Unit 2 ans loads of questions Physics edexcel

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Physics edexcel loads of questions

Transcript of Unit 2 ans loads of questions Physics edexcel

1. D[1] 2. C[1] 3. B[1] 4. A[1] 5. B[1] 6. C[1] 7. B[1] 8. C[1] 9. D[1] 10. (a) D 1(b) WavelengthUse of v = f (1)Use of f = 1/T (1)Answer T = [0.002 s] (1)[give full credit for candidates who do this in 1 stage T =/v ]Example of answerv = ff = 330 / 0.66T = 1/f = 0.66 / 330T = 0.002 s 3[4] 11. Direction of travel of light is water air (1)Angle of incidence is greater than the critical angle (1) 2[2] 12. (a) Transverse waves oscillate in any direction perpendicular to wave direction (1)Longitudinal waves oscillate in one direction only OR parallel to wavedirection. (1)Polarisation reduces wave intensity by limiting oscillations and wavedirection to only one plane OR limiting oscillations to one direction only. (1)(accept vibrations and answers in terms of an example such as a ropepassing through slits) 3 (b) Light source, 2 pieces of polaroid and detector e.g. eye, screen, LED ORlaser, 1 polaroid and detector (1)Rotate one polaroid (1)Intensity of light varies (1) 3[6] 13. Frequency unaltered (1)Wavelength decreases (1)Speed decreases (1) 3[3] 14. (a) Use of sensorEvent happens very quickly OR cannot take readings fast enough (1)Sampling rate: 50+ samples per second (1) 2(b) Initially the temperature is low so current is highResistance of filament increases as temperature increasesCurrent falls to steady value when temperature is constantMaximum heating is when lamp is switched on / when current is highestFilament breaks due to melting caused by temperature rise Max 4[6] 15. The answer must be clear and the answer must be organised in a logicalsequence (QWC It was known that X penetrated (1) It was not known that X rays were harmful (1) Doctors died because of too much exposure (1) Lack of shielding (1) New treatments may have unknown side effects (1) Treatments need to be tested / time allowed for side effects to appear (1) Max 4[4] 16. (a) [1.0 m] (1) 1 (b) Ratio of (5 or 6 / 3 ) 60 (1)Answer [f = 100 Hz] (1) 2[3] New treatments may have unknown side effects (1) Treatments need to be tested / time allowed for side effects to appear (1) Max 4[4]17. Use of sin i / sin r = (1)Use of either 80 or 1.33 (1)[r = 48] (1) 3Example of answersin 80 / sin r = 1.33[r = 48]Both rays refracted towards the normalViolet refracted more than red 2[5] 18. (a) Calculation of energy required by atom (1)Answer [1.8 (eV)] (1)Example of answer:Energy gained by atom = 13.6 eV 3.4 eV = 10.2 eVKE of electron after collision = 12 eV 10.2 eV = 1.8 eV 2 (b) Use of E = hf and c = f(1)Conversion of eV to Joules (1)Answer = [1.22 107 m] (1)Example of answerE = hf and c = f E = hc/= (6.63 1034 J s 3 108 m s2) (10.2 eV 1.6 1019 C)= 1.21 107 m 3[5] 19. The answer must be clear, use an appropriate style and be organised in alogical sequence. (QWC)Reference to I = nqvA (1)For the lampIncreased atomic vibrations reduce the movement of electrons (1)Resistance of lamp increases with temperature (1)For the thermistorIncreased atomic vibrations again reduce movement of electrons (1)But increase in temperature leads to a large increase in n (1)Overall the resistance of the thermistor decreased with increase intemperature. (1) Max 5[5] 20. (a) Diffraction is the change in direction of wave or shape orwavefront (1)when the wave passes an obstacle or gap (1) 2 (b) The energy of the wave is concentrated into a photon (1)One photon gives all its energy to one electron (1) 2 (c) Energy of photon increases as frequency increases OR reference toE = hf (1)Electrons require a certain amount of energy to break free and thiscorresponds to a minimum frequency (1) 2[6] 21. (a) (i) Use of speed = distance over time (1)Distance = 4 cm (1)Answer = [2.7 105 s] (1)Example of answert = 4 cm 1500 m s1t = 2.7 105 s 3 (ii) Use of f = 1/T (1)Answer = [5000 Hz] (1) 2 (iii) Time for pulse to return greater than pulse interval (1)All reflections need to reach transducer before next pulse sent. (1)Will result in an inaccurate image. (1) (Max 2)Need to decrease the frequency of the ultrasound. (1) (Max 3) Max 3 (iv) X-rays damage cells/tissue/foetus/baby but ultrasound doesnot (need reference to both X-rays and ultrasound) (1) 1 (b) The answer must be clear, use an appropriate style and be organisedin a logical sequence (QWC)Doppler shift is the change in frequency of a wave when the source or thereceiver is moving (1)Requirement for a continuous set of waves (1)Two transducers required (one to transmit and one to receive) (1)Change in frequency is directly related to the speed of the blood (1) 4[13] 22. (a) Voltmeter is across resistor should be across cell (1) 1 (b) (i) Plot of graphCheck any three points (award mark if these are correct) (3)Line of best fit 3 (ii) e.m.f. = [1.36 1.44 V] (1) 1 (iii) Attempt to find gradient (1)Answer [0.38 0.42 ] (1) 2 (c) Intercept would twice value above (1) (accept numerical value 2 value(b)(ii))Gradient would be twice value above (1) (accept numerical value 2 value(b)(iii)) 2[9] 23. (a) Diode or LED (1) 1 (b) (i) Use of R = V / I current between 75 and 90 ignoring powers of 10 (1)answer 6.7 8.0 (1)Example of answerR = 0.60 V (85 103) AR = 7.06 2 (ii) Infinite OR very high OR 1 (c) ANY ONERectification / AC to DC / DC supply [not DC appliances]Preventing earth leakageStabilising power outputTo protect componentsA named use of LED if linked to LED as component in (a)(egcalculator display / torch)A voltage controlled switch(Allow current in only one direction) 1[5] 24. (a) Resistivity definitionResistivity = resistance (1) cross sectional area / length (1)= RA/l with symbols defined scores 2/2equation as above without symbols defined scores equation given as R = l/A with symbols defined scores 1/2(1st mark is for linking resistivity to resistance with some other terms) 2(b) (i) Resistance calculationConverts kW to W (1)Use of P = V2/R OR P = VI and V = IR (1)Resistance = 53 (1)Example of answerR = (230 V)2 1000 WR = 53 3 (ii) Length calculationRecall R = l/A (1)Correct substitution of values (1)Length = 6.3 m (accept 6.2 m) (1)ecf value of RExample of answerl = (52.9 1.3 107 m2) (1.1 106 m)l = 6.3 m 3 (iii) Proportion methodIdentifies a smaller diameter is needed (1)Diameter = 0.29 mm (1)ORCalculation methodUse of formula with l = half their value in (b)(ii) (1)Diameter = 0.29 mm (1)(Ecf a wrong formula from part ii for full credit)Example of answerdnew = 0.41 mm 2dnew = 0.29 mm 2[10] B 25. (a) Definition of E.M.F.Energy (conversion) or work done (1)Per unit charge (1)[work done/coulomb 1/2, energy given to a charge 1/2, energygiven to a charge of a coulomb 2/2]OR ORE = W/Q (1) E = P/ISymbols defined (1) Symbolsdefined(E = 1 J/C scores 1) (E = 1 W/A scores 1)((Terminal) potential difference when no current is drawn 1/2) 2(b) (i) Internal resistance calculationAttempt to find current (1)Pd across r = 0.2 V (1)r = 0.36 () (1)[You must follow through the working, I have seen incorrectmethods getting 0.36 ]Example of answerI = 2.8 V 5.0 r = (3.0 2.8) V 0.56 A = 0.36 3 (ii) Combined resistanceUse of parallel resistor formula (1)Resistance = 3.3 [accept 3 1/3 but not 10/3] (1) 2 (iii) Voltmeter reading(ecf bii)Current calculation using 3 V with either 3.3 or 3.7 (1)Total resistance = 3.7 [accept 3.66 to 3.73 ]OR use of V = E Ir (1)Voltmeter reading = 2.7 V (1)ORPotential divider method, ratio of resistors with 3.7 on bottom (1)Multiplied by 3.0 V (1)2.7 V (1)Example of answerRtotal = 3.7 I = 3 V 3.7 = 0.81 AVvoltmeter = 3.3 0.81 A = 2.7 V 3 (c) Ideal voltmeterIdeal voltmeter has infinite resistance OR extremely high resistanceOR highest possible R OR much larger resistance than that ofcomponent it is connected across OR quotes value > 1 M (1)Current through voltmeter is zero (negligible) OR doesnt reduce theresistance of the circuit OR doesnt reduce the p.d. it is meant tobe measuring. (1) 2[12] 26. (a) Circuit diagramPotentiometer correctly connected i.e potential divider circuit (1)Ammeter in series and voltmeter in parallel with bulb (1)(light bulb in series with resistance can score second mark only) 2(b) (i) Graph+I, +V quadrant; curve through origin with decreasing gradient (1)[do not give this mark if curve becomes flat and then starts goingdown i.e. it has a hook]I, V quadrant reasonably accurate rotation of +I,+V quadrant (1) 2 (ii) Shape of graphAs current/voltage increases, temperature of the lamp increases /lamp heats up (1)Leading to increase in resistance of lamp (1)Rate of increase in current decreases OR equal increases in Vlead to smaller increases in I (1)Qowc (1)Ecf if a straight line graph is drawn max 3R constant (1)V I (1)Qowc (1) 4[8] 27. (a) (i) Demonstrating the stationary waveMove microphone between speaker and wall OR perpendicular to wallOR left to right OR towards the wall [could be shown by labelledarrow added to diagram] (1)Oscilloscope/trace shows sequence of maxima and minima (1) 2 (ii) How nodes and antinodes are producedSuperposition/combination/interference/overlapping/crossingof emitted/incident/initial and reflected waves (1)Antinodes: waves (always) in phase OR reference to coincidenceof two compressions/rarefactions/peaks/troughs /maxima/minima,hence constructive interference/reinforcement (1)Nodes: waves (always) in antiphase/exactly out of phase ORcompressions coincide with rarefactions etc, hence destructiveinterference / cancellation (1) 3(iii) Measuring the speed of soundMeasure separation between (adjacent) nodes / antinodes anddouble to get /this is [not between peaks and troughs] (1)Frequency known from/produced by signal generator ORmeasured on CRO / by digital frequency meter (1)Detail on measurement of wavelength OR frequencye.g. measure several [if a number is specified then 3] nodespacings and divide by the number [not one several times]OR measure several (3) periods on CRO and divide by the numberOR adjust cro so only one full wave on screen (1)Use v (allow c) = f 4 (b) (i) Application to concert hallLittle or no sound /amplitudeOR you may be sat at a node (1) (ii) Sensible reasonExamples:Reflected wave not as strong as incident waveOR walls are covered to reduce reflectionsOR waves arrive from elsewhere [reflections/different speakers]OR such positions depend on wavelength / frequency (1) 2[11] 28. (a) Meaning of statement(5.89 1019 J / work function) is the energy needed to remove anelectron [allow electrons] from the (magnesium) surface/plateConsequent markMinimum energy stated or indicated in some way [e.g. at least /or more] (1) 2 (b) (i) Calculation of timeUse of P = IA (1)Use of E = Pt (1)[use of E = IAt scores both marks]Correct answer [210 (s), 2 sig fig minimum, no u.e.] (1)[Reverse argument for calculation leading to either intensity,energy or area gets maximum 2 marks]Example calculation:t = (5.89 1019 J)/(0.035 W m2 8 1020 m2) 3(ii) How wave-particle duality explains immediate photoemissionQOWC (1)Photon energy is hf / depends on frequency / depends on wavelength (1)One/Each photon ejects one/an electron (1)The (photo)electron is ejected at once/immediately (1)[not just