mass, where the first body stops and the second takes over its velocity. Option D, V_2'=2V_2, is zero because V_2=0.

In projectile motion with no air resistance there is no horizontal force, so a_x=0 and the horizontal component of velocity v_x=v_i\cos\theta stays constant throughout the flight. So D is correct.

Option A mixes up the two components: it is the vertical component v_y that is zero at the highest point, while v_x is still v_i\cos\theta there. The horizontal velocity neither increases (B) nor decreases (C), because gravity acts only vertically.

The moment of inertia I=\sum mr^{2} depends on the mass of the body, on how that mass is distributed about the axis, and on where the axis is and how it is oriented (the same body has different I about different axes). It does not depend on how fast the body spins. Option C looks tempting, but moving the axis changes every distance r, so I does change with the orientation of the axis.

Use the first equation of motion with v_i=10\,\mathrm{m\,s^{-1}}, a=2\,\mathrm{m\,s^{-2}} and t=5\,\mathrm s: v_f=v_i+at=10+2\times5=20\,\mathrm{m\,s^{-1}}. So B is correct.

The bus is speeding up, so the final velocity must be greater than 10\,\mathrm{m\,s^{-1}}. Options A, C and D are all smaller than 10\,\mathrm{m\,s^{-1}} and come from dividing instead of multiplying.

Kinetic energy is \tfrac12mv^{2}. For equal masses, \dfrac{K_A}{K_B}=\dfrac{v_A^{2}}{v_B^{2}}=\left(\dfrac{2}{1}\right)^{2}=4:1. Option A (2:1) forgets that the speed is squared.

An incompressible fluid is one whose density stays constant, whatever the pressure (liquids are generally of this kind). Option D (zero density) would mean no fluid at all, and a very high or very small density says nothing about whether the density changes.

Young's modulus is Y=\dfrac{\text{tensile stress}}{\text{tensile strain}}=\dfrac{F/A}{\Delta L/L}, which is option B.

Option A, \dfrac{F\,\Delta L}{L\,A}, equals stress times strain instead of stress divided by strain. Options C and D do not have the form stress over strain.

The equation of continuity, \rho_1A_1v_1=\rho_2A_2v_2 (which becomes A_1v_1=A_2v_2 for an incompressible fluid), says that the mass of fluid entering a pipe in a given time equals the mass leaving it, so it expresses the law of conservation of mass. The tempting option B, energy conservation, is the basis of Bernoulli's equation, not of the equation of continuity.

The equation of continuity \rho_1A_1v_1=\rho_2A_2v_2 says that the mass entering a pipe equals the mass leaving it, so it expresses the law of conservation of mass. Bernoulli's equation is the one based on conservation of energy (option B).

The centripetal force is F_c=\dfrac{mv^{2}}{r}. It is proportional to the square of the speed when m and r stay the same. If v\to2v, then F_c\to\dfrac{m(2v)^{2}}{r}=4\dfrac{mv^{2}}{r}, so the force becomes four times. Option C (double) is the tempting mistake of treating F_c as proportional to v.

Glass is an amorphous solid. Its particles have no long-range regular arrangement and it softens over a range of temperature instead of melting sharply (the book calls glass a super-cooled liquid).

Crystalline (A) and poly-crystalline (C) solids have fully or partly ordered structures. Ductile (D) describes how a solid deforms under stress, not its structure.

Young's modulus is the ratio of tensile stress to tensile strain: Y=\dfrac{F/A}{\Delta L/L}, option D.

Option A is the reciprocal of this (strain over stress). Options B and C are not ratios of stress to strain.

By the first law, Q=\Delta U+W, so \Delta U=Q-W=200-80=120\,\mathrm{J}. The heat added is +200\,\mathrm{J} and the work done by the system is +80\,\mathrm{J}, so the internal energy increases by 120\,\mathrm{J} (option C).

Option B (80\,\mathrm{J}) is just the work. Option D adds the work to the heat instead of subtracting it, and option A has the wrong sign.

A Carnot cycle has four processes: isothermal expansion, adiabatic expansion, isothermal compression and adiabatic compression (option C).

Option A (two) comes from counting only the two kinds of process, isothermal and adiabatic. Each kind occurs twice in the cycle, once as an expansion and once as a compression.

A stationary wave is formed by the superposition of two identical waves of the same speed, amplitude and frequency travelling in opposite directions, for example a wave and its reflection (option C).

Reflection alone (B) only supplies the second wave; the pattern of nodes and antinodes comes from their superposition. Diffraction (A) and refraction (D) are bending of waves.

Light from a star moving away from the Earth is shifted to a lower frequency, towards the red end. This is the red shift (option A), the Doppler effect for light. Hubble observed that all galaxies appeared red-shifted, which showed that the universe is expanding.

Blue shift (B) is the opposite: stars moving towards the Earth. Black and yellow shifts do not exist.

