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OCR Multiple Choice Question on Uncertainty - [Please help]. Watch

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    2. To find the resistivity of a semiconductor, a student makes the following measurements of a cylindrical rod of the material :

    length = 25 ± 1 mm
    diameter = 5.0 ± 0.1 mm
    resistance = 68 ± 1 Ω

    The resistivity is calculated to be 53mΩ m. What is the uncertainty in this result?

    A ± 1mΩ m
    B ± 2mΩ m
    C ± 4mΩ m
    D ± 5mΩ m

    The answer is D, but I am confused how the answer is obtained.

    I converted all of the data above into its respective units.
    length = 25 ± 1x10^3 mmdiameter = 5.0 ± 1x10^3 mmresistance = 68 ± 1000 mΩ

    p = RL/A

    The resistance is ± 1000 * (PI * 1x10^3)^2 / 1x10^3 but no such result was found.
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    (Original post by nwmyname)
    2. To find the resistivity of a semiconductor, a student makes the following measurements of a cylindrical rod of the material :

    length = 25 ± 1 mm
    diameter = 5.0 ± 0.1 mm
    resistance = 68 ± 1 Ω

    The resistivity is calculated to be 53mΩ m. What is the uncertainty in this result?

    A ± 1mΩ m
    B ± 2mΩ m
    C ± 4mΩ m
    D ± 5mΩ m

    The answer is D, but I am confused how the answer is obtained.

    I converted all of the data above into its respective units.
    length = 25 ± 1x10^3 mmdiameter = 5.0 ± 1x10^3 mmresistance = 68 ± 1000 mΩ

    p = RL/A

    The resistance is ± 1000 * (PI * 1x10^3)^2 / 1x10^3 but no such result was found.
    Two things:

    First, the equation you stated is incorrect. Start with:

    R = \frac{\rho L}{A}

    Then re-arrange to make resistivity the subject:

    \rho = \frac{RA}{L}


    Second, a reasonable estimate of the uncertainty in measurements is given by:

    \rm \ uncertainty = \frac{Range}{2} = \frac{(Max - Min)}{2}

    Looking at the resistivity equation, the maximum result is obtained when the numerator variables are at their maximum and when the denominator variables are set to their minimum.

    The minimum result is obtained when the numerator is minimum and the denominator is maximum.

    Which means

    R_{max} = 68 +1 = 69\ \Omega
    R_{min} = 68 - 1= 67\ \Omega


    A_{max} = \pi(\frac{d_{max}}{2})^2 = \pi(\frac{5.1}{2})^2 = 20.43\rm\ mm^2

    A_{min} = \pi(\frac{d_{min}}{2})^2 = \pi(\frac{4.9}{2})^2 = 18.86\rm\ mm^2


    L_{max} = 25 + 1 = 26\rm\ mm
    L_{min} = 25 - 1 = 24\rm\ mm

    Now substitute the max and min values into the resistivity equation, first to get the maximum and then again to get the minimum calculated resistivity. Pay attention to the numerator and denominator.

    \rho_{max} = \frac{A_{max}R_{max}}{L_{min}} = \rm\ \frac{20.43 x 69}{24} = 58.74 m\Omega

    \rho_{min} = \frac{A_{min}R_{min}}{L_{max}} = \rm\ \frac{18.86 x 67}{26} = 48.60 m\Omega


    Finally, substitute the min and max calculated resistivity values into the uncertainty expression:

    \rm \ uncertainty = \frac{(\rho_{max} - \rho{min})}{2} = \frac{(58.74 - 48.60)}{2} = 5.07 m\Omega
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    (Original post by nwmyname)
    L min : isn't it 25 - 1 not 2511?
    Thanks, error corrected.
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    (Original post by nwmyname)
    2. To find the resistivity of a semiconductor, a student makes the following measurements of a cylindrical rod of the material :

    length = 25 ± 1 mm
    diameter = 5.0 ± 0.1 mm
    resistance = 68 ± 1 Ω

    The resistivity is calculated to be 53mΩ m. What is the uncertainty in this result?