photoemission is immediate] 4[9] 29. (a) (i) Condition for reflectionAngle of incidence greater than critical angle [accept i > c] (1) 1 (ii) Description of path of rayAny two from: Ray refracted at A and C Description of direction changes at A and C Total internal reflection at B (1)(1) 2 (b) (i) Things wrong with the diagramAngle of refraction cant be 0 / refracted too much (1)No refraction on emergence from prism (1)[Allow 1 mark for correct reference to partial reflection] 2 (ii) Corrected diagram emergent ray roughly parallel to the rest of the emergent rays (1) direction of refraction first surface correct (1) direction of refraction second surface correct (1) 3[8] 30. (a) Calculation of adaptors inputRecall of: power = IV (1)Correct answer [0.01 A] (1)Example of calculation:power = IVI = P/V = 25 W / 230 V= 0.01 A 2 (b) (i) Explain why VA is a unit of powerPower = voltage current so unit = volt amp (1) 1 (ii) Calculation of efficiency of adaptorUse of efficiency equation (1)Correct answer [24%] (1)Example of calculation:efficiency = (0.6 VA / 2.5 W) 100%= 24 % [0.24] 2 (iii) Reason for efficiency less than 100%Resistance (accept explanations beyond spec, e.g. eddy currents) (1)hence heat loss to surroundings (1) 2 (c) (i) Calculation of chargeRecall of: Q = It (1)Correct answer [4000 C] (1)Example of calculation:Q = It= 0.2 A 6 h= 0.2 A (6 60 60) s= 4000 C (4320 C) 2 (ii) Calculation of work doneRecall of: W = QV OR Recall of W = Pt (1)Correct substitution (1)Correct answer [13 000 J] (1)Example of calculation:W = QVW = 4320 C 3 V [ecf]= 13 000 J (12 960 J)ORW = PtW = 0.6 W 6 hW = 0.6 W (6 60 60) s= 13 000 J 3[12] 31. (a) (i) Add standing waves to diagramsMark for each correct diagram (1)(1) 2(ii) Mark place with largest amplitude of oscillationantinode marked [allow clear indication near centre of wave otherthan an X, allow correct antinode shown on diagrams B or C] (1) 1 (iii) Name of place marked(Displacement) Antinode [allow ecf from (a) (ii)] (1) 1 (b) (i) Calculation of wavelengthCorrect answer [5.6 m]Example of calculation:= 2 2.8 m= 5.6 m (1) 1 (ii) Calculation of frequencyRecall of v = f (1)Correct answer [59 Hz] [ecf] (1)Example of calculation:v = f f = 330 m s1 / 5.6 m= 58.9 Hz 2 (c) (i) Explanation of difference in soundas the room has a standing wave for this frequency / wavelength /it is the fundamental frequency(allow relevant references to resonance) (1) 1 (ii) Suggest another frequency with explanationAppropriate frequency [a multiple of 59 Hz] [ecf] (1)Wavelength 1/2, 1/3 etc (stated or used) (1) 2 (d) Explain change in frequencieswavelengths (of standing waves) bigger / f = v/2l (1)hence frequencies smaller/lower (1) 2[12] 32. (a) Blue light:Wavelength / frequency / (photon) energy 1(b) (i) Frequency:Conversion of either value of eV to JoulesUse of f = E / hCorrect frequency range [4.8 1014 8.2 1014 Hz or range =3.4 1014 Hz][no penalty for rounding errors]eg.2 eV = 2 1.6x 1019 = 3.2 1019 J= 6.63 1034 ff = 4.8 1014 Hz3.4 eV = 3.4 1.6 1019 = 5.4 1019 Jf = 8.2 1014 Hz 3 (ii) Diagrams:Downward arrow from top to bottom levelOn larger energy gap diagram 2 (c) (i) Resistivity drop:Less heating / less energy lost / greater efficiency / lowervoltage needed / less power lost 1 (ii) Resistance:Recall of R = L/AUse of R = L/ACorrect answer [80()] [allow 8084 () for rounding errors]Eg.R = (2 102 5.0 103) / (3.0 103 4.0 104)= 83 3[10] 33. (a) Angles:Normal correctly added to raindrop (by eye)An angle of incidence correctly labelled between normal andincident ray and an angle of refraction correctly labelledbetween normal and refracted ray 2 (b) Angle of refraction:Use of = sin i / sin rCorrect answer [20][allow 2021 to allow for rounding errors]eg.sin r = sin 27/ 1.3r = 20 2 (c) (i) Critical angle:The angle beyond which total internal reflection (of the light)occurs [allow T.I.R] / r = 90 1 (ii) Critical angle calculation:Use of = 1 / sin CCorrect answer [50.3] [allow 50 51]Eg.Sin C = 1/1.3C = 50.3 2 (d) Diagram:i = 35 [allow 33 37]Ray of light shown refracting away from normal on leavingraindropSome internal reflection of ray also shown with i = r [by eye]Reflected ray shown refracting away from the normal as it leavesthe front of the raindrop / angle of refraction correctlycalculated at back surface 4 (e) Refractive index:(Red light has) lower refractive index (than violet light) 1[12] 34. (a) n is (number of) charge carriers per unit volume ornumber density or (number of) charge carriers m3 orcharge carrier density(1)[allow electrons]v is drift speed or average velocity or drift velocity(of the charge carriers) (1)[just speed or velocity scores zero] 2 (b) / A and Q A s or / Cs1 and Q C (1)n m3 (1)A m2 and v m s1 (1)[If no equation written assume order is that of equation] 3(c) (i) n 1 and Q Need all three 1(i) Ratio vA/ vB less than 1 following sensible calculation (1)Ratio = // 0.25 // 1:4 (1)(ratio 4:1 scores 1)[4vA:1vB scores 1] 2[8] 35. (a) Use of P = IV (1)Current in lamp A 2 A (1)[0.5 A scores zero unless 24 = I 12 seen for 1st mark] 2Example of answerI = P V = 24 W 12VI = 2A (b) (i) Voltmeter reading = 12 V (1) 1 (ii) p.d. across R2 = 6 V or their (b)(i) minus 6V (1)Use of R = V/I (1) conditional on first markR2Answer to this part must be consistent withvoltmeter reading and if voltmeter reading is wrongthis part has a max 2. If (b)(i) = 15 V then need to seeIf (b)(i) = 6V or less they are going to score zero for this section. 3 (iii) current through R1 = 5 A (1) ecf answers from (a) 1Example of answerCurrent through R1 = 2 A + 3 A = 5 A (iv) p.d. across R1 = 3 V (1) ecf (15V minus their (b)(i)) 1Example of answerp.d. across R1 = 15 V 12 V = 3 V (v) R11Example of answerR1 = 3 V 5A = 0.6[accept fraction 3/5][9]36. (a) (i) EI (1) 1(ii) I2R (1) 1(iii) I2r (1) 1 (b) EI = I2R + I2r or E = IR + Irecf Must use values (a)(i)-(iii) 1 (c) I for circuit given by Imax = E / r or substitution of5000V into the equation (1)(for safety) need I to be as small as possible (1) 3[7] 37. (a) (i) How we know the speed is constantCrest spacing constant / circular crestsOr wavelength constant / equal wavelength (1)[Accept wavefront for crests][Dont accept wave] 1 (ii) Calculation of speedis 10 mm (1)[Allow 9 to 11]Use of v = f(1)0.40 m s1 (1)[Allow 0.36 to 0.44Allow last two marks for correct calculation from wrong wavelength] 3(40Hz)(10 103 m)= 0.40 m s1 (b) Line X1st constructive interference line below PQ, labelled X (1)[Accept straight lineIgnore other lines provided correct one is clearly labelled X] 1 (c) (i) Superposition along PQConstructive interference / reinforcement / waves of largeramplitude / larger crests and troughs (1)Crests from S1 and S2 coincide / waves are in phase / zero phasedifference / zero path difference (1)Amplitude is the sum of the individual amplitudes (OR twice theamplitude of the separate waves) (1) 3 (ii) TableA constructive (1)B destructive (1) 2[10] 38. (a) Part of spectrumLight / Visible / red (1) 1Calculation of work functionUse of = hc/(1)3.06 1019 [2 sig fig minimum] (1) 2(6.63 1034 J s)(3.00 108 m s1)/(6.5 107 m)= 3.06 1019 J (b) (i) Meaning of stopping potentialMinimum potential difference between C and A / across thephotocell (1)Which reduces current to zero OR stops electrons reaching A /crossing the gap / crossing photocell (1) 2 (ii) Why the graphs are parallelCorrect rearrangement giving Vs = hf/e /e (1)Gradient is h/e which is constant / same for each metal (1)[Second mark can be awarded without the first if norearrangement is given, or if rearranged formula is wrong butdoes represent a linear graph with gradient h/e] 2[7] 39. (a) (i) Calculate maximum currentRecall of P = IV (1)Correct answer [0.49 A] (1)Example of calculation:P = IVI = 5.9 W / 12.0 V= 0.49 A 2 (ii) Show that resistance is about 24 Recall of V = IR (1)Correct answer to 3 s.f. [24.5 ] [no u.e.] (1)Example of calculation:R = 12 V / 0.49 A= 24.5 2 (b) (i) Calculate currentUse of correct circuit resistance (1)Correct answer [0.45 A] (1)Example of calculation:I =V / R= 12 V (24.5 + 2 )= 0.45 A 2 (ii) Calculate powerRecall of P = IV and V = IR (accept P = I2R) (1)or P = RV2Correct answer [5.0 W] (1)Example of calculation:P = I2R= (0.45 A)2 24.5 = 5.0 W 2 (c) Increase in power available to pumpe.g. lower resistance in wire thicker wire, panel nearer to motor (1)(accept relevant answers relating to panels, e.g. more panels) 1[9] 40. (a) (i) Name processRefraction (1) 1 (ii) Explanation of refraction taking placechange in speed / density / wavelength (1) 1 (b) (i) Draw ray from butterfly to fishrefraction shown (1)refraction correct (1) 2 (ii) Explain what is meant by critical angleIdentify the angle as that in the denser medium (1)Indicate that this is max angle for refraction OR total internalreflection occurs beyond this (1)[angles may be described in terms of relevant media] 2(iii) Explain two paths for rays from fish A to fish Bdirect path because no change of medium/refractive index/density (1)(total internal) reflection along other path /angle of incidence > critical angle (1)direct ray correctly drawn with arrow (1)total internal reflection path correctly drawn with arrow (1)[lack of ruler not penalised directly] [arrow penalised once only] 4[10] 41. (a) Ultrasound:High frequency sound / sound above human hearing range / soundabove 20 kHz / sound too high for humans to hear (1) 1 (b) (i) Pulses used:to prevent interference between transmitted and reflected signals /allow time for reflection before next pulse transmitted / to allow forwave to travel to be determined (1) (ii) High pulse rate:Greater accuracy in detection of preys motion / position / continuousmonitoring / more frequent monitoring (1) 2 (c) Size of object:Use of = v/f (1)Correct answer (0.0049 m or 4.9 mm) (1)[accept 0.0048 m or 0.005 m]example: = 340 m s1/ 70000 Hz= 0.0049 m = 4.9 mm (accept 5 mm) 2 (d) Time interval:Use of time = distance / speed (1)Correct answer (2.9 103 s) [allow 3 103 s][allow 1 mark if answer is half the correct value ie. Distance = 0.5mused] (1)example:time = 1 m / 340 m s1= 2.9 103 s 2(e) Effect on frequency:Frequency decreases (1)Greater effect the faster the moth moves / the faster themoth moves the smaller the frequency (1) 2[9] 42. (a) Diffraction diagram:Waves spread out when passing through a gap / past an obstacle(1) stays constant (1) 2 (b) Diagrams:Diagram showing 2 waves in phase (1)Adding to give larger amplitude (1) 2 (c) Information from diffraction pattern:Atomic spacing (similar to )Regular / ordered structureSymmetrical structureDNA is a double helix structure (2) Max 2 (d) Electron behaviour:(Behave) as waves (1) 1[7] 43. (a) (i) Diagram:i and r correctly labelled on diagram (1)i = 25 +/ 2 (1)r = 38 +/ 2 (1)[allow 1 mark if angles measured correctly from interfaceie. i = 65 +/ 2, r = 52 +/ 2] (1) 3 (ii) Refractive index:Use of ga = sin i / sin r [allow ecf] (1)Use of ag = 1/ga (1)example:ga = sin 25 / sin 38 = 0.686ag = 1/ ga = 1.46 2 (b) Ray diagram:Ray added to diagram showing light reflecting at interface withangles equal (by eye) (1) 1(c) Observation:Incident angle > critical angle (1)T.I.R occurs (1) 2 (d) largest angle:sin C = 1/1.46 (allow ecf) (1)C = sin1 (1/1.46) = 43 (1) 