Pearson Edexcel GCSE Physics Paper 2, Higher Tier (1PH0/2H)Paper 2 — every question, answered

The correct symbol for a thermistor is the one that shows a resistor with a diagonal line through it and no arrow, since a thermistor's resistance depends on temperature rather than light, which rules out the light-dependent resistor and diode symbols shown as other options.

At a potential difference of 5.0V the current is 0.26A, so substituting into resistance = potential difference divided by current gives resistance = 5.0 divided by 0.26 = 19.2, which rounds to 19 ohms.

The student's conclusion is not correct, because directly proportional would mean that doubling the potential difference should double the resistance, but the table does not show this: going from 1.0V to 2.0V the resistance only rises from 11 to 14 ohms rather than doubling to 22 ohms, and going from 3.0V to 6.0V it only rises from 17 to 20 ohms rather than doubling to 34 ohms. The resistance does increase as the potential difference increases, but not in the same proportion, which is what you would expect from a filament lamp that gets hotter as more current flows through it.

The correct symbol for a thermistor is the resistor symbol with a diagonal line through it and no arrowhead, which distinguishes it from the light-dependent resistor, variable resistor and diode symbols also shown. As temperature increases, the resistance of this thermistor decreases, and the graph shows this is a non-linear relationship, since the curve is steep at low temperatures and becomes less steep as temperature rises, rather than falling at a constant rate.

Drawing a tangent to the curve exactly at 30°C and forming a right-angled triangle underneath it, the resistance falls by roughly 4.6 kilohms over a temperature range of about 50°C along that tangent, giving a rate of change of resistance with temperature of approximately negative 0.09 kilohms per degree Celsius, since the mark scheme accepts any value between about 0.087 and 0.097 for this gradient. The unit for this rate of change is kilohms per degree Celsius.

One improvement would be to place the thermometer closer to the thermistor itself, since this ensures the thermometer is measuring the same temperature that the thermistor is actually experiencing rather than a slightly different part of the water. For the precision comparison, method 2 gives more precise results than method 1 because the voltmeter and ammeter readings are given to more significant figures and decimal places than the ohmmeter's reading, so the resistance calculated by dividing the voltmeter reading by the ammeter reading can also be stated to more significant figures than the direct ohmmeter reading allows.

The student should add a variable resistor in series with the lamp. Since a variable resistor's resistance can be adjusted smoothly rather than in fixed steps, the share of the fixed supply voltage that appears across the lamp can then also be adjusted smoothly, giving a continuously variable voltage instead of only the whole number settings the power supply itself provides.

Substituting directly into R = V divided by I gives R = 4.5 divided by 0.30 = 15 ohms.

To complete the circuit I connect a voltmeter across the iron wire, in parallel with it, and an ammeter in series with the iron wire, so the ammeter reads the current flowing through the wire and the voltmeter reads the potential difference across just that wire.

To investigate how resistance changes with length, the students would also need a metre rule, so that they can measure and mark out different lengths of the wire to test.

Drawing a straight line of best fit through the plotted points and extending it a little further, the resistance of a 100cm length can be estimated at around 3.0 ohms, following the same roughly straight-line trend shown by the shorter lengths already plotted.

The variable resistor increases the total resistance of the circuit, which keeps the current flowing through the iron wire small. Since a larger current would raise the temperature of the iron wire more, keeping the current small using the variable resistor is what prevents the wire from overheating.

Rearranging V = I x R for resistance gives R = V divided by I, so R = 1.56 divided by 0.45 = 3.47, which rounds to 3.5 ohms.

The lamp in Figure 4 is dimmer than the lamp in Figure 3. Adding a second identical lamp in series increases the total resistance of the circuit, and since the power supply provides the same potential difference as before, this bigger resistance means the current in the circuit is now smaller. With the current reduced, and the supply's potential difference now shared between the two lamps rather than being applied fully across just one, each lamp receives less power than the single lamp did, which is why each one glows more dimly.

Using power = potential difference multiplied by current, power = 4.5 multiplied by 0.30 = 1.35W, which rounds to 1.4W.

Substituting into t = E divided by (I multiplied by V) gives t = 126,000,000 divided by (15.0 multiplied by 400) = 126,000,000 divided by 6000 = 21,000 seconds. Converting this to hours, 21,000 divided by 3600 = 5.8 hours. Since 5.8 hours is less than the claimed 6 hours, the claim is justified.

The earth wire is connected directly to the kettle's metal case, and since metal is a good conductor, once the live wire touches the case, the case becomes connected to the live supply through the earth wire's very low resistance path. Because this earth path has such low resistance, a very large current flows straight down to earth rather than through anyone touching the appliance, and the earth wire keeps the metal case held at earth potential rather than at the dangerous voltage of the live wire, so a person touching the case does not get a shock, since there is no potential difference between the case and earth to drive a current through them.