    A ± 1mΩ m
    B ± 2mΩ m
    C ± 4mΩ m
    D ± 5mΩ m

    The answer is D, but I am confused how the answer is obtained.

    I converted all of the data above into its respective units.
    length = 25 ± 1x10^3 mmdiameter = 5.0 ± 1x10^3 mmresistance = 68 ± 1000 mΩ

    p = RL/A

    The resistance is ± 1000 * (PI * 1x10^3)^2 / 1x10^3 but no such result was found.
    Uberteknik's answer is correct but if you're familiar with A level error calculations there is a faster method of getting the answer.

    First, write out the equation for resistivity:
     \rho = RA/L ~ RD^2/L I wrote A~D^2 because I can't be bothered to write in the factor of \frac{\pi}{4} and I know it's not going to matter.

    Now, the % uncertainty in your answer is just the sum of your % uncertainties:
     \text{percentage uncertainty} = 1\div25 + 2(0.1\div5) + 1\div68 = 0.095
    Multiply by 53mOhm m to get your total : 53\times0.095 = 5\text{mOhm m}

    This is a little bit faster if you're used to the method - calculate all percent uncertainties, add them up (accounting for duplicates, like D^2 here) and multiply back to get absolute uncertainty.
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    (Original post by lerjj)
    Uberteknik's answer is correct but if you're familiar with A level error calculations there is a faster method of getting the answer.

    First, write out the equation for resistivity:
     \rho = RA/L ~ RD^2/L I wrote A~D^2 because I can't be bothered to write in the factor of pi/4 and I know it's not going to matter.

    Now, the % uncertainty in your answer is just the sum of your % uncertainties:
    %unc = 1/25 + 2(0.1/5) + 1/68 = 0.095
    Multiply by 53mOhm m to get your total : 53*0.095 ~ 5m Ohm m

    This is a little bit faster if you're used to the method - calculate all percent uncertainties, add them up (accounting for duplicates, like D^2 here) and multiply back to get absolute uncertainty.
    Your answer seems better.
    Is it possible for you to use latex as well for the remainder of the answer as it may be very useful for others [no offense intended]
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    (Original post by nwmyname)
    Your answer seems better.
    Is it possible for you to use latex as well for the remainder of the answer as it may be very useful for others [no offense intended]
    Done, not too sure it looks better tbh, but it is a bit more readable. All those numbers are still ugly.
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    (Original post by lerjj)
    Uberteknik's answer is correct but if you're familiar with A level error calculations there is a faster method of getting the answer.

    First, write out the equation for resistivity:
     \rho = \frac{RA}{L} = \frac{RD^2}{L} I wrote A as D^2 because I can't be bothered to write in the factor of \frac{\pi}{4} and I know it's not going to matter.

    Now, the % uncertainty in your answer is just the sum of your % uncertainties:
     \text{percentage uncertainty} = 1\div25 + 2(0.1\div5) + 1\div68 = 0.095
    Multiply by 53mOhm m to get your total : 53\times0.095 = 5\text{mOhm m}

    This is a little bit faster if you're used to the method - calculate all percent uncertainties, add them up (accounting for duplicates, like D^2 here) and multiply back to get absolute uncertainty.
    This is better
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    (Original post by lerjj)
    Done, not too sure it looks better tbh, but it is a bit more readable. All those numbers are still ugly.
    Thanks for the alternative solution.

    I very often approach these types of problem from an engineering perspective where tolerances in measurement are de rigueur and so very often stated as skewed limits. i.e. +0.00, -0.03mm; +0.15, -0.07 etc. Also engineering demands a more structured approach such that errata in calculations are easily traceable with an audit trail.

    Your method is good for A-level where there is symmetry with error and one is up against a time limit. In real world situations asymmetry is more often the case.
 
 
 
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