2[10] 44. (i) J C 1 Potential difference (1)(ii) Product of two quantitiesPotential difference (1)(iii) Rate of changecurrent (1)(iv) Base quantitycurrent (1)(for any part if two answers are given score is zero) 4[4] 45. (a) (As temperature of thermistor increases) its resistancedecreases [Do not credit the converse] (1)any TWO(slight) decrease in v (symbol, velocity or drift velocity)Large increase in n increases [accept electrons/charge carriers for n]A, Q and (pd) remain constant (1)(1)[ignore any reference to v staying constant] 3(n constant, cant score mark for 3,4) (b) (i) ammeter reading decreases (1)voltmeter reading unaltered (1)(ii) ammeter is used to indicate temperature (1)(iii) Assumption: ammeter; ideal/ has zero/negligible resistance (1)(Reference to meters is zero mark) 4[7]46. (a) Tungsten filamentQowc (1)I is not (directly) proportional to VTemperature of filament increases/ filament heats up/gets hotter as current/pd increases[accept bulb or lamp but not wire]Links temperature increase to resistance increasestungsten filament does not obey Ohms law/not anOhmic conductor or resistor. (1)(1)(1)Any THREE 4 (b) (i) Reading current from graph 1.5 A (1)answer 5.3 (1)(misread current 0/2)Example of answerV = IRR = 8.0 1.5 = 5.3 2 (ii) Addition of two currents (1)OR use of R = V/I and resistors in parallel formula1.5 + 1.2 = 2.7 A (1)ecf candidates current from above[If you see 2.7 A give 2marks] 2[8] 47. (a) (i) Use of P = V2 / R OR P = IV and V = IR (1)Total R = 4.5 (1) 2Example of answerR = V2 P = 12 V 12 V 32 WR = 4.5 (ii) Use of 1/R = 1/R1 + 1/ R2 OR R = 1/5R (1)[OR find total current, divide that by 5 and use V = IR]Resistance of strip = 22.5 (1)ecf candidates R. 2[common error is to divide by 5 0.9 scores 0/2 butecf to next part gives l = 0.033 m which will then score 3/3](b) R = l / A or correct rearrangement (1)Correct substitution (1)Length = 0.82 m (1)ecf candidates R 3Example of answerl = RA/ = (22.5 4.0 108 m2) 1.1 106 ml = 0.82 m (c) See P = V2 / R OR P = IV leading to increase in currentor decrease in resistance (1)more strips in parallel / material of lower resistivity (1)[not greater conductivity] 2[9] 48. (a) E.M.F. = work done / charge OR energy transferred / charge (1)OR power / current[There is only one mark here and this is consistent withspecification but it must not be Joules or coulombs] 1 (b) (i) Use of V = IR (1)I = 2.0 A (1) 2Example of answerI = V / R = 8.0 V / 4.0 I = 2.0 A (ii) Uses p.d. = 4.0 V (1)r = 2.0 ecf their I (1) 2Example of answerr = V / I = 4.0 V / 2.0 Ar = 2.0 (iii) Use of P = VI // I2R // V2/R (1)P = 16 W ecf their I (1) 2Example of answerP = VI = 8 V 2 AP = 16 W(iv) Uses 4V or 2A 2 or their I r (1)see 5 60 s in an energy equation (1)energy = 2400 J (1) 3Example of answerE = VIt = 4 V 2 A 5 60 sE = 2400 J[10] 49. (a) Experiment[Marks may be earned on diagram or in text]Named light source plus polaroid (OR polariser ORpolarising filter) / Laser / Named light source andsuitable reflector (e.g. bench) (1)2nd Polaroid plus means to detect the transmitted light (1)(i.e. eye OR screen OR LDR OR light detector ORinstruction to e.g. look through polaroids)Rotate one Polaroid [Only award if expt would work] (1)Detected intensity varies / No light when polaroids are at 90 (1)Maxima and minima 90 apart / changes from dark to light every 90 (1)[Use of microwaves, slits or blockers: 0/5Use of filters or diffraction gratings: lose first two marksUse of sunglasses to observe: lose mark 2] 5 (b) Why sound cant be polarisedThey are longitudinal / They are not transverse / Only transversewaves can be polarised / Longitudinal waves cannot be polarised /Because the (*) is parallel to the (**) (1)(*) = vibration OR displacement OR oscillation OR motion of particles(**) = direction of travel OR direction of propagation OR motion ofthe wave OR direction of energy transfer 1[6] 50. (a) (i) Table f2.4 (110)1.2 2200.8 330All wavelengths correct (2)[One or two wavelengths correct gets 1]Both frequencies correct (1)[Accept extra zero following wavelength figure, e.g. 2.40.Accept units written into table, e.g. 2.4 m, 220 Hz] 3 (ii) Why nodesString cannot move / no displacement / zero amplitude /no oscillation / phase change of on reflection / two wavescancel out / two waves are exactly out of phase (1)(OR have phase difference of OR half a cycle) /destructive interference 1 (b) Why waves with more nodes represent higher energiesMore nodes means shorter wavelength (1)Momentum will be larger (1)[OR Allow 1 mark for More nodes means higher frequency and E = hf] 2[6] 51. (a) Why statement correctBlue photon has more energy than red photon (1)Why statement incorrectBlue beam carries less energy per unit area per second / Blue beamcarries less energy per second / Blue beam carries less energy perunit area / Blue beam has lower intensity and intensity = energy per unitarea per secondAdditional explanation[Under correct] Blue has a higher frequency (OR shorter wavelength) /[Under incorrect] Blue beam has fewer photons (1)[Allow reverse statements about Red throughout part a] 3 (b) (i) Meaning of work functionEnergy to remove an electron from the surface (ORmetal OR substance) (1)[Dont accept from the atom. Dont accept electrons.]Minimum energy / Least energy / Energy to just/ without giving the electron any kinetic energy (1) 2 (ii) Calculation of threshold frequencyUse of = hf0 (1)Correct answer [6.00 1014 Hz] (1)e.g.(3.98 1019 J)/(6.63 1034 J s) = 6.00 1014 Hz 2[7]52. (a) Which transitionUse of ()E = hc/ OR ()E = hf and f = c/ (1)Use of 1.6 1019 (1)Correct answer [1.9 eV] (1)C to B / 1.5 to 3.4 (1)[Accept reverse calculations to find wavelengths]e.g.(6.63 1034 J s)(3.00 108 m s1)/(656 109 m)(1.6 1019 J eV1)= 1.9 eV 4 (b) Explanation of absorption lineQOWC (1)Light of this wavelength is absorbed by hydrogen (1)In the outer part of the Sun (OR Suns atmosphere) (1)Absorbed radiation is reemitted in all directions (1)Transition from B to C (OR 3.4 to 1.5) (1) Max 4 (c) Why galaxy recedingWavelength increased (OR stretched) / red shift /frequency decreased 1[9] 53. (a) Describe propagation of longitudinal wavesParticles oscillate / compressions/rarefactions produced (1)oscillation/vibration/displacement parallel to direction of propagation (1) 2 (b) Calculation of wave speedRecall of v = f (1)Correct answer [7.2 km s1] (1)Example of calculation:v = f v = 9 Hz 0.8 km= 7.2 km s1 [7200 m s1] 2(c) Determine if elephants can detect waves more quicklyRecall of v = s / t (1)Correct answer for t in minutes or hours [about 6 minutes] orrelevant comment with 347 s or calculation of tidal wave speed[0.35 km s1] with comment [allow ecf] (1) 2Example of calculation:v = s / tt = 2500 km 7.2 km s1 OR v = 2500 km (2 60 60 s)t = 347 s OR v = 0.35 km s1t = about 6 minutes (stated) / much less than hours / 2 h is 7200 sOR 7.2 km s1 >> 0.35 km s1[6] 54. (a) Meaning of superpositionWhen vibrations/disturbances/waves from 2 or more sources coincideat same position (1)resultant displacement = sum of displacements due to individual waves (1) 2 (b) (i) Explanation of formation of standing wavedescription of combination of incident and reflected waves/waves in opposite directions (1)described as superposition or interference (1)where in phase, constructive interference / antinodesOR where antiphase, destructive interference / nodesOR causes points of constructive and destructive interferenceOR causes nodes and antinodes (1) 3 (ii) Calculate wavelengthIdentify 2 wavelengths (1)Correct answer [2.1 109 m] (1) 2Example of calculation:(NANANANAN) X to Y is 2 = 4.2 109 m 2= 2.1 109 m(iii) Explain termsamplitude maximum displacement (from mean position)(can use diagram with labelled displacement axis) (1)antinode position of maximum amplitudeOR position where waves (always) in phase (1) 2[9] 55. (a) (i) Calculate resistanceRecall of R = V/I (1)Correct answer [8.65 ] (1) 2Example of calculation:R = V/IR = 2.68 V 0.31 A= 8.65 (ii) Show that internal resistance is about 0.4 Recall of relevant formula [V = Ir OR lost volts = ( V) (1) OR = I(R + r)] including emfCorrect answer [0.39 ] [no ue] [allow ecf if = I(R + r)] (1) 2Example of calculation:V = Irr = ( V)/I= (2.8 V 2.68 V)/0.31 A= 0.39 (iii) Comment on match to maximum powerNot matched [ecf for R in (a) (i) and r in (a)(ii)] (1)Max power when internal resistance = load resistance (1) 2 (b) (i) Show that charge is about 14 000 CRecall of Q = It (1)Correct answer [14 400 C] [no ue] (1) 2Example of calculation:Q = It= 2 2 A 60 60 s= 14 400 C (ii) Calculate time for which battery maintains currentUse of Q = It OR use of W = Pt (1)Correct answer [46 450 s or 12.9 h] (1) 2Example of calculation:t = Q/I= 14 400 C / 0.31 A= 46 450 s (c) Explain effect on efficiencyEfficiency = I2R / I2(r + R) / Efficiency depends on R /(r + R) /more heat dissipated in cells / Efficiency is V/ and V decreases (1)so efficiency is less (1)[Must attempt explanation to get 2nd mark] 2[12] 56. (a) Work function:Energy needed for an electron to escape the surface /to be released (from the metal) (1) 1 (b) How current produced:Any 3 from:Photon of light passes energy to an electronIf energy above the work function/frequency above threshold (1)(1)Electron released as a photoelectron / photoelectron released /surface electron released (1)Moving electrons produce a current 3 (c) (i) Intensity of light increased:More electrons released (1) (ii) Frequency of light increased:Electrons gain more (kinetic) energy (1) 2(d) Photon energy:Use of f = v/ or E = hc/ (1)Correct answer for E (4.7 1019 J or 2.96 eV) (1)[allow 3.0 eV] 2Example:f = v/ = 3 108 / 4.2 107 = 7.1 1014 HzE = hf = 4.7 1019 J or 2.96 eVORE = hc/ = 3 108 6.63 1034/ 4.2 107= 4.7 1019 J or 2.96 eV (e) Max kinetic energy:Knowledge that kemax = energy calculated in (d) (1)Correct answer for kemax (0.26 eV or 4.2 1020 J)[allow 0.250.26 eV or 4.1 4.2 1020 J and allow ecf from (d)] (1) 2Example:kemax = 2.96 eV 2.7 eV= 0.26 eV (f) (i) Why current reduced:Many / some electrons will not have enough (kinetic) energyto reach the anode / only electrons with large (kinetic) energywill reach the anode (1) 1 (ii) Stopping potential:eV = () keV = ke / e = 0.26V (1) 1[12] 57. (a) Plane polarised:Vibrations / oscillations (1)in one plane (1)ORdouble-headed arrow diagram (1)with vibrations / oscillations labelled (1) 2 (b) Polarising filter: Intensity goes from maximum to minimum (1) Twice per rotation / after 90 (1) As filter only lets through vibrations in a particular plane (1) 3[marks may be gained from a clearly labelled diagram](c) Response of beetle:Changed direction by 90 / turned through a right-angle (1) 1 (d) No moon:Beetle moves in a random direction / in circles / appears disorientated (1) 1[7] 58. (a) Circuit:Potential divider (1) 1 (b) Relay potential difference:4 V (1) 1Example:5/15 12 = 4V (c) (i) Resistance:Recall of R = L/A (1)Correct substitution of values into formula (1)Correct answer [98()] (1)[allow 97 98 to allow for rounding errors] [no u.e.] 3Example:R = (3.4 102 1.44) / (100 0.05)= 98 (ii) Combined resistance:Use of 1/RTot = 1/R1 + 1/R2 (1)Correct answer for R [4.8] (1)[allow 4.7 4.8 to allow for rounding errors] 21/R = 1/98 + 1/5 (or = 1/100 + 1/5)R = 4.8 (iii) Relay voltage:P.d. across relay with ballast very similar to p.d acrossthe relay alone / p.d. = 3.9 V / p.d. lower (slightly) (1) 1 (iv) Train on track:Relay voltage becomes very small / zero (1) 1(v) Wet ballast:Any two Combined resistance now small / RT = 0.45 Relay voltage now small / V = 0.52 V Relay voltage too small to trigger green light /signal remains red (1)(1) 2[11] 59. Tungsten filament bulb(a) ResistanceUse of P = V2/R or P = VI with V = IR (1)answer 960 (1) 2Example of answerR = (240 V 240 V) 60 WR = 960 (b) Drift speedrearrangement of I = nAvQ (1)Use of Q = 1.6 1019 (C) (1)answer 0.15/0.148 m s1 (1) 3Example of answerv = 0.25 A (3.4 1028 m3 1.6 10 19 C 3.1 1010 m2) (c) ExplanationQowc (1)Any THREE Resistance due to collisions between electrons & ions/atoms/particles (as T increases) ions/atoms/particles have more energy (as T increases) ions/atoms/particles vibrate through largeramplitude /vibrate faster OR amplitude if lattice vibrationincreases. more chance/increased frequency of collision/interactionOR impedes the flow of electrons (1)(1)(1) 4[9] 60. Emf and Internal resistance(a) DerivationE = I (R + r) OR E = IR + Ir (1) 1(b) (i) Correct working (allow even if evidence of working backwards) (1)Example of answerE/I = R + rRearranging R = E/I r (ii) EmfAttempt to use gradient (1)answer 1.5 V (bald answer 1.5 V scores 0/2) (1) 2 (iii) PowerFrom graph find value of 1/I when R = 5 (1)Use of P = I2R (1)answer 0.31 (W) (1) 3Example of answer1/I = 4 A1 I = 0.25 AP = 0.25 A 0.25 A 5 = 0.3125 W (c) GraphIntercept at 2 (ohms) (1)Graph steeper than original (1)Gradient is 3.0 V i.e. line passes through [10, 27-29] [no ecf] (1) 3[10] 61. Potential divider(a) First circuitMiddle terminal MOuter terminals L and K (any order) (1) 1 (b) (i) P.d across lamp.External resistance in circuit is 25 or (20+5) ohms (1)See ratio of resistances (denominator larger) 6.0V (1)OR current = 6/25 Aanswer 4.8 V (1) 3 (ii) AssumptionThe resistance of the ammeter is zero/negligible. (1) 1 (c) Second circuitSee 2 resistors in parallel with supply (1)Supply across ends of variable resistor (10 ) (1)Fixed resistor across one end and slider (consequent mark) (1) 3[8] 62. (a) Why transverse waves can be polarised but not longitudinal waves[Marks can be earned in diagram or text]Transverse waves have * perpendicular to direction of ** (1)* = vibration/displacement/oscillation/motion of particles** = travel/propagation/motion of wave/energy transfer/waveIn a transverse wave, * can be in different planes but polarisationrestricts it to one plane (1)Longitudinal waves have * parallel to ** (1) 3[Dont accept motion for **Diagrams to earn marks must be clearly labelled, but dont insiston a label looking along direction of travel in the usual diagramsto illustrate polarised and unpolarised waves] (b) (i) Effect of Polaroid on intensityIntensity is reduced (OR halved) [not zero] (1)[Accept slightly reduced and greatly reduced]Polaroid stops (OR absorbs) vibrations (OR waves OR light) inone plane/ (1)Polaroid only lets through vibrations (OR waves OR light)in oneplane/Light has been polarised 2 (ii) Effect of rotating PolaroidNo effect (1)[ignore incorrect reasons accompanying statements of effect] 1[6] 63. (a) (i) How the bow causes the wave patternEITHERBow alternately pulls and releases string (or sticks and slips) (1)Creates travelling wave (OR travelling vibration ) (on string) (1)Wave reflects at the end (OR bounces back) (1)Incident and reflected waves (OR waves travelling in opposite (1)directions) superpose (OR interfere OR combine)[Dont accept collide] max 3ORBow alternately pulls and releases string (or sticks and slips) (1)Produces forced oscillation/acts as a driver/exerts periodic force (1)[Dont accept makes it vibrate] At a natural frequency of the string (1)Causing resonance (OR large amplitude oscillation) (1) max 3 (ii) Determination of wavelengthUse of node to node distance = /2 / recognise diagram shows 2] (1)Correct answer [0.4 m] (1) 2e.g. = 2 0.2 m= 0.4 m (iii) Differences between string wave and sound waveAny TWO points from: String wave is transverse, sound wave is longitudinal /can be polarised, cant String wave is stationary (OR standing), sound wave is travelling(OR progressive) / has nodes and antinodes, doesnt /doesnt transmit energy, does The waves have different wavelengths Sound wave is a vibration of the air, not the string (1)(1) 2[Dont accept travel in different directions / can be seen, cant beseen / cant be heard, can be heard / travel at different speedsThe first two marking points require statements about both waves,e.g. not just sound waves are longitudinal] (b) Sketch of the waveformSinusoidal wave with T = 1 ms (1)[Zero crossings correct to within half a small squareAccept a single cycle]Amplitude 1.6 cm (1)[Correct to within half a small square] 2[9] 64. (a) Conditions for observable interferenceAny THREE of: Same type of wave / must overlap (OR superpose) / amplitudelarge enough to detect / fringes sufficiently far apart todistinguish [Only one of these points should be credited] (Approximately) same amplitude (OR intensity) Same frequency (OR wavelength) Constant phase difference (OR coherent OR must come fromthe same source) (1)(1)(1) 3[Accept two or more points appearing on the same lineDont accept must be in phase must be monochromatic must have same speed no other waves present must have similar frequencies answers specific to a particular experimental situation, e.g.comments on slit width or separation](b) (i) Experiment description[Marks may be scored on diagram or in text](Microwave) transmitter, 2 slit barrier and receiver (1)[Inclusion of a screen loses this mark, but ignore a single slit in frontof the transmitter]Barrier, metal sheets (1)[Labels indicating confusion with the light experiment, e.g. slitseparations or widths marked as less than 1 mm, lose this mark]Appropriate movement of receiver relevant to diagram [i.e. move inplane perpendicular to slits along a line parallel to the plane of theslits, or round an arc centred between them] (1) 3 (ii) Finding the wavelengthLocate position P of identified maximum/minimum 1st/2nd/3rd etc. (1)away from centreMeasure distance from each slit to P (1)Difference = OR /2 (consistent with point 1) (1) 3[Accept use of other maxima and corresponding multiple of ][9] 65. (a) [Treat parts (i) and (ii) together. Look for any FIVE of thefollowing points. Each point may appear and be credited in eitherpart (i) or part (ii)](i) Light (OR radiation OR photons) releases electrons from cathode Photon energy is greater than work function / frequency oflight > threshold frequency / flight > fo / wavelength of light isshorter than threshold wavelength / < 0 PD slows down the electrons (OR opposes their motion ORcreates a potential barrier OR means they need energy to crossthe gap) Electrons have a range of energies / With the PD, fewer (ORnot all) have enough (kinetic) energy (OR are fast enough) tocross gap Fewer electrons reach anode / cross the gap (ii) (At or above Vs) no electrons reach the anode / cross the gap Electrons have a maximum kinetic energy / no electrons haveenough energy (OR are fast enough) to crossANY FIVE (1)(1)(1)(1)(1)[Dont worry about whether the candidate is describing the effect ofincreasing the reverse p.d. (as the question actually asks), or simplythe effect of having a reverse p.d.] 5(b) Effects on the stopping potential(i) No change (1)(ii) Increases (1) 2[Ignore incorrect reasons accompanying correct statements of the effect][7] 66. (a) Circuit diagram and explanationammeter and voltmeter shown in series and parallel respectively (1)current measured with ammeter and voltage / p.d. with voltmeter (1) 2 (b) Calculation of resistanceRecall of R = V/I (1)Correct answer [25.0 ] (1) 2Example of calculation:R = V/IR = 3.00 V 0.12 A= 25.0 (c) Calculation of resistanceRecall of P = V2/R (1)Correct answer [29.4 ] (1) 2Example of calculation:P = V2/RR = (230 V)2 1800 WR = 29.4 [Accept calculation of I = 7.8 A (1), calculation of R = 29.4 (1)] (d) Explanation of difference in values of resistanceAt higher voltage value element is at a higher temperature (1)(resistance higher because) increased lattice ion vibrations impedecharge flow (more) (1) 2[8]67. (a) Explain how vapour emits lightelectrons excited to higher energy levels (1)as they fall they emit photons/electromagnetic radiation/waves/energy (1) 2 (b) (i) Meaning of spectral line(when the light is split up) each frequency/wavelength/photon energy isseen as a separate/discrete line (of a different colour) (1) 1 (ii) Calculation of frequencyRecall of v = f (1)Correct answer [f = 5.1 1014 Hz] (1) 2Example of calculation:v = f 3.0 108 m s1 = f 589 109 mf = 5.1 1014 Hz (c) Explanation of different coloursdifferent colours = different freq/wavelengths / photons of differentenergies (1)photon energy/frequency/wavelength depends on difference betweenenergy levels (1)diff atoms have diff energy levels/diff differences in levels (1) 3 (d) Explanation of transverse wavesvariation in E or B-field /oscillations/vibrations/displacementat right angles/perpendicular to direction of travel/propagation[not just motion or movement for both 1st and 3rd part] (1) 1[9] 68. (a) Explanation of maximum or minimumpath difference = 2 125 109 m = 250 109 m (1)= half wavelength /antiphase (1)destructive interference / superposition (1) 3(minimum intensity)(b) Meaning of coherentremains in phase / constant phase relationship (1) 1[4] 69. (a) (i) Energy level diagram: Arrow showing electron moving from lower level to ahigher level (1) Arrow downwards from higher to lower level [mustshow smaller energy change than upward arrow] (1) 2 (ii) Missing energy:Causes a rise in temperature of a named item (1) 1 (iii) Range of energies:Minimum energy when = 400 109 m (1)Use of f = c/ (1)Use of E = hf (1)Correct answer [3.1 eV] (1)[allow 3.0 3.3 eV for rounding errors] [no u.e] 4eg. f = 3 108 / 400 109= 7.5 1014 HzE = hf = 5.0 1019 JE = 3.1 eV (b) Detecting forgeries:Forgery would glow / old painting would not glow (1) 1[8] 70. (a) (i) Critical angle calculation:Use of sin C = 1/da (1)Correct answer [24.4 only acceptable answer] [no u.e] (1) 2eg. Sin C = 1/da = 1/2.42C = 24.4 (ii) Ray diagram:Small angle ray shown passing into air, away from the normal (1)Large angle ray showing T.I.R. with angles equal [by eye] (1) 2(iii) Labelling of angles:An incident angle correctly labelled between normal and rayin diamond (1)An angle of refraction correctly labelled between normaland ray in air (1) 2 (iv) Amount of trapped light:Any 3 of the following: The higher the refractive index the greater the amount oftrapped light The higher the refractive index the lower the critical angle T.I.R occurs at angles greater than the critical angle So, if critical angle is smaller, more light is reflected (1)(1)(1) Max 3 (b) Comment on angle:Lower critical angle so more sparkle (1) 1[10] 71. (a) Graph scale:Log scale (1) 1 (b) (i) Choice of material:Any 2 of the following: (almost) all of the voltage is dropped across the carbon rod gives the greatest speed variation others need to be longer (to