Q 11-01-201JEE MainJEE Main 2020 (6 Sep, Shift 2)Medium
A student measuring the diameter of a pencil of circular cross-section with the help of a vernier scale records the following four readings: $5.50$ mm, $5.55$ mm, $5.45$ mm, $5.65$ mm. The average of these four readings is $5.5375$ mm and the standard deviation of the data is $0.07395$ mm. The average diameter of the pencil should therefore be recorded as:
Answer: (D) $(5.54 \pm 0.07)$ mm
The readings are taken to $0.01$ mm, so the result cannot be quoted more precisely than two decimal places in mm.
Rounding the mean to $5.54$ mm and the uncertainty to the same decimal place, $0.07$ mm:
$$d = (5.54 \pm 0.07)\ \text{mm}$$
Solution by Sreeraj P, M.Sc Physics