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ANSWERS: OCR Physics B(Advancing Physics) G491~ 20th May 2014~ AS Physics Watch

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    Question 9a (ii) is similar to the potential divider question in section A.

    http://www.ocr.org.uk/Images/79820-q...-in-action.pdf
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    (Original post by H0PEL3SS)
    No, I got the same as you, and I know it's wrong. The graph was for the fixed resistor, not the LDR. I think we'll get 1/2 marks from that question.
    It told you v of the LDR. R of the fixed resistor and vin

    So 6-(Vldr)=Vfixedresistor this is then Vout.

    So r2=6x800/2.5 -800 =1120 ohms



    Check: V=IR. 2.5/800=I
    So V= (2.5/800)/3.5=1120


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    (Original post by AlphaNick)
    Hehehe I'm gonna be here next year

    do you all like the OCR B exam board?
    No, don't take physics if the exam board is OCR. From what I'm hearing, you shouldn't take any subject that is OCR.

    If you've got the resources, get a tutor and do another exam board.
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    (Original post by makil)
    what did you guys get for the ostrich question? how much smaller was the mass?
    This is OCR B mate.
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    (Original post by Robbo54)
    Question 9a (ii) is similar to the potential divider question in section A.

    http://www.ocr.org.uk/Images/79820-q...-in-action.pdf
    Can you check mine please?
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    No need to worry. What is done is done.

    I got 100% on all past papers and not a chance did I get 100% raw marks on that, I got at maximum about 55. This was an entirely new style of exam, with roughly 8 marks for not even physics answers. So what does this suggest?

    Very low grade boundaries. It is not like June 2013 that was weird yet accessible, there were many questions which were just not accessible for the majority of candidates, especially the last two of section B. Therefore, an A will be in the region 32-40, without a doubt. It will not be above 40, there is no reason it can be. Therefore 100% will be at maximum 48, probably. Does anybody even know anybody that thinks they got 100% raw marks or did exceptionally well? I certainly don't.
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    (Original post by Mutleybm1996)
    It told you v of the LDR. R of the fixed resistor and vin

    So 6-(Vldr)=Vfixedresistor this is then Vout.

    So r2=6x800/2.5 -800 =1120 ohms



    Check: V=IR. 2.5/800=I
    So V= (2.5/800)/3.5=1120


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    V for fixed resistor (from the graph) is 3.8V.
    We know that it's resistance is 800ohms.
    6-3.8=2.2V for the LDR.
    2.2= R for LDR/(800+ RLDR)x6
    Rearranging and solving gives 463 ohms.
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    (Original post by H0PEL3SS)
    V for fixed resistor (from the graph) is 3.8V.
    We know that it's resistance is 800ohms.
    6-3.8=2.2V for the LDR.
    2.2= R for LDR/(800+ RLDR)x6
    Rearranging and solving gives 463 ohms.
    Graph?


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    (Original post by Mutleybm1996)
    Graph?


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    In the exam? It was for the fixed resistor and the light intensity ( as opposed to the normal one with the LDR and intensity), and at 1000 lux, it gave a p.d of 3.8V.
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    (Original post by H0PEL3SS)
    V for fixed resistor (from the graph) is 3.8V.
    We know that it's resistance is 800ohms.
    6-3.8=2.2V for the LDR.
    2.2= R for LDR/(800+ RLDR)x6
    Rearranging and solving gives 463 ohms.
    How many marks do u think I'll lose for mixing up the resistance of the ldr and fixed resistor and getting like 1380 I think?
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    (Original post by DomStaff)
    No need to worry. What is done is done.

    I got 100% on all past papers and not a chance did I get 100% raw marks on that, I got at maximum about 55. This was an entirely new style of exam, with roughly 8 marks for not even physics answers. So what does this suggest?

    Very low grade boundaries. It is not like June 2013 that was weird yet accessible, there were many questions which were just not accessible for the majority of candidates, especially the last two of section B. Therefore, an A will be in the region 32-40, without a doubt. It will not be above 40, there is no reason it can be. Therefore 100% will be at maximum 48, probably. Does anybody even know anybody that thinks they got 100% raw marks or did exceptionally well? I certainly don't.
    By '100% will be at maximum 48' you mean that 48 marks would round to the full amount/100% of UMS?
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    (Original post by samsamjr)
    How many marks do u think I'll lose for mixing up the resistance of the ldr and fixed resistor and getting like 1380 I think?
    We'll get 1, maybe 2 if we're lucky.
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    why is it not 1380?
    3.8v=LDR volts
    total V=6 therefore fixed must have 6-3.8=2.2v
    fixed has resistance of 800-substitute this value into V=IR
    2.2=Ix800
    therefore I =2.2/800= 2.75x10^-3
    then since current is constant substitute this value again into V=IR for the LDR
    so 3.8=2.75x10^-3 R
    Rearrange for R and you got 1380
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    (Original post by james_1997)
    why is it not 1380?
    3.8v=LDR volts
    total V=6 therefore fixed must have 6-3.8=2.2v
    fixed has resistance of 800-substitute this value into V=IR
    2.2=Ix800
    therefore I =2.2/800= 2.75x10^-3
    then since current is constant substitute this value again into V=IR for the LDR
    so 3.8=2.75x10^-3 R
    Rearrange for R and you got 1380
    It's because the graph where u read 3.8v off is the resitance of the fixed resistor, not the ldr, so if u switch the 2.2 and 3.8 round u get the right answer
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    (Original post by Clai)
    By '100% will be at maximum 48' you mean that 48 marks would round to the full amount/100% of UMS?
    Yes I presume so. To work out 100% UMS, it is double the distance from a B to an A added onto the A. So for January 2010, it was 34 for an A and 29 for B, therefore 44 for 100UMS.