have same resistance as carbon) others need to be thinner (to have same resistance ascarbon) (1)(1) Max 2 (ii) Resistance calculation:Use of R = L/A (1)Correct units used for all terms [all in cm or all in m] (1)Correct answer [1.9 ] (1) 3[allow 1.8 for rounding errors no u.e]eg. R = 1.4 105 0.4 / 3.0 106= 1.9 (iii) Available voltage:X 12 V Y 0 V (1) 1(iv) Effect of connecting wires:Less voltage available for train set as some wasted across wires (1)0.5 is (relatively) large % of total resistance, so effect ishigh / not negligible (1)orCalculation of potential difference available now (1)[9.5 V] [allow 9.5 9.6 V]Significant drop from 12 V (1)Vxy = (Rxy / RTotal) Vsupply = (1.9/ (1.9 + 0.5)) 12 = 9.5 V 2[9] 72. (a) (i) Potential difference = work (done)/(unit) chargeOR Potential difference = Power/current (1) 1 (ii) J = kg m 2 s 2 (1)C = A s or W = J s1 (1)V = kg m2 A1 s3 (1) 3 (b) Converts 2 minutes to 120 seconds (1)Multiplication of VI t or V Q (1)Energy = 1440 J (1) 3Example of answer:Energy = 6.0 V 2.0 A 120 s= 1440 J[7] 73. (a) n = number of charge carriers per unit volume ORn = number of charge carriers m3ORn = charge carrier density (1)v = drift speed/average velocity/drift velocity (of the charge carriers) (1) 2 (b) n is greater in conductors / n less in insulators. (1)[There must be some comparison]larger current flows in a conductor. Dependant on havingreferred to n (1) 2(statement that n large in conductor and so current large max1)(c) (In series), so same current and same n and Q (1)vB greater vA (1)vA/vB = // 0.25 (1) 3[7] 74. (a) pd = 3.6 V (1) 1Example of answer;p.d. = 0.24 A 15 = 3.6 V (b) Calculation of pd across the resistor (1)[6.0 3.6 = 2.4 V]Recall V = IR (1)I1 calculated from their pd / 4 (1)[correct answer is 0.60 A.Common ecf is 6V/4 gives 1.5 A] 3Example of answer:I1 = 2.4 V / 4.0 = 0.6 A (c) Calculation of I2 from I1 0.24 [0.36 A] (1)[allow ecf of their I1. common value = 1.26 A]Substitution V = 3.6 V (1)R = 10 (1) 3[7] 75. (a) (i) ( gradient =) r = 1.95 2 (1)E = 8.9 9 V (1) 2 (ii) I = 2.15 2.17 A (1) 1 (iii) Use of V = IR (1)R = 2.1 2.2 (1) 2 (b) (i) Battery or cell with one or more resistive component (1)Correct placement of voltmeter and ammeter (1) 2 (ii) Vary Re.g. variable resistor, lamps in parallel (1)Record valid readings of current and pd (consequent mark) (1) 2[Do not give these marks if the candidate varies the voltage as well][9] 76. (a) Solar PowerUse of P = Ir2 [no component needed for this mark] (1)Use of cos 40 or sin 50 (with I or A) (1)2.2 [2 sf minimum. No ue] (1) 3e.g. P=1.1 103 W m2 cos 40 (29 103 m)2= 2.2 W (b) EnergyUse of E = Pt (1)1.8 104 J/2.0 104J (1) 2e.g. E = 2.2W (2.5 3600 s)= 2.0 104 J[5] 77. (a) GraphStraight line with positive gradient (1)Starting the straight line on a labelled positive fo (1)[Curved graphs get 0/2. Straight line below axis loses mark 2unless that bit is clearly a construction line.] 2 (b) Work functionFrom the y intercept (1)[Accept if shown on graph]OR Given by gradient f0 (or h f0)[Provided that f0 is markedon their graph, or they say how to get it from the graph]OR Read f and Ek off graph and substitute into Ek = hf [Curved graph can get this mark only by use of hf0 or equation methods.] 1 (c) GradientGradient equals Planck constant (1) 1[Curved graph cant get this mark][4]78. (a) WavelengtheV to J (1)Use of E = hf (1)Use of c = f(1)1.8 1011[2 sf minimum. No ue] (1) 4e.g. f =(1.8 keV ( 69.6 keV)) (103 1.6 1019 J keV1) / 6.6 1034 J s= 1.64 1019 Hz = 3.00 108 m s1/1.64 1019Hz= 1.8 1011 m (b) TypeX rays [Accept gamma rays] (1) 1[5] 79. Meaning of energy levelSpecific allowed energy/energies (of electron in an atom)(1) 1 Meaning of photonQuantum/packet/particle of energy/radiation/light/electromagnetic wave (1) 1 Formula for photon energyE2 E1 (1) 1[Allow E1 + Ephoton = E2] Explanation of photon wavelengthsSame energy change / same energy difference / energy the same (1) 1 Meaning of coherentRemains in phase / constant phase relationship(1) 1 80. Description of soundParticles/molecules/atoms oscillate/vibrate (1)(Oscillations) parallel to/in direction of wave propagation / wavetravel / wave movement [Accept sound for wave] (1)Rarefactions and compressions formed [Accept areas of high and low pressure] (1) 3Meaning of frequencyNumber of oscillations/cycles/waves per second / per unit time (1) 1 Calculation of wavelengthRecall v = f (1)Correct answer [18 m] (1) 2Example of calculationv = f = 330 m s1 18 Hz= 18.3 m[6] 81. Explanation of standing wavesWaves reflected (at the end) (1)Superposition/interference of waves travelling in opposite directions (1)Where in phase, constructive interference/superpositionOR where antiphase, destructive interference/superpositionOR causes points of constructive and destructiveinterference/superposition [Do not penalise here if node/antinode mixed up] (1) 3 Mark node and antinodeBoth marked correctly on diagram (1) 1 Label wavelengthWavelength shown and labelled correctly on diagram (1) 1 Explain appearance of stringAny two from: light flashes twice during each oscillation / strobefrequency twice that of string [accept light or strobe] string seen twice during a cycle idea of persistence of vision (2)max 2Calculate speed of wavesUse of v = T/ (1) Correct answer [57 m s1] (1) 2Example of calculation:v = T/= (1.96 N / 6.0 104 kg m1)= 57.2 m s1[9] 82. Distance to aircraft:Use of distance = speed time(1)Correct answer [7.2(km) / 7200(m) is the only acceptable answer. No u.e.] (1) 2e.g. Distance = speed time = 3 108 24 106= 7.2 km Why pulses are used:Any two of the following: Allow time for pulse to return before next pulse sent To prevent interference/superposition A continuous signal cannot be used for timing Cant transmit / receive at the same time (2) max 2 Doppler shift:Any three of the following Change in frequency/wavelength of the signal [allow specified change,either increase or decrease] Caused by (relative) movement between source and observer[accept movement of aircraft/observer] Size of change relates to the (relative) speed of the aircraft[Allow frequency increasing; do not allow frequency decreasingunless linked to aircraft moving away] Quote v/c = f/f (3) max 3[7]83. Area of wire:Use of A = r2 (1)Correct answer [1.9 107 (m2). Allow 1.9 107 and 2.0 107 (m2)](1) 2[No u.e.]e.g.A = r2 = (2.5 104)2= 1.96 107 m2 Table + graph:Length / Area / 106 m10.00.51.01.52.02.53.13.6 (1)4.0 4.1First 2 points plotted correctly to within 1 mm (1)Rest of points in straight line with origin by eye (1) 3 Resistivity calculation:Drawn through the origin, ignoring first 2 points (1)Recall = R /(L/A) [in any form] (1)Large triangle drawn on graph OR accept the use of a pair of values (1)read from the line[ x> 3 10-6 m1)is required in both cases][x-axis allowed as bottom of triangle]Correct answer [1.2 10-7 m)] (1)[allow 1.1 1.3 10-7( m)] [no u.e.] 4e.g.0.4 / 3.4 106 = 1.2 107 mAnomalous results:Any two of the following: Higher current/lower resistance for shorter lengths/at these points At shorter lengths/at these points wire gets hotter Non-uniform area/diameter Cable / contact resistance Sensitivity of meters Effect on resistance of any of the above (2) max 2[11] 84. Unpolarised and plane polarised light:Correct diagrams showing vibrations in one plane only and in all planes (1)Vibrations/oscillations labelled on diagrams (1) 2 Telescope adaptation:Fit polarising filter / lens [must be lens not lenses] (1)At 90 to polarisation direction to block the moonlight / rotate until 2cuts out moonlight (1)[4] 85. Meaning of plane polarisedOscillations/vibrations/field variations (1)Parallel to one direction, in one plane [allow line with arrow at both ends] (1) 2 Doppler effectDoppler (1)If source/observer have (relative) movement [reflections offvibrating/moving atoms] (1)Waves would be bunched/compressed/stretched or formula quoted[accept diagram] (1)Thus frequency / wavelength changes [accept red /blue shift] (1) 4Frequency about 3 10 14Hz Evidence of use of 1/wavelength = wavenumber (1)laser wavenumber = 9400 or wavelength change =7.69104 (1)New wavenumber = 10700 [or 8100] or conversion of wavelengthchange to m [7.69 106] (1)New wavelength = 935 nm [or 1240 nm]Use of frequency = c / wavelength [in any calculation] (1)f = 3.2 1014 Hz [note answer of 2.8 1014 = 3 , 3.4 1014 = 4](1) 5 Model of lightParticle/photon/quantum model (1)Photon energy must have changed / quote E = hf (1)Energy of atoms must have changed [credit vibrating less/more/faster/slower] (1) 3[14] 86. (a) (i) Lamp brightnessLamp A (1)Larger current through it (at 9.0 V)/greater power (1) 2(at 9.0 V)/smaller resistance (at 9.0 V) (ii) Battery currentAddition of currents (1)Current = 1.88 1.92 A (1) 2 (iii) Total resistanceR = 9 V/1.9 A or use of parallel formula (1)R = 4.6 4.9 (1) 2[full ecf for their current] (b) Lamps in seriesCurrent same in both lamps/current in A reduced from original value (1)Pd across A less than pd across B (1)Lamp A has a lower resistance than lamp B (1)P = VI or P = RI2 (1) Any 2Lamp A will be dimmer than B [conditional on scoring ONE of (1) 1the above marks][9] 87. (a) (i) ResistanceUse of V/I [ignore 10x] (1)3800 (3784 ) (1) 2 (ii) Resistance of thermistorUse TH THRRRVV OR 9V/.74mAROR (1)6.2 V = 0.74 mA RTH8400 [8378 ] [substituting 4000 gives 8857 ie 8900 ] (1)[method 2 substituting 3800 gives 8362 : substituting 4000 gives8162 ] 2 (b) Suggestion and ExplanationThe milliammeter reading increases (1)Thermistor resistance becomes zero /Short circuit (1)Since supply voltage is constant / I = 9.0 V/R (1)ORCircuit resistance reduced 3[7] 88. (a) Definition of E.M.F.Energy (conversion) or work done (1)Per unit charge (1)ORE = W/Q (1)Symbols defined (1)[E = 1J/C scores 1]ORE = P/I (1)Symbols defined (1)[terminal pd when no current drawn or open circuit scores max 1] 2(b) Voltmeter calculationAny attempt to find any current (1)Attempt to calculate pd across 10 resistor (1)5.77 V 2ORPotential divider method; ratio of resistors with 10.4 on the bottom (1)Multiplied by 6.0 V (1)5.77 V (1) 3[For either method, an answer of 0.23 V scores max 1] (c) Second battery addedVoltmeter reading increased (1)Any two of:EMF unchangedTotal resistance reducedcurrent increases or lost volts decreases (2) 3[8] 89. Frequency(a) (i) 1.0(3) 1010 Hz (1) 1 Electromagnetic Spectrum(ii) IR, microwave & radio in correct order above visible (1)UV with either X rays / Gamma rays / both in correct order belowvisible (1) (iii) Wavelength at boundary 1 108 m / 1 109 m (1) 3 Plane polarised(b) (i) Vibrations/oscillations (of electric field/vector) (1)In one direction/plane (of oscillation) (1) 2Description(ii) Diagram showing generator labelled transmitter/generator/source/emitter (1)And suitable detector eg shows how signal is observed by using (1)(micro)ammeter/cro/loudspeaker/computer with interface[Ignore anything drawn between generator and detector but for eachmark do not give credit if a grille etc is attached]To detect max and min (1)(Rotate through) 90 between max and min (1) 4[10] 90. (a) ExplanationQOWC (1)UV/red photon (1) 2EUV > ER/fuv (1)EUV > /fuv > fTH (so electron can break free) (1)One photon absorbed by one electron (1)Both metal