The fuse itself is a thin piece of wire connected inside the plug between the live pin and the wire that goes into the kettle. The temperature this wire reaches depends on the size of the current flowing through it, and the same very large fault current that flows to earth also flows through the thin fuse wire, heating it up rapidly. Once the fuse wire's temperature rises above its melting point, it melts and breaks, which disconnects the live supply to the kettle completely, so the kettle can no longer carry mains voltage anywhere inside it, and the surge is also prevented from damaging the house's own wiring.

The two features work together in sequence: it is the earth wire's low resistance that lets such a large fault current flow in the first place, by giving the current somewhere to go other than through a person, and it is that same surge of current which then heats up and blows the fuse, permanently isolating the kettle from the mains supply.

First, the rate of flow of charge is different: since the current in the immersion heater is 14A, 14 coulombs of charge flow through it every second, but since the current in the kettle's heating element is only 8.3A, only 8.3 coulombs flow through it every second, so charge flows faster in the immersion heater. Second, the direction of charge flow is different: the immersion heater runs from a battery, so charge always flows in the same direction through it, whereas the kettle runs from the mains, so the direction of charge flow in its heating element keeps reversing back and forth many times a second, since mains electricity is alternating current.

Rearranging E = Q x V for charge gives Q = E divided by V, so Q = 126,000,000 divided by 400 = 315,000 coulombs.

With current flowing from P to Q, the magnetic field around the wire forms circles, and the compass needle lines up tangent to that circular field, pointing in the direction shown by option A.

To find the shape of the whole field, the student should mark the direction the compass needle points to at one position on the card, then move the compass to a new position so its tail sits directly over that mark, and record the new direction it points to there. Repeating this process traces out a series of marks that, once joined together, reveal the field forms a complete circle around the wire, and repeating the whole process again starting from a different distance from the wire reveals a family of concentric circles.

Since field lines point away from a north pole and towards a south pole, I draw at least two arrows on the curved lines pointing away from the N end of the magnet and around towards the S end. The field is strongest right at the poles, where the lines are most tightly bunched together, so I mark the X immediately next to the N pole.

Two differences from a uniform field: first, the bar magnet's field lines curve and change direction from point to point, whereas a uniform field's lines are straight and all point the same way everywhere; second, the bar magnet's field lines are close together near the poles and spread further apart away from the magnet, showing the field varies in strength, whereas a uniform field has evenly spaced lines of exactly the same strength throughout.

A uniform magnetic field can be produced in a school laboratory between two flat Magnadur magnets facing each other with opposite poles, or equally inside a long solenoid carrying a steady current.

Sketching the Earth's field, I draw at least two field lines outside the Earth that match the directions shown by the compass needles, curving from one imaginary magnetic pole round to the other with arrows in the correct direction, and I continue at least two of these lines inside the Earth towards the imaginary poles, leaving a gap where the solid Earth itself is drawn. The Earth's magnetic field is generated by its magnetic outer core.

Since both magnets have their N poles facing each other, the two fields repel rather than joining up directly between the poles, so between the two magnets I draw at least four curved field lines that bow outward away from the gap, two curving upward and two curving downward, with none of them crossing straight across the middle. The arrows on these lines point away from each N pole, since field lines always point away from a north pole.

Using Fleming's left-hand rule with the magnetic field running from the N pole to the S pole and the current flowing in the direction shown by the arrow, the force on the wire acts upwards. By Newton's third law, the force the wire pushes back on the magnet with is exactly equal in size and opposite in direction, so the magnet itself experiences an equal force acting downwards.

Rearranging F = B x I x l for current gives I = F divided by (B multiplied by l), and converting the wire's length to metres, I = 0.045 divided by (0.72 multiplied by 0.030) = 0.045 divided by 0.0216 = 2.08, which rounds to 2.1A.

Rearranging F = B x I x l for magnetic flux density gives B = F divided by (I multiplied by l), and converting the current to amps, B = (1.11 x 10^-5) divided by (0.0931 multiplied by 0.600) = (1.11 x 10^-5) divided by 0.05586 = 1.99 x 10^-4, which rounds to 2.0 x 10^-4 T.

Using Fleming's left-hand rule with the field running from the N pole to the S pole and the current flowing in the direction shown, the force on the wire itself acts upwards. By Newton's third law, the wire pushes back on the magnet with an equal and opposite reaction force, so the force on the magnet acts downwards.

Rearranging F = B x I x l for length gives l = F divided by (B multiplied by I), so l = 0.15 divided by (0.50 multiplied by 2.7) = 0.15 divided by 1.35 = 0.111, which rounds to 0.11m.

