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    I have a question on this exam, 9702_w14_qp_23.

    2. a) iii)

    On the mark scheme, we can see the value 0.2 cm for the uncertainty of the oscilloscope. As it is an analogue reading, shouldn't it be half of this number?

    Is it 0.2 because 3.8 cm (on calculating the period) is actually a difference between two readings, each one affected by an uncertainty of 0.1?

    2. b)

    According to the mark scheme, the answer can be 530 +- 30.
    But from the calculations... 526 +- 28... is it acceptable? If not, why should we round it up?

    Thank you.
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Updated: April 11, 2016


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