plate and electron are negative or repel (each other) (1) max 2 (b) (i) Intensity red light increasednothing / no discharge (1)(ii) Intensity of UV increased(Coulombmeter) discharges quicker (1) 2 (c) Max KEUse of E = hc/ (1)conversion of eV to J or vice versa i.e. appropriate use of 1.6 1019 (1)Subtraction hc/ [must use same units] or use of full equation (1)max KE = 2.2 1019J (1) 4[Candidates may convert photon energy to eV leading to max KE = 1.4 eV][10] 91. Explanation increase of resistance with temperatureTemperature increase leads to increased lattice vibrations (1)scattering flowing electrons / increased collisions of electrons. (1) 2Calculation of resistance at 200 CR = V/I [stated or implied] (1)= 7.4 V 0.19 A= 39 (1) 2 Discuss whether results support hypothesisNo. Resistance is not increasing with temperature. (1) 1 Calculation of mains voltageP = V2 R (1)V2 = PR= 1200 W 41 [Mark for rearrangement OR substitution] (1)[Accept 39 41 ] [ecf]V = 220 V (1)[Allow P = I2R (1), 3calculate I = 5.4 A and use in 1200 W = 5.4 A V (1),V= 220 V(1)][8] 92. Explanation of excitedElectrons/atoms gain energy (1)and electrons move to higher (energy) levels (1) 2[Credit may be gained for diagrams in this and the next 3 parts] Explanation of how radiation emitted by mercury atomsElectrons (lose energy as they) drop to lower levels (1)Emit photons / electromagnetic radiation (1) 2 Explanation of why only certain wavelengths are emittedWavelength (of photon) depends one energy (1)Photon energy depends on difference in energy levels (1)Levels discrete / only certain differences / photon energies possible(1) 3(and therefore certain wavelengths) Why phosphor emits different wavelengths to mercuryDifferent energy levels / different differences in energy levels (1) 1 Calculation of chargeQ = It(1)= 0.15 A 20 60s= 180 C (1) 2[10] 93. Why microwaves are reflectedWave is reflected when passing from one medium to another / when density changes / when speed changes (1) 1 Varying amplitudeAny two of the following:Varying differences in density of the two mediums produce different intensities of signal (1)Different distances travelled give different amplitudes (1)Following a reflection there is less energy available (1) Max 2 Varying timeDifferent thicknesses of medium (1) 1What is meant by Doppler shiftChange in frequency/wavelength (1)Caused by movement of a source (1) 2 Changes due to Doppler shiftWavelength increases (1)Frequency decreases (1)[Allow e.c.f. from incorrect wavelength]Any one of the following: Each wave has further to travel than the one before to reach the heart The waves are reflected from the heart at a slower rate (1) 3[9] 94. Resistance calculationUse of R = L/A (1)Substitution R = 1.6 104 0.02/(5 (103) 0.02 (103)) (1)= 32 (1) 3 Total resistanceEither Section 2 = R1 (16 ) OR Section 3 = 31 R1 (10.7 ) (1)Use of RTotal =R1 + R2 + R3 (1)RTotal = 58.7 [55 if 30 used as starting point] (1) 3[ecf if section 3 calculated as R1 = 56 OR 52.5 if 30 usedas starting point] Why thermochromic ink becomes warmCurrent produces heat / reference to I2RORThermal conduction from conductive ink (1) 1[Mark for identifying that the heating effect originates in the conductive ink] Why only thin section transparentThinner / section 1 has more resistance (1)So even a small current will heat it/Power (heating effect) given byI2R / current will heat it more (1) 2[OR opposite argument explaining why thicker section is harder to heat][9] 95. Example of light behaving as a waveAny one of: diffraction refraction interference polarisation (1) 1 What is meant by monochromaticSingle colour / wavelength / frequency (1) 1Completion of graphPoints plotted correctly [1 for each incorrect point] (1) (1)Line of best fit added across graph grid (1) 3WhateV s tells usMaximum (1)Kinetic energy of the electrons / mv2 of electrons (1) 2Threshold frequency for sodiumCorrect reading from graph: 4.3 1014 Hz (1) 1[Accept 4.1 1014 4.7 1014Hz]Work functionf = hf0 = 6.63 1034 J s 4.3 1014 Hz (1)= 2.9 1019J [Allow ecf] (1) 2 Why threshold frequency is needed Electron requires certain amount of energy to escape from surface (1) This energy comes from one photon of light (1) E = hf (1) Max 2[12] 96. Adding angles to diagramCritical angle C correctly labelled (1) 1 Calculation of critical angleUse of = 1/sin C (1)Sin C = 1/1.09C = 66.6 (1) 2 Why black mark not always seenAt (incident) angles greater than the critical angle (1)t.i.r. takes place (so black mark not visible) (1)light does not reach X / X only seen at angles less than C (1) 3[OR opposite argument for why it is seen at angles less than C] Comparison of sugar concentrationLower means greater density (1)Greater density means more sugar (1) 2[8]97. CircuitsBase unit: ampere OR amperes OR amp OR amps (1)Derived quantity: charge OR resistance (1)Derived unit: volt OR volts OR ohm OR ohms (1)Base quantity: current (1) 4[If two answers are given to any of the above, both must be correct to gain the mark][4] 98. (a) Io and Jupiter: Time taken for electrons to reach Jupitert = s/ = (4.2 108 m)/(2.9 107 m s1) = 14.48 sCorrect substitution in = s/t (ignore powers of ten) (1)Answer: 14.48 s, 14.5 s [no ue] (1) 2 (b) Estimate of number of electronsQ = ne = Itn = It/en = (3.0 106 A) (1s)/(1.6 1019 C)Use of ne = It (1)(1.8 2.0) 1025 (1) 2 (c) Current directionFrom Jupiter (to Io) / to Io / to the moon (1) 1[5] 99. (a) p.d. across 4 resistor1.5 (A) 4 ()= 6 V (1) 1 (b) Resistance R2Current through R2 = 0.5 A (1)R2 = 0.5(A)(V) 6R2 = 12 (1) 2[allow ecf their pd across 4 ](c) Resistance R1p.d. across R1 = 12 6 4= 2 V (1)Current through R1 = 2 A (1)R1 = 2(A)V) ( 2 = 1 (1)[allow ecf of pd from (a) if less than 12 V]Alternative methodParallel combination = 3 (1)Circuit resistance = 12(V)/2 (A) = 6 (1)R1 = 6 (3 + 2) = 1 (1) 3[allow ecf of pd from (a) and R from (b)][6] 100. (a) Current in filament lampP = VI or correct rearrangement (1)2 A (1) 2 (b) (i) Sketch graphCorrect shape for their axes (1)IV quadrant showing fair rotational symmetry (1) 2IV(ii) Explanation of shape(As the voltage/p.d. increases), current also increases (1)(As the current increases), temperature of lamp increases (1)(This leads to) an increase in resistance of lamp (1)so equal increases in V lead to smaller increases in I OR rate ofincrease in current decreases OR correct reference to their correct (1) 4gradient[8][If a straight line graph was drawn though the origin then (1)(0)(0)(1) forthe following:V is proportional to Rtherefore the graph has a constant gradient] 101. (a) (i) ReplacementV1 (1) 1 (ii) Explanation[ONE pair of marks]Resistance: resistance of V1 [not just the voltmeter] is much largerthan 100 OR combined resistance of parallel combination is (1)approximately 100 Voltage: p.d. across V1 is much greater than p.d. across 100 OR (1)all 9 V is across V1ORCurrent: no current is flowing in the circuit / very small current (1)Resistance: because V1 has infinite/very large resistance (1)OR(Correct current calculation 0.9 x 10 6 A and) correct pd calculation90 x 10 6 A (1)This is a very small/negligible pd (1) 2 (b) Circuit diagram(i)or equivalent resistor symbol labelled 10 M (1) or equivalent resistor symbol labelled 10 M (1) 2[They must be shown in a correct arrangement with R](ii) Value of R 6 (V): 3 (V) = 10 (M): 5 (M) / Rtotal of parallel combination is 5 (1)M1/5 (M) = 1/10 (M) + 1/R OR some equivalent correct (1)substitution to show workingR = 10 M (1) 3[8] 102. Table 6Wavelength of light in range 390 nm 700 nm (1)Wavelength of gamma 1011 m(1)Source (unstable) nuclei (1)Type of radiation radio (waves) (1)Type of radiation infra red (1)Source Warm objects / hot objects /above 0 K (1)[6] 103. (a) AmplitudeMaximum distance/displacementFrom the mean position / mid point / zero displacement line / (1) 1equilibrium point[If shown on a diagram, at least one full wavelength must be shown,the displacement must be labelled a or amplitude and the zerodisplacement line must be labelled with one of the terms above.] (b) Progressive waveDisplacement at A: 2.0 (cm) [accept 2] (1)Displacement at B: 2.5 (cm) to 2.7 (cm) (1)Displacement at C: 1.5 to 1.7 (cm) (1) 3 Diagram[Minimum] one complete sinusoidal wavelength drawn (1)Peak between A and B [accept on B but not on A] (1)y = 0 (cm) at x = +2.6 cm with EITHER x = +6.2 cm OR x = 1.0 (1)cm 3[7] 104. (a) Transverse wave(Line along which) particles/em field vectors oscillate/vibrate (1)Perpendicular to (1)Direction of travel or of propagation or of energy flow or velocity (1) 3 (b) DifferencesAny two:Standing waves Progressive waves1. store energy 1. transfer energy (1)2. only AN points have max 2. all have the max ampl/displ (1) ampl/displ3. constant (relative) phase 3. variable (relative) phaserelationship relationship (1) Max 2 (c) (i) DropletsFormed at nodes / no net displacement at these points (1) 1(ii) SpeedUse of = f (1)Evidence that wavelength is twice nodenode distance (1)Wavelength = 1.2 (cm) (1)Frequency = 8.0 [8.2 / 8.16] Hz or s1 only (1) 4[10] 105. Photoelectric effect(a) Explanation:Particle theory: one photon (interacts with) one electron (1)Wave theory allows energy to build up, i.e. time delay (1) 2 (b) Explanation:Particle theory: f too low then not enough energy (is released byphoton to knock out an electron) (1)Wave theory: Any frequency beam will produce enough energy (to release an electron, i.e. should emit whatever the frequency) (1) 2[4]106. Description of photonPacket/quantum/particle of energy [accept E = hf for energy] (1) (1)[allow {packet/quantum/particle} of {light/e-m radiation/e-m wave} etc for (1) X] 2[zero marks if error of physics such as particle of light with negative charge]Show that energy to move electron is about 8 10 20J W = QV (1)= 1.6 1019 C 0.48 V= 7.7 1020 J [no ue] (1) 2Calculate efficiency of photon energy conversionEfficiency = (7.7 1020 J 4.0 1019 J) [ecf] (1)= 0.19 or 19 % (1) 2[6] 107. Explanation of pressure nodes or antinodesPressure constant (1)Node as a result (1) 2Relationship between length and wavelengthl = /2 or = 2l (1) 1Calculation of fundamental frequency = 2 0.28 m = 0.56 m [ecf for relationship above] (1)v = f (1)f = v/ = 330 m s1 0.56 m= 590 Hz (1) 3 Calculation of time periodT = 1/f (1)T = 1 590 Hz [ecf]= 0.0017 s (1) 2State another frequency and explain choicee.g. 590 Hz 2 = 1180 Hz (or other multiple) (1)multiple of f0 or correct reference to changed wavelength (1)diagram or description, e.g. N A N A N, of new pattern [ecf for A & N] (1) 3[11]108. Explain zeroing of meterNo resistance when leads touched together/short circuit/calibration forzero error (1) 1Show that resistance is about 70 R = V I (1)= 0.54 V 0.0081 A= 67 [no ue] (1) 2Explain section from passageOther currents/voltages/resistances present (1)change in current changes reading for resistance (1) 2Explain changes in meter reading with temperature increaseIncreased lattice vibrations/vibration of atoms/molecules (1)scattering flowing electrons/more collisions (1)increased resistance/increase meter reading (1) 3[8] 109. Name process of deviationRefraction (1) 1Completion of ray diagramB no deviation of ray (1)A and C refraction of ray away from normal on entering