When the device is used as a loudspeaker, it is an output device that turns an electrical signal into sound. An alternating current representing the sound signal is fed into the coil of wire, which sits inside the permanent magnet's field, and because this current keeps reversing direction it produces a changing magnetic field around the coil that is constantly attracted to and repelled by the magnet's fixed field. This is the motor effect: the alternating current forces the coil, and the movable cone attached to it, to push and pull back and forth repeatedly, with the rate of alternation setting the frequency of the sound produced and the size of the current setting how loud it is.

When the same device is used as a microphone instead, it works in reverse and becomes an input device that takes sound in rather than giving it out. Sound arriving at the cone is vibrations of the air, and these vibrations push and pull the cone, and the coil attached to it, back and forth through the magnet's field. Because the coil is now being physically moved through the field rather than being driven by a current, this is electromagnetic induction rather than the motor effect: the moving coil generates an alternating potential difference across its own wires, producing an alternating current signal that matches the original sound, with louder or higher-pitched sound producing a bigger amplitude or higher frequency in that induced signal.

I would connect the coil of wire across the terminals of the sensitive voltmeter, with the bar magnet held so it can be pushed in and pulled out along the axis of the coil.

To show what affects the direction of the induced pd, I would push the magnet into the coil and then pull it back out again, and note that the voltmeter needle deflects one way for one direction of movement and the opposite way for the other, showing the induced pd reverses direction. To show what affects the size of the induced pd, I would move the magnet into the coil quickly on one trial and slowly on another, and note that the voltmeter gives a much bigger reading for the faster movement than for the slower one, showing that a faster rate of change of the magnetic field induces a bigger potential difference.

As the magnet is repeatedly pushed in and pulled out of the coil, the needle on the meter oscillates, swinging to either side of the centre zero position, first one way and then the other, in response to the magnet's pole entering the coil and then leaving it again. This happens because the changing magnetic field through the coil, caused by the magnet moving in and out, induces an alternating potential difference and so an alternating current in the coil, and the direction of this induced current depends on whether a pole is entering or leaving, which is why the needle swings the opposite way for each motion.

Comparing the two voltmeter readings, the needle deflects further in (b) than in (a) and to the opposite side, so the handle in (a) was turned more slowly than in (b), and in the opposite direction to (b), matching the row that shows slower and opposite.

Once the dynamo is connected to a lamp, turning the handle becomes harder because the induced current now flowing in the coil creates its own magnetic field, and this field interacts with the permanent magnet's field to produce a force that opposes the coil's motion, so the teacher must do extra work to keep turning the handle against that opposing force.

The changing magnetic field in the transformer's core is caused by the alternating current in the primary coil, since an alternating current is constantly changing in size and direction, which produces a magnetic field around the primary coil that is likewise constantly changing.

Rearranging Vp divided by Np equals Vs divided by Ns for the number of secondary turns gives Ns = (Np multiplied by Vs) divided by Vp, so Ns = (2000 multiplied by 15) divided by 230 = 30,000 divided by 230 = 130.4, which rounds to 130 turns.

First I find the secondary voltage using Vp divided by Np equals Vs divided by Ns: Vs = (Ns multiplied by Vp) divided by Np = (400 multiplied by 230) divided by 700 = 131.4V. Since the transformer is 100% efficient, power in equals power out, so Vp multiplied by Ip equals Vs multiplied by Is, giving Ip = (Vs multiplied by Is) divided by Vp = (131.4 multiplied by 1.75) divided by 230 = 230.0 divided by 230 = 1.00A.

At the power station, P, electricity is generated at a moderate voltage, and Q is a step-up transformer that increases this voltage hugely before it enters the transmission line, so that the current flowing along that line for the same delivered power can be made much smaller, since power equals current multiplied by voltage.

R represents the transmission line itself, carrying electricity over a long distance. Because the current in this line has been made so much smaller by stepping up the voltage at Q, far less energy is wasted as heat in the wires, since the power lost to heating a wire depends on the current squared, so a smaller current wastes dramatically less energy; the smaller current also means a smaller potential difference is dropped along the length of the cable, since a wire's own voltage drop depends on the current flowing through it multiplied by its resistance.

At S, a step-down transformer reduces the very high transmission voltage back down to the standard 230V that is safe to use in a house, ready to supply appliances at T. Both Q and S are transformers built from coils of wire wound around a shared iron core, and although real transformers are never perfectly 100% efficient, this step-up-then-step-down arrangement across the whole system minimises the much larger losses that would otherwise occur if electricity were transmitted at its original, lower generating voltage with a correspondingly bigger current.

Rearranging Vp divided by Np equals Vs divided by Ns for the secondary voltage gives Vs = (Ns multiplied by Vp) divided by Np, so Vs = (18 multiplied by 230) divided by 800 = 4140 divided by 800 = 5.175, which rounds to 5.2V.

