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Edexcel Physics Unit 2 "Physics at work" June 2013 Watch

  • View Poll Results: The last question - Does resistance increase or decrease?
    It increases ( using V=IR or some other method)
    70.73%
    It decreases using the 'lattice vibrations' theory
    29.27%

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    (Original post by CWE)
    I thought that it would be because the voltmeter was connected in parallel with the internal resistance the overall resistance would be lower as 1/Rt = 1/R1 + 1/R2 -

    i take it thats wrong?
    TBH doesn't make sense as such but who knows what the mark scheme says !!
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    (Original post by CWE)
    I thought that it would be because the voltmeter was connected in parallel with the internal resistance the overall resistance would be lower as 1/Rt = 1/R1 + 1/R2 -

    i take it thats wrong?
    i think the voltmeter can not connect in parallel with the internal resistance ( since the internal resistance connect in the series with the e.m.f ?? right ? so i think the voltmeter can only connect in parallel with the resistor ) maybe you wrong when you write that "the voltmeter was connected in parallel with the internal resistance" ... but it also maybe I am wrong , iam not sure about this , i think we should need to a person who can write a mark scheme( draft ), so we can check the answers , and discuss this mark scheme
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    Did anyone else notice that there were only about 5 calculation questions?
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    (Original post by jollygood)
    What did u guys do about the ultrasound question? how did u tell if corrosion has occurred or not?
    If the thickness of the metal has been reduced, corrosion has occurred
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    (Original post by orangestarburst)
    Did anyone else notice that there were only about 5 calculation questions?
    Yes I did, bloody annoying. That's the way the papers seem to be going. It's crap because you have to spend longer writing wordy answers and then you have less time to check over. But you also have to spend longer checking over! Sucks. Plus I love maths.

    By the way, what did everyone put for a 3 mark question on defining plane polarised light? 3 marks??!! I think I said oscillates in one plane only, transverse and can be achieved using a polarising filter.
    And two marks for defining a photon? I said it was a particle used to describe light, and that it was a discrete packet of energy or something like that.
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    Anyone tell me the answer to the last one about the bulbs being supllied lower current what would happen to the resistance?

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    (Original post by CharlieTT)
    I said that it had corroded and worked the new thickness to be 3.009cm. Meaning it had corroded by 0.991cm.
    That's exactly what I did too, but everyone keeps telling me you had to divide the time by 2 or something? I thought since it was the reflective pulse time you'd use the time given to you?
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    (Original post by JoshThomas)
    Anyone tell me the answer to the last one about the bulbs being supllied lower current what would happen to the resistance?

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    I said it would increase. I actually tried this on the calculations too, just to make sure it made sense, like, I lowered the current and the resistance did increase
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    (Original post by GCSE-help)
    Yeah you're right on both counts but could you please remind me about the polarisation question? Are you sure it wasn't superposition? I didn't read the question and hastily ticked superposition after seeing the image.

    Also what did you put for the last question?
    I put down superposition too, I think I came across that question in one of the past papers.
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    Someone put the Unit 1 paper on here very quickly this year, hope they do the same with this paper if it's possible!

    I think for the last one the question asked us the effect of low current on the resistance of the bulb, so it wasn't asking for us to use any formulas, but use the fact the temperature would be less due to less current, so the ions in the lattice would have less energy and vibrate less, leading to less electron-ion collisions etc etc, so the resistance would reduce. That's what usually happens when a lamp is operating below it's normal rating, there is a low resistance which could give rise to a huge surge in current momentarily if a significant potential difference is applied
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    Guys. IT WASN'T AS BAD AS JANUARY!! Thank you Edexcel!!

    I think the E.M.F question and the last question were hard, and the rest of the paper was tricky but manageable.

    For 45 and 135 degrees did anyone says that they're perpendicular to each other, thus they let trough completely different oscillations of light?

    I need 91 UMS for an A, can anyone offer a prediction what they tho k an A will be? I think I've dropped a lot of marks oh the last couple of questions 5 marks for E.M.F ugh


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    Think an A* sorry, I'm on my phone!


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    (Original post by jollygood)
    Okay and there wasn't any other hard MCQ apart from these two three. now coming to the questions answers first three were fairly okay. For the resistance one in which the real value was double the calculated value I said that by applying force the wire becomes longer as well as thinner so cross sectional area also decreases and for this reason the ressitance is twice the calculated value. what did u guys put?
    I referred to the equation R=(rho x l)/A and said that R is inversely proportional to A and directly proportional to l.As l increases and A decreases so R would be higher.Am i right?
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    (Original post by jollygood)
    and last past of this question I answered differently. I said that when current is reduced then according to V=IR. V is constant so resistance should increase.incresed resistance would not allow current to flow through.. and I also mentioned that as P=VI so decrease in current would decrease the power dissipation so they will show no light and brightness. do u think it makes some sense?
    I wrote exactly the same thing!
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    You get the paper from my dropbox link https://www.dropbox.com/sh/v6ydqbfrg...ZWR-r9ZOSH?lst sorry its in pics
    Attached Images
     
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    (Original post by Aerosports)
    You get the paper from my dropbox link https://www.dropbox.com/sh/v6ydqbfrg...ZWR-r9ZOSH?lst sorry its in pics
    Thanks a lot bro
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    (Original post by Alexsk)
    I wrote exactly the same thing!
    I wrote the same as you two!


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    (Original post by Aerosports)
    You get the paper from my dropbox link https://www.dropbox.com/sh/v6ydqbfrg...ZWR-r9ZOSH?lst sorry its in pics
    I put violin was that right
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    answers:
    D
    either D or A (i favour D)
    D
    B
    I put C, possibly B
    B
    A
    A
    A

    (3*10^8)/(218.6*10^6)=1.37

    measure the incident, and refraction angles. plot a graph of sini against sinr. the gradient is the refractive index

    Critical angle is the angle at which the light no longer passes through the glass block, but is reflected internally or along the border.

    ref water = 1.33
    r=1/sinc sinc=1/r c=48.6 degrees

    r=pl/a r=4.9*10^-7 * 1 / 2.9*10^-8 = (possibly) 16.9

    as the length increases, the cross-sectional area decreases, and so there is less space for the electrons to move, and they slow down, meaning the resistance increases more than expected


    anyway, that is the first 13 questions. i have to go to school now, but i hope you manage to finish the rest.

    the link to the dropbox is here, but i didnt make it, "Aerosports" above posted it.
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    also who had a graph with a negative gradient http://www.s-cool.co.uk/a-level/asse...resistance.gif
 
 
 
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