hot air region (1)A and C refraction of ray towards normal on leaving hot air region/ (1) 3 Show positions of tree trunksB the same } (1)} [consistent with ray diagram]A and C closer to B } (1) 2Explanation of wobbly appearanceHot air layers rise/density varies/layers uneven (1)Change in the amount of refraction [accept refractive index]/changein direction light comes from (1) 2[8]110. Circuit diagramAmmeter and power source in series (1)Voltmeter in parallel with electrodes (1) 2[Allow both marks if diagram shows an ohmmeter without a powerpack 1 if power pack]Calculation of resistanceUse of area = r2 (1)R = 2.7 103 m 5.0 104 m/A (1)= 172 (171.9 ) (1) 3 Plotting graphAxis drawn with R on y-axis and labelled with units (1)Points plotted correctly [1 for each incorrect] (1)Sensible scale (1)Curve added passing through a minimum of 4 points (1) 4Diameter of holeCorrect reading from graph = 0.23 mm [Allow 0.22 0.26 mm] (1) 1[10] 111. Unpolarised and plane polarised lightMinimum of 2, double-headed arrows indicating more than 1 planeand 1 double-headed arrow indicating 1 plane labelled unpolarisedand polarised (1)Vibrations/oscillations labelled (1) 2 Appearance of screenScreen would look white/bright/no dark bits/light [not dark = 0] (1)ExplanationAs no planes of light prevented from leaving screen/all light getsthrough/all polarised light gets through (1) 2Observations when head is tiltedScreen goes between being bright/no image to image/dark bits (1)Every 90/as the polarising film on the glasses becomes parallel/perpendicular to the plane of polarisation of the light (1) 2Comment on suggestionImage is clear in one eye and not the other (1)If plane of polarisation is horizontal/vertical (1)ORImage is readable in both eyes (1)As the plane of polarisation is not horizontal or vertical (1) 2[8] 112. How sound from speakers can reduce intensity of sound heard by driverAny 6 from: graphs of 2 waveforms, one the inverse of the other graph of sum showing reduced signal noise detected by microphone waveform inverted (electronically) and fed through speaker with (approximately) same amplitude as original noise causing cancellation/destructive superposition error microphone adjusts amplification 6[6] 113. Temperature calculationCurrent = 4.5 103 A (1)p.d. across thermistor is 4.2 V (1)Rthermistor = 930 ecf their current and pd subtraction error (1)Temperature = 32 C 34 C [Allow ecf for accurate reading] (1) 4 Supply doubledAny two from: Current would increase / thermistor warms up / temperature increases Resistance of thermistor would decrease (1) (1) Ratio of p.d.s would changeNo OR voltmeter reading / pd across R more than doubles (1) 3[This mark only awarded if one of the previous two is also given][7] 114. DiagramLabelled wire and a supply (1)Ammeter in series and voltmeter in parallel (1)ORLabelled wire with no supply (1)Ohmmeter across wire (1) 2ReadingsCurrent and potential difference OR resistance ( consistent with diagram) (1)Length of wire (1)Diameter of wire (1) 3 Use of readingsR = V/I OR = RA/l (1)Awareness that A is crosssectional area (may be seen above and credited here) (1)Repetition of calculation OR graphical method (1) 3PrecautionAny two from: Readings of diameter at various places /different orientations Contact errors Zeroing instruments Wire straight when measuring length Wire not heating up / temperature kept constant (1) (1) 2[10] 115. Conductor resistanceR = l/A (1)Correct substitution of data (1)R = 4.3 102 (1) 3 Manufacturers recommendationLarger A has a lower R (1)Energy loss depends on I2R / reduces overheating in wires (1) 2[5]116. Car batteryVoltmeter reading: 12.2 (V) (1) 1EquationTerminal p.d. = 12 V + (5.0 A 0.04 )See 12V (1)See 5.0 A 0.04 (1)Addition of terms (1) 3Wasted powerSee 0.04 + 0.56 OR2.8 V + 0.2 VOR5 x (15 12) W (1)Power = 15 W (1) 2 Efficiency(same current) 12 V / 15 V OR POUT/PIN = 60 W/75 W (1)Efficiency = 0.8/80% Efficiency = 0.8/80% (1) 2 ExplanationAny two from: Starter motor / to start car needs (very) large current I = r RE+ (E and R fixed) rmin Imax (1) (1) (1) 2[10]117. Wavelength0.30 m (1) 1Letter A on graphA at an antinode (1) 1WavespeedUse of = f(1)11(10.8) m s1 (1) 2[allow ecf = 0.15 m ie = 5.4 m s1]Phase relationshipIn phase (1) 1Amplitude2.5 mm (1) 1[6] 118. DiagramOne arrow straight down (from 3.84 to 5.02) (1)Two arrows down (from 3.84 to 4.53, then 4.53 to 5.02) (1) 2Transition TT from 5.02 to 1.85 upwards (1) 1Kinetic energy values and explanation of what has happened to lithium atomin each case0.92 eV (1)Atom stays in 5.02 (eV) level/nothing happens to it (1)0.43 eV (1)Atom excited to 4.53 (eV) level (1) 4Full credit is given to candidates who take the k.e. of the electron to be 0.92 Jafter collision. Any TWO correct energies with correct statement.[7] 119. Value of wavelength= 13.9 cm 0.5 cm (using interpolated sine curve) (1)= 13.4 cm [accept 13.2 to 13.6 cm] (1) 2[12.3 to 12.5 cm for distance using rods (1) ]Value of amplitudePeak to peak = 4.5 cm [Accept 4.3 cm to 4.7 cm] (1)Amplitude = peak to peak= 2.25 cm [Accept 2.15 cm to 2.35 cm] [Allow ecf for 2nd mark if (1) 2first part shown] Calculation of frequencyf = 1/T= 1 2 s= 0.5 Hz (1) 1Explanation of why waves are transverseOscillation/vibration/displacement/disturbance at right angle (1)to direction of propagation/travel of wave (1) 2[Oscillation not in direction of wave (1)] Description of use of machine to illustrate sound waveSound is longitudinal/not transverse (1)with oscillation along the direction of propagation / compressions and rarefactions (1)so model not helpful (1) 3[10] 120. Circuit diagramVariable voltage (1)Includes ammeter and voltmeter (1). in series and parallel respectively (1) 3[No penalty for LED bias] Description of current variation in LEDsInitially, increasing voltage still gives zero currentORCurrent doesnt flow until a specific minimum voltage (1)Current then increases (1).with an increasing rate of increase (1) 3Discussion of whether LEDs obey Ohms lawNo (1)I not proportional to VORR not constant / V/I not constant / R decreases (1) 2 Calculation of resistance of green LED at 1.9 VR = V/I [Stated or implied] (1)= 1.9 V 1.46 103 A= 1300 (1) 2 Calculation of power dissipated by red LED at 1.7 VP = IV [Stated or implied] (1)= 3.89 103 A 1.7 V [do not penalise mA twice]= 6.6 103 W (1) 2[12] 121. Process at ARefraction [Accept dispersion] (1) 1Ray diagramDiagram shows refraction away from normal (1) 1Explanation of condition to stop emergence of red light at BAngle greater than critical angle (1)Correctly identified as angle of incidence [in water] (1) 2 Calculation of wavelength of red light in waterc = f[stated or implied] (1)= 2.2 108 m s1 4.2 1014 Hz= 5.24 107 m (1) 2[6]122. Difference between polarised and unpolarised lightPolarised: vibrations in one plane (at right angles to direction of travel) (1)Unpolarised: vibrations in all planes [NOT 2 planes] (1) 2ORCorrect drawing (1)Vibrations labelled (1) Meaning of advertisement(Light vibrations are) in one plane (1) 1Evidence that glare comprises polarised lightGlare is eliminated, so must be polarised light (1) 1Sunglasses turned through 90Glare would be seen through glasses (1)since they now transmit the reflected polarised light (1) 2[6] 123. ChargeCharge is the current time (1) 1Potential differenceWork done per unit charge [flowing] (1) 1Energy9 V 20 C (1)= 180 J (1) 2[4] 124. Number of electrons(64 109 C) / (1.6 1019 C) = 4.0 1011 electronsUse of n = Q/e (1)Seeing 1.6 1019 C (1)Answer of 4.0 1011 (electrons) (1) 3[Use of a unit is a ue][ve answer: 2/3]Rate of flow(6.4 108 C)/3.8 s = 16.8/17 [nC s1] OR 16.8/17 109 [C s1](6.4) / 3.8 s i.e. use of I = Q/t [Ignore powers of 10] (1)Correct answer [No e.c.f.] [1.7 or 1.68 x 108 or 1.6 108] (1) 2UnitAmp(ere)/A (1) 1[6] 125. Explanation of observationAny two from: LED on reverse bias/R in LED infinite/ LED wrong way round so current is zero /LED does not conduct / very small reversebias current since V = IR V = 0 1K = 0 V (1) (1) 2 Explanation of dimnessRV very large / RV much greater than RLED (1)Current very low / pd across LED very small (not zero) (1) 2Circuit diagramLED in forward bias (1)Variation of pd across LED (1)Voltmeter in parallel with LED alone (1) 3[LED in series with voltmeter 0/3][7] 126. Circuit diagramAmmeter in series with cell and variable resistor (correct symbol) (1)Voltmeter in parallel with cell OR variable resistor (1) 2Power output at point XPower = voltage current (1)= 0.45 V 0.6 A= 0.27 W (1) 2Description of power outputAny three from: Current zero; power output zero/small/low As current increases power output also increases Then (after X ) power decreases Maximum current; power output zero (1) (1) (1) 3[Accept reverse order] e.m.f. of cell0.58 V (1) 1Internal resistanceAttempt to use current" lost volts " OR = V+ IR (1)= A 6 . 00.45V V 58 . 0= 0.217 / 0.2 (1) 2[ecf an emf greater than 0.45 V][10] 127. Statement 1Statement is false (1)Wires in series have same current (1)Use of I = nAe with n and e constant (1) 3[The latter two marks are independent] Statement 2Statement is true (1)Resistors in parallel have same p.d. (1)Use of Power = V2/R leading to R , power (1) 3OR as R , I leading to a lower value of VI3rd mark consequenton second[6]128. Description + diagramDiagram to show:Microwave source/transmitter and detector (not microphone) (1)Transmitter pointing at metal plate/second transmitter from same source (1)Written work to include:Move detector perpendicular to plate/to and fro between /accept ruler on diagram (1)Maxima and minima detected/nodes and antinodes detected (1) 4[Experiments with sound or light or double slit 0/4]ObservationIn phase/constructive interference maximum/antinode (1)Cancel out/out of phase/Antiphase/destructive interference minimum /node (1) 2How to measure wavelength of microwavesDistance between adjacent maxima/antinodes = /2 (1)Measure over a large number of antinodes or nodes (1) 2[8] 129. Incident photon energiesUse of E = hf (1)Use of c = f [ignore 10X errors] (1) e (1)For 320 nm E = 3.9 (eV) and 640 nm E = 1.9 (eV) (1) 4 Photocurrent readingsWork function of Al > 3.9 / energies of the incident photonsOR threshold frequency is greater than incident frequencies (1)For Li (= 2.3 eV / f = 5.6 1014 Hz / = 540 nm hence) a photocurrentat 320 nm but not 640 nm (1)If intensity 5 then photocurrent 5 (1) 3Stopping PotentialKEmax = 4.00/3.88 2.30 = 1.7/1.58 [ignore anything with only e] (1)Vs = 1.7/1.58 V (1) 2[9]130. Wavelength and wave speed calculation= 0.96 m (1)seeing f = 2 their (f = 2.1 Hz) (1) 2Qualitative description(Coil) oscillates / vibrates (1)With SHM / same frequency as wave (their value) (1)Parallel to spring / direction of wave (1) 3[5] 131. Explanation of emission of radiation by hydrogen atomsElectrons excited to higher energy levels (1)as they fall they emit photons / radiation (1) 2[Accept 21 cm line arises from ground state electron changing spinorientation (1) / relative to proton (1)] Why radiation is at specific frequenciesPhoton frequency related to energy / E = hf (1)Energy of photon = energy difference between levels / hf = E1 E2 (1)Energy levels discrete/quantised / only certain