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Edexcel Unit 3 (IAL): Exploring Physics, WPH03 (11th May 2016) Watch

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    Grade boundary will definitely be high.

    How was your graph? I averaged my points and mine was through the origin.

    Resistivity value was around I guess 1.1*10^-6 i think?

    I read the vernier calliper and got the answer as A (dont remember what it was)
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    (Original post by SaadKaleem)
    Grade boundary will definitely be high.

    How was your graph? I averaged my points and mine was through the origin.

    Resistivity value was around I guess 1.1*10^-6 i think?

    I read the vernier calliper and got the answer as A (dont remember what it was)
    Yeah I got around the same for my graph too.

    I messed up on the vernier caliper one and put 10.7 as my answer. Fu*k!

    I hope the grade boundaries arent too high, as I'm almost certain that I've made numerous silly mistakes.
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    when will we have a full mark scheme?
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    (Original post by iWant.A)
    when will we have a full mark scheme?
    An official one? Next year when it gets released.
    In terms of an unofficial one, when someone is bothered enough to come up with an unofficial markscheme.
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    (Original post by SaadKaleem)
    Grade boundary will definitely be high.

    How was your graph? I averaged my points and mine was through the origin.

    Resistivity value was around I guess 1.1*10^-6 i think?

    I read the vernier calliper and got the answer as A (dont remember what it was)
    i got 2.24x10^-6............what was your gradient???????????
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    did anyone get their gradient as 4.5?I made a major error.... i used the diameter on its own as area. how much do you guys think ill get out of 4?For keeping current constant i mentioned using a power pack set at 0.11A. Any chance its correct?
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    (Original post by tejsanghavi)
    did anyone get their gradient as 4.5?I made a major error.... i used the diameter on its own as area. how much do you guys think ill get out of 4?For keeping current constant i mentioned using a power pack set at 0.11A. Any chance its correct?
    To keep current constant, you need to use a variable resistor.
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    (Original post by wima3109)
    i got 2.24x10^-6............what was your gradient???????????
    Around 4 decimal something, don't exactly remember.

    (Original post by tejsanghavi)
    did anyone get their gradient as 4.5?I made a major error.... i used the diameter on its own as area. how much do you guys think ill get out of 4?For keeping current constant i mentioned using a power pack set at 0.11A. Any chance its correct?
    Yes I got around that, probably quoted in 3 s.f though. I think you'll get 1 mark only for quoting the formula.

    For keeping current constant, I mentioned that as you increase the length (The resistance is proportional to it) so you need to increase the voltage with the same ratio, to keep the current constant in the circuit.
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    to keep current constant i put use an EHF power supply
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    so is a variable resistor correct? to keep the current constant
    I wrote the same thing
    and what did u guys write for the last question? why the value is different
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    Change in the temperature in the circuit
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    Also to find the refractive index i wrote that the student must plot a graph of sin i against sin r and the refractive index will be the gradient
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    (Original post by wima3109)
    to keep current constant i put use an EHF power supply
    I mentioned using a power pack set at 0.11A. is it the same thing? any doubt on your answer?
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    (Original post by tejsanghavi)
    did anyone get their gradient as 4.5?I made a major error.... i used the diameter on its own as area. how much do you guys think ill get out of 4?For keeping current constant i mentioned using a power pack set at 0.11A. Any chance its correct?
    I got 4.58 pretty close to you..
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    (Original post by Blazyy)
    To keep current constant, you need to use a variable resistor.
    .
    Yayyy I wrote variable resistor
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    (Original post by iWant.A)
    Also to find the refractive index i wrote that the student must plot a graph of sin i against sin r and the refractive index will be the gradient
    I wrote plot a sin r (y-axis) against sin i (x-axis) and find the inverse of the gradient as the incident angle is what we are changing hence is the independent variable and has to go to the x axis.
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    (Original post by Haamoo7)
    Yeah I got around the same for my graph too.

    I messed up on the vernier caliper one and put 10.7 as my answer. Fu*k!

    I hope the grade boundaries arent too high, as I'm almost certain that I've made numerous silly mistakes.
    For IAL they're constant?

    Not a bad paper overall ;o
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    What was the vernier caliper one?
    and what's the answer to the question where they ask how setup was in timing the can as it rolled down a slope in no.6c)??
    thanks in advance!
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    What did you all write on how to measure the diameter of the wire?
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    Measure the diameter with micrometer and make sure to measure diameter several times around the wire to get an average result
 
 
 
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