JEE Main 2020 · 6 September · evening · Single correct · +4 / −1
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 :
- (a) (5.54 ± 0.07) mm
- (b) (5.5375 ± 0.0740) mm
- (c) (5.5375 ± 0.0739) mm
- (d) (5.538 ± 0.074) mm
Answer and solution
Answer (a) (5.54 ± 0.07) mm
Solution
Physics sir · 4 steps
Your answer cannot be more precise than your readings, so round everything to two decimal places.
- Precision of the data. Every reading is written to two decimal places of a millimetre, like 5.50 mm. So the instrument can only tell you the second decimal place.
- Round the error. The uncertainty must match that precision. 0.07395 mm rounds to 0.07 mm, and this sets the last decimal place you are allowed to report.
- Round the mean. The mean must end at the same decimal place as the error. 5.5375 mm rounds to 5.54 mm. Extra digits from the calculator are not real information.
- So. The diameter is recorded as (5.54 ± 0.07) mm, which is option (a).
The idea to keep. Round the uncertainty to the precision of the measuring instrument first. Then round the mean to the same decimal place as the uncertainty. Never report calculator digits.
Answer from the NTA final answer key. Solution by PrepWiser.