energy differences possible (1) 3 Show that hydrogen frequency corresponds to = 21 cm f = 4.4623 109 = 1.42 109 Hz (1)c = f = 3 108 (1.42 109 Hz) (1)= 0.211 m or 21.1 cm [no up] (1) 3[8] 132. Explanation of assumption that voltmeter does not affect valuesVoltmeter has very high resistance/takes very small current (1) 1Current through X4.8 A 6 = 0.8 AOR 48 V 60 = 0.8A (1) 1Value missing from E7P =IVP = 4.4 A 53 V = 233 W (1) 1Description of appearance of lamp X as lamps switched onGets dimmerfrom table, voltage decreasing / current in X decreasing / power per lamp decreasing (1)So P decreases (1) 3 Formula for cell C6I = / Rtot (1)I = 120 / (15 + B6) (1) 2Effect of internal resistance on powerPower has a maximum value (1)when external resistance = internal resistance (1) 2[10] 133. Fundamental frequency of note440 Hz (1) 1Frequencies of first three overtones880 Hz1320 Hz1760 HzTwo correct frequencies (1)Third correct frequency (1) 2Comment on the patternAny 2 from the following:[Allow ecf]880 Hz = 2 440 Hz1320 Hz = 3 440 Hz1760 Hz = 4 440 Hz1760 Hz = 2 880 Hz (1) (1) 2[OR They are multiples (1) of the fundamental (or similar qualification) (1)][Allow 1 mark for amplitude decreasing with frequency]Measurement of periodExample: 7 cycles takes (0.841 0.825) s [at least 5 cycles] (1)Period = 0.016 s 7= 2.3 103 s[in range 2.2 103 s to 2.4 103 s] (1) 2 Calculation of frequencyf = 1/T (1)= 1 2.2 103 s [Allow ecf]= 454 Hz (1) 2[9] 134. Mark on diagramCorrectly drawn normal (1)Correctly labelled angles to candidates normal (1) 2Show that refractive index of water is about 1.3Angles correctly measured:i = 53 (t 2)r = 39 (t 2) (1)= sin i / sin r = sin 53 / 39= 1.27 [Allow ecf] [Should be to 2 d.p. min] (1) 2Critical angle= 1/sinC (1)so sin C = 1/1.27 so C = 52 [ecf] (1) 2[use of 1.3 gives 50]Explanation of reflection of rayInternal angle of incidence = 39 t1 (1)Compare i with critical angle (1)Valid conclusion as to internal reflection being total/partial (1) 3Refractive indexIt varies with colour (1) 1[10] 135. Explanations(i) Refraction:e.g. bending of wave when travelling from one medium to another [OR change of speed] (1)(ii) Diffraction:e.g. spreading of wave when it goes through a gap (1) 2 Diagram of wavefronts near beachGradual bend in wavefronts (1)Smaller wavelengths (1)Waves bending upwards as they approach shore (1) 3 Diagram of wavefronts in bayConstant wavelength (1)Waves curve (1) 2ExplanationRefraction/diffraction causes waves to bend towards the beach (1) 1[8]136. Measurement neededAny three from: Resistance Distance between probes Effective area/cross sectional area R = AL (1) (1) (1) 3 Equation of line AIntercept = 3.5 (m) (+/ 0.3) (1)Gradient = 1.5 (mm1) (+/ 0.05) (1)So equation is = 1.5 d 3.5 [Or equivalent, e.c.f. allowed] (1) 3 Addition of linePoints correctly plotted (1 for each error, allow square tolerance) (1) (1)Line of best fit drawn (1) 3Best distanceBetween 1.90 and 1.99 km (1) 1[10] 137. UltrasoundUltrasound is very high frequency sound (1)How ultrasound can be usedAny three from: gel between probe and body ultrasound reflects from boundaries between different density materials time taken to reflect gives depth of boundary probe moved around to give extended picture size of reflection gives information on density different (1) (1) (1) 3How reflected ultrasound provides information about heartAny two from: Doppler effect frequency changes when reflected from a moving surface gives speed of heart wall gives heart rate (1) (1) 2[6] 138. Physics principlesRequires 9 V battery:Battery required for electronic circuitry / microphone / speaker (1)Rubberized foam ear cups:Air filled material / material has large surface area (1)Air molecules collide frequently with material (1)Foam deforms plastically/collisions are inelastic (1)Sound converted to heat in material (1) Active noise attenuation:Noise picked up by microphone (1)Feedback signal inverted / 180 out of phase with noise / antiphase (1)Amplified [OR amplitude adjusted] and fed to earphones / speaker (1)Sound generated cancels/superimposes/minimum noise (1)Diagrams of superposing waves showing (approx.) cancellation (1)Amplifier gain automatically adjusted if noise remains (1)Device only works over frequency range 20 800 Hz (1) Max 6 Where does the energy go?Some places will have constructive interference (1)More intense noise (1)Some noise dissipated as heat in air / foam (1)increased kinetic energy of air [OR foam] molecules (1) Max 2[8]139. Resistance calculationsEvidence of 20 for one arm (1)201201 1+ R (1)R = 10 (1) 3 CommentThis combination used instead of a single 10 resistor [or samevalue as before] (1)because a smaller current flows through each resistor/reduce heatingin any one resistor/average out errors in individual resistors (1) 2[5] 140. GraphsDiode:RH quadrant: any curve through origin (1)Graph correct relative to labelled axes (1)LH side: any horizontal line close to axes (1) 3IVL i n e o n o r c l o s e t ov o l t a g e a x i sFilament lampIVRH quadrant:Any curve through origin (1)Curve correct relative to axes (1)LH quadrant:Curve correct relative to RH quadrant (1) 3[Ohmic conductor scores 0/3][6] 141. CircuitAmmeters and two resistors in series (1) 1[1 mark circuit penalty for line through cell or resistor]Cell e.m.fE= 150 x 106 (A) x 40 x 103 ( ) total R (1)Powers of 10 (1) 2E = 6.0 (V) New circuitVoltmeter in parallel with 25 (k ) resistor (1) 1Resistance of voltmeter(Total resistance) = A) ( 10 170V) ( 66 = (35.3 k )(Resistance of ll combination) = 35 15 k= (20 ) [e.c.f. their total resistance]VR1251201+ 1004 5 1VRRV = 100 k[108 kif RT calculated correctly](1)(1)(1) 'Alternative route 1: 3p.d. across 15 k= 2.55 V(p.d. across ll combination = 3.45 V)resistance combination = 20 kRV= 100 k(1)(1)(1)'[7] Alternative route 2: 3p.d. across parallel combination = 3.45 VI through 25 k= 138 ARV= 100 k(1)(1)(1)' 142. Resistance of strain gaugeState R = Al (1)Use of formula (1)x 6 (1)R = 0.13 [ecf their l] (1) 4

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.| 13 . 010 6 . 129m 10 1 . 16 m 10 4 . 2 m 10 9 . 9 3 2 7 2 8 RAlRChange in resistance R = 0.13 0.001 R = 1.3 104 ( ) [no e.c.f.]OR R = 0.02 0.001 R = 2.0 105 0.1% 0.001 (1)Correct number for R (1) 2Drift velocityStretching causes R to increase (1)Any two from: Current will decrease I = nA Q Drift velocity decreases nAe constant (1) (1) 3[9] [For R decreasing, max 1:Any one from: I will increase I = nA Q will increase nAe constant] 143. ExplanationClarity of written communication (1)Wave reflects off bench (1)(Incident and reflected) waves superpose/stationary wave is formed (1)Maxima or antinodes where waves in phase or constructiveinterference occurs (1)Minima or nodes where waves exactly out of phase or destructiveinterference occurs (1) 5 Speed of soundSee a value between 5.0 and 5.6 (cm) (1)Use of = f(1)= 2 spacing (1)320 m s1 to 360 m s1 (1) 4 Explanation of contrastAs height increases, incident wave gets stronger, reflected wave weaker (1)So cancellation is less effective [consequent mark] (1) 2[11]144. Ionisation energy(10.4 eV) (1.6 10 19J eV1) (1)() 1.66 1018(J) (1) 2Kinetic energy0.4 (eV) (1) 1TransitionUse of E = hc/(1)3.9 (eV) (1)Transition is from ()1.6 eV to ()5.5 eV 3[6] 145. Deductions about incident radiations(i) Radiations have same frequency/same wavelength/ same photon energy (1)(ii) Intensity is greater in (a) than in (b) (1) 2 Sketch graph (c)Line of similar shape, starting nearer the origin on negative V axis (1) 1Maximum speedUse of E = hf (1)Subtract 7.2 1019 (J) (1)Equate to m2 (1)3.1 106 ms1 (1) 4[7] 146. WavelengthDistance between two points in phase (1)Distance between successive points in phase (1) 2[May get both marks from suitable diagram]Sunburn more likely from UVUV (photons) have more energy than visible light (photons) (1)Since shorter wavelength / higher frequency (1) 2What happens to atomsMove up energy levels/excitation/ionization (1)Correctly related to electron energy levels (1) 2[6] 147. Emitted pulseGreater amplitude/pulse is larger/taller (1) 1Depth of rail2d = vt = 5100 m s1 4.8 105 s= 0.24 mHence d = 0.12 mReading from graph [4.8 or 48 only] (1)Calculation of 2d [their reading timebase 5 100] (1)Halving their distance (1) 3 Description of traceA reflected peak closer to emitted/now 3 pulses (1)Exact position e.g. 1.6 cm from emitted (1) 2 DiagramShadow region (1)Waves curving round crack (1) 2[8] 148. Resistance in darknessIn the dark R = 4 k(1)so resistance per mm = 4000 /40 mm = 100 (mm1) (1) 2Resistance of 8 mm lengthIn the light R = 200 (1)so resistance of 8 mm strip = (8 mm/40 mm) 200 [= 40 ] (1) 2CalculationsResistance of remainder = 32 mm 100 mm1 = 3200 (1) 1(i) Total resistance = 3240 (1)Current = V/R = 1.2 V/3240 = 3.7 104 A (1)(ii) Across 8 mm, p.d. = IR = 3.7 104 A 40 (1)= 0.015 V (1) 4 Explanation of why current decreasesAny two points from: more of strip is now in the dark greater total resistance I = V/R where V is constant (1) (1) Max 2[11] 149. Error in circuit diagramCell needs to be reversed (1)Any one point from: electrons released from the magnesium copper wire needs to be positive to attract electrons (1) 2 Completion of sentenceUV is made up of particles called photons (1) 1UV and visible light(i) UV has shorter wavelength/higher frequency/higher photonenergy (1)(ii) Both electromagnetic radiation/both transverse waves/samespeed (in vacuum) (1) 2 Explanation of why low intensity UV light produces a currentAny three points from: reference to photons or E = hf frequency > threshold frequency electron must have sufficient energy to be released UV photons have more energy electron is released by ONE photon brighter light just means more photons (1) (1) (1) Max 3 Why current stoppedGlass prevents UV reaching magnesium (1) 1[9] 150. Total internal reflectionAny two points from: from a more dense medium to a less dense medium/high to low refractive index incident angle greater than the critical angle light is reflected not refracted/no light emerges (1) (1) Max 2 Critical angleSin i / sin r = ; gives sin 90/sin C = (1)C = 42 (1) 2DiagramReflection (TIR) at top surface (air gap) (1)Reflection (TIR) at bottom surface and all angles equal by eye (1) 2Path of ray APassing approximately straight through plastic into glass (1)Emerging at glassair surface (1)Refraction away from normal (1) 3 Why there are bright and dark patches on imageBright where refracted/reference to a correct ray A in lower diagram (1)Dark where air gap (produces TIR)/reference to correct top diagram (1) 2[11]151. PolarisationThe (wave) oscillations (1)occur only in one plane (1) 2[OR shown with a suitable diagram]How to measure angle of rotationAny four points from: Polaroid filter at one/both ends with no sugar solution, crossed Polaroids (top and bottom oftube) block out light sugar solution introduced between Polaroids one Polaroid rotated to give new dark view difference in angle between two positions read from scale (1) (1) (1) (1) Max 4 GraphPoints plotted correctly [1 for each incorrect; minimum mark 0] (1) (1)Good best fit line to enable concentration at 38 to be found (1) 3Concentration0.57 ( 0.01) kg l11[10] 152. Expl