    June 2009 36A, 31B, so 46 for 100% UMS.

    Now, after having done all papers, both of those papers are a walk in the park in comparison. No stupid estimation questions or 'suggest a use of this' or 'why do they have the three squares' or that stupid show that question at the end. However, we must take into account that those papers were when the spec was new, so they had less practice, but they had plenty of legacy papers which are almost identical. So, I just don't see a high boundaries. I will discuss it with my teacher tomorrow.
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    (Original post by DomStaff)
    No need to worry. What is done is done.

    I got 100% on all past papers and not a chance did I get 100% raw marks on that, I got at maximum about 55. This was an entirely new style of exam, with roughly 8 marks for not even physics answers. So what does this suggest?

    Very low grade boundaries. It is not like June 2013 that was weird yet accessible, there were many questions which were just not accessible for the majority of candidates, especially the last two of section B. Therefore, an A will be in the region 32-40, without a doubt. It will not be above 40, there is no reason it can be. Therefore 100% will be at maximum 48, probably. Does anybody even know anybody that thinks they got 100% raw marks or did exceptionally well? I certainly don't.
    If I had perhaps 10-20 minutes more, I feel I could have logically worked through most of the 'hard' questions. Perhaps it was the exam pressure, especially the early hard questions which threw me, at home I've already figured out how to do the 'mass of graphene' question in a few minutes.

    On past papers, I averaged around 51/60 (I completed 7). On this paper, I'll be lucky if I get an A.
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    (Original post by DomStaff)
    Yes I presume so. To work out 100% UMS, it is double the distance from a B to an A added onto the A. So for January 2010, it was 34 for an A and 29 for B, therefore 44 for 100UMS.

    June 2009 36A, 31B, so 46 for 100% UMS.

    Now, after having done all papers, both of those papers are a walk in the park in comparison. No stupid estimation questions or 'suggest a use of this' or 'why do they have the three squares' or that stupid show that question at the end. However, we must take into account that those papers were when the spec was new, so they had less practice, but they had plenty of legacy papers which are almost identical. So, I just don't see a high boundaries. I will discuss it with my teacher tomorrow.
    I have seen papers that ask for possible uses, and estimation, but never for 7 marks in total, and usually, the estimation questions are simple, and the possible uses are 2 marks at max. This paper was dreadful and I wouldn't be surprised if the grade boundaries are the lowest they've ever been.
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    (Original post by DomStaff)
    Yes I presume so. To work out 100% UMS, it is double the distance from a B to an A added onto the A. So for January 2010, it was 34 for an A and 29 for B, therefore 44 for 100UMS.

    June 2009 36A, 31B, so 46 for 100% UMS.

    Now, after having done all papers, both of those papers are a walk in the park in comparison. No stupid estimation questions or 'suggest a use of this' or 'why do they have the three squares' or that stupid show that question at the end. However, we must take into account that those papers were when the spec was new, so they had less practice, but they had plenty of legacy papers which are almost identical. So, I just don't see a high boundaries. I will discuss it with my teacher tomorrow.
    Ah good, I'm hoping to apply to Cambridge and I was told they actually ask for your UMS marks, so the higher you can get, the better.
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    I think what's happened is ocr have decided to employ a new exam writer who by the definition of the word is a biatch
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    (Original post by Robbo54)
    If I had perhaps 10-20 minutes more, I feel I could have logically worked through most of the 'hard' questions. Perhaps it was the exam pressure, especially the early hard questions which threw me, at home I've already figured out how to do the 'mass of graphene' question in a few minutes.

    On past papers, I averaged around 51/60 (I completed 7). On this paper, I'll be lucky if I get an A.
    Same. The timing was so difficult on that paper that I barely had any time for the graphene and QR codes.

    Now I realise why my teacher said to do physics papers back-to-front... argh!
 
 
 
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