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    (Original post by TajwarC)
    No. This exact question came up before in a paper (June 14) and the reasoning behind the neutral pion was to conserve charge, except they gave you a list of particles to chose from. I'll be honest here, aside from the obvious charge conservation the only reason I knew it was the neutral pion was because I'd seen the question before

    That's A levels for you haha. You might get the mark, who knows
    Exactly this, they asked us the same thing in June 14 Q1ci - that's the only reason why I knew.
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    (Original post by Ayaz789)
    It varies a lot that the thing!(
    June 15- 41
    June 14- 46
    June 13- 49
    Jan 13- 48
    June 12- 47Jan 12- 47
    Predict the grade boundaries on how you found the paper compared to the past papers and see what you need.
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    (Original post by TheLifelessRobot)
    Predict the grade boundaries on how you found the paper compared to the past papers and see what you need.
    Haha to get 75 Ums ill need 45/70 tbh
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    (Original post by matty9)
    For the conservation to find particle X I put photon, it doesn't have a charge, lepton number = 0, baryon number = 0, doesn't violate anything so would I by any chance get the mark?
    You'll probably get marks for saying charge and lepton number has to be conserved but photons (virtual) are exchange particles for electromagnetic interactions so you wouldn't get the final mark for identifying X. I'd assume it's 3/3 marks for simply stating its a neutral pion because it didn't ask for any explanation.
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    (Original post by BanterBus)
    You'll probably get marks for saying charge and lepton number has to be conserved but photons (virtual) are exchange particles for electromagnetic interactions so you wouldn't get the final mark for identifying X. I'd assume it's 3/3 marks for simply stating its a neutral pion because it didn't ask for any explanation.
    Yeah it didn't ask for an explanation which confused me in the exam because it normally did, I wrote the conservation laws anyway so hopefully should pick up 1 or 2 out of 3
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    Is there an unofficial mark scheme?
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    How many marks was all the heater question
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    (Original post by Imtheish)
    How many marks was all the heater question
    14
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    Honestly didn't think it was too bad, I'd say around 55 ish for an A. The only question that threw me off was the heating element one, specifically part C and D. If the answers in here are anything to go by, I'd say I've got around 60 imo.
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    (Original post by emsieMC)
    I got 3.4 ohms for internal resistance
    Yeah, got that too. Should have full marks for that question if you got that.
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    [QUOTE=Adamwest17;65130195]14[/QUOTE
    Thanks what about the whole question about the internal resistance
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    (Original post by lucabrasi98)
    Last years paper was genuinely disgusting though. It's normally around 55-58 for an A but they had to drop it to 49 because of how many marks were dropped overall. I think it'll go back up to 55-58. There's no year 12s taking it and the paper wasn't that bad.


    smh. I still don't understand why aqa assumed we'd know the last question in the june 2015 paper was a parallel circuit. Or why I was tested on potato conduction.
    Its usually around 53 for an A from what I have seen
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    [QUOTE=Imtheish;65130565]
    (Original post by Adamwest17)
    14[/QUOTE
    Thanks what about the whole question about the internal resistance
    Think 10
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    Also was the peak to peak question singular? I mean if you had a wrong answer there im guessing you wouldnt get further questions wrong?
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    Question 7 Everyone:
    Name:  IMG_1196.jpg
Views: 225
Size:  500.0 KB
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    (Original post by Formless)
    Everyone please correct, suggest and add. Rep is appreciated,
    What i got so far.
    1) isotope - same no. protons, diff no. neutrons
    92*1.6*10^-19/mass no. * 1.67*10^-something I got 3.7 * 10^something 3 electrons removed to 89 fill in -
    mass number stays the same, 93 protons, (0, -1 )for beta minus particle and antineutrino
    Pair production
    Rest Energies of electron and positron must be equal or greater to photon in order to conserve energy. Below energy pair production will not take place
    Show that, using (0.51MeV x 2) in data sheet. = 1.02MeV
    Photon frequency ( 1.6X10^13/h) = 2.4x10^20 Hz
    lepton number conserved (0=0)Kaon decays via weak, so pi 0

    Diodes - 6 marker is June 2009, refer to MS
    Switch wiring to cell so that opposite flow, Or Switch diode so reverse bias.
    Resistance increases, Voltage Decreases. ( i put it so that Res decreases which is wrong as its non-ohmic)

    Heating Element BS
    V peak to Peak is 650V
    4040W, 228V,

    7) Current not flowing, therefore dot on the oscilloscope
    PQ is Emf=6Volts
    current=0.28A,(5/18) I used 5 as it was in parallel with the internal res so voltage has to be 5?.
    Internal resistance = 3.6 Ohms. / 3.8 Due to rounding( Using Emf - V = 1)

    Prediction - 56/55 For an A
    Triple Bumb
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    And the wire resistance question: Name:  IMG_1198.jpg
Views: 164
Size:  498.1 KB
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    (Original post by Boro12345)
    Yeah, got that too. Should have full marks for that question if you got that.
    How did you get 3.4
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    (Original post by Formless)
    Everyone please correct, suggest and add. Rep is appreciated,
    What i got so far.
    1) isotope - same no. protons, diff no. neutrons
    92*1.6*10^-19/mass no. * 1.67*10^-something I got 3.7 * 10^something 3 electrons removed to 89 fill in -
    mass number stays the same, 93 protons, (0, -1 )for beta minus particle and antineutrino
    Pair production
    Rest Energies of electron and positron must be equal or greater to photon in order to conserve energy. Below energy pair production will not take place
    Show that, using (0.51MeV x 2) in data sheet. = 1.02MeV
    Photon frequency ( 1.6X10^13/h) = 2.4x10^20 Hz
    lepton number conserved (0=0)Kaon decays via weak, so pi 0

    Diodes - 6 marker is June 2009, refer to MS
    Switch wiring to cell so that opposite flow, Or Switch diode so reverse bias.
    Resistance increases, Voltage Decreases. ( i put it so that Res decreases which is wrong as its non-ohmic)

    Heating Element BS
    V peak to Peak is 650V
    4040W, 228V,

    7) Current not flowing, therefore dot on the oscilloscope
    PQ is Emf=6Volts
    current=0.28A,(5/18) I used 5 as it was in parallel with the internal res so voltage has to be 5?.
    Internal resistance = 3.6 Ohms. / 3.8 Due to rounding( Using Emf - V = 1)

    Prediction - 56/55 For an A
    How would you get 4040? It was literally (230^2)/12=4400. And even if it was 4040, you would be wrong because you had to use 2 significant figures
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    For the internal resistance I got 3.6 ohms:

    The emf was 6.0V, and the terminal pd was 5.0V (worked out from oscilloscope, work out displacement from P to both Q and R and multiply by two).
    Resistance of the resistor was 18 ohms, so using V=RI you get I=V/R =5/18 =0.27777... A
    Considering the internal resistance, the lost volts were 1V (as emf = terminal pd + lost volts, so lost volts = emf - terminal pd = 6-5 = 1V)
    V=RI and R=V/I =1/0.2777777... =3.6 ohms.

    You can check that this result is true using the emf equation:
    emf = terminal pd + lost volts, so E=IR+Ir
    E=(0.27777...)(18) + (0.2777777...)(3.6)
    E=6.0V [QED]
 
 
 
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