The magnetic force \vec F=q(\vec v\times\vec B) is always perpendicular to the velocity. It therefore changes only the direction of the velocity, never the speed, and acts as a centripetal force: qvB=\dfrac{mv^{2}}{r} gives the circle of radius r=\dfrac{mv}{qB}. The answer is D.

The path is straight (B) only if the particle moves along the field, where the force is zero. It is never parabolic (C), because a force of constant magnitude perpendicular to the velocity does not give a parabola. A (random) is wrong because the path is fixed by v, B, q and m.

The force on a charge is \vec F=q\vec E. For a negative charge q<0, so the force is opposite to the field. The field points from south to north, so the particle accelerates towards the south. The answer is B.

Option A (north) would be correct for a positive charge. East and west are at right angles to the field and the force has no component in those directions.

At the balance point the potentials at the two ends of the galvanometer are equal, so there is no potential difference across it and no current flows. The galvanometer reads zero, which is why the method is called the null method.

Option B (maximum) describes the galvanometer when the bridge is far from balance. A reading of minimum or infinite current would not give a sharp balance point.

The potential difference is the work done per unit charge, V=\dfrac{W}{q}=\dfrac{800\,\mathrm{J}}{4\,\mathrm{C}}=200\,\mathrm{V}, so C is correct.

Option D, 5\times10^{-3}\,\mathrm{V}, comes from dividing the wrong way round (q/W=4/800). Option B (100\,\mathrm{V}) would need a charge of 8\,\mathrm{C}.

The resistor is across the battery, so V=10\,\mathrm{V} and P=\dfrac{V^{2}}{R}=\dfrac{(10)^{2}}{100}=1\,\mathrm{W}. The same answer comes from I=V/R=0.1\,\mathrm{A} and P=VI=1\,\mathrm{W}.

Option C (100\,\mathrm{W}) is V^{2} itself, found by forgetting to divide by R. Option B (10\,\mathrm{W}) would be the answer for a 10\,\Omega resistor.

The tesla is the SI unit of magnetic induction (flux density). It is defined from F=BIL\sin\theta: 1\,\mathrm{T}=1\,\mathrm{N\,A^{-1}\,m^{-1}}.

The gauss (B) is a unit of the same quantity, but in the CGS system, not the SI (1\,\mathrm{T}=10^{4}\,\mathrm{G}). The weber (C) is the unit of magnetic flux, and the newton metre (A) is the unit of work or torque.

Rest energy E=m_0c^{2} with m_0=2.00\,\mathrm{g}=2.00\times10^{-3}\,\mathrm{kg} and c=3.00\times10^{8}\,\mathrm{m\,s^{-1}}:

This is option C. The wrong powers in A (10^{18}) and D (10^{16}) come from forgetting to convert grams to kilograms or from mis-squaring c.

Time dilation means that time between two events is stretched for an object in motion: the moving clock ticks more slowly, so the time interval t=t_0/\sqrt{1-v^{2}/c^{2}} is greater than the proper time t_0. The paper's option A, 'time increases for moving observer', states this as the lengthening of the time interval. Option B is the tempting wrong one: it confuses the slowing of the clock with a shorter interval. Option C (time becomes zero) would need v\to\infty, and D (constant) denies the effect.

From m=m_0/\sqrt{1-v^{2}/c^{2}} the root is less than 1 for any moving body, so the relativistic mass m is greater than the rest mass m_0 (it rises without limit as v\to c). Option C (equal) is the tempting one: it holds only at v=0, or when the speed is so small that the effect cannot be noticed.

An inertial frame is one that is at rest or moves with uniform velocity along a straight line, and in it Newton's first law (law of inertia) holds: a body at rest stays at rest and a body in motion keeps its constant velocity unless a force acts. In a non-inertial (accelerating) frame this fails, since bodies seem to accelerate with no force. Options B and C are tempting but Newton's second and third laws are not what define the frame; D is not a law of motion.

The neutrino is a lepton with zero electric charge, so it is a neutral particle. Neutrinos are not magnetic or charged, and their mass is extremely small (less than 1\,\mathrm{eV}/c^{2} for the electron neutrino), so option C (heavy) is wrong. Option B, charged, is the property of the electron, muon and tau, the charged leptons.

For a solid cylinder (disc) about its central axis the moment of inertia is I=\tfrac12MR^{2} (Figure 4.13 of the book). Substituting,

The wrong options come from slips: 1.5=MR forgets to square R, and MR^{2}=0.45 forgets the factor \tfrac12.

Add the charges of the quarks: uud gives +\tfrac23+\tfrac23-\tfrac13=+1, which is a proton. The neutron is udd (two down quarks and one up quark), charge 0. A meson has only two quarks (a quark and an anti-quark), and a photon is a force carrier, not made of quarks.

Newton's law of gravitation is F=G\dfrac{m_1m_2}{r^{2}}. Make G the subject and put in the dimensions (1 mark for the rearrangement).

