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There is a question which states, find the percentage uncertainty when the measured value of an object is $6\text{cm}$, given that a vernier caliper is used.

$$60.0\text{mm} \pm 0.1\text{mm}$$

Hence we have: $$\frac{0.1}{60.0}\times 100\%=0.167\%$$

However, isn't absolute uncertainty of the instrument, the lowest division divided by $2?$ So if our instrument is the vernier caliper, the lowest division is $0.1\text{mm}$. Therefore, the absolute uncertainty will be then $0.05\text{mm}$, which gives our uncertainty calculation to be:

$$\frac{0.05}{60.0}\times 100\%=0.083\%$$

Furthermore in the markscheme it says:

$$\frac{0.1}{6}\times 100\%=1.67\%$$

At the moment I am so confused to which one is the true answer to this question, could anyone help me understand this?

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    $\begingroup$ 0.083% looks good to me. $\endgroup$ – PM 2Ring May 1 at 16:32
  • $\begingroup$ Ok thank you, so is the mark scheme wrong? $\endgroup$ – Aurora Borealis May 2 at 18:36
  • $\begingroup$ Yes, I believe it's wrong, and you are right. $\endgroup$ – PM 2Ring May 2 at 18:48

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