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The least count of a stop watch is 0.2 s .The time of 20 oscillationsof a pendulum is measured to be 25.What will be the percentage error in measurement of time?

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closed as off-topic by user191954, rob Nov 9 '18 at 18:14

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Total time measurement: $ 25s \pm 0.2s$

Error in one oscillation: $\frac{25s \pm 0.2s}{20} = 12.5s \pm 0.01s$

Hence Percentage error: $\frac{0.01s}{12.5s} \approx 8 \cdot 10^{-4}$

Depending on how exactly your clock works you could argue that all the errors are halved, but the above is the safe bound.

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  • $\begingroup$ If the divisions are every 0.2 seconds then any time between 24.9 and 25.1 seconds will be registered as 25.0 seconds so should it not be $25.0 \pm 0.1$ seconds? $\endgroup$ – Farcher Apr 14 '16 at 16:09
  • $\begingroup$ @Farcher i wasn't quite sure if 0.2 in the question were the error or the spacing (i guess 'least count' suggests the latter as you say), so i put both options. that's what i meant by halving the error in the comment at the end. is that not clear? i could edit and swap that i guess $\endgroup$ – Wolpertinger Apr 14 '16 at 17:46
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    $\begingroup$ It is difficult to decide what to do from the information given as one could say that two observations are needed to get a time value. One at the start of the interval and one at the end so your analysis would fit such a method. I agree with your last sentence. $\endgroup$ – Farcher Apr 14 '16 at 21:21

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