Discussion
Accidental errors are proportional to :
( Where 'n' is number of observations made)
Answer: A
The Board's key is \(1/\sqrt{n}\), the standard result for the accidental error of the mean of n observations.
Accidental (random) errors are as likely to be positive as negative, so they partly cancel. For n independent observations of equal precision, each with standard error \(\sigma\):
\[ \begin{aligned} \sigma_{\text{sum}} &= \sigma\sqrt{n} \\ \sigma_{\text{mean}} &= \frac{\sigma\sqrt{n}}{n} = \frac{\sigma}{\sqrt{n}} \end{aligned} \]
So the accidental error in the average of the readings falls as \(1/\sqrt{n}\), which is why repeating a measurement improves it. The question does not say whether the total or the mean is meant. The error in the total of n observations grows as \(\sqrt{n}\), which is option C, and some books quote that form. The key takes the error of the result, which falls as \(1/\sqrt{n}\).
Option B (n) would suit a systematic error, which adds up in full with every reading. Option D (1/n) would make random error vanish as fast as the readings multiply, which does not happen.
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