I would weigh a number of marbles together on the balance to find their total mass, since using several marbles rather than just one improves the accuracy of the measurement. Then I would lower the marbles fully into the water in the measuring cylinder and record the water level before and after adding them, so the difference between the two readings gives the volume of water displaced, which equals the volume of the marbles. Finally I would calculate the density of the glass by dividing the total mass of the marbles by this volume, using density = mass divided by volume.

Since the graph shows the volume of a fixed 1kg mass of water is smallest at 4°C, the density of water must be at its greatest at exactly 4°C, because density is mass divided by volume and the mass stays fixed. Below 4°C, as temperature rises towards 4°C the volume keeps falling, so density keeps increasing; then above 4°C, as temperature keeps rising the volume starts increasing again, so density decreases.

Reading the measuring cylinder, the water level rises from 490cm3 to 530cm3 once the iron is added, so the lump of iron displaces 530 minus 490 = 40cm3 of water, which is the iron's own volume. Rearranging density = mass divided by volume for mass gives mass = density multiplied by volume, so mass = 7.9 multiplied by 40 = 316g, which rounds to 320g to 2 significant figures.

Because the density of the wood, 0.82g/cm3, is less than the density of water, 1.00g/cm3, the wood would float rather than sink fully like the iron did, so it would only displace a volume of water equal to the part of the wood that is actually submerged, not the wood's whole real volume. This would give a wrongly small displaced volume reading, and so an inaccurate mass calculation.

Substituting directly into density = mass divided by volume gives density = (7.22 x 10^-2) divided by (2.69 x 10^-5) = 2684, which rounds to 2680kg/m3, with the unit kilograms per cubic metre since mass is in kilograms and volume is in cubic metres.

Two sentences the student should have added: first, 'I took a reading of the water level in the measuring cylinder before adding the metal', since without an initial reading there is nothing to compare the new level against; and second, 'I subtracted the first reading from the new reading to find the volume of the metal', since the new level on its own is the volume of the water plus the metal together, not the metal's volume by itself.

When water boils to become steam, the particles end up much further apart than they were in liquid water, which is why steam's density is far less than liquid water's, even though it is made of exactly the same particles.

Sublimation is when a solid changes directly into a gas without ever passing through the liquid state in between, for example solid carbon dioxide turning straight into gas.

The correct term for changing directly from solid to gas is sublimating. Converting aluminium's melting point to kelvin, I add 273 to 660, giving 933K.

In liquid aluminium above 660°C, the particles move around randomly in all directions at a range of different speeds, constantly colliding and sliding past each other, moving noticeably faster overall than the particles did when the aluminium was still a solid.

Rearranging change in thermal energy = mass x specific heat capacity x change in temperature for the temperature rise gives change in temperature = thermal energy divided by (mass multiplied by specific heat capacity), so change in temperature = 84000 divided by (0.25 multiplied by 4200) = 84000 divided by 1050 = 80°C. Since the water finished at boiling point, 100°C, and rose by 80°C to reach that, its temperature before heating must have been 100 minus 80 = 20°C.

Rearranging thermal energy for a change of state = mass x specific latent heat for the specific latent heat gives specific latent heat = thermal energy divided by mass, so specific latent heat = 0.34 divided by 0.15 = 2.27, which rounds to 2.3MJ/kg.

Specific heat capacity is about the energy needed to change a substance's temperature, heating it up or cooling it down, without it changing state, whereas specific latent heat is about the energy needed to change a substance's state altogether, such as melting or boiling it, without its temperature changing at all during that change.

After switching off the power supply, the student should keep stirring the water and keep watching the thermometer for some time afterwards, rather than reading it immediately, because it takes time for the heat that has already been supplied to spread all the way through the water by conduction and convection, so the temperature will keep rising for a while even with the heater off. The student should record the highest, peak temperature the thermometer reaches once it stops climbing any further, reading the scale at eye level to avoid a parallax error.

When the gas is heated, energy is transferred into the particles, which makes them move around with greater average velocities than before. Since kinetic energy is given by kinetic energy = half multiplied by mass multiplied by velocity squared, and the mass of each gas particle stays fixed, this increase in average speed directly means the average kinetic energy of the particles has increased, because kinetic energy depends on velocity squared, so even a modest increase in speed produces a real increase in the particles' kinetic energy.

Because the container's volume stays exactly the same, the faster-moving particles are now confined in the same space but hitting the container's walls with greater force each time they collide, since a faster particle undergoes a bigger change in momentum on each impact. Since pressure is defined as force divided by area, and the wall's area has not changed, this bigger average force per collision on its own would already increase the pressure the gas exerts.

On top of that, the faster particles also cross the fixed-size container and strike its walls more often in a given time, since they are travelling further each second, so the number of collisions per second on each wall increases as well. Both effects work in the same direction together, more force per collision and more collisions per second, which is why heating a gas at constant volume produces a genuinely bigger increase in pressure than either effect on its own.