The unit of G is \mathrm{N\,m^{2}\,kg^{-2}}=\mathrm{m^{3}\,kg^{-1}\,s^{-2}}, which agrees.

By the principle of homogeneity of dimensions, an equation can be correct only if every term has the same dimensions. Here both sides have the dimensions [ML^{-1}T^{-2}], so the equation P=\rho gh is dimensionally homogeneous. (This shows the dimensions are consistent; it cannot fix a numerical constant, which here is 1.)

The dot product of two vectors is \vec A\cdot\vec B=AB\cos\theta, where \theta is the angle between them. It is a scalar. Three properties (1 mark each):

• It is commutative. \vec A\cdot\vec B=\vec B\cdot\vec A, because AB\cos\theta=BA\cos\theta.
• It is zero for perpendicular vectors. For \theta=90^\circ, \cos\theta=0, so \vec A\cdot\vec B=0. Hence \hat\imath\cdot\hat\jmath=\hat\jmath\cdot\hat k=\hat k\cdot\hat\imath=0.
• It is largest for parallel vectors. For \theta=0^\circ, \cos\theta=1, so \vec A\cdot\vec B=AB. In particular \vec A\cdot\vec A=A^{2} and \hat\imath\cdot\hat\imath=\hat\jmath\cdot\hat\jmath=\hat k\cdot\hat k=1. For antiparallel vectors (\theta=180^\circ) it is -AB, the smallest value.

In components, \vec A\cdot\vec B=A_xB_x+A_yB_y+A_zB_z. Work W=\vec F\cdot\vec d and power P=\vec F\cdot\vec v are examples of the dot product.

Entropy is said to measure the degradation of energy: as entropy increases, less of the energy in the system is available to do useful work. In a natural process the energy is not lost, but its ability to do work is reduced.

In a reversible process the entropy is constant and no energy is degraded. Natural processes are irreversible, so energy is always being degraded.

Example. In a car engine the burning of fuel gives some work, and the remaining heat is released into the atmosphere. This heat is no longer available for work, so the energy has been degraded and the entropy has increased.

Timing 10 vibrations together shares the stopwatch's single uncertainty of 0.1\,\mathrm{s} among all of them, so the uncertainty in one period is 10 times smaller than in timing one vibration. The percentage uncertainty is \dfrac{0.01}{1.76}\times100\approx0.57\,\%.

The book calls polymeric solids polycrystalline solids: their structure lies between order and disorder, while an amorphous solid has no regular structure at all.

Explanation. The falling water is a stream of an (almost) incompressible fluid in steady flow, so the equation of continuity applies: the volume of water passing every cross-section per second is the same.

As the water falls under gravity its speed increases (v_2>v_1). For the same flow rate, a larger speed needs a smaller area, so A_2<A_1 and the stream becomes narrower as it comes down. This is the conservation of mass: the same mass of water must pass each cross-section in each second.

The Higgs boson (3 marks). The Higgs boson is a fundamental (elementary) particle that is responsible for the mass of particles.

• Up to 1964 the mass of a particle was thought to be an intrinsic property. In 1964 Peter Higgs proposed the Higgs field, which fills the universe and gives mass to elementary particles. The Higgs boson is the wave (quantum excitation) of that field. It was confirmed at CERN in July 2012, with a mass of 124.97\,\mathrm{GeV}/c^{2}, about 130 times that of a proton.
• How strongly a particle interacts with the Higgs field decides its mass. Quarks interact strongly and gain relatively large mass; electrons interact only slightly, so they are very light; photons do not interact with it and have no mass.
• The Higgs boson has zero spin, zero electric charge and no strong force interaction.

Check: AB=\sqrt{(6\sqrt3)^{2}+6^{2}}=\sqrt{144}=12, and 12\cos30^\circ=6\sqrt3, 12\sin30^\circ=6.

A vector \vec A at angle \theta to the x-axis has rectangular components A_x=A\cos\theta and A_y=A\sin\theta (1 mark).

So the rectangular components of a vector have the same magnitude when the vector makes an angle of 45^\circ with the x-axis (and so also 45^\circ with the y-axis). The same holds in the other quadrants, at 135^\circ, 225^\circ and 315^\circ, where the two components differ only in sign.

Here the first body is much lighter than the second (m_1\ll m_2) and the second is at rest. In the book's notation the velocities before the collision are u_1 and u_2=0 (the paper's V_1 and V_2=0), and after the collision v_1 and v_2. The collision is elastic and in one dimension.

Discussion (1 mark). The lighter body rebounds with the same speed it had, in the opposite direction (v_1=-u_1). The heavier body remains at rest even after the collision (v_2\approx0). A ball thrown at a fixed wall bounces back with the same speed it had before hitting the wall; the wall is the massive body.

Construction. A seismometer has a coil wound on a frame that hangs from springs fixed to the casing. A permanent magnet is fixed to the casing, which rests on the ground, in such a way that the coil can oscillate freely relative to the magnet.